Chemistry Module 6
Chemistry
Module Six: Thermochemistry and rate
Study sheet for the test
G.01: Thermo-chemistry I. Temperature and Thermal Energy
Temperature is a measure of the average kinetic energy of the particles in a sample of matter. The greater the avg. kinetic energy of the particles in a sample of matter, the higher the temperature of that matter. Some temperature scales are Fahrenheit, Celsius and Kelvin.
Thermal energy is the measure of the total kinetic energy in a sample. And Temperature is a measure of the average kinetic energy.
Ex: Which container (the bath tub or the coffee cup) have a temperature that is higher than the other?
The temperature of the coffee cup is higher!
Which sample of water has a greater amount of thermal energy? The bathtub. Even though the avg. kinetic energy of the particles in the coffee cup is higher than the avg.KE of the particles in the bathtub, the thermal energy in the warm bath tub is higher because there are more particles in the tub than the cup.
II. Heat
Heat is the movement, or flow, of thermal energy.
Heat always flows spontaneously from matter at a higher temperature to matter at a lower temperature.
Heat and thermal energy are both measure in units of energy, the SI unit for energy is Joule(J). Another unit of energy that is commonly used is the Calorie. The food calorie means that for every one food calorie that is eaten has enough thermal energy to increase the temperature of 1000 grams of water by 1 degree Celsius.
III. Closed, open and isolated systems
In science, it is important to identify the system being observed or investigated. Some examples of systems are: atom, reaction, cell, ecosystem, universe etc. In chemistry, the system is only defined as the chemical reaction or physical change being examined. a. Isolated systems The total amount of energy and matter contained in an isolated system will remain constant because energy and matter cannot leave or enter.
Ex: Universe, space b. Closed systems
Energy can leave or enter the system but matter cannot.
Ex: earth, beakers c. Open system
Both energy and matter are exchanged freely.
Ex: our body.
IV. Energy in Chemistry
The role of energy in chemical and physical change: virtually every process involves absorption or release of energy.
The law of conservation says that energy remains constant during a reaction
Thermochemistry is the study of the changes in energy that accompany chemical reactions and physical changes.
6.02: Endothermic and Exothermic I. Intro
Endothermic: A chemical or physical change that absorbs energy from the surroundings.
Exothermic: a chemical or physical change that releases energy to the surroundings.
II. Endothermic reactions
The temperature of an object can be used to determine whether heat has moved into or out of an object. * D and boiling are both endothermic processes because both require energy to be absorbed from the surroundings for the process to occur.
Ex: when sweat evaporates from your skin, your skin feels cooler.
This is because the TE from your skin flows into the liquid (sweat), raising the temperature until it evaporates. The TE↓and the skin is cooler. * Photosynthesis involves endothermic reactions that require energy from the sun.
III. Exothermic Reactions
The amount of energy released by an exothermic reaction depends on the amount of reactants as well as the dif. In P.E of the reactants and products
The products have a ↓P.E than the reactants.
Ex: hot packs, ice cubes
IV. Potential energy diagrams
This diagram is used to represent the general change in energy over the course of a reaction. a. Exothermic
In an exothermic reaction, the products have less potential energy than the reactants had at the beginning of the reaction. This is why the potential energy diagram for an exothermic reaction starts at a higher energy value and ends at a lower energy value. The law of conservation of energy is obeyed because the extra energy that does not end up stored in the final products is released to the surroundings in the form of heat, light, or other kinetic energy. b. Endothermic
In an endothermic reaction, the products end up with more stored potential energy than the reactants. This is why energy must be absorbed by the reaction to provide enough energy to form the products. c. Both: Because it always requires energy to break bonds, even exothermic reactions need to take in energy to get the reaction started. Activation energy is the minimum amount of energy required to initiate a chemical reaction.
The Lab of 6.02
H2CO3 -> H2O + CO2 Steel Wool and Vinegar: The cup feels warmer to the touch as the steep wool rusts. Baking Soda and Vinegar: The cup feels cooler to the touch as the baking soda and vinegar react together. The student should also observe bubbling as the reaction takes place. Washing Soda and Water: The cup feels warmer to the touch as the two substances react together. |
Because there was a decrease in temperature in the cup, this reaction is endothermic.
The vinegar and the baking soda seem to absorb energy after it decreased in temperature.
The thermal energy flowed to the cup until it cooled.
(For baking soda and vinegar): The reaction is endothermic because the cup feels cooler as heat is absorbed from the surroundings in order to be used in the reaction. (For steel wool and vinegar or washing soda and water): The reaction is exothermic because the cup feels warm as heat flows out of the reaction to the water, cup, and surroundings.
The chemical reaction that take place when a battery is connected is exothermic. The battery releases energy to the surroundings. The battery flow the energy to the object that it is connected. The reaction that occurs in a battery must be exothermic because energy is given off by the battery to power the machine that is connected to the battery. This means that the stored potential energy of the reactants in the battery is greater than the potential energy of the products formed by the reaction, and the difference between their potential energy values is the amount of energy given off by the battery. They both release heat that is exothermic because the reactants have a higher potential energy than the products. Both will be represented by an exothermic potential energy diagram, and both will have a hill following the initial potential energy of the reactants that represents the activation energy. The difference between the two diagrams is that the firework will have a greater difference between the starting and ending potential energy values because more energy is released by a firework than by a candle.Quiz: |
[06.01]The same amount of heat is removed from 2 kg of water and from 1 kg of water starting at the same temperature. What will happen? ( 1 points)The 1 kg of water will reach the lowest temperature. [06.01]You have 500 grams of water at 80° Celsius. You want to lower the temperature of the water to as close to 60° Celsius as possible. Which of the following would get the 500 grams of water the closest to 60° Celsius? ( 1 points)C) adding 250 grams of 60° Celsius water [06.01]If you drop a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius, what best describes what would occur? (1 points) The heat lost by the metal will equal the heat gained by the water.
[06.02]Which of the following potential energy diagrams represents the process of evaporation?
[06.02]When ammonium nitrate is dissolved in water, energy is absorbed from the surroundings, causing a decrease in the resulting solution's temperature. Which of the following best describes this type of reaction? (1 points)endothermic
[06.02]Which of the following best describes the relationship between products and reactants in an exothermic reaction? ( 1 points)
C) The potential energy of the products is less than the potential energy of the reactants.
[06.02]Which of the following accurately characterizes the process of freezing? (1 points)
A) Because energy is released, freezing is an exothermic process.
6.03: Calorimetry
I. Specific heat capacity
Different substances have dif capacities for storing thermal energy. Dif types of matter require dif amounts of energy to reach the same temperature and they also cool down at a dif rate.
Each substance has its own specific heat capacity.
The specific heat capacity is the quantity of heat that is required to raise the temperature of one gram of the substance by one degree Celsius.
Specific heat values have a unit of joules per degree Celsius per gram (J / (°C × g)).
