Unit 08 Thermodynamics Students Manua
CBSE-i
CLASS
XI
UNIT-8
CHEMISTRY
Thermodynamics
Student's Manual
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
CLASS
CBSE-i
XI
UNIT-8
CHEMISTRY
Thermodynamics
Student's Manual
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
The CBSE-International is grateful for permission to reproduce and/or translate copyright material used in this publication. The acknowledgements have been included wherever appropriate and sources from where the material may be taken are duly mentioned. In case any thing has been missed out, the Board will be pleased to rectify the error at the earliest possible opportunity.
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Preface
Education plays the most important role in acquiring professional and social skills and a positive attitude to face the challenges of life. Curriculum is a comprehensive plan of any educational programme. It is also one of the means of bringing about qualitative improvement in an educational system. The Curriculum initiated by Central
Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content responsive to global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos.
The CBSE introduced the CBSE-i curriculum as a pilot project in few schools situated outside India in 2010 in classes I and IX and extended the programme to classes II, VI and X in the session 2011-12. It is going to be introduced in classes III, VII and for Senior Secondary classes with class XI in the session 2012-13.
The Senior Secondary stage of education decides the course of life of any student. At this stage it becomes extremely important for students to develop the right attitude, a willingness to learn and an understanding of the world around them to be able to take right decisions for their future. The senior secondary curriculum is expected to provide necessary base for the growth of knowledge and skills and thereby enhance a student's potential to face the challenges of global competitiveness. The CBSE-i Senior Secondary Curriculum aims at developing desired professional, managerial and communication skills as per the requirement of the world of work. CBSE-i is for the current session offering curriculum in ten subjects i.e. Physics Chemistry, Biology, Accountancy,
Business-Studies, Economics, Geography, ICT, English, Mathematics I and Mathematics II. Mathematics at two levels caters to the differing needs of students of pure sciences or commerce.
The Curriculum has been designed to nurture multiple intelligences like linguistic or verbal intelligence, logicalmathematical intelligence, spatial intelligence, sports intelligence, musical intelligence, inter-personal intelligence and intra-personal intelligence.
The Core skills are the most significant aspects of a learner's holistic growth and learning curve. The objective of this part of the core of curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This involves trans-disciplinary linkages that would form the core of the learning process.
Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research would be the constituents of this 'Core'.
The CBSE-i Curriculum evolves by building on learning experiences inside the classroom over a period of time.
The Board while addressing the issues of empowerment with the help of the schools' administering this system strongly recommends that practicing teachers become skilful and lifelong learners and also transfer their learning experiences to their peers through the interactive platforms provided by the Board.
The success of this curriculum depends upon its effective implementation and it is expected that the teachers will make efforts to create better facilities, develop linkages with the world of work and foster conducive environment as per recommendations made in the curriculum document.
I appreciate the effort of Dr.Sadhana Parashar, Director (Training), CBSE, Dr. Srijata Das, Education Officer,
CBSE and Ms. Anjali Chhabra, Assistant Education Officer, CBSE and their teams involved in the development of this document.
The CBSE-i website enables all stakeholders to participate in this initiative through the discussion forums. Any further suggestions on improving the portal are always welcome.
Vineet Joshi
Chairman, CBSE
Acknowledgements
Advisory
Shri Vineet Joshi, Chairman, CBSE
Dr. Sadhana Parashar, Director (Training), CBSE
Ideators Classes XI and XII
Prof. A K Bakshi
Dr. N K Sehgal
Prof. Kapil Kapor
Ms. Renu Anand
Dr. Barkatullah Khan
Ms. Avnita Bir
Conceptual Framework
Shri G. Balasubramanian, Former Director (Acad), CBSE
Ms. Abha Adams, Consultant, Step-by-Step School, Noida
Dr. Sadhana Parashar, Director (Training), CBSE
Ms. P Rajeshwari
Ms. Gyatri Khanna
Mrs. Anita Makkar
Prof. Biswajit Nag
Ms. Usha Sharma
Dr. Niti Nandini Chatnani
Dr. Anil K Bali
Dr. Preeti Tewai
Dr. Deeksha Bajpai
Mr. S K Agarwala
Ms. Neeta Rastogi
Dr. Anshu
Dr. Rajesh Hassija
Mr. Mukesh Kumar
Dr. Om Vikas
Material Production Groups: Classes XI-XII
English :
Ms. Gayatri Khanna
Ms. Renu Anand
Ms. P Rajeshwary
Ms. Sandhya Awasthi
Ms. Manna Barua
Ms. Veena Bhasin
Ms. Urmil Guliani
Ms. Sudha Ravi
Mr. Anil Kumar
Ms. Vijaylaxmi Raman
Ms. Neerada Suresh
Ms. Himaal Handoo
Chemistry :
Dr. G S Sodhi
Dr. Vimal Rarh
Dr. Shalini Baxi
Dr. Vinita Arora
Dr. Vandana Soni
Ms. Charu Maini
Ms. Rashmi Sharma
Ms. Kavita Kapoor
Biology :
Dr. Ranjana Saxena
Dr. Neeraja Sood
Dr. P Chitralekha
Ms. Mridula Arora
Ms. Lucy Jad
Ms. Priyanka Choudhury
Ms. Prerna Gosain
Ms. Malini Sridhar
Geography:
Dr. Preeti Tewari
Ms. Rupa Das
Mr. S Fazal Daoud Firdausi
Ms. Neena Phogat
Ms. Sujata Sharma
Ms. Deepa Kapoor
Ms. Bharti Malhotra
Ms. Isha Kaushik
Mr. Riyaz Khan
Physics :
Dr. B. Biswal
Ms. Namarata Alwadhi
Mr. Dhirender Sharma
Ms. Vandana Banga
Mr. Vivek
Economics:
Mr. S K Agarwala
Ms. Ambika Gulati
Ms. Nidhi Singh
Ms. Malti Modi
Ms. Sapna Das
Ms. Ingur Agarwal
Ms. Shankar Kulkarni
Mr. Sandeep Sethi
Mathematics :
Dr. Sushil Kumar
Mrs. Monica Talwar
Mrs. Charu Dureja
Mrs. Seema Juneja
Dr. H L Bhatia
Dr. Sushma Bansal
Mrs. Neeru Aggarwal
Dr. Saroj Khanna
Accountancy :
Mr. S S Sehrawat
Dr. K Mohna
Dr. Balbir Singh
Ms. Bhupendra Kriplani
Ms. Shipra Vaidya
Mr. Sandeep Sethi
Business Studies :
Dr. S K Bhatia
Ms. Meenu Ranjan Arora
Mrs. Shegorika
Mr. Sandeep Sethi
Ms. Usha Sharma
Ms. Komal Bhatia
Ms. Ravisha Aggarwal
ICT :
Mr. Mukesh Kumar
Ms. Nancy Sehgal
Ms. Purvi Srivastava
Ms. Gurpreet Kaur
Coordinators:
Ms. Sugandh Sharma, EO
Dr Rashmi Sethi, EO
Ms. S. Radha Mahalakshmi, EO Mr. Navin Maini, RO (Tech)
Ms. Madhuchhanda, RO (Inn) Shri Al Hilal Ahmed, AEO
Ms. Anjali Chhabra, AEO
Shr. R. P. Singh, AEO
Shri R. P. Sharma,
Consultant (Science)
Ms. Reema Arora,
Consultant (Chemistry)
Mr. Sanjay Sachdeva, SO
Ms. Neelima Sharma,
Consultant (English)
Contents
Preface
Acknowledgement
Syllabus Coverage
1
Learning Outcomes
3
Mind Map
6
Contents
Summary
Students’ Worksheets
7 - 83
84
85 - 119
Common Misconceptions
120
Crossword Puzzle
121
Additional Resource Links
124
UNIT – 8: THERMODYNAMICS
SYLLABUS COVERAGE
8.1
Introduction to thermodynamics
8.2
Thermodynamic Terms
-
System and surrounding
-
Types of system
-
State of a system
-
State function and path function
-
Extensive and intensive properties
-
Reversible and irreversible process
8.3
8.4
8.5
Thermodynamic Quantities
-
Work
-
Heat
First Law of Thermodynamics
-
Internal energy
-
Enthalpy
-
Heat capacity
-
Measurement of ∆U and ∆H
Thermochemistry
-
Enthalpy change in a reaction
-
Endothermic and exothermic reactions
-
Enthalpy changes during phase transformations
-
Standard enthalpy of formation
-
Thermochemical equations
1
8.6
-
Hess‟s law of constant heat summation
-
Enthalpies for different types of reactions
Spontaneity
-
Entropy
-
Second law of thermodynamics
-
Gibb‟s energy change for spontaneous and non-spontaneous processes
-
Criteria for equilibrium
-
Third law of thermodynamics
2
LEARNING OUTCOMES
At the end of this unit students would be able to –
Understand the meaning of the term Thermodynamics.
Differentiate between the terms system and surrounding.
Distinguish between open system, closed system and isolated system.
Understand the meaning of state of a system, state function and path function.
Distinguish between extensive and intensive properties.
Comprehend the concept of thermodynamic reversibility.
Distinguish between reversible and irreversible processes.
Recall the physical interpretation of work.
Understand the work done during the process of expansion and compression
Write the S.I. units of work.
Deduce a mathematical expression to calculate the work done during isothermal expansion and compression of an ideal gas and during free expansion.
Solve numerical for calculating work done during expansion and compression.
Comprehend the difference in the work done during expansion and compression when carried out reversibly and irreversibly.
Understand the concept of heat.
Know the IUPAC sign conventions of work and heat.
Recognize that work and heat are not state functions.
Explain the concept of internal energy.
Show that internal energy is a state function.
State and deduce the mathematical expression for the first law of thermodynamics.
Interpret the first law to explain its consequences.
Solve numerical based on the first law of thermodynamics.
3
Recognize the need for enthalpy function.
Derive a mathematical expression for enthalpy.
Appreciate the physical significance of enthalpy.
Comprehend that Internal energy change is the heat evolved or absorbed at constant volume and Enthalpy change is the heat evolved or absorbed at constant pressure.
Deduce a mathematical relation between ΔH and ΔU and solve numerical.
Explain heat Capacity, molar heat capacity and specific heat capacity.
Differentiate between Cp and Cv.
Derive a mathematical relation between Cp and Cv for an ideal gas.
Know the procedure of measurement of ΔH and ΔU in a calorimeter.
Define reaction enthalpy.
Recognize that heat changes accompany chemical reactions.
Compare exothermic and endothermic reactions.
Define standard enthalpy of reactions.
Examine the enthalpy changes accompanying phase transformations such as fusion, vaporization, sublimation.
Describe and Apply the standard enthalpy of formation values to calculate the enthalpy change for the reaction.
Know the convention for writing Thermochemical equations.
Comprehend the Hess‟s Law of Constant Heat Summation.
Discuss the enthalpies of different types of reactions such as Standard enthalpy of combustion, Enthalpy of atomization, Bond enthalpy, Enthalpy of solution.
Construct a Born-Haber Cycle for determination of lattice enthalpy.
Solve numerical based on Hess‟s Law of Constant Heat Summation.
Understand the limitations of first law of thermodynamics.
Recognize that all natural processes will tend to proceed spontaneously in one direction only.
4
Comprehend with the help of examples that decrease in enthalpy is not a criterion for spontaneity.
Interpret randomness or disorder with the help of examples.
Explain the thermodynamic function, Entropy and write its mathematical expression.
Analyze ΔStotal as the criterion for examining spontaneity.
Examine the statement of second law of thermodynamics.
Understand the need for introducing Gibb‟s free energy and write its mathematical expression.
Derive the criteria for spontaneity given by ΔG.
Interpret the effect of temperature on spontaneity of reaction.
Derive a mathematical expression showing the relation between free energy and equilibrium constant.
Solve numerical based on second law of thermodynamics.
State the third law of thermodynamics.
5
MIND MAP
6
8.1
INTRODUCTION TO THERMODYNAMICS
Warm up
Recall the concept of various states of matter. Try to represent a specific attribute/property of a state of matter. Now ask yourself whether matter can be converted from one form to the other.
Understand the transformation of matter from one form to another for example melting of ice into liquid water and then evaporation into water vapour. This activity can be performed in Chemistry lab by using a Bunsen burner.
Now, interpret your observation. Think, how heat is playing a role in transformation of solid to liquid to gas?
8.1.1.
Origin of the term Thermodynamics
Thermodynamics is the branch of physical science concerned with heat and its relation to other forms of energy and work. The word thermodynamics was coined from two Greek words, thermos (heat) and dynamics (movement).
Thus thermodynamics mean “movement of heat”. It is therefore, a study of heat and its transformation to various other forms of energy, especially mechanical energy.
Thermodynamics is a physical science that involves the study of the changes in the energy of the system when the temperature, pressure or volume of a system, are changed or when a chemical transformation takes place. It deals with the relationship of energy with heat and work and the inter conversion of various forms of energy. Therefore, thermodynamics is sometimes also known as Energetics.
Chemical Thermodynamics is that branch of thermodynamics that deals with energy changes accompanying chemical transformations. It investigates the feasibility of a chemical change and tries to find the answer to the question – Why does a chemical reaction take place?
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8.1.2.
8.1.3.
Characteristics of thermodynamics
It is a study of macroscopic systems, i.e., objects or processes that are of a size that is measurable and observable by the naked eye.
Usefulness of thermodynamics lies in correlating the observable properties of substances with some of the non-observable ones.
It points out the feasibility, extent and direction of changes taking place in a process. These changes may be physical processes or chemical reactions.
It is not necessary to know the nature of the fundamental particles or the mechanism of the phenomenon. The deductions are made using basic laws.
Thermodynamics can be studied by dealing with the energy changes taking place at the macroscopic level without emphasizing the changes that occur at molecular level.
Limitations of thermodynamics
It deals with materials in bulk without any reference to the internal structure and microscopic properties or behaviour.
It deals with stationary or equilibrium states only.
It gives the feasibility, direction and extent of a change but not the time taken for such a change to take place. The rate at which a process undergoes completion is studied under a separate branch of physical chemistry, known as, chemical kinetics.
Student Activity- 1
Objective: To understand the First Law of Thermodynamics, i.e., the heat added to a system is partly used to do work of expansion and the rest is used to increase the internal energy of the system.
Materials Required:
1.
a pack of popcorn kernels
8
2.
a pan with lid
3.
gas stove
Procedure:
1.
Open the pack of popcorn kernels and put all of them in the pan.
2.
Place the pan on the gas stove over low flame and cover it with the lid.
3.
Wait for some time for the pan to get hot. After few minutes, kernels start popping with a pop sound.
4.
Wait for another few minutes for all the kernels to pop.
5.
Put off the gas stove and leave the pan covered with the lid.
Now answer the following questions:
1.
What happened to the kernels when the pan got hot?
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2.
Why do you think kernels start popping up?
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3.
From where the popcorns are getting energy to pop?
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4.
What do you observe happened to the lid of the pan when all the kernels popped simultaneously? Why has it happened?
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5.
Relate your observations with the first law of thermodynamics.
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8.2
THERMODYNAMIC TERMS
System: It is that part of the universe on which theoretical or experimental investigations are carried out. A system is said to be a part of the universe which is separated from the rest of the universe by a real or imaginary well defined boundary. For example, if the effect of pressure on the volume of the gas has to be studied, then the system is the gas taken in a container. Similarly, any solution taken in a beaker is known as the system, when its properties are being studied.
Surroundings: The rest of the universe in the neighbourhood of the system is known as its surroundings. It lies outside the system and is also called the environment of the system. The system can exchange matter or energy or both with its surroundings.
A system and its surrounding are together known as the Universe.
Boundary: Anything that separates the system from the surroundings is known as a boundary. It is also known as the wall of the system. The system interacts with its surroundings across the boundary.
8.2.1.
Classification of systems based on the type of boundary
The system exchanges matter and energy with the surroundings across the boundary. Therefore, the boundary decides the relationship of the system with its surroundings. There can be several categories of systems depending on the nature of boundary present.
1.
Open system: is a system which has a permeable boundary such that both matter and energy can move across the boundary. So, the exchange of matter (mass, m) and energy (U) can take place between the system and the surroundings. For an open system Δm ≠ 0 and ΔU ≠0.
10
8.2.2.
2.
Closed system: is a system which has an impermeable but diathermic boundary such that matter cannot flow across the boundary but energy exchange can take place between the system and the surroundings. For a closed system Δm= 0 and ΔU≠0.
3.
Isolated system: is a system which has an adiabatic boundary, i.e., neither matter nor energy can be exchanged between the system and surroundings. For an isolated system Δm = 0 and ΔU =0.
State of the System
The physical state and other important properties and parameters defining the system are decided by the pressure (p), temperature (T), volume (V) and the amount (number of moles, n) of the system. These four variables define the state of the system. Therefore, these variables are known as state variables.
Thermodynamic process: It is any operation or process that brings about a change in the state of the system. Expansion, compression, heating and cooling are some of the examples of a thermodynamic process. The initial state of the
11
system is defined by the pressure, temperature, volume and amount of the system before the change takes place. The final state of the system is defined by the values of pressure, temperature, volume and amount of the system after the change has taken place. A process is said to have taken place whenever there is a change in any one or more of the state variables. For any state variable Y, if the change in the state variable is large it is represented by
ΔY (delta Y). If the change is infinitesimal (very small) it is represented by dY.
8.2.3.
Classification of processes based on the change in thermodynamic variables Thermodynamic processes are classified into various categories depending upon the thermodynamic properties that are undergoing a change during the process. These categories are as follows:
1.
Isothermal process: It is a process during which the temperature of the system remains constant, that is, ∆T = 0.
2.
Adiabatic process: It is a process during which the system does not exchange heat with the surroundings, that is, q = 0.
3.
Isobaric process: It is a process during which the pressure of the system remains constant, that is, ∆P = 0.
4.
Isochoric process: It is a process during which the volume of the system remains constant, that is, ∆V = 0.
5.
Cyclic process: It is the process during which the system after undergoing a series of changes comes back to the initial state, that is, the series of changes take place in a cyclic manner.
Path: The sequence of steps taken by a system during a thermodynamic process starting from the initial state, through an intermediate state to the final state is known as a path. The path may consist of one or more than one steps.
8.2.4.
State function
A function that depends only on the state of the system and not on the process by which this state is achieved is known as a state function.
12
Thermodynamic properties like internal energy (U), enthalpy (H), entropy (S) and free energy (G and A) change with the change in variables like temperature, pressure, volume and amount of the system. The changes in these properties depend upon the initial and final states of the system. They do not depend upon the path followed during the change. Therefore, these thermodynamic properties are also state properties or state functions.
8.2.5.
Path Function
It is a thermodynamics property that depends upon how the process is carried out, that is, it depends upon the path taken by the system during the thermodynamic process.
Work and heat are path functions. Therefore, there is no such thing as initial work and final work, therefore, w1 and w2 do not mean anything. w is the work done when the system goes from an initial state to the final state.
Similarly, since q is a path function we only talk about the total heat exchange between the system and the surroundings during a process and not the initial heat and the final heat.
8.2.8.
Intensive and extensive properties
Extensive properties are those properties that depend upon the size of the system. They are additive in nature. This means that if the system is divided into small parts then the total value of a property is obtained by adding up the value of that property in all the parts of the system.
Some of the important extensive properties are: mass, volume, amount
(number of moles), energy, enthalpy, entropy, free energy and heat capacity.
Intensive properties are those properties that do not depend upon the size of the system. These properties are not added up to get the total value of a given property of the system. They have the same value in all the parts of the system. Some of the important intensive properties are: pressure, temperature, density, concentration, mole fraction, refractive index, surface tension, viscosity and specific heat capacity.
Let a container containing an ideal gas be divided into four compartments namely A, B, C, D as shown in Figure 1. Let the pressure in each compartment
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be PA, PB PC, PD, respectively. Let the temperature be TA, TB, TC, TD; volume be
VA, VB, VC, VD; and the amount be nA, nB, nC, nD. When the partition is removed without changing the state of the system then, the total volume of the container is
V=VA+ VB+ VC+ VD
The total amount of the gas is n = nA+ nB + nC + nD
However, the total pressure of the container is not equal to the sum of the pressure of each compartment. Similarly, the temperature of the container is also not equal to the sum of the temperatures of each compartment. Thus, it is seen that the volume and amount depend upon the size of the system while temperature and pressure do not depend upon the size of the system.
Another illustration of extensive and intensive properties is of the 1mol dm–3 sodium chloride solution taken in a one litre flask. This solution is divided unequally among four students and each student measures the mass, volume, density, temperature, surface tension and viscosity of the solution. What is the expected result?
All the four students will report almost the same values of density, temperature, surface tension and viscosity of the solution within experimental error. The mass and the volume of the solutions will be different but the volumes of solutions of each student will add up to one litre. Again it is observed that the intensive properties, that is, density, temperature, surface tension and viscosity of the solution do not depend upon the size of the system while the extensive properties depend upon the size.
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List of some extensive and intensive properties
Extensive mass, volume, amount, heat capacity, internal energy, properties enthalpy, entropy, free energy
Intensive pressure, temperature, density, surface tension, properties viscosity, refractive index, molarity, molar volume, specific heat capacity, molar heat capacity, molar energy
8.2.7.
Classification of processes based on the path
Depending upon the path followed by the system there are two types of processes: reversible and irreversible processes.
Reversible process: A process is said to be reversible if the system undergoes a change in the state through a sequence of intermediate states, each one of which is an equilibrium state. In this manner, the system can be restored to the initial state by following the same sequence of steps in the reverse order. That is, a process is said to be reversible if it can proceed in forward and backward directions by a small change in state variables. When the system comes back to the initial state the surroundings are also restored to their initial state. That is, the process is said to be reversible only if the system can retrace its steps without any net change in the system and the surroundings. The process should be reversible in both forward and backward directions. Each reversible process is slow but each slow process may or may not be reversible.
Irreversible process: A process is said to be irreversible if the change takes place without the system being in equilibrium with its surroundings at any stage during the change. The system cannot retrace its path without the help of an external source. Hence, the system can attain its initial state if and only if some change is brought about in the surroundings.
Examples: Most of the natural processes and the spontaneous processes are irreversible processes.
Equilibrium: A system exists in a state of equilibrium if its macroscopic properties such as temperature, pressure, volume and mass remain constant without the help of an external source.
15
Student Activity- 8.2
Objective: To understand the statement of second law of thermodynamics and direction of heat flow.
Materials Required:
1.
metal blocks of the same size and dimensions – 2 No.s
2.
thermometer
Procedure
1.
Place one block of metal in the freezer for about 10 minutes. Before taking it out from the freezer, measure its temperature using a thermometer.
2.
At the same time, immerse the other block in boiling water. Measure its temperature also.
3.
Get the two blocks and place them in contact for 3–5 minutes. Get the temperature of each block.
4.
Record your data in the table below.
Blocks
Temperature (°C)
Before contact
After contact
Block A (hot)
Block B (cold)
Answer the following questions:
1.
What did you observe after the blocks were placed in contact with each other?
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2.
What do you conclude from your observation?
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3.
If we assume that the two blocks are simply interacting with each other and not with the surroundings, how can you relate the heat changes in the two blocks?
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4.
What is the natural direction of heat flow? Can it be reversed?
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8.3
THERMODYNAMIC QUANTITIES
8.3.1. Work
In thermodynamics, work is defined as any quantity that flows across the boundary of a system during a change in its state and is completely convertible into the lifting of a weight in the surroundings. The symbol of work is w. SI unit of work is Joule (J).
It has the following characteristics.
(i)
Work appears only at the boundary of a system.
17
(ii)
Work appears only during a change in state, that is, only during a thermodynamic process.
(iii) Work is manifested by an effect in the surroundings.
(iv) During an adiabatic process, work results in the change in the internal energy of the system. When the work is done by the system, the system uses its internal energy. Therefore, internal energy of the system decreases. When work is done on the system, the internal energy of the system increases.
(v)
Work is a path function, that is, it is an inexact function. If the system is changed from the initial state to the final state by two different processes, then the work done will be different for the two processes.
Mathematical Expression for Work
Mechanical work is the result of displacement of a body when the force is applied on it. It is related to the force (F) and displacement by the following equation. w=F
Δx
When the system does mechanical work, it can be related to the change in the volume of the system.
This can be illustrated by considering a system consisting of a gas taken in a cylinder with a piston at the top. This piston is assumed to be mass less and frictionless; therefore, no work is done to overcome friction when the piston moves up and down. An external pressure (p) is applied on the system. When the gas expands, the piston moves up by a distance Δx. The force is related to the external pressure by the equation
Force = Pressure
area = p
A
Here, A is the area of cross section of the base of the piston. Substituting the expression of force in the above equation of w, w= p
Since, A
A
Δx x = ΔV
(area
distance = volume)
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w = pΔV
….(1)
When the displacement is infinitesimal, i.e., dx the change in volume is represented by dV and the work done is dw. Therefore, dw = pdV
….(2)
Sign Convention in thermodynamics
In thermodynamics studies, the emphasis is on the system. Any change that results in an increase in the energy of the system is taken as positive. During compression, the internal energy of the system increases, therefore, the work done on the system is taken as positive. Similarly, the work done by the system is done at the cost of the internal energy of the system; therefore, the work done by the system is taken as negative. Hence, the Equations (1) and (2) are modified as w = −pΔV
.....for finite change
and dw = − pdV
.....for infinitesimal change
It can be shown that the above formula will give the positive value of w during compression when the work is done on the system and the negative value of w during expansion when the work is done by the system.
Figure 8.1: Compression and expansion of a gas taken in a cylinder
Initially some weights are kept on the piston so that the external pressure is equal to the internal pressure and the piston does not move. Then the weight on the piston is increased to compress the gas and move the piston down. Here, work is done on the gas and it can be calculated using Equation (1).
19
w = –pex ΔV pex is the external pressure acting on the gas. During compression, the volume of the gas decreases, so, ΔV is negative. Therefore, the value of w is positive when the work is done on the gas.
To expand the gas some weights are removed from the piston and the gas expands lifting the piston with the remaining weights. During expansion, work is done by the gas to lift the weights. Volume of the gas increases and so ΔV is positive. Therefore, the value of w is negative when work is done by the gas.
8.3.2. Heat
In thermodynamics heat is defined as a quantity that flows across the boundary of a system, during a change in its state by virtue of the difference in temperature between the system and its surrounding. Heat flows from a region of higher temperature to a region of lower temperature. The symbol of heat is
q. SI Unit of heat is Joule (J).
Heat appears only at the boundary of the system
Heat appears only during the change in state
Heat is manifested by an effect in the surroundings
It is a path function
Sign Convention
The heat absorbed by the system is represented by q. When the system absorbs heat, that is, heat flows from the surroundings to the system, the internal energy of the system increases provided no work is done. Therefore, q is positive. When the system loses heat, that is, heat flows from the system to the surroundings, the internal energy of the system decreases provided no work is done. Therefore, q is negative. q is positive when heat is absorbed by the system q is negative when heat is lost by the system.
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8.3.3. Work done during expansion or compression of an ideal gas
Let the system consist of an ideal gas. The mechanical work on the gas can be done either reversibly or irreversibly. During mechanical work, the pressure and the volume of the gas change. These changes can take place under isothermal conditions or under adiabatic conditions. Therefore, the various processes can be
(i)
Isothermal reversible volume change
(ii)
Isothermal irreversible volume change
(iii)
Adiabatic reversible volume change
(iv)
Adiabatic irreversible volume change
Since work is a path function, it should depend upon the nature of process.
Thus, we shall derive the expressions for work done for all the four processes and it will be proved that the amount of work done in all the four cases will be different. A.
Work involved in reversible process
The reversible work of expansion or compression can be illustrated by the following set up.
An ideal gas is taken in a container fitted with a massless and frictionless piston. On the top of the piston a container with a large number of weights are kept. The weights exert pressure on the gas. This pressure is known as external pressure, Pex. Initially, the external pressure is balanced by the pressure exerted by the gas on the piston, that is, Pin.
Pex = Pin
This state is an equilibrium position. During expansion, the weight on the piston is decreased slightly. As a result the external pressure becomes infinitesimally smaller than Pin by the amount dP. The new external pressure will be P'ex
P'ex = Pin – dP
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That is, the pressure exerted by the gas is infinitesimally different from the new external pressure on the gas. The system is said to be infinitesimally removed from the equilibrium state. To maintain equilibrium, the gas expands infinitesimally and the piston moves up. If the process of decreasing the external pressure infinitesimally is continued, then after a large number of steps, there is a significant change in the volume of the gas. At any stage during the expansion, the internal pressure is practically equal to external pressure,
P'ex. So each step is an equilibrium step. The work done by the gas at each step is (– dw). According to Equation (2) dw is given by the expression,
– dw = P'exdV = (Pin – dP)dV ≈ Pin dV as dPdV is very small it is neglected. Pin keeps on changing after each stage of expansion. During compression the weight on the piston is increased slightly, and the external pressure increases slightly after each addition and becomes greater than the internal pressure. To maintain the equilibrium, the gas compresses and the piston moves down. When all the weights have been replaced, the gas comes back to the original position. The work done during each step of compression will be equal to dw.
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B.
Work involved in irreversible process
When the gas expands or compresses rapidly the process is irreversible in nature. During the expansion, the pressure on the gas is decreased by a finite amount. That is, the new external pressure is less than the pressure exerted by the gas on the piston,
P'ex < Pin
The equilibrium is disturbed and the gas expands without maintaining the equilibrium with the surroundings. The work done by the gas is given by–dw.
–dw = P'exdV
The gas expands against constant external pressure at any given stage of expansion. The total work done by the system (−w) is equal to
−w = P'ex∆V or w = −P'ex∆V
Similarly, during compression, the external pressure on the gas is increased such that the new external pressure is much greater than the pressure exerted by the gas on the piston. So, the gas compresses without maintaining the equilibrium with the surroundings.
The reversible as well as irreversible work done can take place under isothermal as well as adiabatic conditions depending upon the boundary of the container. C.
Work Done During Isothermal Volume Change for an Ideal Gas
During an isothermal process the temperature of the system remains constant.
The system may exchange heat with the surroundings to maintain the temperature. The amount of the gas remains constant.
Reversible process: The process is carried out in an infinite number of steps.
The work done in a step is given by the equation dw = –PexdV
Here, Pex is the external pressure on the gas.
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Let n be the amount of an ideal gas at temperature T. Let the initial and the final states of the gas are as shown:
Initial state
Final state
Pressure
P1
P2
Volume
V1
V2
In the reversible volume change the external pressure differs from the internal pressure by the expression
Pex = (P ± dP). where, +and−signs are meant for the compression and expansion, respectively and P is the internal pressure, i.e., the pressure of the gas.
Substituting this in the expression of dw, we get, dw = −(P ± dP)dV = –PdV −(± dPdV)
Since dPdV is a product of two differentials it may be neglected. Hence, dw = –PdV
Total work done during the process is obtained by integrating the above equation V2
w
dw
PdV
….(3)
V1
Since the pressure is also changing during the process, P cannot be taken out of the integral. However, for an ideal gas,
P = nRT / V w isoth rev ….(4)
V2
(ideal)
V
nRT nV V2
1
nRT dV V
V1
nRTn
2.303nRTog
V2
nRT
dV
V
V1
V2
V1
V2
V1
….(5)
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Note: During the expansion process, V2 > V1. Therefore, log (V2/ V1) is greater than zero. Since n, R and T are always positive; the work done by the ideal gas during isothermal reversible expansion is always negative. During the compression process, V2 < V1. Therefore, log (V2/ V1) is less than zero and the work done is positive.
At a constant temperature for an ideal gas, P1V1 = P2V2, therefore, Equation (5) in terms of pressure becomes
wisoth rev (ideal)
2.303nRTlog
P1
P2
….(6)
Graphical representation of isothermal reversible expansion of a gas
Figure 8.2a illustrates the magnitude of the work involved in one step in which the volume change (= volume of gas at point d – volume of gas at point c) is infinitesimally small against an external pressure of P1. It is obvious that this work is equal to the area of trapezoid abcd. Similarly, the magnitude of the work done by the gas during next step will be the area of the adjacent trapezoid. Thus the total magnitude of the work done by the gas during various steps of a reversible process will be equal to the sum of the areas of all the trapezoids from the initial state A to the final state B (Figure 8.2b). This is equal to the area of the region ABCD.
Work done by an ideal gas under reversible isothermal conditions
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Graphical Representation of Isothermal Reversible Compression
During isothermal compression, the pressure is increased slowly from P2 to P1 in an infinitesimal number of steps and at each stage the volume decreases slightly. According to the Figure 8.3, the change starts from point A (P2, V2) and goes till the point B (P1, V1). The total work done on the gas is equal to the area of the region ABCD.
Figure 8.3: Pressure – volume work done during reversible isothermal compression. Comparison of Isothermal Reversible Work During Expansion and
Compression
On comparing Figures 8.2b and 8.3 it is seen that the magnitude of the work done is same if P1, P2 and V1, V2 in the two figures are same. However, the sign of work done is different.
Work done during isothermal reversible expansion = w = –2.303nRT log
V2
V1
Work done during isothermal reversible compression = w = –2.303nRT log
V2
V1
Note: The work done on an ideal gas to bring it back to its initial state is equal to the work done by the gas during expansion for an isothermal reversible process. Therefore, in a cyclic process from A to B and back from B to A, the total work done is zero.
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Irreversible volume change
In this case pressure may be changed in one step or more than one step. If the volume is changed from V1 to V2 in one step, then the total work done is given by the expression
V2
w
V1
PexdV
If the gas is allowed to expand unrestrictedly, then the external pressure is equal to the final pressure of the gas after expansion. However, the expansion can be done against the external pressure smaller than the final pressure of the gas and the desired expansion may be achieved by a set of stops. If Pex = P2 then the expression of work is given by
V2
w
V1
....(7)
P2 dV
Since P2 is constant we get w isoth irr V2
(ideal)
P2 dV
P2 (V2
....(8)
V1 )
V1
Note: For expansion V2 > V1 and w is negative indicating that the work is done by the gas. For compression V2 < V1 and w is positive indicating that the work is done on the gas.