Water has a higher specific heat capacity than most other substances, 4.18 J / (°C × g).
In contrast, most metals have a specific heat capacity of less than 1.0. It requires a lot of energy to increase the temperature of water, and water takes longer to cool than many other substances. This property of water makes it a useful coolant for hot engines and helps keep climates near large bodies of water more constant.
II. Heat gain or loss
The following equation shows the relationship between specific heat capacity, mass, temperature change, and thermal energy gained or lost. q = m × c × Δt q: The q represents the heat gained or lost by the system, in joules. m: The m in this equation represents the mass of the sample, measured in grams. c: The c represents the specific heat capacity of the substance being heated or cooled. The unit for specific heat capacity is joules per gram per degree Celsius.
Δt: Delta t (Δt) represents the change in temperature, in degrees Celsius, when a substance gains or loses energy. The change in temperature can be determined by subtracting the initial temperature from the final temperature of the substance (tf – ti).
III. Calorimetry
Chemists use a device called a caloriemeter to measure heat absorbed or released by a chemical or physical change.
Calorimetry: the process of measuring the heat absorbed or released by a chemical or physical change.
The simplest form of calorimetry measures the change in temperature as the heat involved in a given process flows into or out a known amount of water.
By measuring the temperature change experienced by the sample of water, the amount of heat involved in the reaction or process can be determined.
Using insulation helps prevent any energy from flowing into or out of the water from outside the calorimeter.
Because polystyrene is a good insulator, sometimes a simple calorimeter can be constructed using polystyrene coffee cups.
These simple “coffee cup calorimeters” are often used in general chemistry laboratories at the high school and college level. If a reaction, such as combustion, is being investigated that cannot be exposed to water, the reaction takes place in a chamber within the calorimeter.
The heat released from the reaction flows into the water that surrounds the reaction chamber, where the temperature change is measured.
PRACTICE:
Order the following objects from heats fastest (requires the least energy to heat) to heats slowest (requires the most energy to heat) when sitting outside in the sun? * iron pot * glass * air * wood * water
Order the different samples of water from heats fastest (requires the least energy to heat) to heats slowest (requires the most energy to heat). * teacup * kettle * bath tub * hot tub * swimming pool
Part Two—Energy Gain/Loss
If we assume that the calorimeter, thermometer, and stirrer have no heat capacity, meaning they do not absorb any thermal energy during the process, the amount of heat gained or lost by the water will be equal to the amount of heat lost or gained by the process being examined.
Because a known amount of water is used in the calorimeter, and we know the specific heat capacity of water, we can use the equation q = m × c × Δt to determine the energy gained or lost by the water. The amount of energy gained or lost by the water is equal to the amount of energy lost or gained by the reaction or process being examined in the calorimeter. This means that the quantity of energy values are the same, but the sign on the values will be opposite because heat is flowing into one and out of the other. qwater/surroundings = -qreaction/system
If the value of q for water is positive (energy was gained), that means that the value of q for the reaction is negative (energy was lost).
If the calculated value of q for water is negative (energy was lost by the water), that means that the value of q for the reaction is positive (energy was absorbed, or gained, by the reaction).
Scientists use various forms of chemistry to examine the direction of the energy flow, as well as the amount of energy flowing into and out of various systems.
Besides examining the amount of energy, in calories or joules, contained in different food items, calorimetry can also be used to compare the usefulness of various possible fuel sources by comparing the amount of energy given off when each fuel is combusted.
Any time scientists want to examine the amount of energy involved in a process, their investigation will probably involve some type of calorimetry.
Example One—Burning Food
A 3.30 gram sample of sucrose (C6H12O6) was burned in a calorimeter. The heat released by the combustion changed the temperature of 850.0 grams of water from 22.3 degrees Celsius to 34.0 degrees Celsius. How much energy, in joules, was released by the burning of the sucrose?
Because we know the mass, specific heat capacity, and temperature change of the water (the surroundings), we can determine the energy gained by the surroundings as the sucrose burned. Remember that the specific heat capacity of water is 4.18 J / (°C × g). qsurroundings = m × c × Δt qsurroundings = 850.0 g × 4.18 J / (°C × g) × (34.0°C − 22.3 °C) qsurroundings = 41,570 J
Remember that the amount of energy absorbed by the surroundings equals the amount of energy given off by the system (qsurroundings = −qsystem). This means that 41,570 joules, or 41.57 kilojoules, were given off when the sucrose was burned (qsystem = −41570 J).
Example Two—Determining Specific Heat
The specific heat capacity is unique to each substance. Objects can be identified by determining its specific heat capacity.
A common laboratory technique for determining the specific heat capacity of an object is to heat the object to a known temperature and then place it in a calorimeter with a given amount of water at a given temperature.
Both the object and the water will end up at the same final temperature, as the object gives off heat to the water. This will give you enough information to determine the specific heat capacity (c) of the object being examined.
Example:
12.5 grams of an unknown substance is heated to 85.0 degrees Celsius and then placed into a calorimeter containing 120.0 grams of water at 23.0 degrees Celsius. If the final temperature reached in the calorimeter is 26.0 degrees Celsius, what is the specific heat of the unknown substance? Remember that the specific heat capacity of water is 4.18 J / (°C × g).
Identify what you know:
Surroundings (water):
Mass (m) = 120.0 g
Specific heat capacity (c) = 4.18 J / (°C × g)
Change in temperature (Δt) = tf − ti = 26.0°C − 23.0°C = 3.0°C
System (unknown substance):
Mass (m) = 12.5 g
Specific heat capacity (c) = unknown (this is what we’re trying to determine)
Change in temperature (Δt) = tf − ti = 26.0°C – 85.0°C = −59.0°C
Notice that the temperature change of the unknown substance is negative, because its temperature decreased over the course of the investigation.
Because we know everything about the surroundings except the energy involved we can solve for the energy (q) gained by the water. qsurroundings = m × c × Δt qsurroundings = 120.0 g × 4.18 J/ (°C × g) × 3.0°C qsurroundings = 1,505 J
If the surrounding water gained 1,505 joules, this means that the unknown substance released 1,505 joules as it cooled in the calorimeter. qsurroundings = −qsystem
This means that qsystem = −1,505 J
Use this information to solve for the specific heat capacity of the unknown substance. qsystem = m × c × Δt
−1505 J = 12.5 g × c × −59.0°C
Solve for c to get: c = c = 2.04 J / (°C × g)
The specific heat capacity of the unknown substance is 2.04 J / (°C × g).
V. Enthalpy change
The enthalpy change of system is represented by the symbol ΔH.
It is equal to the energy flow of heat as long as pressure remains constant.
The sign on the enthalpy change value represents the direction of the heat flow from the perspective of the system. A negative enthalpy change means that energy was lost by the system to the surroundings in an exothermic process. On the other hand, a positive enthalpy change value represents an endothermic process.