Since for an ideal gas, V2 = nRT /P2 and V1 = nRT/P1, the expression of work in terms of pressure is
w isoth irr (ideal)
P2
nRT
P2
nRT
P1
nRT 1
P2
P1
….(9)
Graphical Representation of Isothermal Irreversible Expansion
During an irreversible process, the external pressure is decreased from P1 to P2 and the gas expands from V1 to V2. The total work done is equal to –P2(V2 – V1).
From Figure 8.4, (V2 – V1) = CD and P2 = DE.
Therefore, the magnitude of the work done will be equal to w = ED
CD = area of the region BCDE
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Figure 8.4: Pressure – volume work done by ideal gas under irreversible isothermal conditions
Irreversible Expansion When the Expansion of the Gas is Restricted
In this case, the pressure on the gas is decreased by a finite amount and then the gas is allowed to expand isothermally under an irreversible condition.
However, the expansion is stopped by blocking the movement of the piston with the stoppers as shown in Figure 8.5. As a result, the final pressure of the gas is not equal to the external pressure exerted on the gas. Here, final pressure,
P2 is greater than external pressure. The final volume (V'2) will be less than the situation when the expansion of gas is not stopped (V2).
Figure 8.5: Irreversible expansion when P2 > Pex
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The gas expands against the external pressure. Therefore, the work done by the gas will be equal to w isoth irr (ideal)
V2
Pex dV
Pex (V2
V1 )
V1
It is important to note that the expression of work done during the expansion involves the external pressure and not the final pressure of the gas. However, during the unrestricted expansion the final pressure is equal to the external pressure. The work done by the gas during restricted expansion will be less as compared to the case when the gas is allowed to undergo unrestricted expansion. Graphical Representation of Isothermal Irreversible Compression
The work done on the gas during isothermal irreversible compression is given by the expression, P1(V1–V2). From the Figure 8.6, P1=C'D' and
(V1– V2) = A'D'.
Therefore, the work done on the gas is equal to w = C'D'
A'D' = area of the region A'B'C'D'.
Figure 8.6: Pressure–volume work done during irreversible isothermal compression. 29
Comparison of Isothermal Irreversible Work During Expansion And
Compression
The gas expands from State I to State II irreversibly and then it is compressed irreversibly to State III.
State I (P1,V1, T1 )
isothermal expansion
State II (P2,V2, T1)
isothermal compression
State III (P1,V1, T1 )
As can be seen, State I and State III are identical. Therefore, when the process is complete the system comes back to the initial state. On comparing the magnitude of the work done during irreversible expansion and irreversible compression, it can be seen that the work done on the gas to bring it back to the initial state is different from the work done by the gas during expansion.
From Figures 8.4 and 8.6 area of the region A'B'C'D' > area of the region BCDE
The work done on the gas is greater than the work done by the gas during irreversible process. Comparison of Work Done During Reversible and Irreversible
Expansion
The work done by the gas is the useful work that can be obtained from the gas.
The magnitude of the work done by the gas has been determined graphically in
Figures 2b and 4. On comparing the two figures it is observed that area of the region ABCD > area of the region BCDE.
Therefore, the work done by the gas during reversible expansion is more than work done by the gas during an irreversible expansion.
Work Done at a Constant Volume (Isochoric Process)
When the volume is constant ΔV= 0. The process is isochoric; the gas neither expands nor contracts. The gas does not do any mechanical work.
Work Done At Zero External Pressure
The external pressure is zero if the gas is surrounded by vacuum. Under such conditions the gas undergoes expansion into vacuum. This expansion is known
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as free expansion. Since dw = –Pext dV and Pext = 0, therefore, dw = 0. No work is done by the gas during free expansion.
8.4
FIRST LAW OF THERMODYNAMICS
8.4.1. Laws of Thermodynamics
The laws of thermodynamics are based upon observations. They were numbered in the order in which they were stated. The last law to be given was the zeroth law. Since it is the most fundamental, it should come before first law
(as zero comes before one) it was named zeroth law.
The change in the energy of the system is related to the work done and the heat exchanged between the system and its surroundings. The first law of thermodynamics deals with the relation between the change in the energy and work and heat. Therefore, an understanding of these terms is important before studying the first law of thermodynamics.
8.4.2. Internal Energy (U)
The total energy of the system is due to energy possessed by the constituents of the system such as atoms, ions and molecules as well as their motion and position. The internal energy of the system does not include the kinetic energy due to motion of the system and potential energy due to position of the system.
Energy due to the position of the system is its potential energy and energy due to motion of the system is kinetic energy. These two types of energies constitute external energy of the system.
The energy possessed by all the constituents of the system is the internal energy. It is represented by the symbol U.
Internal energy is the total energy of all the molecules. It includes translational energy (Ut), vibrational energy (Uv), rotational energy (Ur), electronic energy
(Ue), and intermolecular interaction energy (VT)
U = Ut + Uv + Ur + Ue + VT
Internal energy is also known as intrinsic energy.
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Total energy = External energy + Internal energy of system
The SI unit of energy is Joule (J).
1 J = 1 kg m2 s–2 = 107 erg
Other units: calories (cal)
1 cal = 4.184 J electron volt (eV)
1 eV = 1.602 x 10–19 J
1 L atm = 1.013 x 102 J
1 Watt hour = 3.6 x 103 J
In thermodynamic studies, the emphasis is not on the individual energy of the molecules, but the overall contribution of all the molecules to the energy of the system. Therefore, thermodynamics generally deals with the energy at the macroscopic level and not at the microscopic level.
Internal energy is a state function. It depends upon temperature (T), pressure
(p), volume (V) and the amount (number of moles present in the system, n).
Characteristics of Internal Energy
Internal energy is represented by U
The SI unit of U is Joule.
U is an extensive property.
U is a state function
The change in U is equal to the difference in internal energies of the final state and initial state of the system. That is, ΔU = U2 –U1. Here, Δ stands for change.
In a cyclic process, there is no net change in internal energy.
Internal energy of the system can be changed when system exchanges heat with its surroundings or when work is done on the system or by the
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system. The internal energy can also change when the system undergoes physical or chemical process.
The experiments show that the internal energy of the system increases when the work is done on the system or when the heat is absorbed by the system. 8.4.3. Law of Conservation of Energy and First Law of Thermodynamics
According to the law of conservation of energy, energy can neither be created nor destroyed. If it disappears from one form it must appear in another form. This statement is true for those processes also that involve work done and heat transfer. According to the definitions of heat and work done discussed earlier, both heat and work done affect the internal energy of the system. However, they appear only during the change. They are not forms of energy. They are the manner in which energy manifests or shows itself during a change.
When the energy is supplied from outside by heating the system, a part of it is used to do work and the rest remains with the system. Since the total energy should be conserved during any change, the change in internal energy of a system must be related to heat exchanged by the system and work done on/by the system. That is,
ΔU = q + w ….for finite change
….(10)
where ΔU = U2–U1. U1 is the internal energy of initial state and U2 is the internal energy of the final state. q is the heat absorbed by the system and w is the work done on the system
When the process involves infinitesimal change dU = dq +dw
….(11)
Equations (10) and (11) are the respective mathematical forms of the first law of thermodynamics for finite change and infinitesimal change.
In Equation (12), dq and dw do not represent exact differentials. That is, dq ≠ q2 – q1. This is due to the fact that the heat of the initial state and heat of the final state has no meaning. dq represents a small amount of heat absorbed by the system. Similarly, dw is the small amount of work done on the system during the change.
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When only mechanical work, i.e., work of expansion or compression is done then dw = – pdV. Therefore, Equation (11) becomes, dU = dq – pdV
….(12)
Solved Example: An ideal gas absorbs 500 J of heat and its internal energy increases by 260 J. On the basis of first law of thermodynamics explain what happened to rest of the heat.
Solution: Heat absorbed by the system, q = 500 J
The change in internal energy, ∆U = 260 J
According to the first law of thermodynamics
ΔU = q + w
or
w = ΔU – q
w = 260 J – 500 J = – 240 J
Since sign of the value of w is negative, 240 J of work is done by the system. So out of the 500 J of heat absorbed, 240 J was used to do work by the gas and 260 J was used to increase the internal energy of the gas.
8.4.4. Internal energy and heat
When no work is done and the system does not undergo any phase transformation then the heat transfer between the system and its surroundings results in the change in the temperature of the system. As a result, the volume and pressure will also change. The conditions can be maintained in such a way that either the volume or the pressure is kept constant. Heat exchange is also accompanied by the change in energy of the system. This energy change will be different for the two conditions, that is when volume is kept constant and when the pressure is kept constant.
Heat exchange at constant volume
Since, ΔU = q + w
= q –PΔV
When the volume change is constant ΔV = 0, Hence, PΔV is also zero. The heat exchanged, q can be written as qv. Therefore,
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ΔU = qv
This implies that, the change in internal energy of a system is equal to heat exchange between the system and its surroundings at constant volume.
Since internal energy is a state function heat exchange at constant volume is also a state function.
8.4.5. Enthalpy
In the previous section heat exchange at constant volume has been related to the change in internal energy. However, in the actual practice, the experiments are carried out at constant pressure in the laboratory. The energy changes are measured in terms of heat exchange taking place between the system and its surroundings. Since the heat exchange takes place at constant pressure it cannot be equal to the change in internal energy. Therefore, a new energy term is needed that will account for the heat exchange at constant pressure. This new energy term is called enthalpy and the symbol is H. It was earlier also known as heat content. The symbol H is taken from heat content. H is related to the internal energy, pressure and volume by the equation
H = U + PV
….(13)
Since U, P and V are state functions U + PV must also be a state function. Since
U and PV have dimensions of energy, it must also have dimensions of energy.
The change in enthalpy is given by dH. Differential of Equation (13) gives dH = d(U+ PV)
From the first law of thermodynamics, dq = dU + PdV
Therefore,
dH = dq +VdP
....(14)
Heat exchange at constant pressure
When the pressure is constant, dP = 0,
VdP = 0 and dq = dqp.
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Therefore, dH = dqp.
....(15)
For finite change,
∆H = qp where, ∆H = H2 – H1
Therefore, the heat exchanged at constant pressure is equal to the change in enthalpy of the system. Whenever the system exchanges heat at constant pressure (qp), there will be a change in temperature of the system. Some part of heat, qp will be used to increase the internal energy and rest is used up for the change in volume. Hence, change in enthalpy results in change in internal energy as well as change in volume of the system. For solids and liquids the volume change is negligible, therefore, enthalpy and internal energy changes are practically same.
Relationship between ∆H and ∆U for an Ideal Gas
We know, H = U + PV
∆H = ∆U + ∆PV
For an ideal gas, PV = nRT
Therefore,
∆H = ∆U + ∆(nRT)
At constant temperature,
∆H = ∆U + (∆n)RT
Here, ∆n is the change in the amount of the gas. If the temperature changes but the amount of gas is constant then,
∆H = ∆U +nR∆T
Characteristics of Enthalpy
The enthalpy change is equal to the heat exchanged by the system at constant pressure.
It is a state function.
H = U +PV
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For solids and liquids, the enthalpy change (ΔH) is taken as equal to the change in the internal energy (ΔU).
Its units are same as the units of energy.
When the system absorbs heat at a constant pressure, ΔH is positive and the process is known as an endothermic process.
When the system loses heat to the surroundings at a constant pressure,
ΔH is negative and the process is known as an exothermic process.
Since enthalpy is a state function, it depends upon T, V, P and the amount of the system (n).
8.4.6. Heat capacity
Heat capacity is defined as the amount of heat required in order to achieve a unit rise in the temperature of a substance.
C = dq/dT
Unit of heat capacity
heat capacity
heat exchanged temperature change
J
K
SI unit is J K-1.
Heat capacity is an extensive property as it depends upon the size of the system.
Specific heat capacity: It is defined as the heat needed to cause a unit rise in the temperature of a unit mass of a substance. It is denoted by s. Its SI unit is J
K-1 kg-1.
Molar heat capacity: It is defined as the heat needed to cause a unit rise in the temperature of one mole of a substance. Molar heat capacity is denoted by Cm.
Thus,
Molar heat capacity =
Heat capacity of the substance
Amount of the substance
Cm =
C n The SI unit of molar heat capacity is J K-1mol-1.
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Specific heat capacity and molar heat capacity are intensive properties.
When the heat is absorbed at a constant pressure, the heat capacity is known as heat capacity at constant pressure, Cp. It is also known as an isobaric heat capacity. When heat is absorbed at constant volume the heat capacity is known as heat capacity at constant volume, CV. It is also known as isochoric heat capacity.
Relationship Between Cp And Cv For An Ideal Gas
Since, H = U + PV
Therefore, ∆H = ∆U +∆(PV)
For an ideal gas, PV = n RT
Therefore,
∆H = ∆U + ∆(nRT)
Or, ∆H = ∆U + nR ∆T
Using the relations of ∆H and ∆U in terms of specific heat,
Cp∆T = Cv∆T + nR ∆T
Or, Cp = CV + nR
Or, Cp −CV = nR
Dividing the equation by n we get,
Cp,m – CV,m = R
Here, Cp,m = Cp/n and CV,m = CV/n.
8.4.7. Measurement of ∆H and ∆U through Calorimetry
Calorimetry is the measurement of heat. A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured. Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up the surrounding air, which expands and escapes through a tube that leads the air out of the
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calorimeter. When the air is escaping through the copper tube it will also heat up the water outside the tube. The temperature of the water allows for calculating calorie content of the fuel.
Figure 8.7: Set up of a Bomb calorimeter
The bomb, with the known mass of the sample and oxygen, form a closed system - no air escapes during the reaction. The weighted reactant put inside the steel container is then ignited. Energy is released by the combustion and heat flow from this crosses the stainless steel wall, thus raising the temperature of the steel bomb, its contents, and the surrounding water jacket. The temperature change in the water is then accurately measured with a thermometer. This reading is used to calculate the energy given out by the sample burn. After the temperature rise has been measured, the excess pressure in the bomb is released.
Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings) and ignition circuit connected to the bomb. By using stainless steel for the bomb, the reaction will occur with no volume change observed (see figure 8.7).
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Since there is no heat exchange between the calorimeter and surroundings, q = 0 (adiabatic); no work performed, w = 0.
Thus, the total internal energy change ΔU(total) = q + w = 0
Also, total internal energy change ΔU(total) = ΔU(system) + ΔU(surroundings)
=0
Thus, ΔU (system) = - ΔU(surroundings) = - Cv ΔT (since in the entire process volume is constant) where Cv = heat capacity of the bomb
Similarly, when ΔH is to be measured, the heat evolved or absorbed during a reaction is measured, and is known as, heat of reaction or enthalpy of a reaction
ΔrH. In an exothermic process, heat is evolved, and the system loses heat to the surroundings. Thus, qp will be negative and ΔrH will also be negative. On the contrary, in an endothermic process, heat is absorbed, qp will be positive and hence ΔrH will also be positive.
Measurement of Change in Internal energy of the system (∆U)
The change in the internal energy during a chemical reaction is measured using a bomb calorimeter. The reaction is carried out in a container whose volume remains constant. This container is made up of a metal, which is generally steel lined with gold. It is known as the bomb (Figure 8.7). In this experiment, a known mass of the reaction materials are taken in the bomb. The reaction material is the system. The bomb is immersed in a vessel containing a known mass of water. The calorimeter and the water act as its surroundings. The water is stirred by the rotating blades and its initial temperature (T1) is recorded.
As the reaction takes place, there is a change in the chemical energy. If the energy is released during a process, it is transferred to the surroundings as heat. This heat is used to warm water and the calorimeter. When the heat is absorbed during the reaction, the calorimeter gives the heat to the system, so the temperature of the calorimeter decreases. When the temperature of water reaches a constant value, the final reading of thermometer (T2) is noted.
The change in temperature of water and the calorimeter (surroundings)
= ∆T = T2 – T1
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The heat exchanged between the system (reaction) and its surroundings is calculated from the equation q (water + calorimeter) = (m
s + C)
ΔT = q (surroundings)
where, m = mass of water, s = specific heat capacity of water and C = heat capacity of the calorimeter.
The heat liberated in the reaction is calculated as the heat appearing in the surroundings. During the process the volume remains constant so no work is done. Thus,
∆U = q (system) = – q (surroundings) at constant volume
Difficulties in the Experimental Determination of ΔU
If a large quantity of heat is liberated in a reaction then there is a danger of explosion since the experiment is performed at constant volume in a bomb calorimeter. Why? Because the heat liberated during the reaction will make the gaseous products to expand. Since the vessel is closed, the gas cannot expand.
This will cause a build up of high pressure inside the bomb calorimeter. If the walls of the vessel are not strong enough to withstand the high pressure developed, an explosion may take place.
Measurement of Change in Enthalpy of the System (∆H)
In the laboratories since it is easier to maintain a constant pressure as compared to the constant volume, most of the thermochemistry experiments are carried out under constant pressure conditions. The heat exchange is measured using a normal calorimeter that has adiabatic walls. The simple calorimeter is a glass vessel, usually a glass beaker which is well insulated. It is fitted with a glass stirrer and a thermometer. The set up is kept inside a Dewar flask or thermos flask for insulation. The glass vessel (beaker) is kept inside it on a rubber cork.
The reactants are mixed in it and the change in the temperature is measured, using a 0.1oC thermometer. For accurate work, more sensitive devices such as a
Beckmann thermometer, platinum resistance thermometer or thermocouple may be used. The experimental set up is shown in Figure 2. Sometimes the
Dewar flask itself is used as the vessel in which the reaction is carried out. This set up is commonly used in the laboratory for determination of enthalpy reactions involving solutions.
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The enthalpy change is determined by measuring the temperature change before and after the process. The heat gained or lost by the system is manifested by the change in temperature of the solution and the container which are assumed to be the surroundings. It is also assumed that no heat is lost during the exchange of heat between the system and the surroundings.
Therefore,
qsystem = − qsurroundings
During the process if the system gains heat then the surroundings will lose heat and the final temperature recorded by the thermometer will be less than the initial temperature. Similarly, if the system loses heat then the heat lost is gained by the surroundings and the temperature rises. qsurroundings = heat exchanged by the solution + heat exchanged by the calorimeter. Since the process takes place at a constant pressure, the heat exchanged is equal to the enthalpy change. Therefore,
∆Hsystem = − ∆Hsurrounding
∆Hsurrounding = ∆Hsolution + ∆Hcalorimeter
At a constant pressure
∆H = Cp ∆T , here ∆T = T2 −T1
∆Hsolution = Cp (solution)
∆T = m
s
∆T
where, m is the mass of the solution and s is the specific heat capacity of the solution. If the solution is aqueous solution, s is taken as equal to the specific heat capacity of water. For the calorimeter,
∆Hcalorimeter = Cp (calorimeter)
∆T
Therefore, the ∆Hsurrounding becomes equal to
∆Hsurrounding = Cp (solution)
∆T + Cp (calorimeter)
∆T
Since the temperature change is same for the solution and calorimeter,
∆Hsurrounding = {Cp (solution) + Cp (calorimeter)} or ∆Hsystem = −{Cp (solution) + Cp (calorimeter)}
42
∆T
∆T
....(16)
8.4.8.
Heat exchange during isothermal process
According to the first law of thermodynamics dU = dq + dw
For an isothermal process involving an ideal gas, dU = 0. Therefore, dq = –dw and hence, q = – w
For an isothermal reversible volume change for an ideal gas
wisoth rev (ideal)
2.303nRTlog
q = 2.303nRTlog
V2
V1
V2
V1
….(17)
For an isothermal irreversible volume change for an ideal gas, w isoth irr (ideal)
V2
P2 dV
P2 (V2
V1 )
V1
q = P2 (V2
V1 )
….(18)
Here it is assumed that the external pressure on the gas is equal to the final pressure of the gas.
Conclusion
For a reversible isothermal volume change of an ideal gas,
w isoth rev 2.303nRTlog
q = 2.303nRTlog
V2
V1
V2
V
= 2.303nRTlog 1
V1
V2
dU = 0 or ΔU = 0 dH = 0 or ΔH = 0
For an irreversible isothermal volume change of an ideal gas,
43
w isoth irr P2 (V2
V1 )
q = P2(V2 −V1) dU = 0 or ΔU = 0 dH = 0 or ΔH = 0
8.4.9.
Work done during adiabatic volume change for an ideal gas
The process during which there is no exchange of heat between the system and its surrounding is known as an adiabatic process. For a finite change, q = 0
Therefore, from the First law of thermodynamics
ΔU = w
....(19)
For an infinitesimal change, dq = 0, therefore, dU = dw = –PdV
Since dw = dU, the work in adiabatic process is accompanied with the change in energy (hence its temperature) of the system. Let n be the amount of an ideal gas. Let the initial and the final state of the gas be shown as the following
Initial state
Final state
Pressure
P1
P2
Volume
V1
V2
Temperature
T1
T2
Reversible process: The gas expands in an infinite number of steps. The work done during any step of the process is given by dw = –PexdV
….(20)
Hence,
V2
w
dw
Pex dV
V1
For an ideal gas, P is replaced by
nRT
. Therefore,
V
44
w
adiab rev V2
(ideal)
nRT dV V
V1
Since, both T and V change, w cannot be calculated directly by integration.
However, this can be determined by using the expression, dw= dU. dU = CV dT
….(21)
On integrating Equation (21) we get
ΔU
ΔU
U2
U1
Cv
dU
T2
T1
dT
T2
T1
C v dT
C v (T2
T1 )
Hence, the expression for the work is
wadiab
C v (T2 rev (ideal)
T1 )
Note: For the expansion, w is negative, and thus, T2 < T1, i.e., the expansion is accompanied with the decrease in temperature. For compression the reverse is observed. For calculating the enthalpy change, the equation H = U+PV is considered. At a constant pressure, for a finite process the enthalpy change is given by the equation ΔH = ΔU + PΔV = CV ΔT + nRΔT
ΔH = (CV + nR) ΔT
For an ideal gas (CV + nR) = Cp, therefore,
ΔH = CpΔT
....(22)
Process Involving Change in the State of the System
The change in the state of the system can be vaporization, fusion, sublimation and solid-solid transition. These processes take place at constant temperature, therefore, they are isothermal processes. However, the change in internal energy (∆U) and the change in enthalpy (∆H) are not zero as during the change in state of the system internal energy is involved.
The work done during the phase change = w = −P∆V
45
When the state change involves the vapour state then the change in volume is appreciable. Hence the work done is appreciable. However, when only condensed states are involved in the process then the change in volume is very small and hence the work done is negligible.
The heat exchanged during the state change is equal to heat of transition. For solid → liquid, q is equal to heat of fusion. For liquid → vapour, q is equal to heat of vapourization. Since the change in state takes place at constant pressure, the heat exchanged is equal to the enthalpy change for the process. That is, q = qp =∆H
∆U = q + w
∆U = q −P∆V
The change in thermodynamic properties when an ideal gas undergoes volume changes are summarized as follows:
Process
work done(w) Isothermal reversible volume change V nRTn 2
V1
Isothermal irreversible volume change – P2(V2– V1)
heat(q)
change in internal energy(ΔU)
q=–w
V
nRTn 2
V1
q = P2 (V2
V1 )
change in enthalpy (ΔH)
0
0
0
0
Solved Example: One mole of an ideal monoatomic gas at 300 K and 1atm is allowed to expand to the final pressure of 0.1 atm under the conditions:
(i) isothermal reversible (ii) isothermal irreversible against external pressure of
0.1 atm. Calculate work done for each process. R = 8.314 J K-1 mol-1
Solution:
The given data are :
n = 1 mol;
P1 = 1.0 atm
P2 = 0.1 atm
46
T1 = 300 K;
(i)
isothermal reversible process
T1 = T2 = 300 K
wisoth rev 2.303nRTlog
P1
P2
2.303(1mol) (8.314 J K 1mol 1 )(300 K) log
(ii)
1.0 atm
0.1 atm
5744.14 J
Isothermal irreversible process
w isoth irr nRT
isoth wirr P2
P1
1
(1 mol)(8.314 J K 1 mol 1 )(300K)
0.1atm
1
1.0atm
2244.78 J
Solved Example: One mole of an ideal gas is heated at constant pressure from
0°C to 100°C. Calculate w, q, ΔU and ΔH. Given Cp,m = 3.5 R.
Solution: On heating the gas will expand against constant pressure. Therefore, the process is irreversible.
The given data are T1 = 0°C = 273 K ; T2 = 100°C = 373 K ; n = 1mol
The expression of work is w = −Pex(V2 − V1)
For an ideal gas V = nRT / P. Hence w = −Pex {(nRT2/P2) − (nRT1/P1)}
Since the expansion occurs at constant pressure, Pex = P1 = P2. Hence, w = – nR(T2 – T1) = –(1 mol )( 8.314 J K-1mol-1)(373 K – 273 K ) = – 831.4 J
∆U = nCV,m(T2 –T1)
Cp,m – Cv,m = R
Cv,m = Cp,m – R = 3.5 R –R = 2.5 R
∆U = 1 2.5
8.314 JK-1mol-1 (373 K–273 K) = 2078.5 J
q = ∆U–w = 2078.5 J–(–831.4 J) = 2909.9 J
∆H= nCp,m(T2–T1) = (1 mol)(3.5
8.314 J K-1 mol-1)( 373 K–273 K ) = 2909.9 J
47
Solved Example: One mole of an ideal gas is held by a piston at 273 K and under a pressure of 10 kPa. The pressure is suddenly released to 0. 4 kPa and the gas is allowed to expand isothermally. Calculate q, w, ∆U and ∆H for the process. Hint: The change is irreversible.
Solution: Since the gas is expanding against constant pressure the process is irreversible in nature. The work done by an ideal gas under isothermal irreversible conditions is given by the expression
w isoth irr P2 (V2
V1 )
V1 = nRT/P1 and V2 = nRT/P2
w isoth irr nRT
P2
P1
1
The data given are: T = 273 K ; P1 = 10 kPa = 104 N m-2
P2 = 0.4 kPa = 0.4 103 N m-2 ; n = 1mol
w isoth irr (1 mol) (8.314 J K 1 mol 1 ) (273 K)
0.4 10 3 N m
10 4 N m 2
2
1
= –2178.93 J
For an ideal gas under isothermal conditions ∆U = 0 and ∆H = 0
According to first law of thermodynamics
∆U = q + w = 0 or q = –w = 2178.93 J
Thermochemistry is a part of thermodynamics that deals with the changes in energy during a chemical reaction or a physical transformation. Generally, these changes are mentioned in terms of the heat exchange during the process.
Heat is the energy in transit and it flows from a system at a higher temperature to one at lower temperature. It is also defined as the energy transferred across the boundary between a system and its surroundings by the virtue of the temperature difference. By convention, the heat absorbed or evolved during a process is called the heat exchange during the process. Heat exchanges can also take place during the physical or the chemical processes such as the change of state, a chemical reaction or even a simple process such as dissolution. Heat
48
and energy are used interchangeably because heat exchange at a constant volume is equal to the change in the internal energy and the heat exchange at a constant pressure is known as the change in enthalpy. Thus, qv = ΔU and qp = ΔH
Thermochemistry enables us to predict the amount of heat that would be evolved or absorbed in a process without actually performing a tedious set of experiments in the laboratory. We can also calculate the energy changes for the processes that are not feasible experimentally. This is done by taking advantage of the fact that although heat is a path function, the heat exchanges become state functions when the measurements are made under the conditions of constant volume or constant pressure. Under these conditions they are dependent only on the initial and final states of the system and hence can be added, subtracted or multiplied by an integer according to the algebraic laws.
Sign Convention
If the heat is absorbed by the system the process is known as an endothermic process. Since the energy of the system increases when it absorbs heat, ΔU and
ΔH are given positive sign.
If the system loses heat the process is known as an exothermic process. ΔU and
ΔH have negative sign since the energy of the system decreases.
8.5.1. Standard state
The standard state for a substance is defined for the substance in its pure state and the most stable form of that substance at a standard pressure of one bar and a specified temperature. Although the standard state definition is valid for all temperatures, conventionally the data is mentioned for 298.15 K. However, in most cases the temperature for which the standard state is defined is mentioned in the brackets.
For gases the standard state is the one at which pure gas behaves ideally at one bar and a specified temperature. For liquids the standard state is the pure liquid at one bar and specified temperature.
49
For solid it is the pure and most stable crystalline form of aggregation of solid at one bar and specified temperature. For example, graphite is taken as the stable form of carbon and rhombic form is the most stable form for sulphur.
For solution it is one mol dm−3 solution at one bar and the specified temperature. The standard states can be summarized as follows:
S.No.
Phase of the substance Condition
1.
Gas
Pure gas at 1 bar and a specified temperature
2.
Liquid
Pure liquid at 1 bar and a specified temperature
3.
Solid
Pure and stable form of solid at 1 bar and a specified temperature 4.
Solution
One mol dm−3 solution at 1 bar and a specified temperature 8.5.2. Heat of Reaction
Heat of a reaction is defined as the heat exchanged between the system and the surroundings during the complete transformation of reactants to products at a constant temperature and pressure for the given stoichiometrically balanced equation. During a chemical process some bonds break and others are formed so there is a net change in the energy of the system. This results in the heat exchange during a process. This heat exchange can be measured if the energy of initial and final state is known. The heat exchanges are measured either at a constant pressure or at a constant volume.
When the reaction is carried out at a constant volume, the heat exchange is measured in terms of the internal energy change. It is known as the heat of reaction at constant volume, qv or energy of reaction. The symbol is ΔrU. The energy change is given by the equation
ΔrU = Ufinal −Uinitial
When the reaction is carried out at a constant pressure, the heat exchange is measured in terms of the enthalpy change. It is known as heat of reaction at constant pressure, qp or enthalpy of reaction. The symbol is ΔrH.
50
The enthalpy of reaction is given by the equation
ΔrH = Hfinal −Hinitial
Since the measurements are generally made under constant pressure conditions, the heat of reaction is normally referred as enthalpy of reaction.
The enthalpy of reaction is the enthalpy change during a reaction. Since enthalpy is a state function, the change in enthalpy is equal to the difference in the enthalpies of the final state and the initial state. For a general reaction νAA + νBB → νCC + νDD
The reactants are the initial state and the products form the final state assuming that the reaction has been completed. Therefore,
ΔrH = Total enthalpy of products – total enthalpy of reactants
=
H(products)
H(reactant s )
ΔrH = (νCHm,C + νD Hm,D) –(νA Hm,A + νBHm,B)
Here, Hm,A = molar enthalpy of A
Hm,B = molar enthalpy of B
Hm,C = molar enthalpy of C
Hm,D = molar enthalpy of D νA, νB, νC and νD are stoichiometric coefficients of A, B, C and D, respectively in the balanced chemical equation.
For example, during the formation of CO2, one mole of carbon in its solid state reacts with one mole of oxygen gas to form one mole of carbon dioxide. The heat produced at constant pressure and 298 K is 393.5 kJ. This information is represented as,
C(s) + O2(g) → CO2(g)
ΔrH(298 K,1 atm) = −393.5 kJ mol−1
When no pressure is mentioned, it is assumed that the reaction is taking place at atmospheric pressure, that is, 1atm.
When the reactants and the products are in their respective standard states, the enthalpy of reaction is known as standard enthalpy of reaction ΔrHө . If the
51
reaction takes place at 298 K the standard enthalpy of reaction is written as
ΔrHө (298 K)
ΔrHө (298 K) = ΔrH( 298 K, 1 bar)
While defining the heat of a reaction, the nature of the reaction has not been mentioned. Therefore, the term can be used for all the types of reactions.
However, these reactions can be classified into various categories and for each category the heat exchange is given a specific name. Since, these heat exchanges are generally studied at a constant pressure, these will be referred to in terms of enthalpy changes in this chapter. The reactions involving the transformations are reported for 1 mol of the substance to standardize the values of the enthalpy changes.
(a)
Enthalpy of Formation
There are many ways in which a substance can be formed and for each such reaction the heat of reaction is different. In many of the reactions more than one product is formed. The values of the heat of reactions for all the reactions in which a given substance is formed cannot be taken as the heat of formation of that substance. In order to define a single value of heat of formation, the chosen reaction should be such that only one mole of the substance is formed from its constituent elements and no other product is formed. So the enthalpy of formation can be defined as the amount of heat exchanged at constant temperature and pressure during the formation of one mole of the substance from its constituent elements. It is represented by ΔfH. Its unit is kJ mol−1.
For example, the enthalpy of formation of CO2 is equal to the enthalpy change for the following reaction
C(s) + O2(g) → CO2(g)
ΔfH = −393.5 kJ mol−1
Similarly, the enthalpy of formation of water is equal to the enthalpy change for the reaction
H2(g) + ½O2(g) → H2O(ℓ)
ΔfH = −285.8 kJ mol−1
Note that although water is formed during the neutralization process also, it is not taken as the formation reaction of water as it is not formed from its elements in this reaction.
52
Standard enthalpy of formation is defined as the amount of heat exchanged at a given temperature and a pressure of one bar during the formation of one mole of the substance in its standard state from its constituent elements in their respective standard states. It is represented by ΔfHө .
Enthalpy of Formation of an Element
By convention the enthalpy of formation of an element in its standard state is taken as zero at all temperatures. Hence, whether the element is a solid, a liquid or a gas its standard enthalpy of formation is taken as zero. For example, hydrogen gas, solid zinc, rhombic sulphur, liquid bromine and graphite all have ΔfHө (298 K) = 0.
While dealing with the ionic species in the solution, the standard enthalpy of formation of H+ ions is also taken as zero.
Hess’s Law of Constant heat Summation
The total enthalpy change of a reaction is the same regardless of whether the reaction is completed in one step or in several steps. For example for a change from reactant to product that can take place in four steps or a single step, the total enthalpy change will be same.
Single step process
Reactant → Product
ΔH
Multiple step process
Reactant →A
ΔH1
A →B
ΔH2
B→C
ΔH3
C→ Product
ΔH4
According to Hess‟s law
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
This can also be shown by the following diagram
53
Diagram illustrating Hess’s law
The following important rules are derived from thermochemistry
We can reverse any equation and the sign of ΔH is also reversed.
The chemical equations can be added or subtracted. The ΔH of the resulting equation is sum or difference of the ΔH of the chemical equations.
The chemical equation can be multiplied or divided by an integer and the
ΔH is also multiplied or divided by that integer.
The identical terms on both the sides of the equation are cancelled.
These laws help us to theoretically calculate the enthalpy changes of processes which cannot be easily carried out in a laboratory. For example, to determine the enthalpy change for the process
C(graphite)+ ½ O2 (g)→ CO(g)
ΔH
This process when carried out in the laboratory would also generate CO2.
However, the following two processes can be carried out quantitatively.
1.
C(graphite)+ O2 (g)→ CO2(g)
ΔH1 = −393.3 kJ mol−1
2.
CO(g) + ½ O2 (g)→ CO2(g)
ΔH2 = −282.8 kJ mol−1
If the equation 2 is reversed
3.
CO2(g) → CO (g)+ ½ O2 (g)
ΔH3 = − ΔH2
Add Eqs. 1 and 3
54
C(graphite)+ O2 (g) + CO2(g) → CO2(g) + CO (g)+ ½ O2 (g) or C(graphite)+ ½ O2 (g)→ CO(g)
ΔH = ΔH1 + ΔH3 = ΔH1 – ΔH2
ΔH = −393.3 kJ mol−1+ (−282.8 kJ mol−1) = −110.5 kJ mol−1
Hence, the enthalpy change for a reaction that cannot be carried out quantitatively in the laboratory can be theoretically calculated.
Calculation of Enthalpy of a Reaction from Enthalpies of Formation of
Reactants and Products
We have seen that the enthalpy of a reaction is calculated from the molar enthalpies of the products and the reactants. However, ∆rH can also be calculated from the enthalpies of formation values of all the reactants and the products involved in the reaction. Let us consider the following reaction:
(A) CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
By definition,
ΔrHө = Total enthalpy of products – total enthalpy of reactants
=
H θ (products)
H θ (reactants )
ΔrHө = {Hө m(CO2) + 2Hө m(H2O)} – {Hө m(CH4) – 2Hө m(O2)}
The reaction A can be carried out in the following steps:
(i)
C(graphite)+ 2H2 (g) → CH4(g)
∆fH (CH4)
(ii)
O2(g) → O2(g)
∆fH (O2)
(iii) C(graphite)+ O2 (g)→ CO2(g)
∆fH (CO2)
(iv) H2 (g) +½ O2(g)→ H2O(ℓ)
∆fH (H2O)
According to Hess‟s law,
Eq. (A) = {Eq. (iii) + 2Eq.(iv)} –{Eq. (i) + 2Eq. (ii)}
55
Therefore, ∆H(A) = {∆fH (CO2) + 2∆fH (H2O)} −{∆fH (CH4) + 2∆fH (O2)}
Therefore, for a reaction νA A + νBB → νCC + νDD
The enthalpy of reaction can be written as
ΔrH = {νC ΔfHө (C) + νD ΔfHө (D)}−{ νAΔfHө (A) + νB ΔfHө (B)}
....(23)
= Total enthalpy of formation of products – Total enthalpy of formation of reactants Hence, from the tabulated ΔfH values of the substances at a given temperature, the enthalpy of any reaction can be calculated using the above formula. The standard enthalpy of reaction can be calculated from the tabulated values of
ΔfHө .
Relationship Between ΔrU and ΔrH
H = U + PV
For calculating the enthalpy change the equation becomes,
ΔH =ΔU + PΔV
For a chemical reaction the equation becomes
ΔrH = ΔrU + Δr (PV)
....(24)
For the reactions involving solids and liquids the volume change with the change in the pressure is negligible and therefore, the change in PV is negligible in comparison to ΔrU. The Equation (24) gives
ΔrH = ΔrU
For the following reaction involving gases, νAA(g) + νBB(g) → νCC(g) + νDD(g)
V(reactants ) = νA Vm(A) + νB Vm(B)
V(products ) = νC Vm(C) + νD Vm(D)
Δr (PV) = Δr (PνgVm )
56
If all the gaseous species are assumed to behave as an ideal gas, then,
Vm = RT/P
Therefore, Δr (PV) = Δ(νgRT)
At a constant temperature, Δr (νgRT ) = RT Δνg where, Δνgas = νC + νD − νA − νB
Therefore, on substituting the value of Δr (PV) in Equation (24),
ΔrH = ΔrU + RTΔνgas
For heterogeneous reactions, the same rule applies; however, here Δνgas is equal to the change in the stoichiometric number of gaseous species involved in the balanced chemical equation.
Since ΔrH = qp and ΔrU = qv qp = qv + RTΔνgas
....(25)
The Equation (25) gives the relationship between heat exchanges at a constant pressure and at a constant volume for reactions involving gases.
Solved Example: Calculate ΔrU for the following reactions at 1 atm and 298K.
1.
N2(g) + 3H2(g) → 2NH3(g)
ΔrH = −92.38 kJmol−1
2.
C(graphite) + ½O2 → CO(g)
ΔrH = −110.5 kJmol−1
Solution:
From Eq. (3.4), ΔrU = ΔrH − RTΔνgas
R = 8.314 J K−1mol−1
RT = 8.314 J K−1mol−1
1.
T = 298 K
298 K = 2477 J mol−1 = 2.477 kJ mol−1
N2(g) + 3H2(g) → 2NH3(g)
ΔrH = −92.38 kJ mol−1
Δνgas = 2 – (1+3) = −2
ΔrU = −92.38 kJ mol−1 – (− 2
2.
2.477 kJ mol−1 ) = −87.426 kJ mol−1
C(graphite) + ½O2 → CO(g)
ΔrH = −110.5 kJ mol−1
Δνgas =1 – ½ = ½
ΔrU = −110.5kJmol−1 –(½ x 2.477kJ mol−1) = −111.738kJ mol−1
57
(b)
Enthalpy of combustion
It is the heat exchange that takes place when one mole of a substance is burnt completely in the presence of oxygen at a given temperature and pressure. For most of the organic compounds the products of combustion are H 2O and CO2.
Other elements undergo combustion to form their respective oxides. It is denoted by ΔcH and the unit is kJ mol−1. The combustion is always an exothermic process.
For example,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
ΔcHө (298 K)= −890 kJ mol−1
C2H6 (g) +
7
O2(g) → 2 CO2 (g) + 3H2O(ℓ)
2
ΔcHө (298 K)= −1560 kJ mol−1
The enthalpy of combustion is always measured for the combustion of one mole of a compound. Therefore, while writing the balanced chemical equation, the stoichiometric number of the compound undergoing combustion is taken as one and the rest of the reactants and the products are balanced accordingly.
Solved Example: Apply Hess‟s law to calculate the ΔfHө (C4H10) using the following thermodynamic data:
ΔcHө (C4H10,g)= −2878.7 kJ mol−1
ΔfHө (H2O,ℓ) = −285.8kJmol−1
ΔfHө(CO2,g) = −393.5kJmol−1
Solution: The required equation is
4C (graphite) + 5H2 (g) → C4H10(g)
ΔfHө
Given
1.
H2(g) + ½ O2 (g) → H2O(ℓ)
ΔfHө (H2O,l) = −285.8kJmol−1
2.
C(graphite)+ O2 (g) → CO2(g)
ΔfHө (CO2,g) = −393.5kJmol−1
3.
C4H10(g) + 13/2 O2(g) → 4CO2(g) + H2O(ℓ)
58
ΔcHө (C4H10,g)= −2878.7 kJ mol−1
Multiply Eq. 2 by 4 and equation 1 by 5 and add the resulting equations
4.
4C(graphite)+ 4O2 (g) + 5H2(g) + 5/2 O2 (g) → 4CO2(g) + 5H2O(ℓ)
Subtract equation 3 from equation 4
4C(graphite) + 5H2 (g) → C4H10(g)
This is the required equation. Therefore,
ΔfHө (C4H10) = 4
ΔfHө (CO2,g) + 5
ΔfHө (H2O,l) − ΔcHө (C4H10,g)
ΔfHө (C4H10) = 4 mol−1) (−393.5 kJ mol−1 ) + 5
(−285.8 kJ mol−1 ) – (−2878.7 kJ
ΔfHө (C4H10) = (−1574 −1429 + 2878.7) kJ mol−1
ΔfHө (C4H10) = −124.3 kJ mol−1
(c)
Enthalpy of Transition
It is the amount of heat exchanged when one mole of a substance is transformed from one allotropic form to the other allotropic form at a given temperature and pressure. The symbol is ∆trsH.
C(graphite) → C(diamond)
Enthalpy of vaporization
It is the amount of heat absorbed to convert one mole of a liquid to its vapour state at a given temperature and pressure.
H2O(ℓ) → H2O(g)
ΔvapHө (298 K) = 44.0 kJ mol−1
Enthalpy of Fusion
It is the amount of heat absorbed to convert one mole of a solid to its liquid state at a given temperature and pressure.
H2O (s) → H2O(ℓ)
ΔfusHө (298 K)= 6.0 kJ mol−1
59
Enthalpy of Sublimation
It is the amount of heat absorbed to convert one mole of a solid to vapour state at a given temperature and pressure.
ΔsubHө (298 K)= 50.0 kJmol−1
H2O(ℓ) → H2O(g)
Enthalpy of Atomization
It is the amount of heat required to convert one mole of a substance into its constituent atoms in the gaseous state.
C(graphite)
H2(g)
CH4 (g)
atomization
atomization
C(g)
∆aH(C) =716.68 kJmol−1
2H(g)
atomization
∆aH(H)= 436 kJ mol−1
C(g) + 4H(g)
∆aH = 1664 kJ mol−1
For an element in its solid state, the enthalpy of atomization is equal to its enthalpy of sublimation.
Mg(s)
(d)
atomization
ΔaHө = 150.2 kJ mol–1
Mg(g)
Enthalpy of solution
It is the amount of heat exchanged when one mole of a solute is dissolved in a given amount of solvent at a specified temperature and pressure.
When the solvent is water the dissolution of solute X is represented as
X + n H2O → X.n H2O
This process results in the solution with the solute solvent ratio of 1: n
For example,
1.
HCl(g) + 10 H2O(ℓ) → HCl.10H2O(aq)
ΔH = −69.01 kJ mol−1
2.
HCl(g) + 40 H2O(ℓ) → HCl.40H2O(aq)
ΔH = −72.79k J mol−1
3.
HCl(g) + 200 H2O(ℓ) → HCl.200H2O(aq)
ΔH = −73.961 kJ mol−1
60
These values of ΔH show the general dependence of the heat of solution on the amount of the solvent. As more and more solvent is used the value of heat of solution changes. As the amount of the solvent increases the resulting solution becomes more dilute and ultimately it becomes so dilute that further addition of solvent produces no enthalpy change. This solution is known as infinitely dilute solution.
(e)
Bond Energies
Every reaction involves breaking and making of bonds of atoms. If the energies involved in these processes can be determined the net amount of enthalpy change for the reaction ΔrH can be estimated.
Energy is given to the system to break a bond and the energy is released whenever a bond is formed. Therefore, bond formation is an exothermic process while bond dissociation is an endothermic process. For example when one mole of hydrogen gas is formed from its atoms in gaseous state, 436 kJ of energy is released; however, 436 kJ of energy is required to break one mole of H−H bonds. H(g) + H(g) → H2(g)
ΔrH = −436 kJ mol−1
H2(g) → H(g) + H(g)
ΔrH = 436 kJ mol−1
The energy required to break a bond is known as the bond dissociation energy. Since the measurements are made at a constant pressure, the heat required to break the bond is known as the bond dissociation enthalpy and the heat released during bond formation is known as the bond formation enthalpy. NOTE: In our studies, the use of bond dissociation enthalpy is preferred over bond formation enthalpy since it is easier to break the bonds to form gaseous atoms. Measurement of the bond formation enthalpy requires the difficult task of isolating gaseous atoms and then combining them to form the molecules.
In a simple diatomic molecule only one type of bond is broken. However, in a polyatomic molecule the energy required to break a bond will depend upon the presence of other atoms or group of atoms. For example, the bond dissociation enthalpies of the two O−H linkages in the water molecule are different.
61
H−O−H(g) → H(g) + OH(g)
ΔHө (298K) = 498 kJ mol−1
O−H(g) →O(g) + H(g)
∆Hө (298K) = 430 kJ mol−1
Similarly, breaking up of an O−H bond in ethyl alcohol molecule or acetic acid molecule would involve different amount of heat. However, while tabulating the data of enthalpy required for breaking the bond, generally the net average of various values is taken. Therefore, there are two types of enthalpies of dissociation, one is the exact value for a specific bond and the other is the average value for a bond. These are known by different names; bond dissociation enthalpy and bond enthalpy.
Bond dissociation enthalpy: It is the enthalpy required to break a specific bond in a specific molecule at a given pressure. It depends upon the presence of other groups in the molecule.
Bond enthalpy: It is the average of dissociation enthalpies of a given bond in a series of different types of dissociating species.
In the case of diatomic molecules, the bond enthalpy and bond dissociation enthalpy are same.
The convention is to represent it by ε(A−B).
Relationship Between Bond Enthalpy and Bond Energy
The enthalpy of a reaction is related to the internal energy change of that reaction by the equation
ΔrH = ΔrU + RTΔνgas
Similarly, for the dissociation of A−B bond
A−B(g) → A(g) + B(g) the bond enthalpy is related to the bond energy by the equation ε (A−B) = ΔU(A−B) + RTΔνgas
For this reaction, Δνgas= 2−1 =1 and at 298K RTΔνgas = 2.48 kJ mol−1
This value is almost negligible compared to bond enthalpy values. So the bond enthalpy is approximately equal to the bond energy and the two are used interchangeably. 62
Calculation of Bond Enthalpy
In a simple molecule such as methane all the bonds are identical. It is assumed that the enthalpy required to break each bond is equal and is equal to bond enthalpy of C−H bond. So, the total enthalpy required to break all the bonds and form gaseous atoms is equal to four times the bond enthalpy of C−H bond.
That is 4 ∆H(C−H) = ∆H for the reaction
CH4(g) → C(g) + 4H(g)
∆H
∆H for this reaction can be calculated using the following data: sublimation C(g)
1.
C(graphite)
2.
½H2(g)
3.
C(graphite) + 2H2(g) → CH4(g)
dissociation
∆H1 = 716.68 kJ mol−1
H(g)
∆H2 = 217.97 kJ mol−1
∆H3 = −74.75 kJ mol−1
Using the laws of thermochemistry these equations are modified to get the required value.
Multiplying Eq. 2 by 4 gives
4.
2H2(g)
dissociation
4 H(g)
∆H4 = 4
∆H2
Reversing Eq. 3 we get
5.
CH4(g) → C(graphite) + 2H2(g)
∆H5 = −∆H3
Adding Equations (1), (4) and (5) gives the required equation
CH4(g) → C(g) + 4H(g)
Therefore, ∆H = ∆H1 + ∆H4 + ∆H5
∆H = ∆H1 + 4
∆H2 −∆H3
∆H = 716.68 kJ mol−1 + 4
217.97 kJmol−1 –(−74.75 kJmol−1)
= (716.68 + 871.88 + 74.75) kJmol−1
= 1663.31 kJ mol−1
H ε(C−H) =
4
1663.31 kj mol
4
1
= 415.82 kJ mol−1
63
Since C−H bond is present in other molecules also, the average value is calculated by taking into account C−H bonds in different molecules. The bond enthalpy of C−H bond is taken as 413 kJ mol−1.
Similarly, the C−C, C=C and C≡C bond enthalpies can be calculated by using the enthalpies of formation of ethane, ethene and ethyne and bond enthalpy of
C−H bond. Thus the bond enthalpy data of polyatomic molecules can also be tabulated. Applications of bond enthalpies
1.
Determination of Enthalpy Changes Accompanying a Chemical
Reaction:
Bond enthalpies can be used for determining the enthalpies of reaction.
∆rH = Total bond enthalpies of reactants −Total bond enthalpies of products
...(26)
During a reaction some bonds of the reactants dissociate and bonds of the products are formed. The enthalpy of a reaction is the total enthalpy change during bond formation and bond dissociation. The enthalpy change during bond dissociation is equal to the bond enthalpy. Since the bond formation is reverse of bond dissociation, the enthalpy change during bond formation is negative of bond enthalpy. Therefore, the total bond enthalpies of the products are subtracted from the total bond enthalpies of the reactants to get the enthalpy of reaction.
Solved Example: For the following reaction, calculate the amount of energy released during bond formation:
C2H4(g)+H2(g) → C2H6(g)
Or
HH
H-C=C-H (g) + H-H(g)
H H
H-C-C-H (g)
HH
64
Solution: The bonds being broken on the reactant side are C=C and H−H bonds in ethene and hydrogen gas, respectively and bond formed are C−H bonds and C−C. Given, ε (C=C) = 610 kJ mol−1, ε (H−H) = 436 kJ mol−1, ε (C−C) = 348 kJ mol−1, ε (C−H) = 413 kJ mol−1,
Total enthalpy needed to break the bonds = Total bond enthalpies of reactants
= ε (C=C) + ε (H−H) = (610 + 436) kJmol−1 = 1046 kJmol−1
Enthalpy released during the formation of bonds = Total bond enthalpies of products = ε (C−C)+ 2 ε (C−H) = (348 +2
413) kJmol−1 = 1174 kJmol−1
∆rH = { ε (C=C) + ε (H−H) } –{ ε (C−C) + 2 ε (C−H)}
= 1046 kJmol−1 −1174 kJmol−1
∆rH = −128 kJmol−1
2.
Determination of enthalpy of formation:
If the reaction involves the formation of a compound from its elements then the bond enthalpy data can be used to calculate the enthalpy of formation of the compound by the formula
∆fHө = Total enthalpy of atomization of the reactants – Bond enthalpies of products ...(27)
Solved Example: In the following reaction, 2C(graphite)+3H2(g)+½O2(g)
→C2H5OH(ℓ) calculate the enthalpy of formation of ethanol.
Solution: The steps involved are
2C(graphite)
3H2(g)
atomization
atomization
2C(g)
∆rH=2 ∆aH(C) =716.68 kJmol−1
6H(g)
∆rH=3∆aH(H)= 436 kJ mol−1
½O2 (g) → O(g)
∆rH=½∆aH(O) = 498.4 kJmol−1
65
HH
2C(g) + 6H(g) + O(g) → H-C-C-O-H (I)
HH
Enthalpy change during bond breakage = Total enthalpy of atomization
= 3∆aH(H) + 2∆aH(C) + ½∆aH(O)
=3
436 kJ mol−1 + 2 716.68 kJmol−1 + 249.2 kJmol−1
= 2990.88 kJmol−1
The bond enthalpy values are ε(C−C) = 348 kJ mol−1, ε (C−H) = 413 kJ mol−1, ε(C−O) = 351 kJ mol−1 ε (O−H)= 363 kJ mol−1
The enthalpy released during the formation of bonds in C2H5OH = Bond enthalpy of C2H5OH = ε (C−C) + 5 ε(C−H)+ ε (C−O) + ε (O−H)
= (348 + 5 413 + 351 + 363) kJ mol−1 = 3227 kJ mol−1
∆fH (C2H5OH) = Total enthalpy of atomization – Bond enthalpy
= (2990.38 – 3227) kJ mol−1 = −236.52 kJmol−1
IMPORTANT: The drawback of this method is that it will give same value of
∆fHө for all the isomers of a compound which is not correct.
Solved Example: Calculate ΔrU for the following reactions at 1 atm and 298K.
C(graphite) + O2 → CO2 (g)
ΔrH = −393.5 kJmol−1
Ans. ΔrU = ΔrH − RTΔνgas
R = 8.314 J K−1mol−1
RT = 8.314 J K−1mol−1
T = 298 K
298 K = 2477 J mol−1 = 2.477 kJ mol−1
C(graphite) + O2 → CO2 (g)
ΔrH = −393.5 kJ mol−1
Δνgas =1 –1=0
ΔrU = ΔrH = −393.5 kJ mol−1
66
6.6
Spontaneity
Some processes take place on their own without any help from the external agencies. Such processes are known as spontaneous processes. For example, expansion of gas kept in an open container and cooling of hot tea kept in a cup take place spontaneously, while air can never compress itself and fill a balloon on its own. Similarly a body at room temperature can never take heat from the surroundings and become hot. These processes can take place with the help of an external agency. Such processes that do not take place on their own but require the help of an external agency are known as non-spontaneous processes.
8.6.1. Limitations of the first law of thermodynamics
The first law of thermodynamics is about conservation of energy. It is very useful in calculating energy changes taking place during various processes.
However, it gives no idea about the direction of energy flow. The first law of thermodynamics is unable to predict whether a process will be spontaneous or nonspontaneous. To make this prediction we have to look into the most important property of the system which is energy. It is a known fact that energy flows from a high energy state to a low energy state. The system having lower energy is more stable than the system having higher energy. Then the question that can be asked is whether all the processes in which energy of the system is decreasing are always spontaneous? Are all exothermic processes always spontaneous? It is observed that it is not necessary that all spontaneous processes are exothermic. Various cases are known where a spontaneous process is endothermic. For example, a gas expands in vacuum spontaneously although there is no energy change. Similarly, water evaporates spontaneously into water vapour but the process is endothermic in nature.
There is another reason why an exothermic process cannot be taken as the sole criterion of spontaneity and that is the law of conservation of energy. If the energy of the system decreases during a process then the energy of the surrounding must increase since the total energy must be conserved. So if the system undergoes an exothermic process spontaneously then the change in the surroundings should also be spontaneous although the energy of the surrounding will increase.
67
The second law of thermodynamics decides the direction of a spontaneous process. It introduces a new concept of entropy that governs the criterion of spontaneity. 8.6.2. Various statements of the Second Law of Thermodynamics
The second law of thermodynamics has been stated in various ways by different scientists. Although these statements are different but each one of them predicts the direction in which a change will take place spontaneously.
1.
Heat cannot pass spontaneously from a colder to a warmer body.
(Clausius)
2.
No change is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.(Kelvin)
3.
It is impossible to transfer heat from a colder system to a warmer system without other simultaneous changes occurring in the two systems or in their environment.(P.S. Epstein)
4.
Every system which is left to itself will on the average change towards a condition of maximum probability.
5.
The state of maximum entropy is the most stable state of an isolated system. (Enrico Fermi)
6.
In any irreversible process the total entropy of all bodies concerned is increased. 7.
In an adiabatic process the entropy either increases or remains unchanged.
8.
Entropy is times arrow. (A. Edington)
9.
There is general tendency in nature for energy to pass from more available form to less available form.
10.
The entropys
8.6.3. Entropy
Whenever a change occurs, the internal energy is distributed into various forms. It has been found that those processes are spontaneous in which the energy has been distributed into a more dispersed form, such that is there is
68
more chaotic or random distribution of energy. This random distribution of energy makes it less useful for doing work. For example, when a ball bounces it loses some energy at each bounce. The energy lost is either in the form of frictional energy or it is given to the floor molecules as thermal energy. The total energy is conserved but the energy is in the distributed state and hence is of less use.
A new concept of entropy was introduced by Clausius as a measurement of randomness or chaos (also referred to as disorder). The word entropy has been made from energy (en) and trope (tropy) which is for chaos. The symbol is S. It was given in honour of famous scientist Sadi Carnot.
According to one of the statements of second law of thermodynamics, heat cannot be completely converted into work. This has been attributed to entropy and it can be explained by taking the example of a gas kept in a cylinder with a piston that can move without friction.
When the gas kept in the cylinder is heated its kinetic energy increases. The gas molecules move faster. They collide with each other and with the walls of the cylinder. When they collide with the piston, it is pushed forward and the gas expands. This results in useful work done by the gas. Now, consider a hypothetical condition when all the gas molecules move in ordered manner towards the piston. Then each molecule will strike the piston and contribute to the total work done by the gas as shown in Figure 8.7. Under this situation the entire heat absorbed can be converted into work. However, in reality the molecules are moving randomly in all directions. Therefore, all the molecules do not strike the piston. Hence, all the collisions of the gas molecules do not lead to expansion. Since only a fraction of collisions result in useful work, only a fraction of heat absorbed is converted into useful work and the rest is the unavailable heat, that is, the heat that is not used for doing work. The fraction of heat used to do useful work decreases and the amount of unavailable energy increases as the randomness of the gas molecules increases.
69
Figure 8.7 a and b. Conversion of heat into work when the collisions are ordered and when the collisions are random
Mathematical form of entropy
The change in entropy is defined as
dS
dq rev
T
Note that while calculating the entropy change, the heat transfer only during the reversible processes is considered. However, the entropy change for an irreversible process is not related to the heat transfer during an irreversible process. q is a path function. It can be shown that the entropy is a state function.
Characteristics of Entropy
Entropy is a measure of randomness. More is the randomness greater is the entropy of the system. Entropy can also be related to unavailable energy.
For a given substance, the vapour phase has maximum randomness and solid phase has least randomness. Therefore,
Svapour > Sliquid > Ssolid
Entropy is a state function. Therefore, ΔS = Sfinal – Sinitial = S2 – S1
70
It depends upon temperature, pressure, volume and amount.
Entropy is an extensive property.
dS ≠ dqirr / T
Entropy increases with increase in temperature.
The entropy increases as the volume of the system increases.
Units of entropy dS dq rev
T
J
K
The SI unit of entropy is J K-1
For molar entropy, the unit is J K-1mol-1
When heat is measured in calories, the unit is cal K-1
Relationship between internal energy and entropy dU = dq + dw
....First law of thermodynamics
dU = dq –PdV
For reversible process, dq = dqrev = TdS dU = TdS –PdV
....(28)
This is also known as the combined form of first law and second law for reversible processes. This expression indicates that the internal energy depends upon the change in entropy and volume.
Relationship between enthalpy and entropy
H = U + PV
Differentiating the equation we get dH = dU + PdV + VdP
....(29)
From the combined form of the first law and the second law of thermodynamics 71
dU = TdS – PdV
Substituting the above expression in Equation (29) we get dH = TdS –PdV + PdV + VdP dH = TdS + VdP
....(30)
Entropy change for an isolated system
According to the first law of thermodynamics, dUrev = dqrev + dwrev and dUirr = dqirr + dwirr
If the initial states and final states of the system are same for both the processes, then dUrev = dU irr dqrev + dwrev = dqirr + dwirr
....(31)
dwrev< dwirr
Therefore,
dqrev
T
or
dqrev > dqirr
dqirr
....(32)
T
dq rev
= dS
T
Therefore,
Since,
dS >
dS >
dq irr
T
....(33)
dqirr =dU − dwirr
dU dw irr
T
....(34)
For an isolated system dq = 0, dU = 0. Therefore, dw = 0 dqrev = 0
(dS)rev = 0
dqirr = 0 (dS) irr >0 or dS(isolated) ≥ 0
....(35)
72
Although entropy is a state function, it appears that the quantity dS is different for reversible and irreversible processes for an isolated system. This can be explained if the nature of the isolated system is considered. An isolated system includes the system as well as its immediate surroundings. Therefore, the entropy change for an isolated system includes the entropy change for the system as well as its surroundings. dS(isolated) = dS(system) + dS(surroundings)
Since, Universe = System + Surroundings dS(isolated) = dS (universe)
For the state functions, the initial state and the final state of the system are considered and not that of the surroundings. Since the initial states and final states of the system are same for both the processes, the entropy change for the system is same for reversible and irreversible process. The changes in the surroundings are different for reversible and irreversible processes so the entropy changes for the surroundings will be different. As a result, the total entropy change for the universe will be different for reversible and irreversible process. dSrev(system) = dSirr(system)
....(36)
For a reversible process,
....(37)
dSrev(system) = and dqrev(surroundings) = – dqrev(system)
Therefore,
dSrev(surrounding) =
....(38)
dSrev(universe) = dSrev(system) + dSrev(surrounding)
Substituting the values of Equations (37) and (38) in the above equation, dSrev(universe) =
In view of Equation (38) the above equation becomes
73
dSrev(universe) =
For an irreversible process,
Heat absorbed by the system = dqirr(system)
From Equation (33), dSirr(system)>
....(39)
Since the surroundings are large, the addition or removal of a small amount of heat in the surrounding is taken as a reversible process. Therefore,
ΔSirr(surrounding) =
....(40)
The entropy change for the universe is given by the equation, dSirr(universe) = dSirr(system) + dSirr(surrounding)
On substituting the value of ΔSirr(surrounding) from Equation(40) in the above equation, dSirr(universe) = dSirr(system)
....(41)
From Equations (39) and (41) it is seen that dSirr(universe) > 0
8.6.4. Free Energy Function
One of the purposes of studying thermodynamics is to predict the feasibility of a process, that is, to predict the direction in which a process will be spontaneous. We have seen that the sign of internal energy alone cannot be the criterion for determining the spontaneity of a process. The second law of thermodynamics introduced the concept of entropy. Whenever a process occurs spontaneously it is an irreversible process. It is observed that for such a process the entropy change of the universe is positive. This implies that to predict the spontaneity of a process it is important to determine the entropy change of the universe. This involves measurement in entropy change of the system as well as that of surroundings. The problem arises in the measurement of entropy change of the surroundings since the surroundings are not clearly defined. For most of the spontaneous processes it is observed that the system either tries to attain the state of minimum energy or maximum entropy. So it is necessary
74
to define a new function which takes into account both the properties. Gibbs free energy (G) is a function that takes into account the composite effect of energy and entropy change of the system.
Gibbs free Energy or Gibbs Energy as criterion of spontaneity
The symbol of Gibbs energy is G. It is related to the enthalpy and entropy by the relation
G = H – TS
....(42)
Since H, T and S are state functions; therefore, G must also be a state function.
Differential of Equation (42) is dG = dH – TdS – SdT
If the temperature is constant dT = 0, therefore, dG = dH – TdS
....(43)
At constant pressure, dH = dq. Therefore,
(dG)T, P = dq – TdS
....(44)
Let us see how Gibbs energy can be used as criterion of spontaneity dS =
dS >
Whether the process is reversible or irreversible
TdS = dqrev
Therefore, (dG)T, P = dq –dqrev
For a reversible process each step is an equilibrium step and dq = dqrev.
Therefore,
(dG)T, P = dqrev – dqrev = 0
So for the equilibrium state, dG = 0
For an irreversible process dq = dqirr .Therefore,
(dG)T, P (irr) = dqirr – dqrev
We know that dqrev > dqirr, therefore,
75
dqirr –dqrev < 0 and (dG)T, P < 0
...(45)
Since all the spontaneous processes are irreversible processes, for all spontaneous process taking place at constant temperature and pressure the following relation is valid dG < 0
Characteristics of Gibbs free Energy
G is a state function
For a spontaneous process the change in free energy is always negative.
For equilibrium state dG = 0
For non-spontaneous process dG >0
The SI unit of G is J
For finite change at constant temperature and pressure
ΔG = ΔH – TΔS
....(46)
Here ΔG is the free energy change in the system
ΔH is the enthalpy change in the system
ΔS is the entropy change in the system
The process will be spontaneous, that is the change in free energy of the system, ΔG will be negative if either of the following conditions is true
ΔH < 0 and ΔS >0
ΔH > 0 but ΔH < TΔS
ΔH< 0 and ΔS<0 but ΔH > TΔS
When ΔH > 0 and ΔS < 0, the process will always be non-spontaneous.
This can be elaborated as follows:
On the basis of the enthalpy change there are two types of processes
76
(i)
Exothermic process for which the enthalpy change is negative, that is,
ΔH < 0
(ii)
Endothermic process for which the enthalpy change is positive, that is,
ΔH > 0
Each type of the process is also accompanied by the change in entropy ΔS which is positive when the entropy increases during a process and negative when the entropy decreases during a process. Therefore, there are the following four cases:
(i)
Exothermic process, ΔH < 0
(a)
ΔS > 0, which is the entropy increases during the process. Therefore,
TΔS > 0
ΔH −TΔS < 0
Therefore ΔG is always negative and the process is always spontaneous. (b)
ΔS < 0, which is the entropy decreases during the process. Therefore,
T ΔS < 0
ΔH −T ΔS will be less than zero and ΔG will be negative when
∣ΔH ∣ >∣TΔS∣
That is, the process will be spontaneous when ΔH is less than TΔS
(ii)
Endothermic process, ΔH > 0
(a)
ΔS > 0, which is the entropy increases during the process. Therefore,
TΔS > 0.
The process will be spontaneous when
ΔH −T ΔS < 0 or ∣ΔH ∣ <∣TΔS∣
(b)
ΔS < 0, which is the entropy decreases during the process. Therefore,
TΔS < 0
77
ΔH−TΔS will always be positive, that is ΔG > 0. Therefore the process is always non-spontaneous.