-ΔH = exothermic (-q)
As heat leaves the system, the enthalpy of the system decreases. This is why an exothermic process has a negative enthalpy change value.
Decrease in energy of the system = -q/-ΔH
+ΔH = endothermic (+q)
As heat enters the system, the enthalpy of the system increases. This is why an endothermic process has a positive enthalpy change value.
Increase in energy of the system = +q/+ΔH
The enthalpy change of reaction or process is usually expressed in kilojoules per mole (kJ/mol). This means that enthalpy change can be used as a conversion factor to determine the energy gained or lost when a certain amount of reactants are used.
CH4 + O2 → CO2 + 2 H2O ΔH = −890 kJ/mol
The enthalpy change value of this reaction tells us that for every one mole of methane (CH4) burned, 890 kilojoules are given off.
If you burned 38.5 grams of methane, how many kilojoules of energy would you expect to be given off?
The enthalpy change can be used as a conversion factor between moles of methane and kilojoules of energy.
38.5 g CH4 × × = 2,135 kJ given off 1. Calculate the number of joules released when 72.5 grams of water at 95.0 degrees Celsius cools to a final temperature of 28.0 degrees Celsius.
Check your answer: q = m × c × Δt q = 72.5 g × 4.18 J / (°C × g) × (28.0°C − 95.0°C) q = −20,300 J (20,300 joules (or 20.3 kilojoules) of energy would be released) 1. A 120.0 gram sample of metal at 75.0 degrees Celsius is added to 150.0 g of water at 15.0 degrees Celsius. The temperature of the water rises to 18.3 degrees Celsius. What is the specific heat of the metal?
Check your answer: qsurroundings = m × c × Δt qsurroundings = 150.0 g × 4.18 J / (°C × g) × (18.3°C − 15.0°C) qsurroundings = 2,069 J gained qsurroundings = −qsystem
−2,069 J = 120.0 g × c × (18.3°C − 75.0°C) c = 0.304 J / (°C × g) 1. The combustion of isooctane fuel (C8H18) has an enthalpy change of −5,460 kJ/mol. How many kilojoules of energy will be released when 2.0 liters of isooctane is burned, if the density of the fuel is 0.692 g/mL?
Check your answer:
2.0 L C8H18 × × × × = 66,112 kJ released
Lab:
Calculations:
Show your work and write a short explanation with each calculation.
Part I: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
Metal | Aluminum | Mass of metal | 27.776 g | Volume of water | 25.9 ml | Initial temperature of water | 25.3 C | Temperature of hot water and metal in hot water bath | 100.5 C | Final temperature | 38.9 C | m × c × ΔT of water= -(m × c × ΔT of metal)
(25.9 ) x (4.18J/g*C) x (38.9 – 25.3)
= 1472.36 2. Calculate the specific heat of the metal. m × c × ΔT of water= -(m × c × ΔT of metal)
(25.9 ) x (4.18J/g*C) x (38.9 – 25.3)
= 1472.36
-1472.36= 27.776g x C x (38.9 C-100.5)
= 0.8605 J / (g × °C)
Part II: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C), and the density of water is 1.00 g/mL. Metal | Metal B | Mass of metal | 25.605 g | Initial temperature of water | 25.3 C | Temperature of hot water and metal in hot water | 100.3 C | Final temperature | 32.2 C |
m × c × ΔT of water= -(m × c × ΔT of metal)
25.9g x (4.18J/g*C) x (32.2-25.3) 2. Calculate the specific heat of the metal.
-747.0078J = 25.605g x C x (32.2-100.3)
= 0.428 J / (g × °C)
Conclusion:
1. Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in Part I of the lab.
Known specific heat values—Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C
0.8605/.900 x100 =95.6 %
So the percent error is 4.4% 1. Using the experimental specific heat capacity value that you determined in Part II of the lab, what is the most probable identity of the metal that you examined?
= 0.428 J / (g × °C)
It most likely a Nickel because the value is closest to the specific heat capacity~ 2. Assuming that is the identity of the metal, determine the percent error of your calculated specific heat capacity value. Metal | Specific Heat Capacity | Nickel | 0.440 | Tin | 0.210 | Silver | 0.237 | Magnesium | 0.140 | Calcium | 0.650 | Mercury | 0.140 | * Metal you examined: Metal B * Experimental specific heat capacity: 0.428 J / (g × °C) * Metal identity: Nickel * Known specific heat capacity value: 0.440 J / (g × °C)
0.428/0.0.440= 97.3%
The percent error is 2.7%. 3. In complete sentences, describe three sources of experimental error that could occur with this type of calorimetry lab. Explain, in detail, the effect that each specific error would have on the calculated specific heat capacity values.
The experimental error could occur when I round some of the values to the nearest hundredth or tenth. By this way, it is not exact.
Some questions from 6.02 and 6.03:
When ammonium nitrate is dissolved in water, energy is absorbed from the surroundings, causing a decrease in the resulting solution's temperature. Which of the following best describes this type of reaction? – endothermic
Which of the following best describes the relationship between products and reactants in an exothermic reaction?
Potential energy of the products is less than the potential energy of the reactants.
Which of the following accurately characterizes the process of freezing? Because energy is released, freezing is an exothermic process.
6.04 Enthalpy, Entropy and Free energy
I. Intro
A spontaneous change is a change in a system that proceeds without a net input of energy from an outside source. Spontaneous reaction still needs some activation energy to get it started, and the reaction itself may be fast or slow. II. Enthalpy * You would not be surprised to see a ball roll down a hill unassisted, but you might be surprised if it turned around and rolled back up to the top of the hill without anyone giving it a push. * From most of our experiences with the world around us, we have learned that processes that lead to a lower energy state tend to occur. * The ball rolls on its own from the top of the hill to the bottom because it has a more stable, lower energy state at the bottom of the hill. Some of the potential energy that the ball had at the top of the hill was released as kinetic energy as it rolled.
Many of the chemical reactions that occur naturally are exothermic, meaning that energy is released and the final products are a t a lower energy state than the original reactants.
Although the enthalpy change of a process can help us predict spontaneity, enthalpy change is not the only factor that affects spontaneity. Ex: A solid only melts spontaneously at temperatures above its melting point. Whether a process will proceed spontaneously can depend on temperature or pressure, and also on a property called entropy.
III. Entropy
Entropy is a measure of the disorder or molecular randomness, of a system.
Entropy is represented(ΔS) that occurs over the course of a chemical reaction or process.
Phase changes – an example of entropy change. (As a solid, the particles are packed closely with limited movement. As a liquid, it moves randomly but they are still condensed and packed relatively close. As a gas, the particles are spread far apart and moving randomly at greater speeds. These gas particles have more positions and arrangements available to them than the particles in the solid state, that is why entropy of a gas is greater than the entropy of a solid.