Variation of G with Temperature and Pressure
G = H –TS
....(47)
Differential of this equation gives dG = dH –TdS – SdT
...(48)
According to combined form of first law and second law of thermodynamics, for reversible process dH = TdS + VdP
Substituting the value in equation we get dG = TdS +VdP –TdS –SdT or dG = – SdT +VdP
....(49)
Various thermodynamic relations between U, H, S, T, V and P
1.
dU = TdS –PdV
2.
dH = TdS + VdP
3.
dG = – SdT + VdP
4.
dA = – SdT –PdV
Solved example: Calculate H° and S° for the following reaction:
NH 4 NO 3 (s) + H2 O() NH+4 (aq) + NO 3- (aq)
Use the results of this calculation to determine the value of Go for this reaction at 25o C, and explain why NH4NO3 spontaneously dissolves is water at room temperature. 78
Given data:
Compound
Hfo(kJ/mol)
NH4NO3(s)
-365.56
151.08
NH +4 (aq)
-132.51
113.4
NO -3 (aq)
-205.0
146.4
Solution: Ho =
S°(J/mol-K)
o
o
H f (products) -
H f (reactants)
= [1 mol NH4 132.51 kJ/mol + 1 mol NO -3 -205.0 kJ/mol] - [1 mol NH4NO3
-365.56 kJ/mol]
= 28.05 kJ
The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction:
So =
o
S (products) -
o
S (reactants)
= [1 mol NH4 113.4 J/mol-K + 1 mol NO -3
146.4 J/mol-K] - [1 mol NH4NO3
151.08 J/mol-K]
= 108.7 J/K
To decide whether NH4NO3 should dissolve in water at 25o C we have to compare the Ho and T So to see which is larger. Before we can do this, we have to convert the temperature from oC to kelvin:
TK = 25o C + 273.15 = 298.15 K
We also have to recognize that the units of Ho for this reaction are kilojoules and the units of So are joules per kelvin. At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis. Perhaps the easiest way of doing this is to convert Ho to joules. We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term:
Go = Ho - T So
= 28,050 J - (298.15 K
108.7 J/K)
79
= 28,050 J - 32,410 J
= - 4360 J
At 25o C, the standard-state free energy for this reaction is negative because the entropy term at this temperature is larger than the enthalpy term:
Go = -4.4 kJ
The reaction is therefore spontaneous at room temperature.
8.6.5. Third law of Thermodynamics
The entropy at absolute zero is determined with the help of third law of thermodynamics which states that:
Entropy of perfectly crystalline substance is zero at absolute zero.
This statement has a logical explanation. The best organized physical state of a system is that of a single crystal, since in this state each particle has definite arrangement in space with respect to other particles. At very low temperatures only vibrational motion of the particle (atom or molecule) remains and at absolute zero all the particles vibrate with lowest possible energy. Therefore, the arrangement is most ordered. For a perfectly crystalline substance there is only one possible arrangement of particles, so there is no randomness. Hence, the entropy is zero at absolute zero.
Characteristics of absolute entropy
Entropy of perfectly crystalline substance is zero at absolute zero.
Substances that are not perfectly crystalline have positive entropy at absolute zero.
At higher temperatures the absolute entropy of all the substances is positive.
Entropy depends upon mass of the molecules. Heavier molecules have greater entropy.
Entropy depends upon the number of atoms in a molecule. More are the number of atoms in a molecule greater is the entropy.
Entropy depends
Sgas > Sliquid > Ssolid
upon
the
80
physical
state
of
the
substance.
Solved Example: Arrange the diatomic molecules: F2, C 2, Br2 and I2, in the increasing order of entropy at room temperature.
Solution: The increasing order of entropy is I2 < Br2 < F2 < Cl2
Since iodine is solid at room temperature it has least entropy followed by bromine which is in liquid state and so has greater entropy than iodine.
Chlorine and fluorine both are gases but chlorine has greater molecular mass than fluorine so chlorine has maximum entropy in case of halogens.
8.6.6. Equilibrium Constant from the Free Energy Change
Standard-State Free Energy of a Reaction
What does the value of Go tell us about the following reaction?
N2(g) + 3 H2(g)
2 NH3(g)
Go = -32.96 kJ
By definition, the value of Go for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard-state conditions.
Go therefore describes this reaction only when all three components are present at 1 atm pressure.
The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium. The fact that in the above example, Go is negative for this reaction at 25oC means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of Go for a reaction tells us how far the standard state is from equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium.
At equilibrium, G of a reaction is zero. Therefore, if we know the standard state free energy change, Go, for a chemical process at some temperature T, we can calculate the equilibrium constant for the process at that temperature using the relationship between Go and Keq.
0 = Go- RTℓn Keq
81
Or, Go = –RTℓn Keq
Rearrangement gives
ΔG o or = K eq
RT
In Keq = –
e
ΔG o/RT
Once the value of Go is known to us, all the other thermodynamic properties can also be calculated with the help of the thermodynamic relations derived in the presious sections.
Solved Example: Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (Ka) at for formic acid: Compound
Gfo(kJ/mol)
HCO2(aq)
-372.3
H+(aq)
0.00
HCO -2 (aq)
-351.0
Solution: We can start by writing the equation that corresponds to the aciddissociation equilibrium for formic acid:
HCO2H(aq)
H+(aq) + HCO -2 (aq)
We then calculate the value of
Go =
G of (products) -
= [1 mol H+
Go for this reaction:
G of (reactants)
0.00 kJ/mol + 1 mol HCO -2
-351.0 kJ/mol] - [1 mol HCO2H
–372.3 kJ/mol]
= 21.3 kJ
We now use the relationship between Go and the equilibrium constant for the reaction: Go = - RTℓn K and solve for the natural log of the equilibrium constant:
In K =
82
ΔG o
RT
Substituting the known value of Go, R, and T into this equation gives the following result:
In K =
21.300 J/mol
(8.314 J/mol k)(298K)
8.60
We can now calculate the value of the equilibrium constant:
K = e-8.60 = 1.8
10-4
The value of Ka obtained from this calculation agrees with the table value for formic acid, within experimental error.
83
SUMMARY
Energy changes in physical or chemical processes can be studied through a branch of science called as Thermodynamics. These changes can be studied quantitatively which helps in providing useful predictions. The universe may be divided as system and surrounding. These processes lead to conversion evolution or absorption of heat or q, which may be partially converted into work which is denoted as (w). First law of thermodynamics ΔU = q+w; wherein ΔU, change in internal energy, depends on initial and final states, q and w depend on path. ΔU is a state function while q and w are not the state functions. In expansion of gases, work can be measured.
At constant volume, w=0, then ΔU = qv. Another state function is enthalpy. Heat changes at constant pressure can give Enthalpy change.
Change of phase- vaporization, sublimation etc. (enthalpy changes) occurs at constant temperature. Enthalpy changes are always positive. Various enthalpy changes as
Enthalpy of formation, combustion etc. is always positive.
First law of thermodynamics does not guide us about the direction of chemical reactions
i.e., what is the driving force of a chemical reaction. Another state function is Entropy or
S. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive.
Chemical reactions are generally carried at constant pressure, it is defined by another state function i.e. Gibbs energy. G which is related to entropy and enthalpy changes of the system of the equation.
84
Students’ Worksheets
Student Worksheet- 1
1.
Why is it important to study thermodynamics? Search the web and list three simple applications of thermodynamics.
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2.
In a thermodynamic process, one form of energy is transformed into another form.
Identify the initial and the final forms of energy in the following processes.
Thermodynamic Process
Initial form of energy Final form of energy A battery running on a fuel cell
Tuning fork after jerk
A car running on petrol
Speaking on a telephone
Producing fire by rubbing stones
Making a toasted bread in a toaster
3.
Heat flows from one body to another (usually from hotter body to colder body) until equilibrium is reached. Identify the direction of heat flow in the following cases. 85
Heat flows
From
To
When our body is sweating!
When ice cube is added to a glass of water at room temperature!
When we keep water in a clay pot out in summers 4.
What is the role thermodynamics in chemistry?
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5.
What are the limitations of thermodynamics? Which branch of science can overcome these limitations and complement the findings of thermodynamics?
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6.
Thermodynamics is the study of macroscopic objects, i.e, the objects visible to naked human eye! On the contrary, quantum mechanics is the study of microscopic objects, i.e., the objects NOT visible to naked human eye!
a) Identify macroscopic and microscopic objects from the following.
86
A glass of water, a metal block, an electron moving in an orbital, a gas enclosed in a container, sodium and chloride ions dissolved in water, two hydrogen atoms colliding with each other, a matchstick.
Macroscopic system
Microscopic system
b) Based on the above answer, formulate suitable definitions for macroscopic and microscopic systems.
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c) What properties (or variables) do you think can be used define macroscopic systems? ____________________________________________________________________________
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7.
The word thermodynamics stems from two Greek words meaning
a)
conservation of heat
b) interactions of heat
c)
study of heat
d) movement of heat
87
8.
Which of the following thermodynamics? a)
laws of
physics
becomes
the foundation
of
Newton‟s laws of motion
b) Law of conservation of energy
c)
Law of universal gravitation
d) Law of conservation of momentum
9.
When heat is added to a system, all of the following may happen EXCEPT
a)
increase in internal energy
b) decrease in the system‟s temperature
c)
external work is done by the system
d) increase in the pressure in the system
10.
Thermodynamics focuses on the macroscopic details of systems. Which of the following is NOT used to describe a system macroscopically?
a)
Temperature
b) Pressure
c)
Work
d) Spin
11.
Which of the following is TRUE about thermodynamics?
a)
It is based on conservation principle.
b) It deals with energy.
c)
It discusses direction of heat movements.
d) All of the above
12.
What happens when you get inside an air-conditioned room after staying under the sun for sometime?
a)
You feel warm because heat flows from your body to the room
88
b) You feel warm because heat flows from the room to your body
c)
You feel cold because heat flows from your body to the room
d) You feel cold because heat flows from the room to your body
13. Heat energy can change into
a)
light only
b) work only
c)
chemical energy only
d) any form of energy
14.
Which of the following best describes thermodynamics?
a)
It is the study of the hotness and coldness of a body
b) It is the study of the atomic and molecular nature of matter
c)
It is the study of energy and its transformations
d) It is the study of the interaction between heat and temperature
15.
Which of the following statement is true about thermodynamics?
a)
Thermodynamics will help in predicting whether a physical or chemical change is possible under given conditions.
b) Thermodynamics only deals with the initial and final states of the system and is not helpful in evolving the mechanism of the process.
c)
The rate of a reaction can be evolved from thermodynamics.
d) The mechanism of a reaction can be evolved from thermodynamics
i.
I only
ii.
I & II
iii.
I, II & III
iv.
I & IV
89
Student Worksheet- 2
1.
Define system and surroundings in thermodynamic terms. How the two are separated? Identify the system and the surroundings in the following cases.
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Example
Topic of investigation A cup of tea
To measure the heat absorbed by the cup walls as well as heat given out in the air
Water inside a thermos flask* To measure the temperature of water after adding some ice
An inflated football To measure the volume occupied by the ball inflated with air
System
Surroundings
Boundary
(real or imaginary) * assume that no heat or mass can be exchanged from thermos flask to the surroundings 90
2.
Differentiate between an open, a closed and an isolated system. Identify the type of the system in the examples given in question number 1.
Type of system Definition
Example from
Q. No. 1
Open
Closed
Isolated
3.
What do you understand by the state of a thermodynamic system? What are state variables? Which variables may be used to define the state of the systems given below? System
Variables
A glass of lemonade
A balloon filled with air An ice cube
<
4.
Define a state function and a path function. Put „√‟ for the state functions or „Χ‟ for path functions against the given thermodynamic properties.
Function
Definition
State function Path function Enthalpy
Heat
Temperature
Internal energy 91
Volume
Pressure
Entropy
Work
Free
energy
5.
Define intensive and extensive properties. Put „I‟ for intensive property or „E‟ for extensive property against the given thermodynamic properties.
Property
Definition
Intensive property Extensive property Internal
Energy
Enthalpy Density Refractive
Index
Pressure Resistance Entropy
Surface
Tension
6.
Volume
Specific
Volume
Hof
Electric
Field
Mass
Free
Energy
Temperature
Normality
Molality
Resistivity
Polarization Magnetic Magnetization
Field
An extensive property can become an intensive property if mass or size of the system is fixed! For example, volume is an extensive property, but molar volume is an intensive property as volume of one mole of the system will always be constant.
List some more examples where an extensive property can become an intensive property. ________________________________________________________________________________
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92
7.
Define a thermodynamic process. What do you understand by reversible and irreversible process? Which of them is always in thermodynamic equilibrium?
8.
As the name suggests, a reversible process means a process which can be reversed without causing any change, elsewhere!
Of the following processes, which of them do you think are reversible?
Process
Reversible or irreversible
Stretching of a rubber band within its elastic limit Stretching of a rubber band beyond its elastic limit A gas enclosed in a cylinder with frictionless pulley is compressed by putting a 100 g of weight on the pulley
9.
State the zeroth law of thermodynamics? Why do you think it is numbered as
„Zero‟ in the sequence of thermodynamic laws?
10.
A system is said to be in thermodynamic equilibrium if it is
11.
a)
in thermal equilibrium with the surroundings (temperature is the same for the whole system and surroundings)
b)
in mechanical equilibrium (no mechanical work is done on any part of the system or between the system and the surroundings)
c)
in chemical equilibrium (the composition or concentration remains fixed and definite) d)
all of the above
A closed system is that which exchanges ______ with the surroundings.
a)
Neither matter nor energy
b)
Energy only
c)
Matter only
d)
Energy and matter
93
12.
13.
14.
15.
16.
A process in which the driving force is infinitesimally greater than the opposing force is known as
a)
a reversible process
b)
an adiabatic process
c)
a cyclic process
d)
an irreversible process
An irreversible process is one which
a)
goes from the initial to the final state in finite number of steps and in finite time b)
is not in equilibrium at all stages
c)
has a driving force much greater than the opposing force
d)
all of the above
A thermos flask, when opened
a)
becomes an isolated system from an open system
b)
becomes an open system from an isolated system
c)
becomes a closed system
d)
remains an isolated system
Most of the processes occurring in nature are
a)
Isochoric
b)
Adiabatic
c)
Isobaric
d)
all of the above
A process in which the apparatus is completely insulated is called an
a)
isothermal process
b)
adiabatic process
94
17.
c)
isochoric process
d)
isobaric process
The human body is
a)
an open system
b)
a closed system
c)
an isolated system
d)
not a system
18.
What do you understand by a thermodynamic equilibrium?
19.
What is a reversible process in thermodynamics?
95
Student Worksheet - 3
1.
What do you understand by „work‟ in thermodynamic terms? Give an example.
What will be the sign („+‟ or „-„) of work if it is said to be done on the system?
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2.
The work done by an ideal gas enclosed in a cylinder with a frictionless piston can be obtained by calculating the area under the P-V curve.
In the following P-V curve for an isothermal expansion of a given amount of gas from state A to a state B, draw the area under the curve that represents
A)
Work done reversibly B)
work done
irreversibly in a single step
Work done in reversible isothermal process 3.
Work done in an irreversible isothermal process
Compare the work done in an isothermal process carried out reversibly and irreversibly, which process gives maximum work and why? Do you think this maximum amount of work is practically achievable?
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96
________________________________________________________________________
4.
Show that work done cannot be a state function.
Step - I
Consider a system going from state A to state C in a PV diagram via two routes : a)
A → B → C and
b) A → D → C
Step - II
Calculate the work done in going from A→B
Step - III
Calculate the work done in going from B→C
Step - IV
Calculate the total work done in going from
A→B→C
Step - V
Calculate the work done in going from A→D
Step - VI
Calculate the work done in going from D→C
Step - VII
Calculate the total work done in going from
A→D→C
Step VIII
Compare the total work done in going from A → C from both the routes and infer your result
97
5.
Give the expression for calculating work done in a reversible isothermal process and an irreversible isothermal process.
One mole of an ideal gas at 300 K occupies a volume of 10 L in a cylinder fitted with a frictionless piston. It is expanded isothermally to a final volume of 20 L.
Calculate w and q if the expansion is carried out (i) reversibly, (ii) against a constant external pressure of 0.1 atm. and (iii) against a zero external pressure (i.e., against vacuum).
Answer
Expression
Work done in a reversible isothermal process
Work done in an irreversible isothermal process
Work done against a zero external pressure wrev =
wirrev =
wirrev =
Calculation
Result (with proper units)
6.
Three moles of an ideal gas expand from one litre to a volume of ten litres at 27 oC.
What is the maximum work done?
(Hint: The maximum work is obtained in a reversible and isothermal process)
Formula used
Conversion
of units
Calculation
w nRT In
T = 27 + 273 = 300 K
w nRT In
= 3 mol
= x 3.303 RT log
2.303
8.314 J K-1 mol-1
Result (with proper units)
98
300 K
log
= 17234.4 J
7.
10 moles of an ideal gas expands isothermally and reversibly from a volume of 5 L to 50 L at 25°C. What is the maximum work done?
Formula used
Conversion of units Calculation
Result (with proper units)
8.
Calculate the amount of work obtain in isothermal reversible expansion of 20 g of
Ar. At 27°C from a pressure of 4 atm to 1 atm.
Formula used
Conversion of units Calculation
Result (with proper units)
9.
Define „heat‟. What is the symbol used to denote heat? What will be the sign of quantity of heat absorbed by the system?
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99
10.
Show that heat is also a path function.
Step – I
Consider an isothermal process going from state A to state C in a P-V diagram via two routes :
a) A → B → C and
b) A → D → C
Step – II
Calculate the heat absorbed in going from A→B
Step – III
Calculate the heat absorbed in going from B→C
Step – IV
Calculate the total heat absorbed in going from A → B → C
Step – V
Calculate the heat absorbed in going from A→D
Step – VI
Calculate the heat absorbed in going from D→C
Step – VII Calculate the total heat absorbed in going from A → D → C
Step VIII
11.
Compare the total heat absorbed in going from A → C from both the routes and infer your result
An isothermal expansion may be represented as in the figure below. Calculate the work done by the system and heat absorbed by the system when a gas expands against a constant external pressure of 1 atm from a volume of 10 L to 20 L.
100
Work
Heat
Formula used
Calculation
Result (with proper units)
12.
Two moles of H2 expands isothermally and reversibly to ten times its volume at
25°C. Assuming ideal behavior for the gas, calculate the work done and the heat absorbed during the process. ( R = 8.314 JK-1 mol-1)
Formula used
Conversion of units Calculation
Result (with proper units)
13.
What causes the lid of a pot to move up when the water in it starts boiling?
a)
The heated air inside the pot compresses
b) The heated air inside the pot expands
c)
The boiling water compresses
d) The pressure inside the pot decreases
14.
A system goes from state (i) P1, V to 2P1, V. Another system goes from state
(ii) P1, V to P1, 2V. The work done in the two cases is
101
a)
(i) zero (ii) zero
b) (i) zero (ii) P1 V
c)
(i) P1 V (ii) zero
d) (i) P1 V (ii) P1 V
15.
The units of heat are
a)
J m-1
b) J m
c)
Cal
d) N m-2
16.
Heat added to a system is equal to
a)
a change in its internal kinetic energy
b) a change in its internal potential energy
c)
work done by it
d) sum of above all the three
17.
The area inside a closed curve on P-V diagram represents
a)
the condition of a system
b) work done on the system
c)
work done in a cyclic process
d) a thermodynamic process
18.
What will be the work done when 2 mole of an ideal gas are compressed reversibly from 10.00 bar to 50.00 bar at a constant temperature of 300 K.
a)
-14.01 kJ
b) 8.02 kJ
c)
4.02 kJ
102
d) -18.02 kJ
19.
An ideal gas occupies 1.25 dm3 at a pressure of 6 bar. If the gas is compressed isothermally at a constant pressure pext, to attain a volume of 0.500 dm3, the minimum possible value of pext is
a)
10 bar
b) 15 bar
c)
20 bar
d) 30 bar
20.
What will be the maximum amount of work done when 1 mol of an ideal gas is expanded from 2.0 dm3 to 4.0 dm3 at 27oC?
a)
7.78 kJ
b) -1.73 kJ
c)
-4.78 kJ
d) +11.73 kJ
103
Student Worksheet- 4
1.
What is meant by the term „Internal Energy‟ of the system? Is it a measurable property? What are the various forms of energy that gives rise to Internal energy of the system?
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2.
State first law of thermodynamics. Give examples from daily life in support of this statement. (You may search internet for finding some interesting examples!)
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3.
The mathematical statement of first law is ∆U = q + w. Following this relation predict how the internal energy of the system is increased or decreased?
Process
Sign convention
Work done on the system
Work done by the system
Heat absorbed by the system
Heat given out by the system
104
Effect on ∆U (increase/ decrease) 4.
Discuss the significance of the mathematical expression of the first law where the heat absorbed by a system is related to the internal energy and work done by the system. ________________________________________________________________________________
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________________________________________________________________________________
________________________________________________________________________________
5.
How will you infer from first law of thermodynamics that ∆U is a state function although q and w are both path functions?
(Hint: In a cyclic process, the system returns to its initial state! Can changing the path create or destroy energy? Can ∆U be different if different paths are followed?)
Example
Cyclic process
Case – I
(If ∆U is a path function )
Cyclic Process 1
Case – II (If
∆U is a path function )
Cyclic Process 2
Case – III (If
∆U is a state function )
Cyclic Process 1
Change in internal energy(∆U)
(+)
A (P1,V1) → B
(P2,V2) → A
(P1,V1)
(-)
A (P1,V1) → C
(P3,V3) → A
(P1,V1)
zero
A (P1,V1) → B
(P2,V2) → A
(P1,V1)
105
Effect on the surroundings Violation / conformation of law of conservation of energy Inference
6.
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
Name the two most common modes by which system and surroundings exchange their energies.
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________________________________________________________________________________
7.
What is the main limitation of the first law of thermodynamics?
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________________________________________________________________________________
8.
A sample of an ideal gas is allowed to expand isothermally.
(a)
Does the gas do work?
___________________________________________________________________________
(b)
Does the system exchange heat with the surroundings?
___________________________________________________________________________
(c)
What happens to the temperature of the gas?
___________________________________________________________________________
106
(d) What is the change in internal energy?
___________________________________________________________________________
9.
The densities of ice and water at 0oC are 0.90 g cm-3 and 0.98 g cm-3, respectively.
The latent heat of ice is 80 cal g-1. Calculate the change in internal energy when 5 kg of ice melts.
Formula used
Conversion of units Calculation
Result (with proper units)
10.
Define „enthalpy‟ of the system. What was the necessity to introduce the term enthalpy over internal energy for measuring the energy changes in the system?
Give the mathematical expression for enthalpy.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
11.
Using the thermodynamic relationships: ∆U = q + w ; w = - P∆V ; H = E + PV and
PV = nRT, derive the following relationships:
a)
∆U = qv
b) ∆H = qp
107
c)
∆H = ∆U + (∆n)RT
To Derive
∆U = qV
12.
H = qP
∆H = ∆U + (∆n)RT
Why are the heat changes reported, usually as enthalpy changes and not internal energy changes?
_________________________________________________________________________
________________________________________________________________________________
_______________________________________________________________________________
13.
What are the conditions under which ∆H becomes equal to ∆U?
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________________________________________________________________________________
________________________________________________________________________________
14.
A sample of argon gas at 1atm pressure and 27oC expands reversibly and isothermally from 1.25 dm3 to 2.50 dm3. Calculate w, q, ∆U and ∆H enthalpy change in this process.
Formula used
Conversion of units Calculation
Result (with proper units)
108
15.
a)
A cooking gas cylinder is assumed to contain 11.2 kg of isobutane. If a family needs 15000 kJ of energy per day for cooking, how long would the cylinder last? b)
Assuming that 30% of the gas is wasted due to incomplete combustion, how long would the cylinder last? The combustion of butane is given by
C4H10(g) +
O2(g) → 4CO2(g) + 5H2O(ℓ)
H = -2658 kJ mol-1
(Hints:C4H10(g) +
O2(g) → 4CO2(g) + 5H2O(ℓ)
H = -2658 kJ mol-1
58 g of C4H10 gives 2658 kJ of energy.
Calculate for 11.2 kg of C4H10)
Formula used
Conversion of units
Calculation
Result (with proper units) 16.
When a boy mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter the temperature of the resultant solution increases from 21.0 oC to
27.5oC. Assuming that the calorimeter absorbs only a negligible quantity of heat, the total volume of the solution is 100 mL, its density is 1.0 g mol -1 and its specific heat 4.18 J g-1.Calculate
a)
the heat change during mixing
b) the enthalpy change during the reaction
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O
(Hints: Qmixing = mxsxt))
109
Formula used
Conversion of units
Calculation
Result (with proper units) 17.
In the diagrams (i) to (iv) of figure, variation of volume by changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas will be
a)
positive in all the cases (i) to (iv) positive in cases (i), (ii) and (iii) but zero in case (iv)
b) negative in cases (i), (ii) and (iii) but zero in case (iv)
c)
18.
zero in all the four cases.
State which of the following will decrease the total energy content of the system.
a)
Work done by the system
b) Heat transformed from the surroundings
c)
Work done on the system
d) Decrease in volume of the system
19.
The change in the internal energy of a system
a)
is an extensive property
b) is a thermodynamic parameter
110
c)
is zero in a cyclic process
d) all of the above
20.
When an ideal gas expands at constant temperature, the change in internal energy is a)
positive
b) negative
c)
zero
d) equal to the work done by the gas
21.
Enthalpy of a system is
a)
an extensive property
b) ∆U + p∆V
c)
qp
d) all of the above
22.
For the reaction
3H2O(ℓ)
vapHm rU 3H2O(g)
= 10k cal mol–1 at 500 K and 1 atm pressure.
of the given reaction is
a) 27 kcal
b) -27 kcal
c) 13.5 kcal
d) 54 kcal
111
Student Worksheet- 5
1.
What is the basic difference between standard enthalpy of formation and standard enthalpy of reaction? Illustrate with suitable examples.
Standard enthalpy of formation
Definition
Example
Standard enthalpy of reaction
Definition
Example
2.
What is value of standard enthalpy of formation of an element in its most stable state? Is it a measurable property?
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112
3.
Diamond is an elementary form of carbon, yet its standard heat of formation is not taken as zero. Why?
________________________________________________________________________________
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________________________________________________________________________________
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4.
What are endothermic and exothermic reactions? What is the sign of ∆H for these reactions? ________________________________________________________________________________
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________________________________________________________________________________
5.
The standard enthalpy of formation of H2O( ) and H2O(g) are -285.8 and -241.8 kJ mol-1 respectively. What is the heat of vaporization of water per gram at 25 oC and
1 atm?
Desired chemical equation Calculation
Result (with proper units) 6.
From the data given below, at 298 K for the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O( ), calculate the enthalpy of formation of CH4(g) at 298 K.
Enthalpy of reaction = -890.5 kJ
113
Enthalpy of formation of CO2(g) = -393.5 kJ mol-1
Enthalpy of formation of H2O( ) = -286.0 kJ mol-1
Formula used
Calculation
Result (with proper units)
7.
The
standard
heats
of
formation
of CH4(g),
CO2(g)
and
H2O( )
are
-76.2 kJ mol-1, -398.8 kJmol-1 and -241.6 kJ mol-1 respectively. Calculate the amount of heat evolved by burning 1 m3 of methane measured at normal conditions.
Desired chemical equation Calculation
Result (with proper units)
8.
Calculate the standard enthalpy change for the following reaction
2H2O2( ) → 2H2O ( ) + O2(g)
When the standard enthalpy of formation of various compounds are
= -188 kJ mol-1
= -286 kJ mol-1
Identify whether the reaction is exothermic or endothermic
114
Desired chemical equation Calculation
Result (with proper units)
9.
Calculate ΔH° for the reaction N2H4( ) + O2 (g) → N2(g) + 2 H2O( ) from the reactions given below.
ΔH° (kJ)
(i) 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(ℓ)
ؘ1011.4
(ii) N2O(g) + 3 H2(g) → N2H4( ) + H2O(ℓ)
-317.25
(iii)
2 NH3(g) + 1/2 O2(g) → N2H4(ℓ) + H2O(ℓ)
-142.98
(iv)
H2(g) + 1/2 O2(g) → H2O( )
-285.83
Desired chemical equation Calculation
Result (with proper units) 10.
In the production of cement calcium carbonate, CaCO3, is converted to calcium oxide, CaO, according to the following balanced chemical equation.
CaCO3 (s) → CaO (s) + CO2 (g)
115
i. From the given standard molar enthalpies of formation, calculate ΔH° for this reaction ΔHf° (CaO, s) = -635.1 kJ mol-1; ΔHf° (CO2, g) = -303.5 kJ mol-1; ΔHf°
(CaCO3, s) = -1206.9 kJ mol-1
Desired chemical equation Calculation
Result (with proper units) ii. Is this an endothermic or exothermic reaction?
_____________________________________________________________________
_____________________________________________________________________ iii. CaCO3 actually exists in two forms. The calculation you did in part a was for the conversion of the form of CaCO3 known as calcite to CaO. If ΔH for the conversion of aragonite (the other form of CaCO3) to CaO is +178.1 kJ mol-1, what is ΔH for the conversion of calcite to aragonite?
Desired chemical equation Calculation
Result (with proper units)
116
11.
Calculate the bond energy of HC , if the bond energies of H - H and C - C are
433 and 242 kJ mol-1 respectively and the heat of formation of HC is -92 kJ mol-1.
Desired chemical equation Calculation
Result (with proper units) 12.
Compute the enthalpy of formation of liquid methanol in kJ mol-1 using the following data:
∆vapH (CH3OH) = 38 kJ mol-1
∆fH (H, g) = 218 kJ mol-1
∆fH (C, g) = 715 kJ mol-1
∆fH (O, g) = 249 kJ mol-1
∆HC-H = 415 kJ mol-1
∆H C-O = 356 kJ mol-1
∆H O-H = 463 kJ mol-1
Desired chemical equation Calculation
Result (with proper units) 117
13.
∆H for the given reaction is -150 kJ. Calculate the enthalpy of the A
C bond.
[Given, bond enthalpies of C-H = 414 kJ mol-1, H-H = 435 kJ mol-1, C-N = 293 kJ mol-1, N-H = 369 kJ mol-1]
Formula used
Calculation
Result (with proper units)
14.
15.
16.
Exothermic reactions are those
a)
which absorb heat at high temperatures
b)
in which heat is evolved
c)
which absorb heat at low temperatures and evolve heat at high temperatures
d)
where heat is evolved only at high temperatures
For endothermic reactions the sign given for heat exchanged is
a)
positive
b)
negative
c)
positive at high temperatures and negative at low temperatures
d)
positive at high temperatures and negative at low temperatures
Heat of combustion of carbon (graphite) is the same as
a)
heat of combustion of carbon (diamond)
118
b)
heat of formation of carbon dioxide
c)
both (a) and (b)
d)
neither (a) nor (b)
[Hint] The standard state of carbon is taken as carbon (graphite) and not carbon (diamond).
119
SOME COMMON MISCONCEPTIONS dq and dw where ever they are mentioned do not mean change in heat or change in work. dq stands for small amount of heat exchanged by the system. dw denotes small amount of work done.
Standard state of a system is defined for the pure substance at 1 bar pressure and a specified temperature and not at 1 bar pressure and 273.15 K temperature.
For gases the standard state is the one at which pure gas behaves ideally at one bar and a specified temperature. For liquids the standard state is the pure liquid at one bar and specified temperature. For solids it is the pure and most stable crystalline form of aggregation of solid at one bar and specified temperature.
120
CROSSWORD PUZZLE
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
121
ACROSS
Hints
1
Metal is a better---of heat than wood
7
---can neither be created nor destroyed.
8
Things that are cooled will---in length
9
System + surroundings
11
Change of state from liquid to gas
14
---reactions require energy in order to take place
16
Study of motion of heat and work
17
Device used to measure temperature
18
---processes take place on their own.
DOWN
Hints
2
Specific heat capacity is the energy needed to raise temperature of a substance by ---degree Celsius
3
Isothermal process takes place at constant---
4
Process taking place at constant volume
5
At equilibrium, change in Gibbs free energy
6
A measure of disordered energy in the system
10
Boundary or wall which does not allow heat to pass through
12
Lowest temperature known
13
Temperature scale with 100 degrees between freezing and boiling
15
Another word for freezing
122
ANSWERS
C
O
N
D
U
C
N
E
N
E
R
G
Y
T
O
E
I
Z
E
M
S
E
N
P
U
N
E
I
V
C
E
N
D
V
A
O
S
R
O
E
C
R
H
A
T
I
O
N
A
N
A
O
D
P
Y
U
I
A
C
B
I
T
U
E
S
Z
E
T
H
E
R
M
I
E
H
E
R
M
O
D
Y
N
M
A
F
T
U
I
C
C
S
I
O
R
M
O
M
E
T
E
R
N
T
A
N
E
O
U
S
R
O
A
T
R
S
O
C
O
I
U
P
R
R
S
S
T
T
L
H
P
R
C
B
L
T
E
R
123
N
ADDITIONAL RESOURCE LINKS:
Smith, J.M.; Van Ness, H.C., Abbott, M.M. (2005). Introduction to Chemical
Engineering Thermodynamics.
Introductory Thermodynamics: i.UIC; ii. U Pittsburgh; iii. Kids' Almanac; iv. Eden
Prairie High School; v. School of Champions; vi. Scitoys vii. THERMO spoken here
Engineering Thermodynamics: i.A Graphical Approach /
124
CENTRAL BOARD OF SECONDARY EDUCATION
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
CLASS
XI
UNIT-8
CHEMISTRY
Thermodynamics
Student's Manual
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
CLASS
CBSE-i
XI
UNIT-8
CHEMISTRY
Thermodynamics
Student's Manual
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
The CBSE-International is grateful for permission to reproduce and/or translate copyright material used in this publication. The acknowledgements have been included wherever appropriate and sources from where the material may be taken are duly mentioned. In case any thing has been missed out, the Board will be pleased to rectify the error at the earliest possible opportunity.
All Rights of these documents are reserved. No part of this publication may be reproduced, printed or transmitted in any form without the prior permission of the CBSE-i. This material is meant for the use of schools who are a part of the CBSE-International only.
Preface
Education plays the most important role in acquiring professional and social skills and a positive attitude to face the challenges of life. Curriculum is a comprehensive plan of any educational programme. It is also one of the means of bringing about qualitative improvement in an educational system. The Curriculum initiated by Central
Board of Secondary Education -International (CBSE-i) is a progressive step in making the educational content responsive to global needs. It signifies the emergence of a fresh thought process in imparting a curriculum which would restore the independence of the learner to pursue the learning process in harmony with the existing personal, social and cultural ethos.
The CBSE introduced the CBSE-i curriculum as a pilot project in few schools situated outside India in 2010 in classes I and IX and extended the programme to classes II, VI and X in the session 2011-12. It is going to be introduced in classes III, VII and for Senior Secondary classes with class XI in the session 2012-13.
The Senior Secondary stage of education decides the course of life of any student. At this stage it becomes extremely important for students to develop the right attitude, a willingness to learn and an understanding of the world around them to be able to take right decisions for their future. The senior secondary curriculum is expected to provide necessary base for the growth of knowledge and skills and thereby enhance a student's potential to face the challenges of global competitiveness. The CBSE-i Senior Secondary Curriculum aims at developing desired professional, managerial and communication skills as per the requirement of the world of work. CBSE-i is for the current session offering curriculum in ten subjects i.e. Physics Chemistry, Biology, Accountancy,
Business-Studies, Economics, Geography, ICT, English, Mathematics I and Mathematics II. Mathematics at two levels caters to the differing needs of students of pure sciences or commerce.
The Curriculum has been designed to nurture multiple intelligences like linguistic or verbal intelligence, logicalmathematical intelligence, spatial intelligence, sports intelligence, musical intelligence, inter-personal intelligence and intra-personal intelligence.
The Core skills are the most significant aspects of a learner's holistic growth and learning curve. The objective of this part of the core of curriculum is to scaffold the learning experiences and to relate tacit knowledge with formal knowledge. This involves trans-disciplinary linkages that would form the core of the learning process.
Perspectives, SEWA (Social Empowerment through Work and Action), Life Skills and Research would be the constituents of this 'Core'.
The CBSE-i Curriculum evolves by building on learning experiences inside the classroom over a period of time.
The Board while addressing the issues of empowerment with the help of the schools' administering this system strongly recommends that practicing teachers become skilful and lifelong learners and also transfer their learning experiences to their peers through the interactive platforms provided by the Board.
The success of this curriculum depends upon its effective implementation and it is expected that the teachers will make efforts to create better facilities, develop linkages with the world of work and foster conducive environment as per recommendations made in the curriculum document.
I appreciate the effort of Dr.Sadhana Parashar, Director (Training), CBSE, Dr. Srijata Das, Education Officer,
CBSE and Ms. Anjali Chhabra, Assistant Education Officer, CBSE and their teams involved in the development of this document.
The CBSE-i website enables all stakeholders to participate in this initiative through the discussion forums. Any further suggestions on improving the portal are always welcome.
Vineet Joshi
Chairman, CBSE
Acknowledgements
Advisory
Shri Vineet Joshi, Chairman, CBSE
Dr. Sadhana Parashar, Director (Training), CBSE
Ideators Classes XI and XII
Prof. A K Bakshi
Dr. N K Sehgal
Prof. Kapil Kapor
Ms. Renu Anand
Dr. Barkatullah Khan
Ms. Avnita Bir
Conceptual Framework
Shri G. Balasubramanian, Former Director (Acad), CBSE
Ms. Abha Adams, Consultant, Step-by-Step School, Noida
Dr. Sadhana Parashar, Director (Training), CBSE
Ms. P Rajeshwari
Ms. Gyatri Khanna
Mrs. Anita Makkar
Prof. Biswajit Nag
Ms. Usha Sharma
Dr. Niti Nandini Chatnani
Dr. Anil K Bali
Dr. Preeti Tewai
Dr. Deeksha Bajpai
Mr. S K Agarwala
Ms. Neeta Rastogi
Dr. Anshu
Dr. Rajesh Hassija
Mr. Mukesh Kumar
Dr. Om Vikas
Material Production Groups: Classes XI-XII
English :
Ms. Gayatri Khanna
Ms. Renu Anand
Ms. P Rajeshwary
Ms. Sandhya Awasthi
Ms. Manna Barua
Ms. Veena Bhasin
Ms. Urmil Guliani
Ms. Sudha Ravi
Mr. Anil Kumar
Ms. Vijaylaxmi Raman
Ms. Neerada Suresh
Ms. Himaal Handoo
Chemistry :
Dr. G S Sodhi
Dr. Vimal Rarh
Dr. Shalini Baxi
Dr. Vinita Arora
Dr. Vandana Soni
Ms. Charu Maini
Ms. Rashmi Sharma
Ms. Kavita Kapoor
Biology :
Dr. Ranjana Saxena
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Dr. P Chitralekha
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Ms. Malini Sridhar
Geography:
Dr. Preeti Tewari
Ms. Rupa Das
Mr. S Fazal Daoud Firdausi
Ms. Neena Phogat
Ms. Sujata Sharma
Ms. Deepa Kapoor
Ms. Bharti Malhotra
Ms. Isha Kaushik
Mr. Riyaz Khan
Physics :
Dr. B. Biswal
Ms. Namarata Alwadhi
Mr. Dhirender Sharma
Ms. Vandana Banga
Mr. Vivek
Economics:
Mr. S K Agarwala
Ms. Ambika Gulati
Ms. Nidhi Singh
Ms. Malti Modi
Ms. Sapna Das
Ms. Ingur Agarwal
Ms. Shankar Kulkarni
Mr. Sandeep Sethi
Mathematics :
Dr. Sushil Kumar
Mrs. Monica Talwar
Mrs. Charu Dureja
Mrs. Seema Juneja
Dr. H L Bhatia
Dr. Sushma Bansal
Mrs. Neeru Aggarwal
Dr. Saroj Khanna
Accountancy :
Mr. S S Sehrawat
Dr. K Mohna
Dr. Balbir Singh
Ms. Bhupendra Kriplani
Ms. Shipra Vaidya
Mr. Sandeep Sethi
Business Studies :
Dr. S K Bhatia
Ms. Meenu Ranjan Arora
Mrs. Shegorika
Mr. Sandeep Sethi
Ms. Usha Sharma
Ms. Komal Bhatia
Ms. Ravisha Aggarwal
ICT :
Mr. Mukesh Kumar
Ms. Nancy Sehgal
Ms. Purvi Srivastava
Ms. Gurpreet Kaur
Coordinators:
Ms. Sugandh Sharma, EO
Dr Rashmi Sethi, EO
Ms. S. Radha Mahalakshmi, EO Mr. Navin Maini, RO (Tech)
Ms. Madhuchhanda, RO (Inn) Shri Al Hilal Ahmed, AEO
Ms. Anjali Chhabra, AEO
Shr. R. P. Singh, AEO
Shri R. P. Sharma,
Consultant (Science)
Ms. Reema Arora,
Consultant (Chemistry)
Mr. Sanjay Sachdeva, SO
Ms. Neelima Sharma,
Consultant (English)
Contents
Preface
Acknowledgement
Syllabus Coverage
1
Learning Outcomes
3
Mind Map
6
Contents
Summary
Students’ Worksheets
7 - 83
84
85 - 119
Common Misconceptions
120
Crossword Puzzle
121
Additional Resource Links
124
UNIT – 8: THERMODYNAMICS
SYLLABUS COVERAGE
8.1
Introduction to thermodynamics
8.2
Thermodynamic Terms
-
System and surrounding
-
Types of system
-
State of a system
-
State function and path function
-
Extensive and intensive properties
-
Reversible and irreversible process
8.3
8.4
8.5
Thermodynamic Quantities
-
Work
-
Heat
First Law of Thermodynamics
-
Internal energy
-
Enthalpy
-
Heat capacity
-
Measurement of ∆U and ∆H
Thermochemistry
-
Enthalpy change in a reaction
-
Endothermic and exothermic reactions
-
Enthalpy changes during phase transformations
-
Standard enthalpy of formation
-
Thermochemical equations
1
8.6
-
Hess‟s law of constant heat summation
-
Enthalpies for different types of reactions
Spontaneity
-
Entropy
-
Second law of thermodynamics
-
Gibb‟s energy change for spontaneous and non-spontaneous processes
-
Criteria for equilibrium
-
Third law of thermodynamics
2
LEARNING OUTCOMES
At the end of this unit students would be able to –
Understand the meaning of the term Thermodynamics.
Differentiate between the terms system and surrounding.
Distinguish between open system, closed system and isolated system.
Understand the meaning of state of a system, state function and path function.
Distinguish between extensive and intensive properties.
Comprehend the concept of thermodynamic reversibility.
Distinguish between reversible and irreversible processes.
Recall the physical interpretation of work.
Understand the work done during the process of expansion and compression
Write the S.I. units of work.
Deduce a mathematical expression to calculate the work done during isothermal expansion and compression of an ideal gas and during free expansion.
Solve numerical for calculating work done during expansion and compression.
Comprehend the difference in the work done during expansion and compression when carried out reversibly and irreversibly.
Understand the concept of heat.
Know the IUPAC sign conventions of work and heat.
Recognize that work and heat are not state functions.
Explain the concept of internal energy.
Show that internal energy is a state function.
State and deduce the mathematical expression for the first law of thermodynamics.
Interpret the first law to explain its consequences.
Solve numerical based on the first law of thermodynamics.
3
Recognize the need for enthalpy function.
Derive a mathematical expression for enthalpy.
Appreciate the physical significance of enthalpy.
Comprehend that Internal energy change is the heat evolved or absorbed at constant volume and Enthalpy change is the heat evolved or absorbed at constant pressure.
Deduce a mathematical relation between ΔH and ΔU and solve numerical.
Explain heat Capacity, molar heat capacity and specific heat capacity.
Differentiate between Cp and Cv.
Derive a mathematical relation between Cp and Cv for an ideal gas.
Know the procedure of measurement of ΔH and ΔU in a calorimeter.
Define reaction enthalpy.
Recognize that heat changes accompany chemical reactions.
Compare exothermic and endothermic reactions.
Define standard enthalpy of reactions.
Examine the enthalpy changes accompanying phase transformations such as fusion, vaporization, sublimation.
Describe and Apply the standard enthalpy of formation values to calculate the enthalpy change for the reaction.
Know the convention for writing Thermochemical equations.
Comprehend the Hess‟s Law of Constant Heat Summation.
Discuss the enthalpies of different types of reactions such as Standard enthalpy of combustion, Enthalpy of atomization, Bond enthalpy, Enthalpy of solution.
Construct a Born-Haber Cycle for determination of lattice enthalpy.
Solve numerical based on Hess‟s Law of Constant Heat Summation.
Understand the limitations of first law of thermodynamics.
Recognize that all natural processes will tend to proceed spontaneously in one direction only.
4
Comprehend with the help of examples that decrease in enthalpy is not a criterion for spontaneity.
Interpret randomness or disorder with the help of examples.
Explain the thermodynamic function, Entropy and write its mathematical expression.
Analyze ΔStotal as the criterion for examining spontaneity.
Examine the statement of second law of thermodynamics.
Understand the need for introducing Gibb‟s free energy and write its mathematical expression.
Derive the criteria for spontaneity given by ΔG.
Interpret the effect of temperature on spontaneity of reaction.
Derive a mathematical expression showing the relation between free energy and equilibrium constant.
Solve numerical based on second law of thermodynamics.
State the third law of thermodynamics.
5
MIND MAP
6
8.1
INTRODUCTION TO THERMODYNAMICS
Warm up
Recall the concept of various states of matter. Try to represent a specific attribute/property of a state of matter. Now ask yourself whether matter can be converted from one form to the other.
Understand the transformation of matter from one form to another for example melting of ice into liquid water and then evaporation into water vapour. This activity can be performed in Chemistry lab by using a Bunsen burner.
Now, interpret your observation. Think, how heat is playing a role in transformation of solid to liquid to gas?
8.1.1.
Origin of the term Thermodynamics
Thermodynamics is the branch of physical science concerned with heat and its relation to other forms of energy and work. The word thermodynamics was coined from two Greek words, thermos (heat) and dynamics (movement).
Thus thermodynamics mean “movement of heat”. It is therefore, a study of heat and its transformation to various other forms of energy, especially mechanical energy.
Thermodynamics is a physical science that involves the study of the changes in the energy of the system when the temperature, pressure or volume of a system, are changed or when a chemical transformation takes place. It deals with the relationship of energy with heat and work and the inter conversion of various forms of energy. Therefore, thermodynamics is sometimes also known as Energetics.
Chemical Thermodynamics is that branch of thermodynamics that deals with energy changes accompanying chemical transformations. It investigates the feasibility of a chemical change and tries to find the answer to the question – Why does a chemical reaction take place?
7
8.1.2.
8.1.3.
Characteristics of thermodynamics
It is a study of macroscopic systems, i.e., objects or processes that are of a size that is measurable and observable by the naked eye.
Usefulness of thermodynamics lies in correlating the observable properties of substances with some of the non-observable ones.
It points out the feasibility, extent and direction of changes taking place in a process. These changes may be physical processes or chemical reactions.
It is not necessary to know the nature of the fundamental particles or the mechanism of the phenomenon. The deductions are made using basic laws.
Thermodynamics can be studied by dealing with the energy changes taking place at the macroscopic level without emphasizing the changes that occur at molecular level.
Limitations of thermodynamics
It deals with materials in bulk without any reference to the internal structure and microscopic properties or behaviour.
It deals with stationary or equilibrium states only.
It gives the feasibility, direction and extent of a change but not the time taken for such a change to take place. The rate at which a process undergoes completion is studied under a separate branch of physical chemistry, known as, chemical kinetics.
Student Activity- 1
Objective: To understand the First Law of Thermodynamics, i.e., the heat added to a system is partly used to do work of expansion and the rest is used to increase the internal energy of the system.
Materials Required:
1.
a pack of popcorn kernels
8
2.
a pan with lid
3.
gas stove
Procedure:
1.
Open the pack of popcorn kernels and put all of them in the pan.
2.
Place the pan on the gas stove over low flame and cover it with the lid.
3.
Wait for some time for the pan to get hot. After few minutes, kernels start popping with a pop sound.
4.
Wait for another few minutes for all the kernels to pop.
5.
Put off the gas stove and leave the pan covered with the lid.
Now answer the following questions:
1.
What happened to the kernels when the pan got hot?
_________________________________________________________________________
_________________________________________________________________________
2.
Why do you think kernels start popping up?
_________________________________________________________________________
_________________________________________________________________________
3.
From where the popcorns are getting energy to pop?
_________________________________________________________________________
_________________________________________________________________________
4.
What do you observe happened to the lid of the pan when all the kernels popped simultaneously? Why has it happened?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
9
5.
Relate your observations with the first law of thermodynamics.
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
8.2
THERMODYNAMIC TERMS
System: It is that part of the universe on which theoretical or experimental investigations are carried out. A system is said to be a part of the universe which is separated from the rest of the universe by a real or imaginary well defined boundary. For example, if the effect of pressure on the volume of the gas has to be studied, then the system is the gas taken in a container. Similarly, any solution taken in a beaker is known as the system, when its properties are being studied.
Surroundings: The rest of the universe in the neighbourhood of the system is known as its surroundings. It lies outside the system and is also called the environment of the system. The system can exchange matter or energy or both with its surroundings.
A system and its surrounding are together known as the Universe.
Boundary: Anything that separates the system from the surroundings is known as a boundary. It is also known as the wall of the system. The system interacts with its surroundings across the boundary.
8.2.1.
Classification of systems based on the type of boundary
The system exchanges matter and energy with the surroundings across the boundary. Therefore, the boundary decides the relationship of the system with its surroundings. There can be several categories of systems depending on the nature of boundary present.
1.
Open system: is a system which has a permeable boundary such that both matter and energy can move across the boundary. So, the exchange of matter (mass, m) and energy (U) can take place between the system and the surroundings. For an open system Δm ≠ 0 and ΔU ≠0.
10
8.2.2.
2.
Closed system: is a system which has an impermeable but diathermic boundary such that matter cannot flow across the boundary but energy exchange can take place between the system and the surroundings. For a closed system Δm= 0 and ΔU≠0.
3.
Isolated system: is a system which has an adiabatic boundary, i.e., neither matter nor energy can be exchanged between the system and surroundings. For an isolated system Δm = 0 and ΔU =0.
State of the System
The physical state and other important properties and parameters defining the system are decided by the pressure (p), temperature (T), volume (V) and the amount (number of moles, n) of the system. These four variables define the state of the system. Therefore, these variables are known as state variables.
Thermodynamic process: It is any operation or process that brings about a change in the state of the system. Expansion, compression, heating and cooling are some of the examples of a thermodynamic process. The initial state of the
11
system is defined by the pressure, temperature, volume and amount of the system before the change takes place. The final state of the system is defined by the values of pressure, temperature, volume and amount of the system after the change has taken place. A process is said to have taken place whenever there is a change in any one or more of the state variables. For any state variable Y, if the change in the state variable is large it is represented by
ΔY (delta Y). If the change is infinitesimal (very small) it is represented by dY.
8.2.3.
Classification of processes based on the change in thermodynamic variables Thermodynamic processes are classified into various categories depending upon the thermodynamic properties that are undergoing a change during the process. These categories are as follows:
1.
Isothermal process: It is a process during which the temperature of the system remains constant, that is, ∆T = 0.
2.
Adiabatic process: It is a process during which the system does not exchange heat with the surroundings, that is, q = 0.
3.
Isobaric process: It is a process during which the pressure of the system remains constant, that is, ∆P = 0.
4.
Isochoric process: It is a process during which the volume of the system remains constant, that is, ∆V = 0.
5.
Cyclic process: It is the process during which the system after undergoing a series of changes comes back to the initial state, that is, the series of changes take place in a cyclic manner.
Path: The sequence of steps taken by a system during a thermodynamic process starting from the initial state, through an intermediate state to the final state is known as a path. The path may consist of one or more than one steps.
8.2.4.
State function
A function that depends only on the state of the system and not on the process by which this state is achieved is known as a state function.
12
Thermodynamic properties like internal energy (U), enthalpy (H), entropy (S) and free energy (G and A) change with the change in variables like temperature, pressure, volume and amount of the system. The changes in these properties depend upon the initial and final states of the system. They do not depend upon the path followed during the change. Therefore, these thermodynamic properties are also state properties or state functions.
8.2.5.
Path Function
It is a thermodynamics property that depends upon how the process is carried out, that is, it depends upon the path taken by the system during the thermodynamic process.
Work and heat are path functions. Therefore, there is no such thing as initial work and final work, therefore, w1 and w2 do not mean anything. w is the work done when the system goes from an initial state to the final state.
Similarly, since q is a path function we only talk about the total heat exchange between the system and the surroundings during a process and not the initial heat and the final heat.
8.2.8.
Intensive and extensive properties
Extensive properties are those properties that depend upon the size of the system. They are additive in nature. This means that if the system is divided into small parts then the total value of a property is obtained by adding up the value of that property in all the parts of the system.
Some of the important extensive properties are: mass, volume, amount
(number of moles), energy, enthalpy, entropy, free energy and heat capacity.
Intensive properties are those properties that do not depend upon the size of the system. These properties are not added up to get the total value of a given property of the system. They have the same value in all the parts of the system. Some of the important intensive properties are: pressure, temperature, density, concentration, mole fraction, refractive index, surface tension, viscosity and specific heat capacity.
Let a container containing an ideal gas be divided into four compartments namely A, B, C, D as shown in Figure 1. Let the pressure in each compartment
13
be PA, PB PC, PD, respectively. Let the temperature be TA, TB, TC, TD; volume be
VA, VB, VC, VD; and the amount be nA, nB, nC, nD. When the partition is removed without changing the state of the system then, the total volume of the container is
V=VA+ VB+ VC+ VD
The total amount of the gas is n = nA+ nB + nC + nD
However, the total pressure of the container is not equal to the sum of the pressure of each compartment. Similarly, the temperature of the container is also not equal to the sum of the temperatures of each compartment. Thus, it is seen that the volume and amount depend upon the size of the system while temperature and pressure do not depend upon the size of the system.
Another illustration of extensive and intensive properties is of the 1mol dm–3 sodium chloride solution taken in a one litre flask. This solution is divided unequally among four students and each student measures the mass, volume, density, temperature, surface tension and viscosity of the solution. What is the expected result?
All the four students will report almost the same values of density, temperature, surface tension and viscosity of the solution within experimental error. The mass and the volume of the solutions will be different but the volumes of solutions of each student will add up to one litre. Again it is observed that the intensive properties, that is, density, temperature, surface tension and viscosity of the solution do not depend upon the size of the system while the extensive properties depend upon the size.
14
List of some extensive and intensive properties
Extensive mass, volume, amount, heat capacity, internal energy, properties enthalpy, entropy, free energy
Intensive pressure, temperature, density, surface tension, properties viscosity, refractive index, molarity, molar volume, specific heat capacity, molar heat capacity, molar energy
8.2.7.
Classification of processes based on the path
Depending upon the path followed by the system there are two types of processes: reversible and irreversible processes.
Reversible process: A process is said to be reversible if the system undergoes a change in the state through a sequence of intermediate states, each one of which is an equilibrium state. In this manner, the system can be restored to the initial state by following the same sequence of steps in the reverse order. That is, a process is said to be reversible if it can proceed in forward and backward directions by a small change in state variables. When the system comes back to the initial state the surroundings are also restored to their initial state. That is, the process is said to be reversible only if the system can retrace its steps without any net change in the system and the surroundings. The process should be reversible in both forward and backward directions. Each reversible process is slow but each slow process may or may not be reversible.
Irreversible process: A process is said to be irreversible if the change takes place without the system being in equilibrium with its surroundings at any stage during the change. The system cannot retrace its path without the help of an external source. Hence, the system can attain its initial state if and only if some change is brought about in the surroundings.
Examples: Most of the natural processes and the spontaneous processes are irreversible processes.
Equilibrium: A system exists in a state of equilibrium if its macroscopic properties such as temperature, pressure, volume and mass remain constant without the help of an external source.
15
Student Activity- 8.2
Objective: To understand the statement of second law of thermodynamics and direction of heat flow.
Materials Required:
1.
metal blocks of the same size and dimensions – 2 No.s
2.
thermometer
Procedure
1.
Place one block of metal in the freezer for about 10 minutes. Before taking it out from the freezer, measure its temperature using a thermometer.
2.
At the same time, immerse the other block in boiling water. Measure its temperature also.
3.
Get the two blocks and place them in contact for 3–5 minutes. Get the temperature of each block.
4.
Record your data in the table below.
Blocks
Temperature (°C)
Before contact
After contact
Block A (hot)
Block B (cold)
Answer the following questions:
1.
What did you observe after the blocks were placed in contact with each other?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
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2.
What do you conclude from your observation?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
3.
If we assume that the two blocks are simply interacting with each other and not with the surroundings, how can you relate the heat changes in the two blocks?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
4.
What is the natural direction of heat flow? Can it be reversed?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
8.3
THERMODYNAMIC QUANTITIES
8.3.1. Work
In thermodynamics, work is defined as any quantity that flows across the boundary of a system during a change in its state and is completely convertible into the lifting of a weight in the surroundings. The symbol of work is w. SI unit of work is Joule (J).
It has the following characteristics.
(i)
Work appears only at the boundary of a system.
17
(ii)
Work appears only during a change in state, that is, only during a thermodynamic process.
(iii) Work is manifested by an effect in the surroundings.
(iv) During an adiabatic process, work results in the change in the internal energy of the system. When the work is done by the system, the system uses its internal energy. Therefore, internal energy of the system decreases. When work is done on the system, the internal energy of the system increases.
(v)
Work is a path function, that is, it is an inexact function. If the system is changed from the initial state to the final state by two different processes, then the work done will be different for the two processes.
Mathematical Expression for Work
Mechanical work is the result of displacement of a body when the force is applied on it. It is related to the force (F) and displacement by the following equation. w=F
Δx
When the system does mechanical work, it can be related to the change in the volume of the system.
This can be illustrated by considering a system consisting of a gas taken in a cylinder with a piston at the top. This piston is assumed to be mass less and frictionless; therefore, no work is done to overcome friction when the piston moves up and down. An external pressure (p) is applied on the system. When the gas expands, the piston moves up by a distance Δx. The force is related to the external pressure by the equation
Force = Pressure
area = p
A
Here, A is the area of cross section of the base of the piston. Substituting the expression of force in the above equation of w, w= p
Since, A
A
Δx x = ΔV
(area
distance = volume)
18
w = pΔV
….(1)
When the displacement is infinitesimal, i.e., dx the change in volume is represented by dV and the work done is dw. Therefore, dw = pdV
….(2)
Sign Convention in thermodynamics
In thermodynamics studies, the emphasis is on the system. Any change that results in an increase in the energy of the system is taken as positive. During compression, the internal energy of the system increases, therefore, the work done on the system is taken as positive. Similarly, the work done by the system is done at the cost of the internal energy of the system; therefore, the work done by the system is taken as negative. Hence, the Equations (1) and (2) are modified as w = −pΔV
.....for finite change
and dw = − pdV
.....for infinitesimal change
It can be shown that the above formula will give the positive value of w during compression when the work is done on the system and the negative value of w during expansion when the work is done by the system.
Figure 8.1: Compression and expansion of a gas taken in a cylinder
Initially some weights are kept on the piston so that the external pressure is equal to the internal pressure and the piston does not move. Then the weight on the piston is increased to compress the gas and move the piston down. Here, work is done on the gas and it can be calculated using Equation (1).
19
w = –pex ΔV pex is the external pressure acting on the gas. During compression, the volume of the gas decreases, so, ΔV is negative. Therefore, the value of w is positive when the work is done on the gas.
To expand the gas some weights are removed from the piston and the gas expands lifting the piston with the remaining weights. During expansion, work is done by the gas to lift the weights. Volume of the gas increases and so ΔV is positive. Therefore, the value of w is negative when work is done by the gas.
8.3.2. Heat
In thermodynamics heat is defined as a quantity that flows across the boundary of a system, during a change in its state by virtue of the difference in temperature between the system and its surrounding. Heat flows from a region of higher temperature to a region of lower temperature. The symbol of heat is
q. SI Unit of heat is Joule (J).
Heat appears only at the boundary of the system
Heat appears only during the change in state
Heat is manifested by an effect in the surroundings
It is a path function
Sign Convention
The heat absorbed by the system is represented by q. When the system absorbs heat, that is, heat flows from the surroundings to the system, the internal energy of the system increases provided no work is done. Therefore, q is positive. When the system loses heat, that is, heat flows from the system to the surroundings, the internal energy of the system decreases provided no work is done. Therefore, q is negative. q is positive when heat is absorbed by the system q is negative when heat is lost by the system.
20
8.3.3. Work done during expansion or compression of an ideal gas
Let the system consist of an ideal gas. The mechanical work on the gas can be done either reversibly or irreversibly. During mechanical work, the pressure and the volume of the gas change. These changes can take place under isothermal conditions or under adiabatic conditions. Therefore, the various processes can be
(i)
Isothermal reversible volume change
(ii)
Isothermal irreversible volume change
(iii)
Adiabatic reversible volume change
(iv)
Adiabatic irreversible volume change
Since work is a path function, it should depend upon the nature of process.
Thus, we shall derive the expressions for work done for all the four processes and it will be proved that the amount of work done in all the four cases will be different. A.
Work involved in reversible process
The reversible work of expansion or compression can be illustrated by the following set up.
An ideal gas is taken in a container fitted with a massless and frictionless piston. On the top of the piston a container with a large number of weights are kept. The weights exert pressure on the gas. This pressure is known as external pressure, Pex. Initially, the external pressure is balanced by the pressure exerted by the gas on the piston, that is, Pin.
Pex = Pin
This state is an equilibrium position. During expansion, the weight on the piston is decreased slightly. As a result the external pressure becomes infinitesimally smaller than Pin by the amount dP. The new external pressure will be P'ex
P'ex = Pin – dP
21
That is, the pressure exerted by the gas is infinitesimally different from the new external pressure on the gas. The system is said to be infinitesimally removed from the equilibrium state. To maintain equilibrium, the gas expands infinitesimally and the piston moves up. If the process of decreasing the external pressure infinitesimally is continued, then after a large number of steps, there is a significant change in the volume of the gas. At any stage during the expansion, the internal pressure is practically equal to external pressure,
P'ex. So each step is an equilibrium step. The work done by the gas at each step is (– dw). According to Equation (2) dw is given by the expression,
– dw = P'exdV = (Pin – dP)dV ≈ Pin dV as dPdV is very small it is neglected. Pin keeps on changing after each stage of expansion. During compression the weight on the piston is increased slightly, and the external pressure increases slightly after each addition and becomes greater than the internal pressure. To maintain the equilibrium, the gas compresses and the piston moves down. When all the weights have been replaced, the gas comes back to the original position. The work done during each step of compression will be equal to dw.
22
B.
Work involved in irreversible process
When the gas expands or compresses rapidly the process is irreversible in nature. During the expansion, the pressure on the gas is decreased by a finite amount. That is, the new external pressure is less than the pressure exerted by the gas on the piston,
P'ex < Pin
The equilibrium is disturbed and the gas expands without maintaining the equilibrium with the surroundings. The work done by the gas is given by–dw.
–dw = P'exdV
The gas expands against constant external pressure at any given stage of expansion. The total work done by the system (−w) is equal to
−w = P'ex∆V or w = −P'ex∆V
Similarly, during compression, the external pressure on the gas is increased such that the new external pressure is much greater than the pressure exerted by the gas on the piston. So, the gas compresses without maintaining the equilibrium with the surroundings.
The reversible as well as irreversible work done can take place under isothermal as well as adiabatic conditions depending upon the boundary of the container. C.
Work Done During Isothermal Volume Change for an Ideal Gas
During an isothermal process the temperature of the system remains constant.
The system may exchange heat with the surroundings to maintain the temperature. The amount of the gas remains constant.
Reversible process: The process is carried out in an infinite number of steps.
The work done in a step is given by the equation dw = –PexdV
Here, Pex is the external pressure on the gas.
23
Let n be the amount of an ideal gas at temperature T. Let the initial and the final states of the gas are as shown:
Initial state
Final state
Pressure
P1
P2
Volume
V1
V2
In the reversible volume change the external pressure differs from the internal pressure by the expression
Pex = (P ± dP). where, +and−signs are meant for the compression and expansion, respectively and P is the internal pressure, i.e., the pressure of the gas.
Substituting this in the expression of dw, we get, dw = −(P ± dP)dV = –PdV −(± dPdV)
Since dPdV is a product of two differentials it may be neglected. Hence, dw = –PdV
Total work done during the process is obtained by integrating the above equation V2
w
dw
PdV
….(3)
V1
Since the pressure is also changing during the process, P cannot be taken out of the integral. However, for an ideal gas,
P = nRT / V w isoth rev ….(4)
V2
(ideal)
V
nRT nV V2
1
nRT dV V
V1
nRTn
2.303nRTog
V2
nRT
dV
V
V1
V2
V1
V2
V1
….(5)
24
Note: During the expansion process, V2 > V1. Therefore, log (V2/ V1) is greater than zero. Since n, R and T are always positive; the work done by the ideal gas during isothermal reversible expansion is always negative. During the compression process, V2 < V1. Therefore, log (V2/ V1) is less than zero and the work done is positive.
At a constant temperature for an ideal gas, P1V1 = P2V2, therefore, Equation (5) in terms of pressure becomes
wisoth rev (ideal)
2.303nRTlog
P1
P2
….(6)
Graphical representation of isothermal reversible expansion of a gas
Figure 8.2a illustrates the magnitude of the work involved in one step in which the volume change (= volume of gas at point d – volume of gas at point c) is infinitesimally small against an external pressure of P1. It is obvious that this work is equal to the area of trapezoid abcd. Similarly, the magnitude of the work done by the gas during next step will be the area of the adjacent trapezoid. Thus the total magnitude of the work done by the gas during various steps of a reversible process will be equal to the sum of the areas of all the trapezoids from the initial state A to the final state B (Figure 8.2b). This is equal to the area of the region ABCD.
Work done by an ideal gas under reversible isothermal conditions
25
Graphical Representation of Isothermal Reversible Compression
During isothermal compression, the pressure is increased slowly from P2 to P1 in an infinitesimal number of steps and at each stage the volume decreases slightly. According to the Figure 8.3, the change starts from point A (P2, V2) and goes till the point B (P1, V1). The total work done on the gas is equal to the area of the region ABCD.
Figure 8.3: Pressure – volume work done during reversible isothermal compression. Comparison of Isothermal Reversible Work During Expansion and
Compression
On comparing Figures 8.2b and 8.3 it is seen that the magnitude of the work done is same if P1, P2 and V1, V2 in the two figures are same. However, the sign of work done is different.
Work done during isothermal reversible expansion = w = –2.303nRT log
V2
V1
Work done during isothermal reversible compression = w = –2.303nRT log
V2
V1
Note: The work done on an ideal gas to bring it back to its initial state is equal to the work done by the gas during expansion for an isothermal reversible process. Therefore, in a cyclic process from A to B and back from B to A, the total work done is zero.
26
Irreversible volume change
In this case pressure may be changed in one step or more than one step. If the volume is changed from V1 to V2 in one step, then the total work done is given by the expression
V2
w
V1
PexdV
If the gas is allowed to expand unrestrictedly, then the external pressure is equal to the final pressure of the gas after expansion. However, the expansion can be done against the external pressure smaller than the final pressure of the gas and the desired expansion may be achieved by a set of stops. If Pex = P2 then the expression of work is given by
V2
w
V1
....(7)
P2 dV
Since P2 is constant we get w isoth irr V2
(ideal)
P2 dV
P2 (V2
....(8)
V1 )
V1
Note: For expansion V2 > V1 and w is negative indicating that the work is done by the gas. For compression V2 < V1 and w is positive indicating that the work is done on the gas.
Since for an ideal gas, V2 = nRT /P2 and V1 = nRT/P1, the expression of work in terms of pressure is
w isoth irr (ideal)
P2
nRT
P2
nRT
P1
nRT 1
P2
P1
….(9)
Graphical Representation of Isothermal Irreversible Expansion
During an irreversible process, the external pressure is decreased from P1 to P2 and the gas expands from V1 to V2. The total work done is equal to –P2(V2 – V1).