The table below shows a relative comparison of entropy between the common states of matter. State of substance | Relative Entropy (S) | Gas | highest S | Aqueous
(dissolved particles in solution) | high S | Liquid | medium S | Solid | lowest S | 1. Entropy increases as a substance changes to a less condensed state (solid to liquid or liquid to gas) or when the moles of gaseous products are greater than the moles of gaseous reactants in a process. This is because an increase in the motion of particles means an increase in disorder. 2. Entropy increases when substances mix or something is dissolved because mixing particles together increases the disorder. 3. Entropy increases when the total number of moles of products is greater than the number of moles of reactants because a greater number of particles lead to a greater chance of disorder. 4. The entropy of any sample increases when temperature increases because of the increase in kinetic energy and movement. 5. Entropy is greater for particles with a greater mass. For example, one mole of Cl2 has a greater mass than one mole of F2, so the entropy of the chlorine sample is greater. 6. Entropy is greater for weakly bonded compounds than for strongly bonded compounds. For example, the carbon atoms in graphite have weaker bonds than the carbon atoms in diamonds, so the entropy is greater for the sample of graphite. 7. Entropy is greater for compounds with greater complexity (greater number of atoms, greater mass, etc.).
Determine if each of the following changes would result in an increase or decrease in entropy (disorder). 1. Freezing water into ice.
Check your answer:
Decrease in entropy. The change in state of a liquid to a solid is a decrease in entropy (disorder) because the particles in a solid are arranged in a more orderly way than when they are randomly moving around as a liquid. 1. Dissolving salt in water.
Check your answer:
Increase in entropy. The dissolving process involves mixing two substances together, which increases the disorder of the system. 1. Evaporating rubbing alcohol.
Check your answer:
Increase in entropy. The change in state of a liquid to a gas is an increase in entropy (disorder) because the particles of a gas are moving faster and forming more random arrangements than they are in the more condensed liquid state. 1. The reaction of carbon dioxide gas with solid calcium oxide to produce solid calcium carbonate.
CO2 (g) + CaO (s) → CaCO3 (s)
Check your answer:
Decrease in entropy. Over the course of this reaction, one mole of gas and one mole of solid are converted to one mole of solid. As a general rule, the side of the equation with more moles of gas has a greater amount of entropy (disorder). Also, the total number of moles decreases over the course of the reaction, another indication of a decrease in entropy. 1. The decomposition of solid potassium chlorate to produce solid potassium chloride and oxygen gas.
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
Check your answer:
Increase in entropy. Over the course of this reaction, two moles of solid decomposes into two moles of solid and three moles of gas. The presence of gas particles in the products indicates greater entropy (disorder). The fact that there are more moles of products than moles of reactants is a second indicator that entropy increased over the course of this reaction. V. Free energy
Entropy, enthalpy and temperature have an effect on the spontaneity of a chemical process. These properties can be combined in an equation that helps determine if a physical or chemical change is spontaneous at a given temperature.
ΔG = ΔH − (T × ΔS)
The temperature used here is Kelvin. (Positive value at all times)
The free energy change (ΔG) of a process is a function of temperature, enthalpy change, and entropy change
At a constant temperature and pressure, a process is spontaneous in the direction that gives a negative change in free energy (−ΔG)
Chemists can determine values for entropy change, enthalpy change, and temperature and plug them into the equation above to determine if a reaction or process is spontaneous under those conditions. In addition, the relationship given by the equation above is useful to make general predictions of the spontaneity of a process.
Refer to the table below to see if reactions with various combinations of entropy, enthalpy, and temperature will be spontaneous. Enthalpy Change | Entropy Change | Spontaneous Reaction? | Exothermic (-ΔH) | Increase (+ΔS) | Yes, -ΔG at any temperature | Exothermic (-ΔH) | Decrease (-ΔS) | Only at low temps, if |T ΔS| < |ΔH| | Endothermic (+ΔH) | Increase (+ΔS) | Only at high temps, if |T ΔS| > |ΔH| | Endothermic (+ΔH) | Decrease (-ΔS) | No, +ΔG at any temperature | 1. The melting of ice has an enthalpy value of ΔH = +333 kJ/kg. What is true about the spontaneity of this process?
Spontaneous at high temperatures
The melting of ice has a positive entropy change (melting from a solid to a liquid is always a positive entropy change) and a positive enthalpy change (ΔH = +333 kJ/kg). For this combination of entropy and enthalpy, the reaction will be spontaneous (-ΔG) at high temperatures. 1. The combustion of ethane gas has a negative enthalpy change and a positive entropy change.
Spontaneous at all temperatures
A negative enthalpy change and a positive entropy change are both “preferred” by nature, so this process is spontaneous at any temperature. 1. The process of freezing methane (CH4 (l) → CH4 (s)) has a negative enthalpy change (ΔH = -58.4 kJ/kg).
Spontaneous at low temperatures
The freezing of methane has a negative entropy change (freezing from a liquid to a solid is always a negative entropy change) and a negative enthalpy change (ΔH = -58.4 kJ/kg). For this combination of entropy and enthalpy, the reaction will be spontaneous (-ΔG) at low temperatures.
6.05: Rate and Collision theory
I. intro
The area of chemistry that studies reaction rate is Chemical Kinetics. The measure of the rate of a reaction involves measuring the change in something (amount, progress, motion etc) measured by ↑ or↓ concentration or amount of a reactant/ product.
II. Collision Theory
Chemists believe that for compounds, atoms, or ions to react successfully, the particles must collide together. This is the basis for a scientific theory known simply as the collision theory
This theory is based on the idea that the rate of a reaction is determined by the number of successful collisions that occur over a given amount of time. A successful collision is one that results in the making of the product(s).
The two factors are sufficient energy and correct orientation. These must be true for a given collsion to successfully result in the production of products. A. Sufficient Energy
The collision between reactants must have enough energy for the reactants’ valence energy levels to penetrate each other. This interaction between valence energy levels allows the electrons to rearrange to form new bonds.
A Maxwell-Boltzmann distribution graph.
The y-axis of the graph is labeled “number of particles,” increasing as you go up the axis, and the x-axis is labeled (energy), increasing left to right. The shape of the graph is like that of a bell or hill, showing that the majority of the particles have a kinetic energy near the average kinetic energy of the sample, while some particles have more energy (far right of the graph) and some particles have less energy (far left of the graph).
A Maxwell-Boltzmann distribution graph.
The y-axis of the graph is labeled “number of particles,” increasing as you go up the axis, and the x-axis is labeled (energy), increasing left to right. The shape of the graph is like that of a bell or hill, showing that the particles in this area of the curve do not have enough energy to react when they collide.
However, the particles to the right of the activation energy value are the only particles in the sample with enough energy to possibly react when they collide. B. Correct Orientation
When reactants come in contact with each other during their random movement, the orientation of their collision will determine if the collision is successful. The atoms from each reactant that will bond together to form the new products must come in contact with each other if a bond is going to form between them.