From Figure 8.4, (V2 – V1) = CD and P2 = DE.
Therefore, the magnitude of the work done will be equal to w = ED
CD = area of the region BCDE
27
Figure 8.4: Pressure – volume work done by ideal gas under irreversible isothermal conditions
Irreversible Expansion When the Expansion of the Gas is Restricted
In this case, the pressure on the gas is decreased by a finite amount and then the gas is allowed to expand isothermally under an irreversible condition.
However, the expansion is stopped by blocking the movement of the piston with the stoppers as shown in Figure 8.5. As a result, the final pressure of the gas is not equal to the external pressure exerted on the gas. Here, final pressure,
P2 is greater than external pressure. The final volume (V'2) will be less than the situation when the expansion of gas is not stopped (V2).
Figure 8.5: Irreversible expansion when P2 > Pex
28
The gas expands against the external pressure. Therefore, the work done by the gas will be equal to w isoth irr (ideal)
V2
Pex dV
Pex (V2
V1 )
V1
It is important to note that the expression of work done during the expansion involves the external pressure and not the final pressure of the gas. However, during the unrestricted expansion the final pressure is equal to the external pressure. The work done by the gas during restricted expansion will be less as compared to the case when the gas is allowed to undergo unrestricted expansion. Graphical Representation of Isothermal Irreversible Compression
The work done on the gas during isothermal irreversible compression is given by the expression, P1(V1–V2). From the Figure 8.6, P1=C'D' and
(V1– V2) = A'D'.
Therefore, the work done on the gas is equal to w = C'D'
A'D' = area of the region A'B'C'D'.
Figure 8.6: Pressure–volume work done during irreversible isothermal compression. 29
Comparison of Isothermal Irreversible Work During Expansion And
Compression
The gas expands from State I to State II irreversibly and then it is compressed irreversibly to State III.
State I (P1,V1, T1 )
isothermal expansion
State II (P2,V2, T1)
isothermal compression
State III (P1,V1, T1 )
As can be seen, State I and State III are identical. Therefore, when the process is complete the system comes back to the initial state. On comparing the magnitude of the work done during irreversible expansion and irreversible compression, it can be seen that the work done on the gas to bring it back to the initial state is different from the work done by the gas during expansion.
From Figures 8.4 and 8.6 area of the region A'B'C'D' > area of the region BCDE
The work done on the gas is greater than the work done by the gas during irreversible process. Comparison of Work Done During Reversible and Irreversible
Expansion
The work done by the gas is the useful work that can be obtained from the gas.
The magnitude of the work done by the gas has been determined graphically in
Figures 2b and 4. On comparing the two figures it is observed that area of the region ABCD > area of the region BCDE.
Therefore, the work done by the gas during reversible expansion is more than work done by the gas during an irreversible expansion.
Work Done at a Constant Volume (Isochoric Process)
When the volume is constant ΔV= 0. The process is isochoric; the gas neither expands nor contracts. The gas does not do any mechanical work.
Work Done At Zero External Pressure
The external pressure is zero if the gas is surrounded by vacuum. Under such conditions the gas undergoes expansion into vacuum. This expansion is known
30
as free expansion. Since dw = –Pext dV and Pext = 0, therefore, dw = 0. No work is done by the gas during free expansion.
8.4
FIRST LAW OF THERMODYNAMICS
8.4.1. Laws of Thermodynamics
The laws of thermodynamics are based upon observations. They were numbered in the order in which they were stated. The last law to be given was the zeroth law. Since it is the most fundamental, it should come before first law
(as zero comes before one) it was named zeroth law.
The change in the energy of the system is related to the work done and the heat exchanged between the system and its surroundings. The first law of thermodynamics deals with the relation between the change in the energy and work and heat. Therefore, an understanding of these terms is important before studying the first law of thermodynamics.
8.4.2. Internal Energy (U)
The total energy of the system is due to energy possessed by the constituents of the system such as atoms, ions and molecules as well as their motion and position. The internal energy of the system does not include the kinetic energy due to motion of the system and potential energy due to position of the system.
Energy due to the position of the system is its potential energy and energy due to motion of the system is kinetic energy. These two types of energies constitute external energy of the system.
The energy possessed by all the constituents of the system is the internal energy. It is represented by the symbol U.
Internal energy is the total energy of all the molecules. It includes translational energy (Ut), vibrational energy (Uv), rotational energy (Ur), electronic energy
(Ue), and intermolecular interaction energy (VT)
U = Ut + Uv + Ur + Ue + VT
Internal energy is also known as intrinsic energy.
31
Total energy = External energy + Internal energy of system
The SI unit of energy is Joule (J).
1 J = 1 kg m2 s–2 = 107 erg
Other units: calories (cal)
1 cal = 4.184 J electron volt (eV)
1 eV = 1.602 x 10–19 J
1 L atm = 1.013 x 102 J
1 Watt hour = 3.6 x 103 J
In thermodynamic studies, the emphasis is not on the individual energy of the molecules, but the overall contribution of all the molecules to the energy of the system. Therefore, thermodynamics generally deals with the energy at the macroscopic level and not at the microscopic level.
Internal energy is a state function. It depends upon temperature (T), pressure
(p), volume (V) and the amount (number of moles present in the system, n).
Characteristics of Internal Energy
Internal energy is represented by U
The SI unit of U is Joule.
U is an extensive property.
U is a state function
The change in U is equal to the difference in internal energies of the final state and initial state of the system. That is, ΔU = U2 –U1. Here, Δ stands for change.
In a cyclic process, there is no net change in internal energy.
Internal energy of the system can be changed when system exchanges heat with its surroundings or when work is done on the system or by the
32
system. The internal energy can also change when the system undergoes physical or chemical process.
The experiments show that the internal energy of the system increases when the work is done on the system or when the heat is absorbed by the system. 8.4.3. Law of Conservation of Energy and First Law of Thermodynamics
According to the law of conservation of energy, energy can neither be created nor destroyed. If it disappears from one form it must appear in another form. This statement is true for those processes also that involve work done and heat transfer. According to the definitions of heat and work done discussed earlier, both heat and work done affect the internal energy of the system. However, they appear only during the change. They are not forms of energy. They are the manner in which energy manifests or shows itself during a change.
When the energy is supplied from outside by heating the system, a part of it is used to do work and the rest remains with the system. Since the total energy should be conserved during any change, the change in internal energy of a system must be related to heat exchanged by the system and work done on/by the system. That is,
ΔU = q + w ….for finite change
….(10)
where ΔU = U2–U1. U1 is the internal energy of initial state and U2 is the internal energy of the final state. q is the heat absorbed by the system and w is the work done on the system
When the process involves infinitesimal change dU = dq +dw
….(11)
Equations (10) and (11) are the respective mathematical forms of the first law of thermodynamics for finite change and infinitesimal change.
In Equation (12), dq and dw do not represent exact differentials. That is, dq ≠ q2 – q1. This is due to the fact that the heat of the initial state and heat of the final state has no meaning. dq represents a small amount of heat absorbed by the system. Similarly, dw is the small amount of work done on the system during the change.
33
When only mechanical work, i.e., work of expansion or compression is done then dw = – pdV. Therefore, Equation (11) becomes, dU = dq – pdV
….(12)
Solved Example: An ideal gas absorbs 500 J of heat and its internal energy increases by 260 J. On the basis of first law of thermodynamics explain what happened to rest of the heat.
Solution: Heat absorbed by the system, q = 500 J
The change in internal energy, ∆U = 260 J
According to the first law of thermodynamics
ΔU = q + w
or
w = ΔU – q
w = 260 J – 500 J = – 240 J
Since sign of the value of w is negative, 240 J of work is done by the system. So out of the 500 J of heat absorbed, 240 J was used to do work by the gas and 260 J was used to increase the internal energy of the gas.
8.4.4. Internal energy and heat
When no work is done and the system does not undergo any phase transformation then the heat transfer between the system and its surroundings results in the change in the temperature of the system. As a result, the volume and pressure will also change. The conditions can be maintained in such a way that either the volume or the pressure is kept constant. Heat exchange is also accompanied by the change in energy of the system. This energy change will be different for the two conditions, that is when volume is kept constant and when the pressure is kept constant.
Heat exchange at constant volume
Since, ΔU = q + w
= q –PΔV
When the volume change is constant ΔV = 0, Hence, PΔV is also zero. The heat exchanged, q can be written as qv. Therefore,
34
ΔU = qv
This implies that, the change in internal energy of a system is equal to heat exchange between the system and its surroundings at constant volume.
Since internal energy is a state function heat exchange at constant volume is also a state function.
8.4.5. Enthalpy
In the previous section heat exchange at constant volume has been related to the change in internal energy. However, in the actual practice, the experiments are carried out at constant pressure in the laboratory. The energy changes are measured in terms of heat exchange taking place between the system and its surroundings. Since the heat exchange takes place at constant pressure it cannot be equal to the change in internal energy. Therefore, a new energy term is needed that will account for the heat exchange at constant pressure. This new energy term is called enthalpy and the symbol is H. It was earlier also known as heat content. The symbol H is taken from heat content. H is related to the internal energy, pressure and volume by the equation
H = U + PV
….(13)
Since U, P and V are state functions U + PV must also be a state function. Since
U and PV have dimensions of energy, it must also have dimensions of energy.
The change in enthalpy is given by dH. Differential of Equation (13) gives dH = d(U+ PV)
From the first law of thermodynamics, dq = dU + PdV
Therefore,
dH = dq +VdP
....(14)
Heat exchange at constant pressure
When the pressure is constant, dP = 0,
VdP = 0 and dq = dqp.
35
Therefore, dH = dqp.
....(15)
For finite change,
∆H = qp where, ∆H = H2 – H1
Therefore, the heat exchanged at constant pressure is equal to the change in enthalpy of the system. Whenever the system exchanges heat at constant pressure (qp), there will be a change in temperature of the system. Some part of heat, qp will be used to increase the internal energy and rest is used up for the change in volume. Hence, change in enthalpy results in change in internal energy as well as change in volume of the system. For solids and liquids the volume change is negligible, therefore, enthalpy and internal energy changes are practically same.
Relationship between ∆H and ∆U for an Ideal Gas
We know, H = U + PV
∆H = ∆U + ∆PV
For an ideal gas, PV = nRT
Therefore,
∆H = ∆U + ∆(nRT)
At constant temperature,
∆H = ∆U + (∆n)RT
Here, ∆n is the change in the amount of the gas. If the temperature changes but the amount of gas is constant then,
∆H = ∆U +nR∆T
Characteristics of Enthalpy
The enthalpy change is equal to the heat exchanged by the system at constant pressure.
It is a state function.
H = U +PV
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For solids and liquids, the enthalpy change (ΔH) is taken as equal to the change in the internal energy (ΔU).
Its units are same as the units of energy.
When the system absorbs heat at a constant pressure, ΔH is positive and the process is known as an endothermic process.
When the system loses heat to the surroundings at a constant pressure,
ΔH is negative and the process is known as an exothermic process.
Since enthalpy is a state function, it depends upon T, V, P and the amount of the system (n).
8.4.6. Heat capacity
Heat capacity is defined as the amount of heat required in order to achieve a unit rise in the temperature of a substance.
C = dq/dT
Unit of heat capacity
heat capacity
heat exchanged temperature change
J
K
SI unit is J K-1.
Heat capacity is an extensive property as it depends upon the size of the system.
Specific heat capacity: It is defined as the heat needed to cause a unit rise in the temperature of a unit mass of a substance. It is denoted by s. Its SI unit is J
K-1 kg-1.
Molar heat capacity: It is defined as the heat needed to cause a unit rise in the temperature of one mole of a substance. Molar heat capacity is denoted by Cm.
Thus,
Molar heat capacity =
Heat capacity of the substance
Amount of the substance
Cm =
C n The SI unit of molar heat capacity is J K-1mol-1.
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Specific heat capacity and molar heat capacity are intensive properties.
When the heat is absorbed at a constant pressure, the heat capacity is known as heat capacity at constant pressure, Cp. It is also known as an isobaric heat capacity. When heat is absorbed at constant volume the heat capacity is known as heat capacity at constant volume, CV. It is also known as isochoric heat capacity.
Relationship Between Cp And Cv For An Ideal Gas
Since, H = U + PV
Therefore, ∆H = ∆U +∆(PV)
For an ideal gas, PV = n RT
Therefore,
∆H = ∆U + ∆(nRT)
Or, ∆H = ∆U + nR ∆T
Using the relations of ∆H and ∆U in terms of specific heat,
Cp∆T = Cv∆T + nR ∆T
Or, Cp = CV + nR
Or, Cp −CV = nR
Dividing the equation by n we get,
Cp,m – CV,m = R
Here, Cp,m = Cp/n and CV,m = CV/n.
8.4.7. Measurement of ∆H and ∆U through Calorimetry
Calorimetry is the measurement of heat. A bomb calorimeter is a type of constant-volume calorimeter used in measuring the heat of combustion of a particular reaction. Bomb calorimeters have to withstand the large pressure within the calorimeter as the reaction is being measured. Electrical energy is used to ignite the fuel; as the fuel is burning, it will heat up the surrounding air, which expands and escapes through a tube that leads the air out of the
38
calorimeter. When the air is escaping through the copper tube it will also heat up the water outside the tube. The temperature of the water allows for calculating calorie content of the fuel.
Figure 8.7: Set up of a Bomb calorimeter
The bomb, with the known mass of the sample and oxygen, form a closed system - no air escapes during the reaction. The weighted reactant put inside the steel container is then ignited. Energy is released by the combustion and heat flow from this crosses the stainless steel wall, thus raising the temperature of the steel bomb, its contents, and the surrounding water jacket. The temperature change in the water is then accurately measured with a thermometer. This reading is used to calculate the energy given out by the sample burn. After the temperature rise has been measured, the excess pressure in the bomb is released.
Basically, a bomb calorimeter consists of a small cup to contain the sample, oxygen, a stainless steel bomb, water, a stirrer, a thermometer, the dewar or insulating container (to prevent heat flow from the calorimeter to the surroundings) and ignition circuit connected to the bomb. By using stainless steel for the bomb, the reaction will occur with no volume change observed (see figure 8.7).
39
Since there is no heat exchange between the calorimeter and surroundings, q = 0 (adiabatic); no work performed, w = 0.
Thus, the total internal energy change ΔU(total) = q + w = 0
Also, total internal energy change ΔU(total) = ΔU(system) + ΔU(surroundings)
=0
Thus, ΔU (system) = - ΔU(surroundings) = - Cv ΔT (since in the entire process volume is constant) where Cv = heat capacity of the bomb
Similarly, when ΔH is to be measured, the heat evolved or absorbed during a reaction is measured, and is known as, heat of reaction or enthalpy of a reaction
ΔrH. In an exothermic process, heat is evolved, and the system loses heat to the surroundings. Thus, qp will be negative and ΔrH will also be negative. On the contrary, in an endothermic process, heat is absorbed, qp will be positive and hence ΔrH will also be positive.
Measurement of Change in Internal energy of the system (∆U)
The change in the internal energy during a chemical reaction is measured using a bomb calorimeter. The reaction is carried out in a container whose volume remains constant. This container is made up of a metal, which is generally steel lined with gold. It is known as the bomb (Figure 8.7). In this experiment, a known mass of the reaction materials are taken in the bomb. The reaction material is the system. The bomb is immersed in a vessel containing a known mass of water. The calorimeter and the water act as its surroundings. The water is stirred by the rotating blades and its initial temperature (T1) is recorded.
As the reaction takes place, there is a change in the chemical energy. If the energy is released during a process, it is transferred to the surroundings as heat. This heat is used to warm water and the calorimeter. When the heat is absorbed during the reaction, the calorimeter gives the heat to the system, so the temperature of the calorimeter decreases. When the temperature of water reaches a constant value, the final reading of thermometer (T2) is noted.
The change in temperature of water and the calorimeter (surroundings)
= ∆T = T2 – T1
40
The heat exchanged between the system (reaction) and its surroundings is calculated from the equation q (water + calorimeter) = (m
s + C)
ΔT = q (surroundings)
where, m = mass of water, s = specific heat capacity of water and C = heat capacity of the calorimeter.
The heat liberated in the reaction is calculated as the heat appearing in the surroundings. During the process the volume remains constant so no work is done. Thus,
∆U = q (system) = – q (surroundings) at constant volume
Difficulties in the Experimental Determination of ΔU
If a large quantity of heat is liberated in a reaction then there is a danger of explosion since the experiment is performed at constant volume in a bomb calorimeter. Why? Because the heat liberated during the reaction will make the gaseous products to expand. Since the vessel is closed, the gas cannot expand.
This will cause a build up of high pressure inside the bomb calorimeter. If the walls of the vessel are not strong enough to withstand the high pressure developed, an explosion may take place.
Measurement of Change in Enthalpy of the System (∆H)
In the laboratories since it is easier to maintain a constant pressure as compared to the constant volume, most of the thermochemistry experiments are carried out under constant pressure conditions. The heat exchange is measured using a normal calorimeter that has adiabatic walls. The simple calorimeter is a glass vessel, usually a glass beaker which is well insulated. It is fitted with a glass stirrer and a thermometer. The set up is kept inside a Dewar flask or thermos flask for insulation. The glass vessel (beaker) is kept inside it on a rubber cork.
The reactants are mixed in it and the change in the temperature is measured, using a 0.1oC thermometer. For accurate work, more sensitive devices such as a
Beckmann thermometer, platinum resistance thermometer or thermocouple may be used. The experimental set up is shown in Figure 2. Sometimes the
Dewar flask itself is used as the vessel in which the reaction is carried out. This set up is commonly used in the laboratory for determination of enthalpy reactions involving solutions.
41
The enthalpy change is determined by measuring the temperature change before and after the process. The heat gained or lost by the system is manifested by the change in temperature of the solution and the container which are assumed to be the surroundings. It is also assumed that no heat is lost during the exchange of heat between the system and the surroundings.
Therefore,
qsystem = − qsurroundings
During the process if the system gains heat then the surroundings will lose heat and the final temperature recorded by the thermometer will be less than the initial temperature. Similarly, if the system loses heat then the heat lost is gained by the surroundings and the temperature rises. qsurroundings = heat exchanged by the solution + heat exchanged by the calorimeter. Since the process takes place at a constant pressure, the heat exchanged is equal to the enthalpy change. Therefore,
∆Hsystem = − ∆Hsurrounding
∆Hsurrounding = ∆Hsolution + ∆Hcalorimeter
At a constant pressure
∆H = Cp ∆T , here ∆T = T2 −T1
∆Hsolution = Cp (solution)
∆T = m
s
∆T
where, m is the mass of the solution and s is the specific heat capacity of the solution. If the solution is aqueous solution, s is taken as equal to the specific heat capacity of water. For the calorimeter,
∆Hcalorimeter = Cp (calorimeter)
∆T
Therefore, the ∆Hsurrounding becomes equal to
∆Hsurrounding = Cp (solution)
∆T + Cp (calorimeter)
∆T
Since the temperature change is same for the solution and calorimeter,
∆Hsurrounding = {Cp (solution) + Cp (calorimeter)} or ∆Hsystem = −{Cp (solution) + Cp (calorimeter)}
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∆T
∆T
....(16)
8.4.8.
Heat exchange during isothermal process
According to the first law of thermodynamics dU = dq + dw
For an isothermal process involving an ideal gas, dU = 0. Therefore, dq = –dw and hence, q = – w
For an isothermal reversible volume change for an ideal gas
wisoth rev (ideal)
2.303nRTlog
q = 2.303nRTlog
V2
V1
V2
V1
….(17)
For an isothermal irreversible volume change for an ideal gas, w isoth irr (ideal)
V2
P2 dV
P2 (V2
V1 )
V1
q = P2 (V2
V1 )
….(18)
Here it is assumed that the external pressure on the gas is equal to the final pressure of the gas.
Conclusion
For a reversible isothermal volume change of an ideal gas,
w isoth rev 2.303nRTlog
q = 2.303nRTlog
V2
V1
V2
V
= 2.303nRTlog 1
V1
V2
dU = 0 or ΔU = 0 dH = 0 or ΔH = 0
For an irreversible isothermal volume change of an ideal gas,
43
w isoth irr P2 (V2
V1 )
q = P2(V2 −V1) dU = 0 or ΔU = 0 dH = 0 or ΔH = 0
8.4.9.
Work done during adiabatic volume change for an ideal gas
The process during which there is no exchange of heat between the system and its surrounding is known as an adiabatic process. For a finite change, q = 0
Therefore, from the First law of thermodynamics
ΔU = w
....(19)
For an infinitesimal change, dq = 0, therefore, dU = dw = –PdV
Since dw = dU, the work in adiabatic process is accompanied with the change in energy (hence its temperature) of the system. Let n be the amount of an ideal gas. Let the initial and the final state of the gas be shown as the following
Initial state
Final state
Pressure
P1
P2
Volume
V1
V2
Temperature
T1
T2
Reversible process: The gas expands in an infinite number of steps. The work done during any step of the process is given by dw = –PexdV
….(20)
Hence,
V2
w
dw
Pex dV
V1
For an ideal gas, P is replaced by
nRT
. Therefore,
V
44
w
adiab rev V2
(ideal)
nRT dV V
V1
Since, both T and V change, w cannot be calculated directly by integration.
However, this can be determined by using the expression, dw= dU. dU = CV dT
….(21)
On integrating Equation (21) we get
ΔU
ΔU
U2
U1
Cv
dU
T2
T1
dT
T2
T1
C v dT
C v (T2
T1 )
Hence, the expression for the work is
wadiab
C v (T2 rev (ideal)
T1 )
Note: For the expansion, w is negative, and thus, T2 < T1, i.e., the expansion is accompanied with the decrease in temperature. For compression the reverse is observed. For calculating the enthalpy change, the equation H = U+PV is considered. At a constant pressure, for a finite process the enthalpy change is given by the equation ΔH = ΔU + PΔV = CV ΔT + nRΔT
ΔH = (CV + nR) ΔT
For an ideal gas (CV + nR) = Cp, therefore,
ΔH = CpΔT
....(22)
Process Involving Change in the State of the System
The change in the state of the system can be vaporization, fusion, sublimation and solid-solid transition. These processes take place at constant temperature, therefore, they are isothermal processes. However, the change in internal energy (∆U) and the change in enthalpy (∆H) are not zero as during the change in state of the system internal energy is involved.
The work done during the phase change = w = −P∆V
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When the state change involves the vapour state then the change in volume is appreciable. Hence the work done is appreciable. However, when only condensed states are involved in the process then the change in volume is very small and hence the work done is negligible.
The heat exchanged during the state change is equal to heat of transition. For solid → liquid, q is equal to heat of fusion. For liquid → vapour, q is equal to heat of vapourization. Since the change in state takes place at constant pressure, the heat exchanged is equal to the enthalpy change for the process. That is, q = qp =∆H
∆U = q + w
∆U = q −P∆V
The change in thermodynamic properties when an ideal gas undergoes volume changes are summarized as follows:
Process
work done(w) Isothermal reversible volume change V nRTn 2
V1
Isothermal irreversible volume change – P2(V2– V1)
heat(q)
change in internal energy(ΔU)
q=–w
V
nRTn 2
V1
q = P2 (V2
V1 )
change in enthalpy (ΔH)
0
0
0
0
Solved Example: One mole of an ideal monoatomic gas at 300 K and 1atm is allowed to expand to the final pressure of 0.1 atm under the conditions:
(i) isothermal reversible (ii) isothermal irreversible against external pressure of
0.1 atm. Calculate work done for each process. R = 8.314 J K-1 mol-1
Solution:
The given data are :
n = 1 mol;
P1 = 1.0 atm
P2 = 0.1 atm
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T1 = 300 K;
(i)
isothermal reversible process
T1 = T2 = 300 K
wisoth rev 2.303nRTlog
P1
P2
2.303(1mol) (8.314 J K 1mol 1 )(300 K) log
(ii)
1.0 atm
0.1 atm
5744.14 J
Isothermal irreversible process
w isoth irr nRT
isoth wirr P2
P1
1
(1 mol)(8.314 J K 1 mol 1 )(300K)
0.1atm
1
1.0atm
2244.78 J
Solved Example: One mole of an ideal gas is heated at constant pressure from
0°C to 100°C. Calculate w, q, ΔU and ΔH. Given Cp,m = 3.5 R.
Solution: On heating the gas will expand against constant pressure. Therefore, the process is irreversible.
The given data are T1 = 0°C = 273 K ; T2 = 100°C = 373 K ; n = 1mol
The expression of work is w = −Pex(V2 − V1)
For an ideal gas V = nRT / P. Hence w = −Pex {(nRT2/P2) − (nRT1/P1)}
Since the expansion occurs at constant pressure, Pex = P1 = P2. Hence, w = – nR(T2 – T1) = –(1 mol )( 8.314 J K-1mol-1)(373 K – 273 K ) = – 831.4 J
∆U = nCV,m(T2 –T1)
Cp,m – Cv,m = R
Cv,m = Cp,m – R = 3.5 R –R = 2.5 R
∆U = 1 2.5
8.314 JK-1mol-1 (373 K–273 K) = 2078.5 J
q = ∆U–w = 2078.5 J–(–831.4 J) = 2909.9 J
∆H= nCp,m(T2–T1) = (1 mol)(3.5
8.314 J K-1 mol-1)( 373 K–273 K ) = 2909.9 J
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Solved Example: One mole of an ideal gas is held by a piston at 273 K and under a pressure of 10 kPa. The pressure is suddenly released to 0. 4 kPa and the gas is allowed to expand isothermally. Calculate q, w, ∆U and ∆H for the process. Hint: The change is irreversible.
Solution: Since the gas is expanding against constant pressure the process is irreversible in nature. The work done by an ideal gas under isothermal irreversible conditions is given by the expression
w isoth irr P2 (V2
V1 )
V1 = nRT/P1 and V2 = nRT/P2
w isoth irr nRT
P2
P1
1
The data given are: T = 273 K ; P1 = 10 kPa = 104 N m-2
P2 = 0.4 kPa = 0.4 103 N m-2 ; n = 1mol
w isoth irr (1 mol) (8.314 J K 1 mol 1 ) (273 K)
0.4 10 3 N m
10 4 N m 2
2
1
= –2178.93 J
For an ideal gas under isothermal conditions ∆U = 0 and ∆H = 0
According to first law of thermodynamics
∆U = q + w = 0 or q = –w = 2178.93 J
Thermochemistry is a part of thermodynamics that deals with the changes in energy during a chemical reaction or a physical transformation. Generally, these changes are mentioned in terms of the heat exchange during the process.
Heat is the energy in transit and it flows from a system at a higher temperature to one at lower temperature. It is also defined as the energy transferred across the boundary between a system and its surroundings by the virtue of the temperature difference. By convention, the heat absorbed or evolved during a process is called the heat exchange during the process. Heat exchanges can also take place during the physical or the chemical processes such as the change of state, a chemical reaction or even a simple process such as dissolution. Heat
48
and energy are used interchangeably because heat exchange at a constant volume is equal to the change in the internal energy and the heat exchange at a constant pressure is known as the change in enthalpy. Thus, qv = ΔU and qp = ΔH
Thermochemistry enables us to predict the amount of heat that would be evolved or absorbed in a process without actually performing a tedious set of experiments in the laboratory. We can also calculate the energy changes for the processes that are not feasible experimentally. This is done by taking advantage of the fact that although heat is a path function, the heat exchanges become state functions when the measurements are made under the conditions of constant volume or constant pressure. Under these conditions they are dependent only on the initial and final states of the system and hence can be added, subtracted or multiplied by an integer according to the algebraic laws.
Sign Convention
If the heat is absorbed by the system the process is known as an endothermic process. Since the energy of the system increases when it absorbs heat, ΔU and
ΔH are given positive sign.
If the system loses heat the process is known as an exothermic process. ΔU and
ΔH have negative sign since the energy of the system decreases.
8.5.1. Standard state
The standard state for a substance is defined for the substance in its pure state and the most stable form of that substance at a standard pressure of one bar and a specified temperature. Although the standard state definition is valid for all temperatures, conventionally the data is mentioned for 298.15 K. However, in most cases the temperature for which the standard state is defined is mentioned in the brackets.
For gases the standard state is the one at which pure gas behaves ideally at one bar and a specified temperature. For liquids the standard state is the pure liquid at one bar and specified temperature.
49
For solid it is the pure and most stable crystalline form of aggregation of solid at one bar and specified temperature. For example, graphite is taken as the stable form of carbon and rhombic form is the most stable form for sulphur.
For solution it is one mol dm−3 solution at one bar and the specified temperature. The standard states can be summarized as follows:
S.No.
Phase of the substance Condition
1.
Gas
Pure gas at 1 bar and a specified temperature
2.
Liquid
Pure liquid at 1 bar and a specified temperature
3.
Solid
Pure and stable form of solid at 1 bar and a specified temperature 4.
Solution
One mol dm−3 solution at 1 bar and a specified temperature 8.5.2. Heat of Reaction
Heat of a reaction is defined as the heat exchanged between the system and the surroundings during the complete transformation of reactants to products at a constant temperature and pressure for the given stoichiometrically balanced equation. During a chemical process some bonds break and others are formed so there is a net change in the energy of the system. This results in the heat exchange during a process. This heat exchange can be measured if the energy of initial and final state is known. The heat exchanges are measured either at a constant pressure or at a constant volume.
When the reaction is carried out at a constant volume, the heat exchange is measured in terms of the internal energy change. It is known as the heat of reaction at constant volume, qv or energy of reaction. The symbol is ΔrU. The energy change is given by the equation
ΔrU = Ufinal −Uinitial
When the reaction is carried out at a constant pressure, the heat exchange is measured in terms of the enthalpy change. It is known as heat of reaction at constant pressure, qp or enthalpy of reaction. The symbol is ΔrH.
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The enthalpy of reaction is given by the equation
ΔrH = Hfinal −Hinitial
Since the measurements are generally made under constant pressure conditions, the heat of reaction is normally referred as enthalpy of reaction.
The enthalpy of reaction is the enthalpy change during a reaction. Since enthalpy is a state function, the change in enthalpy is equal to the difference in the enthalpies of the final state and the initial state. For a general reaction νAA + νBB → νCC + νDD
The reactants are the initial state and the products form the final state assuming that the reaction has been completed. Therefore,
ΔrH = Total enthalpy of products – total enthalpy of reactants
=
H(products)
H(reactant s )
ΔrH = (νCHm,C + νD Hm,D) –(νA Hm,A + νBHm,B)
Here, Hm,A = molar enthalpy of A
Hm,B = molar enthalpy of B
Hm,C = molar enthalpy of C
Hm,D = molar enthalpy of D νA, νB, νC and νD are stoichiometric coefficients of A, B, C and D, respectively in the balanced chemical equation.
For example, during the formation of CO2, one mole of carbon in its solid state reacts with one mole of oxygen gas to form one mole of carbon dioxide. The heat produced at constant pressure and 298 K is 393.5 kJ. This information is represented as,
C(s) + O2(g) → CO2(g)
ΔrH(298 K,1 atm) = −393.5 kJ mol−1
When no pressure is mentioned, it is assumed that the reaction is taking place at atmospheric pressure, that is, 1atm.
When the reactants and the products are in their respective standard states, the enthalpy of reaction is known as standard enthalpy of reaction ΔrHө . If the
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reaction takes place at 298 K the standard enthalpy of reaction is written as
ΔrHө (298 K)
ΔrHө (298 K) = ΔrH( 298 K, 1 bar)
While defining the heat of a reaction, the nature of the reaction has not been mentioned. Therefore, the term can be used for all the types of reactions.
However, these reactions can be classified into various categories and for each category the heat exchange is given a specific name. Since, these heat exchanges are generally studied at a constant pressure, these will be referred to in terms of enthalpy changes in this chapter. The reactions involving the transformations are reported for 1 mol of the substance to standardize the values of the enthalpy changes.
(a)
Enthalpy of Formation
There are many ways in which a substance can be formed and for each such reaction the heat of reaction is different. In many of the reactions more than one product is formed. The values of the heat of reactions for all the reactions in which a given substance is formed cannot be taken as the heat of formation of that substance. In order to define a single value of heat of formation, the chosen reaction should be such that only one mole of the substance is formed from its constituent elements and no other product is formed. So the enthalpy of formation can be defined as the amount of heat exchanged at constant temperature and pressure during the formation of one mole of the substance from its constituent elements. It is represented by ΔfH. Its unit is kJ mol−1.
For example, the enthalpy of formation of CO2 is equal to the enthalpy change for the following reaction
C(s) + O2(g) → CO2(g)
ΔfH = −393.5 kJ mol−1
Similarly, the enthalpy of formation of water is equal to the enthalpy change for the reaction
H2(g) + ½O2(g) → H2O(ℓ)
ΔfH = −285.8 kJ mol−1
Note that although water is formed during the neutralization process also, it is not taken as the formation reaction of water as it is not formed from its elements in this reaction.
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Standard enthalpy of formation is defined as the amount of heat exchanged at a given temperature and a pressure of one bar during the formation of one mole of the substance in its standard state from its constituent elements in their respective standard states. It is represented by ΔfHө .
Enthalpy of Formation of an Element
By convention the enthalpy of formation of an element in its standard state is taken as zero at all temperatures. Hence, whether the element is a solid, a liquid or a gas its standard enthalpy of formation is taken as zero. For example, hydrogen gas, solid zinc, rhombic sulphur, liquid bromine and graphite all have ΔfHө (298 K) = 0.
While dealing with the ionic species in the solution, the standard enthalpy of formation of H+ ions is also taken as zero.
Hess’s Law of Constant heat Summation
The total enthalpy change of a reaction is the same regardless of whether the reaction is completed in one step or in several steps. For example for a change from reactant to product that can take place in four steps or a single step, the total enthalpy change will be same.