The more complex a reaction is, the slower the rate of reaction. For example, if more than two reactants need to collide together at the same time with correct orientation the reaction time would be slow.
For the reaction between ethene (CH2CH2) and hydrogen chloride (HCl) to occur, the double bond between the carbon atoms in ethene must be broken.
The double bond is converted to a single bond, allowing the carbon atoms to each bond with one of the atoms from the hydrogen chloride.
© 2010 FLVS
The reaction can only happen if the hydrogen end of the hydrogen chloride compound collides with the double bond between the carbon atoms of the ethene.
Any other orientation will not be successful in producing the new product; the two reactants will just “bounce” off of each other.
06.05 Rate and Collision Theory: Factors Affecting Reaction Rate—Text Version
For most reactions, there is no practical way to have an effect on the orientation of particles in collisions. However, there are ways that chemists can increase the number of collisions that have sufficient energy to successfully produce the products.
Concentration and Rate
For many reactions involving liquids or gases, increasing the concentration of the reactants increases the rate of reaction. In a few cases, increasing the concentration of one of the reactants may have little noticeable effect of the rate.
In reactions that involve two or more reactants, increasing the concentration of the reactants will increase the rate of the reaction. According to the collision theory, a reaction between two different particles can only happen if those particles collide together. If the concentration of reactants is higher, meaning there are more moles of reactants in a given volume, the chances of the particles coming in contact increases. As the number of collisions between reactants increases, the number of successful collisions in a given amount of time will also increase.
The greater the concentration of solute particles in the solution, the greater the chance of the solute particles colliding with the surface of the solid reactant.
Be careful: We cannot assume that by doubling the concentration of one of the reactants you will double the rate of the reaction. The amount by which an increase in concentration increases the rate of a reaction depends on many different properties of the reaction.
Temperature and Rate
Because temperature is a measure of average kinetic energy, we know that the temperature of a system will affect the number of collisions that have sufficient energy to react. You have experienced the effects of temperature on rate if you have ever cooked something on the stove top or in the oven.
You know that increasing the temperature at which you cook your dinner increases the rate of the cooking process. This can be a great time saver, but you need to be careful that you don’t end up burning your food!
Increasing the temperature of a system increases the rate of a reaction. This direct relationship between temperature and rate is due to two reasons, both related to the collision theory. 1. Increasing the temperature of a system increases the average kinetic energy of the particles. This means more collisions between the reactants will have enough energy to react successfully. 2. Increasing the temperature of a system increases the kinetic energy of the particles, meaning that the particles are moving faster. This means that there are more collisions per minute at a higher temperature than there are in the same system at a lower temperature. Having more collisions in a given amount of time means there are more opportunities for collisions to meet both requirements for success.
As you can see on the Maxwell-Boltzmann distribution curve below, a slight increase of temperature changes the distribution of energy in a sample. Although a slight increase in temperature (T2) may only increase the average kinetic energy by a small amount, there is a significant increase to the number of particles that meet and exceed the activation energy requirement for a given reaction.
Catalysts and Rate
We have seen that an increase in temperature increases the rate of a reaction; if a reaction does not happen fast enough at normal temperatures, raising the temperature can help get the job done in less time. However, sometimes it is not possible to raise the temperature when we need a reaction to occur at a faster rate. In industrial settings, scientists need to keep safety and cost in mind, and so sometimes cannot exceed a certain temperature. In your body, living cells can only survive within a narrow temperature range.
Raising the temperature within the cells to speed up the necessary chemical reactions would result in damage to the very cells that the reactions support.
So, how do we get the complicated biochemical reactions in your body to occur at a sufficient rate to keep you alive without raising the temperature to a point that would kill the cells you are trying to support? Thankfully, your body contains a variety of enzymes that speed up the rate of these reactions without the need to raise the temperature. Almost every important biological reaction occurs in living organisms with the assistance of an enzyme.
Enzymes are biological catalysts. A catalyst is a substance that speeds up a reaction without being consumed by the reaction. Catalysts are important in many biological and industrial reactions where the speed of a reaction is important but temperature cannot exceed a certain range. So, how do catalysts manage to speed up a reaction without adding heat to increase the temperature?
If temperature cannot be increased to meet the activation energy requirement of a given reaction, another way to speed up the reaction is to provide an alternative reaction pathway with a lower activation energy requirement. As you can see in the potential energy diagram below, a catalyst provides a different pathway between reactants and products that requires a lower amount of activation energy.
Imagine you are hiking through the woods and there are two paths that will take you to your destination. The main path requires you to climb a steeper hill than the alternate path. Both paths take you to the same exact destination; the alternative path just requires less energy to get there. That is how a catalyst works: It provides an alternative path to the same final products but requires less activation energy to get there.
The lower activation energy requirement of this alternative reaction pathway results in a larger percentage of the given collisions between reactants that will be successful and produce products. Notice in the Maxwell-Boltzmann distribution curve below that the lower activation energy requirement of the catalyzed pathway allows many more particles to meet the energy requirement for a successful collision without any increase in temperature.
Did You Know?
Some substances, called inhibitors, actually decrease the rate of catalyzed chemical reactions. They do this by interfering with the catalysts’ ability to participate in the reaction. Many biological poisons are inhibitors that interfere with the enzymes in your body.
Speeding up a Reaction
If all the students in a math class are about to take a very tough exam, what are the possible ways to increase the percentage of the class that successfully passes the exam with a grade of C or higher? If all the students work hard studying and practicing for the test, they would probably see an increase in the number of students who pass the exam successfully.
Another option is for the teacher to lower the standard for a C on the grading scale. If your teacher changed the grading scale to make 55 percent the requirement for a C grade, it would be easier for more students to earn a C or higher with less effort on their part.
This story gives an analogy for the two main ways we can increase the rate of a chemical reaction.
Just as increasing the effort made by the students will result in more students successfully meeting the grade requirement of a C, increasing the temperature of a system allows more particle collisions to meet the energy requirement for a successful collision.
On the other hand, lowering the standard for a passing grade can also increase the number of students who successfully make the grade. This is how a catalyst works. Lowering the activation energy requirement of a reaction allows more collisions to be successful without changing any properties of the collisions themselves.
Although using a catalyst to lower the energy standard in a chemical reaction is a common practice in chemistry, do not expect your teachers to lower their standards for your success anytime soon.
Some practice: he reaction shown below has a positive enthalpy change and a negative entropy change.
2C (s) + 2H2 (g) C2H4 (g) The reaction will not be spontaneous at any temperature.
Which of the following conditions will result in a reaction that is spontaneous only at low temperatures? negative enthalpy change and negative entropy change .
Compare the following words. Which word best describes a scientific model? Analogy
The collision theory states that the rate of a chemical reaction is determined by collisions occurring at the molecular level. Which of the following will best help us visualize collision theory? A scientific model
Scientific models help predict experimental results. The model of an ideal gas is useful because it approximates the behavior of most gas molecules.
Catalytic converters on automobiles use the metals rhodium and platinum as catalysts to convert harmful gases to carbon dioxide, nitrogen, and water. Which answer best explains why rhodium and platinum do not have to be continuously replaced in the system?
The metals speed up the reactions but are not used up in the reaction.
Quiz: Which of the following changes will always be true for a spontaneous reaction? - G
Which of the following changes has a decrease in entropy? 3Fe (s) + 2O2 (g) Fe3O4 (g)
Which of the following processes involves an increase in entropy? melting gold bars
Which of the following combinations will result in a reaction that is never spontaneous? positive enthalpy change and negative entropy change
Describe in detail what you expect for the changes in enthalpy, entropy, and free energy when a sample of liquid evaporates. How does temperature affect these changes?
For a sample of liquid to evaporate, heat is needed to allow the liquid to become gas.
As the temperature rises, the kinetic energy affects the liquid to have higher entropy.
Also, when the sample changes to a less condense state such as liquid to gas, the entropy is high.
The enthalpy of the sample would be endothermic (+ triangle H) because it first absorbed the heat energy. So the free energy would be only present at high temperatures.
+ΔH (enthalpy change) +ΔS (entropy change) + or - ΔG (free energy change) depending on temperature This process will be spontaneous (have a negative ΔG) at high temperatures. 1. What factors did you investigate in your procedure, and why did you choose to compare these two factors?
I checked if the amount of water and temperature would affect the rate of reaction. I learned that increased temperature means higher kinetic energy, which would affect the movement of the particles in the mixture. I was just really curious if the amount of water would have anything with dissolving the tablet. 2. What other factors did you need to control during your investigation? Explain how you controlled each one in your procedure.
The water need to be controlled during the investigation. So I used filtered water for each procedure (except the ones that I had to heat up).
The tablet had to be the same size for each experiment. 3. What is your prediction about the results of each trial in your lab? Explain your predictions based on your knowledge of the dissolving process, collision theory, and reaction rates.
I hypothesize that the hotter water would have a faster rate of reaction than the one with the room temperature water. This will occur because the kinetic energy will cause the particles to move faster, causing them to collide more easily. I also hypothesize that the large amount of water would be cause the tablet to dissolve faster because the collision theory states that some factors such as the change in amount, progress, motion etc, can affect the rate of reaction. 4. Explain the collision theory, in your own words, and what is necessary for a collision to be successful.
According to the theory, if the molecules in the reaction are allowed to collide more, the faster the reaction rate would be. In order for a collision to be successful, sufficient energy and orientation must be true for the collision to be successful. 5. A specific catalyst was not provided for this reaction, but catalysts are useful for increasing the rate of many slow reactions. In your own words, give a detailed explanation of how catalysts can increase the rate of a reaction or process.
A catalyst can speed up the reaction rate, by allowing the reactants and products to collide to only require a lower amount of activation energy.
Module Six: Thermochemistry and rate
Study sheet for the test
G.01: Thermo-chemistry I. Temperature and Thermal Energy
Temperature is a measure of the average kinetic energy of the particles in a sample of matter. The greater the avg. kinetic energy of the particles in a sample of matter, the higher the temperature of that matter. Some temperature scales are Fahrenheit, Celsius and Kelvin.
Thermal energy is the measure of the total kinetic energy in a sample. And Temperature is a measure of the average kinetic energy.
Ex: Which container (the bath tub or the coffee cup) have a temperature that is higher than the other?
The temperature of the coffee cup is higher!
Which sample of water has a greater amount of thermal energy? The bathtub. Even though the avg. kinetic energy of the particles in the coffee cup is higher than the avg.KE of the particles in the bathtub, the thermal energy in the warm bath tub is higher because there are more particles in the tub than the cup.
II. Heat
Heat is the movement, or flow, of thermal energy.
Heat always flows spontaneously from matter at a higher temperature to matter at a lower temperature.
Heat and thermal energy are both measure in units of energy, the SI unit for energy is Joule(J). Another unit of energy that is commonly used is the Calorie. The food calorie means that for every one food calorie that is eaten has enough thermal energy to increase the temperature of 1000 grams of water by 1 degree Celsius.
III. Closed, open and isolated systems
In science, it is important to identify the system being observed or investigated. Some examples of systems are: atom, reaction, cell, ecosystem, universe etc. In chemistry, the system is only defined as the chemical reaction or physical change being examined. a. Isolated systems The total amount of energy and matter contained in an isolated system will remain constant because energy and matter cannot leave or enter.
Ex: Universe, space b. Closed systems
Energy can leave or enter the system but matter cannot.
Ex: earth, beakers c. Open system
Both energy and matter are exchanged freely.
Ex: our body.
IV. Energy in Chemistry
The role of energy in chemical and physical change: virtually every process involves absorption or release of energy.
The law of conservation says that energy remains constant during a reaction
Thermochemistry is the study of the changes in energy that accompany chemical reactions and physical changes.
6.02: Endothermic and Exothermic I. Intro
Endothermic: A chemical or physical change that absorbs energy from the surroundings.
Exothermic: a chemical or physical change that releases energy to the surroundings.
II. Endothermic reactions
The temperature of an object can be used to determine whether heat has moved into or out of an object. * D and boiling are both endothermic processes because both require energy to be absorbed from the surroundings for the process to occur.
Ex: when sweat evaporates from your skin, your skin feels cooler.
This is because the TE from your skin flows into the liquid (sweat), raising the temperature until it evaporates. The TE↓and the skin is cooler. * Photosynthesis involves endothermic reactions that require energy from the sun.
III. Exothermic Reactions
The amount of energy released by an exothermic reaction depends on the amount of reactants as well as the dif. In P.E of the reactants and products
The products have a ↓P.E than the reactants.
Ex: hot packs, ice cubes
IV. Potential energy diagrams
This diagram is used to represent the general change in energy over the course of a reaction. a. Exothermic
In an exothermic reaction, the products have less potential energy than the reactants had at the beginning of the reaction. This is why the potential energy diagram for an exothermic reaction starts at a higher energy value and ends at a lower energy value. The law of conservation of energy is obeyed because the extra energy that does not end up stored in the final products is released to the surroundings in the form of heat, light, or other kinetic energy. b. Endothermic
In an endothermic reaction, the products end up with more stored potential energy than the reactants. This is why energy must be absorbed by the reaction to provide enough energy to form the products. c. Both: Because it always requires energy to break bonds, even exothermic reactions need to take in energy to get the reaction started. Activation energy is the minimum amount of energy required to initiate a chemical reaction.