Single step process
Reactant → Product
ΔH
Multiple step process
Reactant →A
ΔH1
A →B
ΔH2
B→C
ΔH3
C→ Product
ΔH4
According to Hess‟s law
ΔH = ΔH1 + ΔH2 + ΔH3 + ΔH4
This can also be shown by the following diagram
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Diagram illustrating Hess’s law
The following important rules are derived from thermochemistry
We can reverse any equation and the sign of ΔH is also reversed.
The chemical equations can be added or subtracted. The ΔH of the resulting equation is sum or difference of the ΔH of the chemical equations.
The chemical equation can be multiplied or divided by an integer and the
ΔH is also multiplied or divided by that integer.
The identical terms on both the sides of the equation are cancelled.
These laws help us to theoretically calculate the enthalpy changes of processes which cannot be easily carried out in a laboratory. For example, to determine the enthalpy change for the process
C(graphite)+ ½ O2 (g)→ CO(g)
ΔH
This process when carried out in the laboratory would also generate CO2.
However, the following two processes can be carried out quantitatively.
1.
C(graphite)+ O2 (g)→ CO2(g)
ΔH1 = −393.3 kJ mol−1
2.
CO(g) + ½ O2 (g)→ CO2(g)
ΔH2 = −282.8 kJ mol−1
If the equation 2 is reversed
3.
CO2(g) → CO (g)+ ½ O2 (g)
ΔH3 = − ΔH2
Add Eqs. 1 and 3
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C(graphite)+ O2 (g) + CO2(g) → CO2(g) + CO (g)+ ½ O2 (g) or C(graphite)+ ½ O2 (g)→ CO(g)
ΔH = ΔH1 + ΔH3 = ΔH1 – ΔH2
ΔH = −393.3 kJ mol−1+ (−282.8 kJ mol−1) = −110.5 kJ mol−1
Hence, the enthalpy change for a reaction that cannot be carried out quantitatively in the laboratory can be theoretically calculated.
Calculation of Enthalpy of a Reaction from Enthalpies of Formation of
Reactants and Products
We have seen that the enthalpy of a reaction is calculated from the molar enthalpies of the products and the reactants. However, ∆rH can also be calculated from the enthalpies of formation values of all the reactants and the products involved in the reaction. Let us consider the following reaction:
(A) CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
By definition,
ΔrHө = Total enthalpy of products – total enthalpy of reactants
=
H θ (products)
H θ (reactants )
ΔrHө = {Hө m(CO2) + 2Hө m(H2O)} – {Hө m(CH4) – 2Hө m(O2)}
The reaction A can be carried out in the following steps:
(i)
C(graphite)+ 2H2 (g) → CH4(g)
∆fH (CH4)
(ii)
O2(g) → O2(g)
∆fH (O2)
(iii) C(graphite)+ O2 (g)→ CO2(g)
∆fH (CO2)
(iv) H2 (g) +½ O2(g)→ H2O(ℓ)
∆fH (H2O)
According to Hess‟s law,
Eq. (A) = {Eq. (iii) + 2Eq.(iv)} –{Eq. (i) + 2Eq. (ii)}
55
Therefore, ∆H(A) = {∆fH (CO2) + 2∆fH (H2O)} −{∆fH (CH4) + 2∆fH (O2)}
Therefore, for a reaction νA A + νBB → νCC + νDD
The enthalpy of reaction can be written as
ΔrH = {νC ΔfHө (C) + νD ΔfHө (D)}−{ νAΔfHө (A) + νB ΔfHө (B)}
....(23)
= Total enthalpy of formation of products – Total enthalpy of formation of reactants Hence, from the tabulated ΔfH values of the substances at a given temperature, the enthalpy of any reaction can be calculated using the above formula. The standard enthalpy of reaction can be calculated from the tabulated values of
ΔfHө .
Relationship Between ΔrU and ΔrH
H = U + PV
For calculating the enthalpy change the equation becomes,
ΔH =ΔU + PΔV
For a chemical reaction the equation becomes
ΔrH = ΔrU + Δr (PV)
....(24)
For the reactions involving solids and liquids the volume change with the change in the pressure is negligible and therefore, the change in PV is negligible in comparison to ΔrU. The Equation (24) gives
ΔrH = ΔrU
For the following reaction involving gases, νAA(g) + νBB(g) → νCC(g) + νDD(g)
V(reactants ) = νA Vm(A) + νB Vm(B)
V(products ) = νC Vm(C) + νD Vm(D)
Δr (PV) = Δr (PνgVm )
56
If all the gaseous species are assumed to behave as an ideal gas, then,
Vm = RT/P
Therefore, Δr (PV) = Δ(νgRT)
At a constant temperature, Δr (νgRT ) = RT Δνg where, Δνgas = νC + νD − νA − νB
Therefore, on substituting the value of Δr (PV) in Equation (24),
ΔrH = ΔrU + RTΔνgas
For heterogeneous reactions, the same rule applies; however, here Δνgas is equal to the change in the stoichiometric number of gaseous species involved in the balanced chemical equation.
Since ΔrH = qp and ΔrU = qv qp = qv + RTΔνgas
....(25)
The Equation (25) gives the relationship between heat exchanges at a constant pressure and at a constant volume for reactions involving gases.
Solved Example: Calculate ΔrU for the following reactions at 1 atm and 298K.
1.
N2(g) + 3H2(g) → 2NH3(g)
ΔrH = −92.38 kJmol−1
2.
C(graphite) + ½O2 → CO(g)
ΔrH = −110.5 kJmol−1
Solution:
From Eq. (3.4), ΔrU = ΔrH − RTΔνgas
R = 8.314 J K−1mol−1
RT = 8.314 J K−1mol−1
1.
T = 298 K
298 K = 2477 J mol−1 = 2.477 kJ mol−1
N2(g) + 3H2(g) → 2NH3(g)
ΔrH = −92.38 kJ mol−1
Δνgas = 2 – (1+3) = −2
ΔrU = −92.38 kJ mol−1 – (− 2
2.
2.477 kJ mol−1 ) = −87.426 kJ mol−1
C(graphite) + ½O2 → CO(g)
ΔrH = −110.5 kJ mol−1
Δνgas =1 – ½ = ½
ΔrU = −110.5kJmol−1 –(½ x 2.477kJ mol−1) = −111.738kJ mol−1
57
(b)
Enthalpy of combustion
It is the heat exchange that takes place when one mole of a substance is burnt completely in the presence of oxygen at a given temperature and pressure. For most of the organic compounds the products of combustion are H 2O and CO2.
Other elements undergo combustion to form their respective oxides. It is denoted by ΔcH and the unit is kJ mol−1. The combustion is always an exothermic process.
For example,
CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ)
ΔcHө (298 K)= −890 kJ mol−1
C2H6 (g) +
7
O2(g) → 2 CO2 (g) + 3H2O(ℓ)
2
ΔcHө (298 K)= −1560 kJ mol−1
The enthalpy of combustion is always measured for the combustion of one mole of a compound. Therefore, while writing the balanced chemical equation, the stoichiometric number of the compound undergoing combustion is taken as one and the rest of the reactants and the products are balanced accordingly.
Solved Example: Apply Hess‟s law to calculate the ΔfHө (C4H10) using the following thermodynamic data:
ΔcHө (C4H10,g)= −2878.7 kJ mol−1
ΔfHө (H2O,ℓ) = −285.8kJmol−1
ΔfHө(CO2,g) = −393.5kJmol−1
Solution: The required equation is
4C (graphite) + 5H2 (g) → C4H10(g)
ΔfHө
Given
1.
H2(g) + ½ O2 (g) → H2O(ℓ)
ΔfHө (H2O,l) = −285.8kJmol−1
2.
C(graphite)+ O2 (g) → CO2(g)
ΔfHө (CO2,g) = −393.5kJmol−1
3.
C4H10(g) + 13/2 O2(g) → 4CO2(g) + H2O(ℓ)
58
ΔcHө (C4H10,g)= −2878.7 kJ mol−1
Multiply Eq. 2 by 4 and equation 1 by 5 and add the resulting equations
4.
4C(graphite)+ 4O2 (g) + 5H2(g) + 5/2 O2 (g) → 4CO2(g) + 5H2O(ℓ)
Subtract equation 3 from equation 4
4C(graphite) + 5H2 (g) → C4H10(g)
This is the required equation. Therefore,
ΔfHө (C4H10) = 4
ΔfHө (CO2,g) + 5
ΔfHө (H2O,l) − ΔcHө (C4H10,g)
ΔfHө (C4H10) = 4 mol−1) (−393.5 kJ mol−1 ) + 5
(−285.8 kJ mol−1 ) – (−2878.7 kJ
ΔfHө (C4H10) = (−1574 −1429 + 2878.7) kJ mol−1
ΔfHө (C4H10) = −124.3 kJ mol−1
(c)
Enthalpy of Transition
It is the amount of heat exchanged when one mole of a substance is transformed from one allotropic form to the other allotropic form at a given temperature and pressure. The symbol is ∆trsH.
C(graphite) → C(diamond)
Enthalpy of vaporization
It is the amount of heat absorbed to convert one mole of a liquid to its vapour state at a given temperature and pressure.
H2O(ℓ) → H2O(g)
ΔvapHө (298 K) = 44.0 kJ mol−1
Enthalpy of Fusion
It is the amount of heat absorbed to convert one mole of a solid to its liquid state at a given temperature and pressure.
H2O (s) → H2O(ℓ)
ΔfusHө (298 K)= 6.0 kJ mol−1
59
Enthalpy of Sublimation
It is the amount of heat absorbed to convert one mole of a solid to vapour state at a given temperature and pressure.
ΔsubHө (298 K)= 50.0 kJmol−1
H2O(ℓ) → H2O(g)
Enthalpy of Atomization
It is the amount of heat required to convert one mole of a substance into its constituent atoms in the gaseous state.
C(graphite)
H2(g)
CH4 (g)
atomization
atomization
C(g)
∆aH(C) =716.68 kJmol−1
2H(g)
atomization
∆aH(H)= 436 kJ mol−1
C(g) + 4H(g)
∆aH = 1664 kJ mol−1
For an element in its solid state, the enthalpy of atomization is equal to its enthalpy of sublimation.
Mg(s)
(d)
atomization
ΔaHө = 150.2 kJ mol–1
Mg(g)
Enthalpy of solution
It is the amount of heat exchanged when one mole of a solute is dissolved in a given amount of solvent at a specified temperature and pressure.
When the solvent is water the dissolution of solute X is represented as
X + n H2O → X.n H2O
This process results in the solution with the solute solvent ratio of 1: n
For example,
1.
HCl(g) + 10 H2O(ℓ) → HCl.10H2O(aq)
ΔH = −69.01 kJ mol−1
2.
HCl(g) + 40 H2O(ℓ) → HCl.40H2O(aq)
ΔH = −72.79k J mol−1
3.
HCl(g) + 200 H2O(ℓ) → HCl.200H2O(aq)
ΔH = −73.961 kJ mol−1
60
These values of ΔH show the general dependence of the heat of solution on the amount of the solvent. As more and more solvent is used the value of heat of solution changes. As the amount of the solvent increases the resulting solution becomes more dilute and ultimately it becomes so dilute that further addition of solvent produces no enthalpy change. This solution is known as infinitely dilute solution.
(e)
Bond Energies
Every reaction involves breaking and making of bonds of atoms. If the energies involved in these processes can be determined the net amount of enthalpy change for the reaction ΔrH can be estimated.
Energy is given to the system to break a bond and the energy is released whenever a bond is formed. Therefore, bond formation is an exothermic process while bond dissociation is an endothermic process. For example when one mole of hydrogen gas is formed from its atoms in gaseous state, 436 kJ of energy is released; however, 436 kJ of energy is required to break one mole of H−H bonds. H(g) + H(g) → H2(g)
ΔrH = −436 kJ mol−1
H2(g) → H(g) + H(g)
ΔrH = 436 kJ mol−1
The energy required to break a bond is known as the bond dissociation energy. Since the measurements are made at a constant pressure, the heat required to break the bond is known as the bond dissociation enthalpy and the heat released during bond formation is known as the bond formation enthalpy. NOTE: In our studies, the use of bond dissociation enthalpy is preferred over bond formation enthalpy since it is easier to break the bonds to form gaseous atoms. Measurement of the bond formation enthalpy requires the difficult task of isolating gaseous atoms and then combining them to form the molecules.
In a simple diatomic molecule only one type of bond is broken. However, in a polyatomic molecule the energy required to break a bond will depend upon the presence of other atoms or group of atoms. For example, the bond dissociation enthalpies of the two O−H linkages in the water molecule are different.
61
H−O−H(g) → H(g) + OH(g)
ΔHө (298K) = 498 kJ mol−1
O−H(g) →O(g) + H(g)
∆Hө (298K) = 430 kJ mol−1
Similarly, breaking up of an O−H bond in ethyl alcohol molecule or acetic acid molecule would involve different amount of heat. However, while tabulating the data of enthalpy required for breaking the bond, generally the net average of various values is taken. Therefore, there are two types of enthalpies of dissociation, one is the exact value for a specific bond and the other is the average value for a bond. These are known by different names; bond dissociation enthalpy and bond enthalpy.
Bond dissociation enthalpy: It is the enthalpy required to break a specific bond in a specific molecule at a given pressure. It depends upon the presence of other groups in the molecule.
Bond enthalpy: It is the average of dissociation enthalpies of a given bond in a series of different types of dissociating species.
In the case of diatomic molecules, the bond enthalpy and bond dissociation enthalpy are same.
The convention is to represent it by ε(A−B).
Relationship Between Bond Enthalpy and Bond Energy
The enthalpy of a reaction is related to the internal energy change of that reaction by the equation
ΔrH = ΔrU + RTΔνgas
Similarly, for the dissociation of A−B bond
A−B(g) → A(g) + B(g) the bond enthalpy is related to the bond energy by the equation ε (A−B) = ΔU(A−B) + RTΔνgas
For this reaction, Δνgas= 2−1 =1 and at 298K RTΔνgas = 2.48 kJ mol−1
This value is almost negligible compared to bond enthalpy values. So the bond enthalpy is approximately equal to the bond energy and the two are used interchangeably. 62
Calculation of Bond Enthalpy
In a simple molecule such as methane all the bonds are identical. It is assumed that the enthalpy required to break each bond is equal and is equal to bond enthalpy of C−H bond. So, the total enthalpy required to break all the bonds and form gaseous atoms is equal to four times the bond enthalpy of C−H bond.
That is 4 ∆H(C−H) = ∆H for the reaction
CH4(g) → C(g) + 4H(g)
∆H
∆H for this reaction can be calculated using the following data: sublimation C(g)
1.
C(graphite)
2.
½H2(g)
3.
C(graphite) + 2H2(g) → CH4(g)
dissociation
∆H1 = 716.68 kJ mol−1
H(g)
∆H2 = 217.97 kJ mol−1
∆H3 = −74.75 kJ mol−1
Using the laws of thermochemistry these equations are modified to get the required value.
Multiplying Eq. 2 by 4 gives
4.
2H2(g)
dissociation
4 H(g)
∆H4 = 4
∆H2
Reversing Eq. 3 we get
5.
CH4(g) → C(graphite) + 2H2(g)
∆H5 = −∆H3
Adding Equations (1), (4) and (5) gives the required equation
CH4(g) → C(g) + 4H(g)
Therefore, ∆H = ∆H1 + ∆H4 + ∆H5
∆H = ∆H1 + 4
∆H2 −∆H3
∆H = 716.68 kJ mol−1 + 4
217.97 kJmol−1 –(−74.75 kJmol−1)
= (716.68 + 871.88 + 74.75) kJmol−1
= 1663.31 kJ mol−1
H ε(C−H) =
4
1663.31 kj mol
4
1
= 415.82 kJ mol−1
63
Since C−H bond is present in other molecules also, the average value is calculated by taking into account C−H bonds in different molecules. The bond enthalpy of C−H bond is taken as 413 kJ mol−1.
Similarly, the C−C, C=C and C≡C bond enthalpies can be calculated by using the enthalpies of formation of ethane, ethene and ethyne and bond enthalpy of
C−H bond. Thus the bond enthalpy data of polyatomic molecules can also be tabulated. Applications of bond enthalpies
1.
Determination of Enthalpy Changes Accompanying a Chemical
Reaction:
Bond enthalpies can be used for determining the enthalpies of reaction.
∆rH = Total bond enthalpies of reactants −Total bond enthalpies of products
...(26)
During a reaction some bonds of the reactants dissociate and bonds of the products are formed. The enthalpy of a reaction is the total enthalpy change during bond formation and bond dissociation. The enthalpy change during bond dissociation is equal to the bond enthalpy. Since the bond formation is reverse of bond dissociation, the enthalpy change during bond formation is negative of bond enthalpy. Therefore, the total bond enthalpies of the products are subtracted from the total bond enthalpies of the reactants to get the enthalpy of reaction.
Solved Example: For the following reaction, calculate the amount of energy released during bond formation:
C2H4(g)+H2(g) → C2H6(g)
Or
HH
H-C=C-H (g) + H-H(g)
H H
H-C-C-H (g)
HH
64
Solution: The bonds being broken on the reactant side are C=C and H−H bonds in ethene and hydrogen gas, respectively and bond formed are C−H bonds and C−C. Given, ε (C=C) = 610 kJ mol−1, ε (H−H) = 436 kJ mol−1, ε (C−C) = 348 kJ mol−1, ε (C−H) = 413 kJ mol−1,
Total enthalpy needed to break the bonds = Total bond enthalpies of reactants
= ε (C=C) + ε (H−H) = (610 + 436) kJmol−1 = 1046 kJmol−1
Enthalpy released during the formation of bonds = Total bond enthalpies of products = ε (C−C)+ 2 ε (C−H) = (348 +2
413) kJmol−1 = 1174 kJmol−1
∆rH = { ε (C=C) + ε (H−H) } –{ ε (C−C) + 2 ε (C−H)}
= 1046 kJmol−1 −1174 kJmol−1
∆rH = −128 kJmol−1
2.
Determination of enthalpy of formation:
If the reaction involves the formation of a compound from its elements then the bond enthalpy data can be used to calculate the enthalpy of formation of the compound by the formula
∆fHө = Total enthalpy of atomization of the reactants – Bond enthalpies of products ...(27)
Solved Example: In the following reaction, 2C(graphite)+3H2(g)+½O2(g)
→C2H5OH(ℓ) calculate the enthalpy of formation of ethanol.
Solution: The steps involved are
2C(graphite)
3H2(g)
atomization
atomization
2C(g)
∆rH=2 ∆aH(C) =716.68 kJmol−1
6H(g)
∆rH=3∆aH(H)= 436 kJ mol−1
½O2 (g) → O(g)
∆rH=½∆aH(O) = 498.4 kJmol−1
65
HH
2C(g) + 6H(g) + O(g) → H-C-C-O-H (I)
HH
Enthalpy change during bond breakage = Total enthalpy of atomization
= 3∆aH(H) + 2∆aH(C) + ½∆aH(O)
=3
436 kJ mol−1 + 2 716.68 kJmol−1 + 249.2 kJmol−1
= 2990.88 kJmol−1
The bond enthalpy values are ε(C−C) = 348 kJ mol−1, ε (C−H) = 413 kJ mol−1, ε(C−O) = 351 kJ mol−1 ε (O−H)= 363 kJ mol−1
The enthalpy released during the formation of bonds in C2H5OH = Bond enthalpy of C2H5OH = ε (C−C) + 5 ε(C−H)+ ε (C−O) + ε (O−H)
= (348 + 5 413 + 351 + 363) kJ mol−1 = 3227 kJ mol−1
∆fH (C2H5OH) = Total enthalpy of atomization – Bond enthalpy
= (2990.38 – 3227) kJ mol−1 = −236.52 kJmol−1
IMPORTANT: The drawback of this method is that it will give same value of
∆fHө for all the isomers of a compound which is not correct.
Solved Example: Calculate ΔrU for the following reactions at 1 atm and 298K.
C(graphite) + O2 → CO2 (g)
ΔrH = −393.5 kJmol−1
Ans. ΔrU = ΔrH − RTΔνgas
R = 8.314 J K−1mol−1
RT = 8.314 J K−1mol−1
T = 298 K
298 K = 2477 J mol−1 = 2.477 kJ mol−1
C(graphite) + O2 → CO2 (g)
ΔrH = −393.5 kJ mol−1
Δνgas =1 –1=0
ΔrU = ΔrH = −393.5 kJ mol−1
66
6.6
Spontaneity
Some processes take place on their own without any help from the external agencies. Such processes are known as spontaneous processes. For example, expansion of gas kept in an open container and cooling of hot tea kept in a cup take place spontaneously, while air can never compress itself and fill a balloon on its own. Similarly a body at room temperature can never take heat from the surroundings and become hot. These processes can take place with the help of an external agency. Such processes that do not take place on their own but require the help of an external agency are known as non-spontaneous processes.
8.6.1. Limitations of the first law of thermodynamics
The first law of thermodynamics is about conservation of energy. It is very useful in calculating energy changes taking place during various processes.
However, it gives no idea about the direction of energy flow. The first law of thermodynamics is unable to predict whether a process will be spontaneous or nonspontaneous. To make this prediction we have to look into the most important property of the system which is energy. It is a known fact that energy flows from a high energy state to a low energy state. The system having lower energy is more stable than the system having higher energy. Then the question that can be asked is whether all the processes in which energy of the system is decreasing are always spontaneous? Are all exothermic processes always spontaneous? It is observed that it is not necessary that all spontaneous processes are exothermic. Various cases are known where a spontaneous process is endothermic. For example, a gas expands in vacuum spontaneously although there is no energy change. Similarly, water evaporates spontaneously into water vapour but the process is endothermic in nature.
There is another reason why an exothermic process cannot be taken as the sole criterion of spontaneity and that is the law of conservation of energy. If the energy of the system decreases during a process then the energy of the surrounding must increase since the total energy must be conserved. So if the system undergoes an exothermic process spontaneously then the change in the surroundings should also be spontaneous although the energy of the surrounding will increase.
67
The second law of thermodynamics decides the direction of a spontaneous process. It introduces a new concept of entropy that governs the criterion of spontaneity. 8.6.2. Various statements of the Second Law of Thermodynamics
The second law of thermodynamics has been stated in various ways by different scientists. Although these statements are different but each one of them predicts the direction in which a change will take place spontaneously.
1.
Heat cannot pass spontaneously from a colder to a warmer body.
(Clausius)
2.
No change is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work.(Kelvin)
3.
It is impossible to transfer heat from a colder system to a warmer system without other simultaneous changes occurring in the two systems or in their environment.(P.S. Epstein)
4.
Every system which is left to itself will on the average change towards a condition of maximum probability.
5.
The state of maximum entropy is the most stable state of an isolated system. (Enrico Fermi)
6.
In any irreversible process the total entropy of all bodies concerned is increased. 7.
In an adiabatic process the entropy either increases or remains unchanged.
8.
Entropy is times arrow. (A. Edington)
9.
There is general tendency in nature for energy to pass from more available form to less available form.
10.
The entropys
8.6.3. Entropy
Whenever a change occurs, the internal energy is distributed into various forms. It has been found that those processes are spontaneous in which the energy has been distributed into a more dispersed form, such that is there is
68
more chaotic or random distribution of energy. This random distribution of energy makes it less useful for doing work. For example, when a ball bounces it loses some energy at each bounce. The energy lost is either in the form of frictional energy or it is given to the floor molecules as thermal energy. The total energy is conserved but the energy is in the distributed state and hence is of less use.
A new concept of entropy was introduced by Clausius as a measurement of randomness or chaos (also referred to as disorder). The word entropy has been made from energy (en) and trope (tropy) which is for chaos. The symbol is S. It was given in honour of famous scientist Sadi Carnot.
According to one of the statements of second law of thermodynamics, heat cannot be completely converted into work. This has been attributed to entropy and it can be explained by taking the example of a gas kept in a cylinder with a piston that can move without friction.
When the gas kept in the cylinder is heated its kinetic energy increases. The gas molecules move faster. They collide with each other and with the walls of the cylinder. When they collide with the piston, it is pushed forward and the gas expands. This results in useful work done by the gas. Now, consider a hypothetical condition when all the gas molecules move in ordered manner towards the piston. Then each molecule will strike the piston and contribute to the total work done by the gas as shown in Figure 8.7. Under this situation the entire heat absorbed can be converted into work. However, in reality the molecules are moving randomly in all directions. Therefore, all the molecules do not strike the piston. Hence, all the collisions of the gas molecules do not lead to expansion. Since only a fraction of collisions result in useful work, only a fraction of heat absorbed is converted into useful work and the rest is the unavailable heat, that is, the heat that is not used for doing work. The fraction of heat used to do useful work decreases and the amount of unavailable energy increases as the randomness of the gas molecules increases.
69
Figure 8.7 a and b. Conversion of heat into work when the collisions are ordered and when the collisions are random
Mathematical form of entropy
The change in entropy is defined as
dS
dq rev
T
Note that while calculating the entropy change, the heat transfer only during the reversible processes is considered. However, the entropy change for an irreversible process is not related to the heat transfer during an irreversible process. q is a path function. It can be shown that the entropy is a state function.
Characteristics of Entropy
Entropy is a measure of randomness. More is the randomness greater is the entropy of the system. Entropy can also be related to unavailable energy.
For a given substance, the vapour phase has maximum randomness and solid phase has least randomness. Therefore,
Svapour > Sliquid > Ssolid
Entropy is a state function. Therefore, ΔS = Sfinal – Sinitial = S2 – S1
70
It depends upon temperature, pressure, volume and amount.
Entropy is an extensive property.
dS ≠ dqirr / T
Entropy increases with increase in temperature.
The entropy increases as the volume of the system increases.
Units of entropy dS dq rev
T
J
K
The SI unit of entropy is J K-1
For molar entropy, the unit is J K-1mol-1
When heat is measured in calories, the unit is cal K-1
Relationship between internal energy and entropy dU = dq + dw
....First law of thermodynamics
dU = dq –PdV
For reversible process, dq = dqrev = TdS dU = TdS –PdV
....(28)
This is also known as the combined form of first law and second law for reversible processes. This expression indicates that the internal energy depends upon the change in entropy and volume.
Relationship between enthalpy and entropy
H = U + PV
Differentiating the equation we get dH = dU + PdV + VdP
....(29)
From the combined form of the first law and the second law of thermodynamics 71
dU = TdS – PdV
Substituting the above expression in Equation (29) we get dH = TdS –PdV + PdV + VdP dH = TdS + VdP
....(30)
Entropy change for an isolated system
According to the first law of thermodynamics, dUrev = dqrev + dwrev and dUirr = dqirr + dwirr
If the initial states and final states of the system are same for both the processes, then dUrev = dU irr dqrev + dwrev = dqirr + dwirr
....(31)
dwrev< dwirr
Therefore,
dqrev
T
or
dqrev > dqirr
dqirr
....(32)
T
dq rev
= dS
T
Therefore,
Since,
dS >
dS >
dq irr
T
....(33)
dqirr =dU − dwirr
dU dw irr
T
....(34)
For an isolated system dq = 0, dU = 0. Therefore, dw = 0 dqrev = 0
(dS)rev = 0
dqirr = 0 (dS) irr >0 or dS(isolated) ≥ 0
....(35)
72
Although entropy is a state function, it appears that the quantity dS is different for reversible and irreversible processes for an isolated system. This can be explained if the nature of the isolated system is considered. An isolated system includes the system as well as its immediate surroundings. Therefore, the entropy change for an isolated system includes the entropy change for the system as well as its surroundings. dS(isolated) = dS(system) + dS(surroundings)
Since, Universe = System + Surroundings dS(isolated) = dS (universe)
For the state functions, the initial state and the final state of the system are considered and not that of the surroundings. Since the initial states and final states of the system are same for both the processes, the entropy change for the system is same for reversible and irreversible process. The changes in the surroundings are different for reversible and irreversible processes so the entropy changes for the surroundings will be different. As a result, the total entropy change for the universe will be different for reversible and irreversible process. dSrev(system) = dSirr(system)
....(36)
For a reversible process,
....(37)
dSrev(system) = and dqrev(surroundings) = – dqrev(system)
Therefore,
dSrev(surrounding) =
....(38)
dSrev(universe) = dSrev(system) + dSrev(surrounding)
Substituting the values of Equations (37) and (38) in the above equation, dSrev(universe) =
In view of Equation (38) the above equation becomes
73
dSrev(universe) =
For an irreversible process,
Heat absorbed by the system = dqirr(system)
From Equation (33), dSirr(system)>
....(39)
Since the surroundings are large, the addition or removal of a small amount of heat in the surrounding is taken as a reversible process. Therefore,
ΔSirr(surrounding) =
....(40)
The entropy change for the universe is given by the equation, dSirr(universe) = dSirr(system) + dSirr(surrounding)
On substituting the value of ΔSirr(surrounding) from Equation(40) in the above equation, dSirr(universe) = dSirr(system)
....(41)
From Equations (39) and (41) it is seen that dSirr(universe) > 0
8.6.4. Free Energy Function
One of the purposes of studying thermodynamics is to predict the feasibility of a process, that is, to predict the direction in which a process will be spontaneous. We have seen that the sign of internal energy alone cannot be the criterion for determining the spontaneity of a process. The second law of thermodynamics introduced the concept of entropy. Whenever a process occurs spontaneously it is an irreversible process. It is observed that for such a process the entropy change of the universe is positive. This implies that to predict the spontaneity of a process it is important to determine the entropy change of the universe. This involves measurement in entropy change of the system as well as that of surroundings. The problem arises in the measurement of entropy change of the surroundings since the surroundings are not clearly defined. For most of the spontaneous processes it is observed that the system either tries to attain the state of minimum energy or maximum entropy. So it is necessary
74
to define a new function which takes into account both the properties. Gibbs free energy (G) is a function that takes into account the composite effect of energy and entropy change of the system.
Gibbs free Energy or Gibbs Energy as criterion of spontaneity
The symbol of Gibbs energy is G. It is related to the enthalpy and entropy by the relation
G = H – TS
....(42)
Since H, T and S are state functions; therefore, G must also be a state function.
Differential of Equation (42) is dG = dH – TdS – SdT
If the temperature is constant dT = 0, therefore, dG = dH – TdS
....(43)
At constant pressure, dH = dq. Therefore,
(dG)T, P = dq – TdS
....(44)
Let us see how Gibbs energy can be used as criterion of spontaneity dS =
dS >
Whether the process is reversible or irreversible
TdS = dqrev
Therefore, (dG)T, P = dq –dqrev
For a reversible process each step is an equilibrium step and dq = dqrev.
Therefore,
(dG)T, P = dqrev – dqrev = 0
So for the equilibrium state, dG = 0
For an irreversible process dq = dqirr .Therefore,
(dG)T, P (irr) = dqirr – dqrev
We know that dqrev > dqirr, therefore,
75
dqirr –dqrev < 0 and (dG)T, P < 0
...(45)
Since all the spontaneous processes are irreversible processes, for all spontaneous process taking place at constant temperature and pressure the following relation is valid dG < 0
Characteristics of Gibbs free Energy
G is a state function
For a spontaneous process the change in free energy is always negative.
For equilibrium state dG = 0
For non-spontaneous process dG >0
The SI unit of G is J
For finite change at constant temperature and pressure
ΔG = ΔH – TΔS
....(46)
Here ΔG is the free energy change in the system
ΔH is the enthalpy change in the system
ΔS is the entropy change in the system
The process will be spontaneous, that is the change in free energy of the system, ΔG will be negative if either of the following conditions is true
ΔH < 0 and ΔS >0
ΔH > 0 but ΔH < TΔS
ΔH< 0 and ΔS<0 but ΔH > TΔS
When ΔH > 0 and ΔS < 0, the process will always be non-spontaneous.
This can be elaborated as follows:
On the basis of the enthalpy change there are two types of processes
76
(i)
Exothermic process for which the enthalpy change is negative, that is,
ΔH < 0
(ii)
Endothermic process for which the enthalpy change is positive, that is,
ΔH > 0
Each type of the process is also accompanied by the change in entropy ΔS which is positive when the entropy increases during a process and negative when the entropy decreases during a process. Therefore, there are the following four cases:
(i)
Exothermic process, ΔH < 0
(a)
ΔS > 0, which is the entropy increases during the process. Therefore,
TΔS > 0
ΔH −TΔS < 0
Therefore ΔG is always negative and the process is always spontaneous. (b)
ΔS < 0, which is the entropy decreases during the process. Therefore,
T ΔS < 0
ΔH −T ΔS will be less than zero and ΔG will be negative when
∣ΔH ∣ >∣TΔS∣
That is, the process will be spontaneous when ΔH is less than TΔS
(ii)
Endothermic process, ΔH > 0
(a)
ΔS > 0, which is the entropy increases during the process. Therefore,
TΔS > 0.
The process will be spontaneous when
ΔH −T ΔS < 0 or ∣ΔH ∣ <∣TΔS∣
(b)
ΔS < 0, which is the entropy decreases during the process. Therefore,
TΔS < 0
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ΔH−TΔS will always be positive, that is ΔG > 0. Therefore the process is always non-spontaneous.