The Lab of 6.02
H2CO3 -> H2O + CO2 Steel Wool and Vinegar: The cup feels warmer to the touch as the steep wool rusts. Baking Soda and Vinegar: The cup feels cooler to the touch as the baking soda and vinegar react together. The student should also observe bubbling as the reaction takes place. Washing Soda and Water: The cup feels warmer to the touch as the two substances react together. |
Because there was a decrease in temperature in the cup, this reaction is endothermic.
The vinegar and the baking soda seem to absorb energy after it decreased in temperature.
The thermal energy flowed to the cup until it cooled.
(For baking soda and vinegar): The reaction is endothermic because the cup feels cooler as heat is absorbed from the surroundings in order to be used in the reaction. (For steel wool and vinegar or washing soda and water): The reaction is exothermic because the cup feels warm as heat flows out of the reaction to the water, cup, and surroundings.
The chemical reaction that take place when a battery is connected is exothermic. The battery releases energy to the surroundings. The battery flow the energy to the object that it is connected. The reaction that occurs in a battery must be exothermic because energy is given off by the battery to power the machine that is connected to the battery. This means that the stored potential energy of the reactants in the battery is greater than the potential energy of the products formed by the reaction, and the difference between their potential energy values is the amount of energy given off by the battery. They both release heat that is exothermic because the reactants have a higher potential energy than the products. Both will be represented by an exothermic potential energy diagram, and both will have a hill following the initial potential energy of the reactants that represents the activation energy. The difference between the two diagrams is that the firework will have a greater difference between the starting and ending potential energy values because more energy is released by a firework than by a candle.Quiz: |
[06.01]The same amount of heat is removed from 2 kg of water and from 1 kg of water starting at the same temperature. What will happen? ( 1 points)The 1 kg of water will reach the lowest temperature. [06.01]You have 500 grams of water at 80° Celsius. You want to lower the temperature of the water to as close to 60° Celsius as possible. Which of the following would get the 500 grams of water the closest to 60° Celsius? ( 1 points)C) adding 250 grams of 60° Celsius water [06.01]If you drop a 50 gram piece of metal with a temperature of 125° Celsius into 1000 grams of water at 20° Celsius, what best describes what would occur? (1 points) The heat lost by the metal will equal the heat gained by the water.
[06.02]Which of the following potential energy diagrams represents the process of evaporation?
[06.02]When ammonium nitrate is dissolved in water, energy is absorbed from the surroundings, causing a decrease in the resulting solution's temperature. Which of the following best describes this type of reaction? (1 points)endothermic
[06.02]Which of the following best describes the relationship between products and reactants in an exothermic reaction? ( 1 points)
C) The potential energy of the products is less than the potential energy of the reactants.
[06.02]Which of the following accurately characterizes the process of freezing? (1 points)
A) Because energy is released, freezing is an exothermic process.
6.03: Calorimetry
I. Specific heat capacity
Different substances have dif capacities for storing thermal energy. Dif types of matter require dif amounts of energy to reach the same temperature and they also cool down at a dif rate.
Each substance has its own specific heat capacity.
The specific heat capacity is the quantity of heat that is required to raise the temperature of one gram of the substance by one degree Celsius.
Specific heat values have a unit of joules per degree Celsius per gram (J / (°C × g)).
Water has a higher specific heat capacity than most other substances, 4.18 J / (°C × g).
In contrast, most metals have a specific heat capacity of less than 1.0. It requires a lot of energy to increase the temperature of water, and water takes longer to cool than many other substances. This property of water makes it a useful coolant for hot engines and helps keep climates near large bodies of water more constant.
II. Heat gain or loss
The following equation shows the relationship between specific heat capacity, mass, temperature change, and thermal energy gained or lost. q = m × c × Δt q: The q represents the heat gained or lost by the system, in joules. m: The m in this equation represents the mass of the sample, measured in grams. c: The c represents the specific heat capacity of the substance being heated or cooled. The unit for specific heat capacity is joules per gram per degree Celsius.
Δt: Delta t (Δt) represents the change in temperature, in degrees Celsius, when a substance gains or loses energy. The change in temperature can be determined by subtracting the initial temperature from the final temperature of the substance (tf – ti).
III. Calorimetry
Chemists use a device called a caloriemeter to measure heat absorbed or released by a chemical or physical change.
Calorimetry: the process of measuring the heat absorbed or released by a chemical or physical change.
The simplest form of calorimetry measures the change in temperature as the heat involved in a given process flows into or out a known amount of water.
By measuring the temperature change experienced by the sample of water, the amount of heat involved in the reaction or process can be determined.
Using insulation helps prevent any energy from flowing into or out of the water from outside the calorimeter.
Because polystyrene is a good insulator, sometimes a simple calorimeter can be constructed using polystyrene coffee cups.
These simple “coffee cup calorimeters” are often used in general chemistry laboratories at the high school and college level. If a reaction, such as combustion, is being investigated that cannot be exposed to water, the reaction takes place in a chamber within the calorimeter.
The heat released from the reaction flows into the water that surrounds the reaction chamber, where the temperature change is measured.
PRACTICE:
Order the following objects from heats fastest (requires the least energy to heat) to heats slowest (requires the most energy to heat) when sitting outside in the sun? * iron pot * glass * air * wood * water
Order the different samples of water from heats fastest (requires the least energy to heat) to heats slowest (requires the most energy to heat). * teacup * kettle * bath tub * hot tub * swimming pool
Part Two—Energy Gain/Loss
If we assume that the calorimeter, thermometer, and stirrer have no heat capacity, meaning they do not absorb any thermal energy during the process, the amount of heat gained or lost by the water will be equal to the amount of heat lost or gained by the process being examined.
Because a known amount of water is used in the calorimeter, and we know the specific heat capacity of water, we can use the equation q = m × c × Δt to determine the energy gained or lost by the water. The amount of energy gained or lost by the water is equal to the amount of energy lost or gained by the reaction or process being examined in the calorimeter. This means that the quantity of energy values are the same, but the sign on the values will be opposite because heat is flowing into one and out of the other. qwater/surroundings = -qreaction/system
If the value of q for water is positive (energy was gained), that means that the value of q for the reaction is negative (energy was lost).
If the calculated value of q for water is negative (energy was lost by the water), that means that the value of q for the reaction is positive (energy was absorbed, or gained, by the reaction).
Scientists use various forms of chemistry to examine the direction of the energy flow, as well as the amount of energy flowing into and out of various systems.
Besides examining the amount of energy, in calories or joules, contained in different food items, calorimetry can also be used to compare the usefulness of various possible fuel sources by comparing the amount of energy given off when each fuel is combusted.
Any time scientists want to examine the amount of energy involved in a process, their investigation will probably involve some type of calorimetry.
Example One—Burning Food
A 3.30 gram sample of sucrose (C6H12O6) was burned in a calorimeter. The heat released by the combustion changed the temperature of 850.0 grams of water from 22.3 degrees Celsius to 34.0 degrees Celsius. How much energy, in joules, was released by the burning of the sucrose?