Variation of G with Temperature and Pressure
G = H –TS
....(47)
Differential of this equation gives dG = dH –TdS – SdT
...(48)
According to combined form of first law and second law of thermodynamics, for reversible process dH = TdS + VdP
Substituting the value in equation we get dG = TdS +VdP –TdS –SdT or dG = – SdT +VdP
....(49)
Various thermodynamic relations between U, H, S, T, V and P
1.
dU = TdS –PdV
2.
dH = TdS + VdP
3.
dG = – SdT + VdP
4.
dA = – SdT –PdV
Solved example: Calculate H° and S° for the following reaction:
NH 4 NO 3 (s) + H2 O() NH+4 (aq) + NO 3- (aq)
Use the results of this calculation to determine the value of Go for this reaction at 25o C, and explain why NH4NO3 spontaneously dissolves is water at room temperature. 78
Given data:
Compound
Hfo(kJ/mol)
NH4NO3(s)
-365.56
151.08
NH +4 (aq)
-132.51
113.4
NO -3 (aq)
-205.0
146.4
Solution: Ho =
S°(J/mol-K)
o
o
H f (products) -
H f (reactants)
= [1 mol NH4 132.51 kJ/mol + 1 mol NO -3 -205.0 kJ/mol] - [1 mol NH4NO3
-365.56 kJ/mol]
= 28.05 kJ
The reaction leads to a significant increase in the disorder of the system, however, and is therefore favored by the entropy of reaction:
So =
o
S (products) -
o
S (reactants)
= [1 mol NH4 113.4 J/mol-K + 1 mol NO -3
146.4 J/mol-K] - [1 mol NH4NO3
151.08 J/mol-K]
= 108.7 J/K
To decide whether NH4NO3 should dissolve in water at 25o C we have to compare the Ho and T So to see which is larger. Before we can do this, we have to convert the temperature from oC to kelvin:
TK = 25o C + 273.15 = 298.15 K
We also have to recognize that the units of Ho for this reaction are kilojoules and the units of So are joules per kelvin. At some point in this calculation, we therefore have to convert these quantites to a consistent set of untis. Perhaps the easiest way of doing this is to convert Ho to joules. We then multiply the entropy term by the absolute temperature and subtract this quantity from the enthalpy term:
Go = Ho - T So
= 28,050 J - (298.15 K
108.7 J/K)
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= 28,050 J - 32,410 J
= - 4360 J
At 25o C, the standard-state free energy for this reaction is negative because the entropy term at this temperature is larger than the enthalpy term:
Go = -4.4 kJ
The reaction is therefore spontaneous at room temperature.
8.6.5. Third law of Thermodynamics
The entropy at absolute zero is determined with the help of third law of thermodynamics which states that:
Entropy of perfectly crystalline substance is zero at absolute zero.
This statement has a logical explanation. The best organized physical state of a system is that of a single crystal, since in this state each particle has definite arrangement in space with respect to other particles. At very low temperatures only vibrational motion of the particle (atom or molecule) remains and at absolute zero all the particles vibrate with lowest possible energy. Therefore, the arrangement is most ordered. For a perfectly crystalline substance there is only one possible arrangement of particles, so there is no randomness. Hence, the entropy is zero at absolute zero.
Characteristics of absolute entropy
Entropy of perfectly crystalline substance is zero at absolute zero.
Substances that are not perfectly crystalline have positive entropy at absolute zero.
At higher temperatures the absolute entropy of all the substances is positive.
Entropy depends upon mass of the molecules. Heavier molecules have greater entropy.
Entropy depends upon the number of atoms in a molecule. More are the number of atoms in a molecule greater is the entropy.
Entropy depends
Sgas > Sliquid > Ssolid
upon
the
80
physical
state
of
the
substance.
Solved Example: Arrange the diatomic molecules: F2, C 2, Br2 and I2, in the increasing order of entropy at room temperature.
Solution: The increasing order of entropy is I2 < Br2 < F2 < Cl2
Since iodine is solid at room temperature it has least entropy followed by bromine which is in liquid state and so has greater entropy than iodine.
Chlorine and fluorine both are gases but chlorine has greater molecular mass than fluorine so chlorine has maximum entropy in case of halogens.
8.6.6. Equilibrium Constant from the Free Energy Change
Standard-State Free Energy of a Reaction
What does the value of Go tell us about the following reaction?
N2(g) + 3 H2(g)
2 NH3(g)
Go = -32.96 kJ
By definition, the value of Go for a reaction measures the difference between the free energies of the reactants and products when all components of the reaction are present at standard-state conditions.
Go therefore describes this reaction only when all three components are present at 1 atm pressure.
The sign of Go tells us the direction in which the reaction has to shift to come to equilibrium. The fact that in the above example, Go is negative for this reaction at 25oC means that a system under standard-state conditions at this temperature would have to shift to the right, converting some of the reactants into products, before it can reach equilibrium. The magnitude of Go for a reaction tells us how far the standard state is from equilibrium. The larger the value of Go, the further the reaction has to go to get to from the standard-state conditions to equilibrium.
At equilibrium, G of a reaction is zero. Therefore, if we know the standard state free energy change, Go, for a chemical process at some temperature T, we can calculate the equilibrium constant for the process at that temperature using the relationship between Go and Keq.
0 = Go- RTℓn Keq
81
Or, Go = –RTℓn Keq
Rearrangement gives
ΔG o or = K eq
RT
In Keq = –
e
ΔG o/RT
Once the value of Go is known to us, all the other thermodynamic properties can also be calculated with the help of the thermodynamic relations derived in the presious sections.
Solved Example: Use the following standard-state free energy of formation data to calculate the acid-dissociation equilibrium constant (Ka) at for formic acid: Compound
Gfo(kJ/mol)
HCO2(aq)
-372.3
H+(aq)
0.00
HCO -2 (aq)
-351.0
Solution: We can start by writing the equation that corresponds to the aciddissociation equilibrium for formic acid:
HCO2H(aq)
H+(aq) + HCO -2 (aq)
We then calculate the value of
Go =
G of (products) -
= [1 mol H+
Go for this reaction:
G of (reactants)
0.00 kJ/mol + 1 mol HCO -2
-351.0 kJ/mol] - [1 mol HCO2H
–372.3 kJ/mol]
= 21.3 kJ
We now use the relationship between Go and the equilibrium constant for the reaction: Go = - RTℓn K and solve for the natural log of the equilibrium constant:
In K =
82
ΔG o
RT
Substituting the known value of Go, R, and T into this equation gives the following result:
In K =
21.300 J/mol
(8.314 J/mol k)(298K)
8.60
We can now calculate the value of the equilibrium constant:
K = e-8.60 = 1.8
10-4
The value of Ka obtained from this calculation agrees with the table value for formic acid, within experimental error.
83
SUMMARY
Energy changes in physical or chemical processes can be studied through a branch of science called as Thermodynamics. These changes can be studied quantitatively which helps in providing useful predictions. The universe may be divided as system and surrounding. These processes lead to conversion evolution or absorption of heat or q, which may be partially converted into work which is denoted as (w). First law of thermodynamics ΔU = q+w; wherein ΔU, change in internal energy, depends on initial and final states, q and w depend on path. ΔU is a state function while q and w are not the state functions. In expansion of gases, work can be measured.
At constant volume, w=0, then ΔU = qv. Another state function is enthalpy. Heat changes at constant pressure can give Enthalpy change.
Change of phase- vaporization, sublimation etc. (enthalpy changes) occurs at constant temperature. Enthalpy changes are always positive. Various enthalpy changes as
Enthalpy of formation, combustion etc. is always positive.
First law of thermodynamics does not guide us about the direction of chemical reactions
i.e., what is the driving force of a chemical reaction. Another state function is Entropy or
S. Entropy is a measure of disorder or randomness. For a spontaneous change, total entropy change is positive.
Chemical reactions are generally carried at constant pressure, it is defined by another state function i.e. Gibbs energy. G which is related to entropy and enthalpy changes of the system of the equation.
84
Students’ Worksheets
Student Worksheet- 1
1.
Why is it important to study thermodynamics? Search the web and list three simple applications of thermodynamics.
_________________________________________________________________________
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2.
In a thermodynamic process, one form of energy is transformed into another form.
Identify the initial and the final forms of energy in the following processes.
Thermodynamic Process
Initial form of energy Final form of energy A battery running on a fuel cell
Tuning fork after jerk
A car running on petrol
Speaking on a telephone
Producing fire by rubbing stones
Making a toasted bread in a toaster
3.
Heat flows from one body to another (usually from hotter body to colder body) until equilibrium is reached. Identify the direction of heat flow in the following cases. 85
Heat flows
From
To
When our body is sweating!
When ice cube is added to a glass of water at room temperature!
When we keep water in a clay pot out in summers 4.
What is the role thermodynamics in chemistry?
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
_________________________________________________________________________
5.
What are the limitations of thermodynamics? Which branch of science can overcome these limitations and complement the findings of thermodynamics?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
6.
Thermodynamics is the study of macroscopic objects, i.e, the objects visible to naked human eye! On the contrary, quantum mechanics is the study of microscopic objects, i.e., the objects NOT visible to naked human eye!
a) Identify macroscopic and microscopic objects from the following.
86
A glass of water, a metal block, an electron moving in an orbital, a gas enclosed in a container, sodium and chloride ions dissolved in water, two hydrogen atoms colliding with each other, a matchstick.
Macroscopic system
Microscopic system
b) Based on the above answer, formulate suitable definitions for macroscopic and microscopic systems.
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
c) What properties (or variables) do you think can be used define macroscopic systems? ____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
____________________________________________________________________________
7.
The word thermodynamics stems from two Greek words meaning
a)
conservation of heat
b) interactions of heat
c)
study of heat
d) movement of heat
87
8.
Which of the following thermodynamics? a)
laws of
physics
becomes
the foundation
of
Newton‟s laws of motion
b) Law of conservation of energy
c)
Law of universal gravitation
d) Law of conservation of momentum
9.
When heat is added to a system, all of the following may happen EXCEPT
a)
increase in internal energy
b) decrease in the system‟s temperature
c)
external work is done by the system
d) increase in the pressure in the system
10.
Thermodynamics focuses on the macroscopic details of systems. Which of the following is NOT used to describe a system macroscopically?
a)
Temperature
b) Pressure
c)
Work
d) Spin
11.
Which of the following is TRUE about thermodynamics?
a)
It is based on conservation principle.
b) It deals with energy.
c)
It discusses direction of heat movements.
d) All of the above
12.
What happens when you get inside an air-conditioned room after staying under the sun for sometime?
a)
You feel warm because heat flows from your body to the room
88
b) You feel warm because heat flows from the room to your body
c)
You feel cold because heat flows from your body to the room
d) You feel cold because heat flows from the room to your body
13. Heat energy can change into
a)
light only
b) work only
c)
chemical energy only
d) any form of energy
14.
Which of the following best describes thermodynamics?
a)
It is the study of the hotness and coldness of a body
b) It is the study of the atomic and molecular nature of matter
c)
It is the study of energy and its transformations
d) It is the study of the interaction between heat and temperature
15.
Which of the following statement is true about thermodynamics?
a)
Thermodynamics will help in predicting whether a physical or chemical change is possible under given conditions.
b) Thermodynamics only deals with the initial and final states of the system and is not helpful in evolving the mechanism of the process.
c)
The rate of a reaction can be evolved from thermodynamics.
d) The mechanism of a reaction can be evolved from thermodynamics
i.
I only
ii.
I & II
iii.
I, II & III
iv.
I & IV
89
Student Worksheet- 2
1.
Define system and surroundings in thermodynamic terms. How the two are separated? Identify the system and the surroundings in the following cases.
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________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
Example
Topic of investigation A cup of tea
To measure the heat absorbed by the cup walls as well as heat given out in the air
Water inside a thermos flask* To measure the temperature of water after adding some ice
An inflated football To measure the volume occupied by the ball inflated with air
System
Surroundings
Boundary
(real or imaginary) * assume that no heat or mass can be exchanged from thermos flask to the surroundings 90
2.
Differentiate between an open, a closed and an isolated system. Identify the type of the system in the examples given in question number 1.
Type of system Definition
Example from
Q. No. 1
Open
Closed
Isolated
3.
What do you understand by the state of a thermodynamic system? What are state variables? Which variables may be used to define the state of the systems given below? System
Variables
A glass of lemonade
A balloon filled with air An ice cube
<
4.
Define a state function and a path function. Put „√‟ for the state functions or „Χ‟ for path functions against the given thermodynamic properties.
Function
Definition
State function Path function Enthalpy
Heat
Temperature
Internal energy 91
Volume
Pressure
Entropy
Work
Free
energy
5.
Define intensive and extensive properties. Put „I‟ for intensive property or „E‟ for extensive property against the given thermodynamic properties.
Property
Definition
Intensive property Extensive property Internal
Energy
Enthalpy Density Refractive
Index
Pressure Resistance Entropy
Surface
Tension
6.
Volume
Specific
Volume
Hof
Electric
Field
Mass
Free
Energy
Temperature
Normality
Molality
Resistivity
Polarization Magnetic Magnetization
Field
An extensive property can become an intensive property if mass or size of the system is fixed! For example, volume is an extensive property, but molar volume is an intensive property as volume of one mole of the system will always be constant.
List some more examples where an extensive property can become an intensive property. ________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
92
7.
Define a thermodynamic process. What do you understand by reversible and irreversible process? Which of them is always in thermodynamic equilibrium?
8.
As the name suggests, a reversible process means a process which can be reversed without causing any change, elsewhere!
Of the following processes, which of them do you think are reversible?
Process
Reversible or irreversible
Stretching of a rubber band within its elastic limit Stretching of a rubber band beyond its elastic limit A gas enclosed in a cylinder with frictionless pulley is compressed by putting a 100 g of weight on the pulley
9.
State the zeroth law of thermodynamics? Why do you think it is numbered as
„Zero‟ in the sequence of thermodynamic laws?
10.
A system is said to be in thermodynamic equilibrium if it is
11.
a)
in thermal equilibrium with the surroundings (temperature is the same for the whole system and surroundings)
b)
in mechanical equilibrium (no mechanical work is done on any part of the system or between the system and the surroundings)
c)
in chemical equilibrium (the composition or concentration remains fixed and definite) d)
all of the above
A closed system is that which exchanges ______ with the surroundings.
a)
Neither matter nor energy
b)
Energy only
c)
Matter only
d)
Energy and matter
93
12.
13.
14.
15.
16.
A process in which the driving force is infinitesimally greater than the opposing force is known as
a)
a reversible process
b)
an adiabatic process
c)
a cyclic process
d)
an irreversible process
An irreversible process is one which
a)
goes from the initial to the final state in finite number of steps and in finite time b)
is not in equilibrium at all stages
c)
has a driving force much greater than the opposing force
d)
all of the above
A thermos flask, when opened
a)
becomes an isolated system from an open system
b)
becomes an open system from an isolated system
c)
becomes a closed system
d)
remains an isolated system
Most of the processes occurring in nature are
a)
Isochoric
b)
Adiabatic
c)
Isobaric
d)
all of the above
A process in which the apparatus is completely insulated is called an
a)
isothermal process
b)
adiabatic process
94
17.
c)
isochoric process
d)
isobaric process
The human body is
a)
an open system
b)
a closed system
c)
an isolated system
d)
not a system
18.
What do you understand by a thermodynamic equilibrium?
19.
What is a reversible process in thermodynamics?
95
Student Worksheet - 3
1.
What do you understand by „work‟ in thermodynamic terms? Give an example.
What will be the sign („+‟ or „-„) of work if it is said to be done on the system?
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2.
The work done by an ideal gas enclosed in a cylinder with a frictionless piston can be obtained by calculating the area under the P-V curve.
In the following P-V curve for an isothermal expansion of a given amount of gas from state A to a state B, draw the area under the curve that represents
A)
Work done reversibly B)
work done
irreversibly in a single step
Work done in reversible isothermal process 3.
Work done in an irreversible isothermal process
Compare the work done in an isothermal process carried out reversibly and irreversibly, which process gives maximum work and why? Do you think this maximum amount of work is practically achievable?
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96
________________________________________________________________________
4.
Show that work done cannot be a state function.
Step - I
Consider a system going from state A to state C in a PV diagram via two routes : a)
A → B → C and
b) A → D → C
Step - II
Calculate the work done in going from A→B
Step - III
Calculate the work done in going from B→C
Step - IV
Calculate the total work done in going from
A→B→C
Step - V
Calculate the work done in going from A→D
Step - VI
Calculate the work done in going from D→C
Step - VII
Calculate the total work done in going from
A→D→C
Step VIII
Compare the total work done in going from A → C from both the routes and infer your result
97
5.
Give the expression for calculating work done in a reversible isothermal process and an irreversible isothermal process.
One mole of an ideal gas at 300 K occupies a volume of 10 L in a cylinder fitted with a frictionless piston. It is expanded isothermally to a final volume of 20 L.
Calculate w and q if the expansion is carried out (i) reversibly, (ii) against a constant external pressure of 0.1 atm. and (iii) against a zero external pressure (i.e., against vacuum).
Answer
Expression
Work done in a reversible isothermal process
Work done in an irreversible isothermal process
Work done against a zero external pressure wrev =
wirrev =
wirrev =
Calculation
Result (with proper units)
6.
Three moles of an ideal gas expand from one litre to a volume of ten litres at 27 oC.
What is the maximum work done?
(Hint: The maximum work is obtained in a reversible and isothermal process)
Formula used
Conversion
of units
Calculation
w nRT In
T = 27 + 273 = 300 K
w nRT In
= 3 mol
= x 3.303 RT log
2.303
8.314 J K-1 mol-1
Result (with proper units)
98
300 K
log
= 17234.4 J
7.
10 moles of an ideal gas expands isothermally and reversibly from a volume of 5 L to 50 L at 25°C. What is the maximum work done?
Formula used
Conversion of units Calculation
Result (with proper units)
8.
Calculate the amount of work obtain in isothermal reversible expansion of 20 g of
Ar. At 27°C from a pressure of 4 atm to 1 atm.
Formula used
Conversion of units Calculation
Result (with proper units)
9.
Define „heat‟. What is the symbol used to denote heat? What will be the sign of quantity of heat absorbed by the system?
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99
10.
Show that heat is also a path function.
Step – I
Consider an isothermal process going from state A to state C in a P-V diagram via two routes :
a) A → B → C and
b) A → D → C
Step – II
Calculate the heat absorbed in going from A→B
Step – III
Calculate the heat absorbed in going from B→C
Step – IV
Calculate the total heat absorbed in going from A → B → C
Step – V
Calculate the heat absorbed in going from A→D
Step – VI
Calculate the heat absorbed in going from D→C
Step – VII Calculate the total heat absorbed in going from A → D → C
Step VIII
11.
Compare the total heat absorbed in going from A → C from both the routes and infer your result
An isothermal expansion may be represented as in the figure below. Calculate the work done by the system and heat absorbed by the system when a gas expands against a constant external pressure of 1 atm from a volume of 10 L to 20 L.
100
Work
Heat
Formula used
Calculation
Result (with proper units)
12.
Two moles of H2 expands isothermally and reversibly to ten times its volume at
25°C. Assuming ideal behavior for the gas, calculate the work done and the heat absorbed during the process. ( R = 8.314 JK-1 mol-1)
Formula used
Conversion of units Calculation
Result (with proper units)
13.
What causes the lid of a pot to move up when the water in it starts boiling?
a)
The heated air inside the pot compresses
b) The heated air inside the pot expands
c)
The boiling water compresses
d) The pressure inside the pot decreases
14.
A system goes from state (i) P1, V to 2P1, V. Another system goes from state
(ii) P1, V to P1, 2V. The work done in the two cases is
101
a)
(i) zero (ii) zero
b) (i) zero (ii) P1 V
c)
(i) P1 V (ii) zero
d) (i) P1 V (ii) P1 V
15.
The units of heat are
a)
J m-1
b) J m
c)
Cal
d) N m-2
16.
Heat added to a system is equal to
a)
a change in its internal kinetic energy
b) a change in its internal potential energy
c)
work done by it
d) sum of above all the three
17.
The area inside a closed curve on P-V diagram represents
a)
the condition of a system
b) work done on the system
c)
work done in a cyclic process
d) a thermodynamic process
18.
What will be the work done when 2 mole of an ideal gas are compressed reversibly from 10.00 bar to 50.00 bar at a constant temperature of 300 K.
a)
-14.01 kJ
b) 8.02 kJ
c)
4.02 kJ
102
d) -18.02 kJ
19.
An ideal gas occupies 1.25 dm3 at a pressure of 6 bar. If the gas is compressed isothermally at a constant pressure pext, to attain a volume of 0.500 dm3, the minimum possible value of pext is
a)
10 bar
b) 15 bar
c)
20 bar
d) 30 bar
20.
What will be the maximum amount of work done when 1 mol of an ideal gas is expanded from 2.0 dm3 to 4.0 dm3 at 27oC?
a)
7.78 kJ
b) -1.73 kJ
c)
-4.78 kJ
d) +11.73 kJ
103
Student Worksheet- 4
1.
What is meant by the term „Internal Energy‟ of the system? Is it a measurable property? What are the various forms of energy that gives rise to Internal energy of the system?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
2.
State first law of thermodynamics. Give examples from daily life in support of this statement. (You may search internet for finding some interesting examples!)
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
3.
The mathematical statement of first law is ∆U = q + w. Following this relation predict how the internal energy of the system is increased or decreased?
Process
Sign convention
Work done on the system
Work done by the system
Heat absorbed by the system
Heat given out by the system
104
Effect on ∆U (increase/ decrease) 4.
Discuss the significance of the mathematical expression of the first law where the heat absorbed by a system is related to the internal energy and work done by the system. ________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
5.
How will you infer from first law of thermodynamics that ∆U is a state function although q and w are both path functions?
(Hint: In a cyclic process, the system returns to its initial state! Can changing the path create or destroy energy? Can ∆U be different if different paths are followed?)
Example
Cyclic process
Case – I
(If ∆U is a path function )
Cyclic Process 1
Case – II (If
∆U is a path function )
Cyclic Process 2
Case – III (If
∆U is a state function )
Cyclic Process 1
Change in internal energy(∆U)
(+)
A (P1,V1) → B
(P2,V2) → A
(P1,V1)
(-)
A (P1,V1) → C
(P3,V3) → A
(P1,V1)
zero
A (P1,V1) → B
(P2,V2) → A
(P1,V1)
105
Effect on the surroundings Violation / conformation of law of conservation of energy Inference
6.
___________________________________________________________
___________________________________________________________
___________________________________________________________
___________________________________________________________
Name the two most common modes by which system and surroundings exchange their energies.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
7.
What is the main limitation of the first law of thermodynamics?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
8.
A sample of an ideal gas is allowed to expand isothermally.
(a)
Does the gas do work?
___________________________________________________________________________
(b)
Does the system exchange heat with the surroundings?
___________________________________________________________________________
(c)
What happens to the temperature of the gas?
___________________________________________________________________________
106
(d) What is the change in internal energy?
___________________________________________________________________________
9.
The densities of ice and water at 0oC are 0.90 g cm-3 and 0.98 g cm-3, respectively.
The latent heat of ice is 80 cal g-1. Calculate the change in internal energy when 5 kg of ice melts.
Formula used
Conversion of units Calculation
Result (with proper units)
10.
Define „enthalpy‟ of the system. What was the necessity to introduce the term enthalpy over internal energy for measuring the energy changes in the system?
Give the mathematical expression for enthalpy.
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
11.
Using the thermodynamic relationships: ∆U = q + w ; w = - P∆V ; H = E + PV and
PV = nRT, derive the following relationships:
a)
∆U = qv
b) ∆H = qp
107
c)
∆H = ∆U + (∆n)RT
To Derive
∆U = qV
12.
H = qP
∆H = ∆U + (∆n)RT
Why are the heat changes reported, usually as enthalpy changes and not internal energy changes?
_________________________________________________________________________
________________________________________________________________________________
_______________________________________________________________________________
13.
What are the conditions under which ∆H becomes equal to ∆U?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
14.
A sample of argon gas at 1atm pressure and 27oC expands reversibly and isothermally from 1.25 dm3 to 2.50 dm3. Calculate w, q, ∆U and ∆H enthalpy change in this process.
Formula used
Conversion of units Calculation
Result (with proper units)
108
15.
a)
A cooking gas cylinder is assumed to contain 11.2 kg of isobutane. If a family needs 15000 kJ of energy per day for cooking, how long would the cylinder last? b)
Assuming that 30% of the gas is wasted due to incomplete combustion, how long would the cylinder last? The combustion of butane is given by
C4H10(g) +
O2(g) → 4CO2(g) + 5H2O(ℓ)
H = -2658 kJ mol-1
(Hints:C4H10(g) +
O2(g) → 4CO2(g) + 5H2O(ℓ)
H = -2658 kJ mol-1
58 g of C4H10 gives 2658 kJ of energy.
Calculate for 11.2 kg of C4H10)
Formula used
Conversion of units
Calculation
Result (with proper units) 16.
When a boy mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee cup calorimeter the temperature of the resultant solution increases from 21.0 oC to
27.5oC. Assuming that the calorimeter absorbs only a negligible quantity of heat, the total volume of the solution is 100 mL, its density is 1.0 g mol -1 and its specific heat 4.18 J g-1.Calculate
a)
the heat change during mixing
b) the enthalpy change during the reaction
HCl(aq) + NaOH(aq) → NaCl(aq) + H2O
(Hints: Qmixing = mxsxt))
109
Formula used
Conversion of units
Calculation
Result (with proper units) 17.
In the diagrams (i) to (iv) of figure, variation of volume by changing pressure is shown. A gas is taken along the path ABCD. The change in internal energy of the gas will be
a)
positive in all the cases (i) to (iv) positive in cases (i), (ii) and (iii) but zero in case (iv)
b) negative in cases (i), (ii) and (iii) but zero in case (iv)
c)
18.
zero in all the four cases.
State which of the following will decrease the total energy content of the system.
a)
Work done by the system
b) Heat transformed from the surroundings
c)
Work done on the system
d) Decrease in volume of the system
19.
The change in the internal energy of a system
a)
is an extensive property
b) is a thermodynamic parameter
110
c)
is zero in a cyclic process
d) all of the above
20.
When an ideal gas expands at constant temperature, the change in internal energy is a)
positive
b) negative
c)
zero
d) equal to the work done by the gas
21.
Enthalpy of a system is
a)
an extensive property
b) ∆U + p∆V
c)
qp
d) all of the above
22.
For the reaction
3H2O(ℓ)
vapHm rU 3H2O(g)
= 10k cal mol–1 at 500 K and 1 atm pressure.
of the given reaction is
a) 27 kcal
b) -27 kcal
c) 13.5 kcal
d) 54 kcal
111
Student Worksheet- 5
1.
What is the basic difference between standard enthalpy of formation and standard enthalpy of reaction? Illustrate with suitable examples.
Standard enthalpy of formation
Definition
Example
Standard enthalpy of reaction
Definition
Example
2.
What is value of standard enthalpy of formation of an element in its most stable state? Is it a measurable property?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
112
3.
Diamond is an elementary form of carbon, yet its standard heat of formation is not taken as zero. Why?
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
4.
What are endothermic and exothermic reactions? What is the sign of ∆H for these reactions? ________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
________________________________________________________________________________
5.
The standard enthalpy of formation of H2O( ) and H2O(g) are -285.8 and -241.8 kJ mol-1 respectively. What is the heat of vaporization of water per gram at 25 oC and
1 atm?
Desired chemical equation Calculation
Result (with proper units) 6.
From the data given below, at 298 K for the reaction:
CH4(g) + 2O2(g) → CO2(g) + 2H2O( ), calculate the enthalpy of formation of CH4(g) at 298 K.
Enthalpy of reaction = -890.5 kJ
113
Enthalpy of formation of CO2(g) = -393.5 kJ mol-1
Enthalpy of formation of H2O( ) = -286.0 kJ mol-1
Formula used
Calculation
Result (with proper units)
7.
The
standard
heats
of
formation
of CH4(g),
CO2(g)
and
H2O( )
are
-76.2 kJ mol-1, -398.8 kJmol-1 and -241.6 kJ mol-1 respectively. Calculate the amount of heat evolved by burning 1 m3 of methane measured at normal conditions.
Desired chemical equation Calculation
Result (with proper units)
8.
Calculate the standard enthalpy change for the following reaction
2H2O2( ) → 2H2O ( ) + O2(g)
When the standard enthalpy of formation of various compounds are
= -188 kJ mol-1
= -286 kJ mol-1
Identify whether the reaction is exothermic or endothermic
114
Desired chemical equation Calculation
Result (with proper units)
9.
Calculate ΔH° for the reaction N2H4( ) + O2 (g) → N2(g) + 2 H2O( ) from the reactions given below.
ΔH° (kJ)
(i) 2 NH3(g) + 3 N2O(g) → 4 N2(g) + 3 H2O(ℓ)
ؘ1011.4
(ii) N2O(g) + 3 H2(g) → N2H4( ) + H2O(ℓ)
-317.25
(iii)
2 NH3(g) + 1/2 O2(g) → N2H4(ℓ) + H2O(ℓ)
-142.98
(iv)
H2(g) + 1/2 O2(g) → H2O( )
-285.83
Desired chemical equation Calculation
Result (with proper units) 10.
In the production of cement calcium carbonate, CaCO3, is converted to calcium oxide, CaO, according to the following balanced chemical equation.
CaCO3 (s) → CaO (s) + CO2 (g)
115
i. From the given standard molar enthalpies of formation, calculate ΔH° for this reaction ΔHf° (CaO, s) = -635.1 kJ mol-1; ΔHf° (CO2, g) = -303.5 kJ mol-1; ΔHf°
(CaCO3, s) = -1206.9 kJ mol-1
Desired chemical equation Calculation
Result (with proper units) ii. Is this an endothermic or exothermic reaction?
_____________________________________________________________________
_____________________________________________________________________ iii. CaCO3 actually exists in two forms. The calculation you did in part a was for the conversion of the form of CaCO3 known as calcite to CaO. If ΔH for the conversion of aragonite (the other form of CaCO3) to CaO is +178.1 kJ mol-1, what is ΔH for the conversion of calcite to aragonite?
Desired chemical equation Calculation
Result (with proper units)
116
11.
Calculate the bond energy of HC , if the bond energies of H - H and C - C are
433 and 242 kJ mol-1 respectively and the heat of formation of HC is -92 kJ mol-1.
Desired chemical equation Calculation
Result (with proper units) 12.
Compute the enthalpy of formation of liquid methanol in kJ mol-1 using the following data:
∆vapH (CH3OH) = 38 kJ mol-1
∆fH (H, g) = 218 kJ mol-1
∆fH (C, g) = 715 kJ mol-1
∆fH (O, g) = 249 kJ mol-1
∆HC-H = 415 kJ mol-1
∆H C-O = 356 kJ mol-1
∆H O-H = 463 kJ mol-1
Desired chemical equation Calculation
Result (with proper units) 117
13.
∆H for the given reaction is -150 kJ. Calculate the enthalpy of the A
C bond.
[Given, bond enthalpies of C-H = 414 kJ mol-1, H-H = 435 kJ mol-1, C-N = 293 kJ mol-1, N-H = 369 kJ mol-1]
Formula used
Calculation
Result (with proper units)
14.
15.
16.
Exothermic reactions are those
a)
which absorb heat at high temperatures
b)
in which heat is evolved
c)
which absorb heat at low temperatures and evolve heat at high temperatures
d)
where heat is evolved only at high temperatures
For endothermic reactions the sign given for heat exchanged is
a)
positive
b)
negative
c)
positive at high temperatures and negative at low temperatures
d)
positive at high temperatures and negative at low temperatures
Heat of combustion of carbon (graphite) is the same as
a)
heat of combustion of carbon (diamond)
118
b)
heat of formation of carbon dioxide
c)
both (a) and (b)
d)
neither (a) nor (b)
[Hint] The standard state of carbon is taken as carbon (graphite) and not carbon (diamond).
119
SOME COMMON MISCONCEPTIONS dq and dw where ever they are mentioned do not mean change in heat or change in work. dq stands for small amount of heat exchanged by the system. dw denotes small amount of work done.
Standard state of a system is defined for the pure substance at 1 bar pressure and a specified temperature and not at 1 bar pressure and 273.15 K temperature.
For gases the standard state is the one at which pure gas behaves ideally at one bar and a specified temperature. For liquids the standard state is the pure liquid at one bar and specified temperature. For solids it is the pure and most stable crystalline form of aggregation of solid at one bar and specified temperature.
120
CROSSWORD PUZZLE
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
121
ACROSS
Hints
1
Metal is a better---of heat than wood
7
---can neither be created nor destroyed.
8
Things that are cooled will---in length
9
System + surroundings
11
Change of state from liquid to gas
14
---reactions require energy in order to take place
16
Study of motion of heat and work
17
Device used to measure temperature
18
---processes take place on their own.
DOWN
Hints
2
Specific heat capacity is the energy needed to raise temperature of a substance by ---degree Celsius
3
Isothermal process takes place at constant---
4
Process taking place at constant volume
5
At equilibrium, change in Gibbs free energy
6
A measure of disordered energy in the system
10
Boundary or wall which does not allow heat to pass through
12
Lowest temperature known
13
Temperature scale with 100 degrees between freezing and boiling
15
Another word for freezing
122
ANSWERS
C
O
N
D
U
C
N
E
N
E
R
G
Y
T
O
E
I
Z
E
M
S
E
N
P
U
N
E
I
V
C
E
N
D
V
A
O
S
R
O
E
C
R
H
A
T
I
O
N
A
N
A
O
D
P
Y
U
I
A
C
B
I
T
U
E
S
Z
E
T
H
E
R
M
I
E
H
E
R
M
O
D
Y
N
M
A
F
T
U
I
C
C
S
I
O
R
M
O
M
E
T
E
R
N
T
A
N
E
O
U
S
R
O
A
T
R
S
O
C
O
I
U
P
R
R
S
S
T
T
L
H
P
R
C
B
L
T
E
R
123
N
ADDITIONAL RESOURCE LINKS:
Smith, J.M.; Van Ness, H.C., Abbott, M.M. (2005). Introduction to Chemical
Engineering Thermodynamics.
Introductory Thermodynamics: i.UIC; ii. U Pittsburgh; iii. Kids' Almanac; iv. Eden
Prairie High School; v. School of Champions; vi. Scitoys vii. THERMO spoken here
Engineering Thermodynamics: i.A Graphical Approach /
124
CENTRAL BOARD OF SECONDARY EDUCATION
Shiksha Kendra, 2, Community Centre, Preet Vihar, Delhi-110 092 India
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