Because we know the mass, specific heat capacity, and temperature change of the water (the surroundings), we can determine the energy gained by the surroundings as the sucrose burned. Remember that the specific heat capacity of water is 4.18 J / (°C × g). qsurroundings = m × c × Δt qsurroundings = 850.0 g × 4.18 J / (°C × g) × (34.0°C − 22.3 °C) qsurroundings = 41,570 J
Remember that the amount of energy absorbed by the surroundings equals the amount of energy given off by the system (qsurroundings = −qsystem). This means that 41,570 joules, or 41.57 kilojoules, were given off when the sucrose was burned (qsystem = −41570 J).
Example Two—Determining Specific Heat
The specific heat capacity is unique to each substance. Objects can be identified by determining its specific heat capacity.
A common laboratory technique for determining the specific heat capacity of an object is to heat the object to a known temperature and then place it in a calorimeter with a given amount of water at a given temperature.
Both the object and the water will end up at the same final temperature, as the object gives off heat to the water. This will give you enough information to determine the specific heat capacity (c) of the object being examined.
Example:
12.5 grams of an unknown substance is heated to 85.0 degrees Celsius and then placed into a calorimeter containing 120.0 grams of water at 23.0 degrees Celsius. If the final temperature reached in the calorimeter is 26.0 degrees Celsius, what is the specific heat of the unknown substance? Remember that the specific heat capacity of water is 4.18 J / (°C × g).
Identify what you know:
Surroundings (water):
Mass (m) = 120.0 g
Specific heat capacity (c) = 4.18 J / (°C × g)
Change in temperature (Δt) = tf − ti = 26.0°C − 23.0°C = 3.0°C
System (unknown substance):
Mass (m) = 12.5 g
Specific heat capacity (c) = unknown (this is what we’re trying to determine)
Change in temperature (Δt) = tf − ti = 26.0°C – 85.0°C = −59.0°C
Notice that the temperature change of the unknown substance is negative, because its temperature decreased over the course of the investigation.
Because we know everything about the surroundings except the energy involved we can solve for the energy (q) gained by the water. qsurroundings = m × c × Δt qsurroundings = 120.0 g × 4.18 J/ (°C × g) × 3.0°C qsurroundings = 1,505 J
If the surrounding water gained 1,505 joules, this means that the unknown substance released 1,505 joules as it cooled in the calorimeter. qsurroundings = −qsystem
This means that qsystem = −1,505 J
Use this information to solve for the specific heat capacity of the unknown substance. qsystem = m × c × Δt
−1505 J = 12.5 g × c × −59.0°C
Solve for c to get: c = c = 2.04 J / (°C × g)
The specific heat capacity of the unknown substance is 2.04 J / (°C × g).
V. Enthalpy change
The enthalpy change of system is represented by the symbol ΔH.
It is equal to the energy flow of heat as long as pressure remains constant.
The sign on the enthalpy change value represents the direction of the heat flow from the perspective of the system. A negative enthalpy change means that energy was lost by the system to the surroundings in an exothermic process. On the other hand, a positive enthalpy change value represents an endothermic process.
-ΔH = exothermic (-q)
As heat leaves the system, the enthalpy of the system decreases. This is why an exothermic process has a negative enthalpy change value.
Decrease in energy of the system = -q/-ΔH
+ΔH = endothermic (+q)
As heat enters the system, the enthalpy of the system increases. This is why an endothermic process has a positive enthalpy change value.
Increase in energy of the system = +q/+ΔH
The enthalpy change of reaction or process is usually expressed in kilojoules per mole (kJ/mol). This means that enthalpy change can be used as a conversion factor to determine the energy gained or lost when a certain amount of reactants are used.
CH4 + O2 → CO2 + 2 H2O ΔH = −890 kJ/mol
The enthalpy change value of this reaction tells us that for every one mole of methane (CH4) burned, 890 kilojoules are given off.
If you burned 38.5 grams of methane, how many kilojoules of energy would you expect to be given off?
The enthalpy change can be used as a conversion factor between moles of methane and kilojoules of energy.
38.5 g CH4 × × = 2,135 kJ given off 1. Calculate the number of joules released when 72.5 grams of water at 95.0 degrees Celsius cools to a final temperature of 28.0 degrees Celsius.
Check your answer: q = m × c × Δt q = 72.5 g × 4.18 J / (°C × g) × (28.0°C − 95.0°C) q = −20,300 J (20,300 joules (or 20.3 kilojoules) of energy would be released) 1. A 120.0 gram sample of metal at 75.0 degrees Celsius is added to 150.0 g of water at 15.0 degrees Celsius. The temperature of the water rises to 18.3 degrees Celsius. What is the specific heat of the metal?
Check your answer: qsurroundings = m × c × Δt qsurroundings = 150.0 g × 4.18 J / (°C × g) × (18.3°C − 15.0°C) qsurroundings = 2,069 J gained qsurroundings = −qsystem
−2,069 J = 120.0 g × c × (18.3°C − 75.0°C) c = 0.304 J / (°C × g) 1. The combustion of isooctane fuel (C8H18) has an enthalpy change of −5,460 kJ/mol. How many kilojoules of energy will be released when 2.0 liters of isooctane is burned, if the density of the fuel is 0.692 g/mL?
Check your answer:
2.0 L C8H18 × × × × = 66,112 kJ released
Lab:
Calculations:
Show your work and write a short explanation with each calculation.
Part I: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL.
Metal | Aluminum | Mass of metal | 27.776 g | Volume of water | 25.9 ml | Initial temperature of water | 25.3 C | Temperature of hot water and metal in hot water bath | 100.5 C | Final temperature | 38.9 C | m × c × ΔT of water= -(m × c × ΔT of metal)
(25.9 ) x (4.18J/g*C) x (38.9 – 25.3)
= 1472.36 2. Calculate the specific heat of the metal. m × c × ΔT of water= -(m × c × ΔT of metal)
(25.9 ) x (4.18J/g*C) x (38.9 – 25.3)
= 1472.36
-1472.36= 27.776g x C x (38.9 C-100.5)
= 0.8605 J / (g × °C)
Part II: 1. Calculate the energy change (q) of the surroundings (water). We can assume that the specific heat capacity of water is 4.18 J / (g × °C), and the density of water is 1.00 g/mL. Metal | Metal B | Mass of metal | 25.605 g | Initial temperature of water | 25.3 C | Temperature of hot water and metal in hot water | 100.3 C | Final temperature | 32.2 C |
m × c × ΔT of water= -(m × c × ΔT of metal)
25.9g x (4.18J/g*C) x (32.2-25.3) 2. Calculate the specific heat of the metal.
-747.0078J = 25.605g x C x (32.2-100.3)
= 0.428 J / (g × °C)
Conclusion:
1. Use the given specific heat capacity values below to calculate the percent error of the experimental specific heat capacity that you determined in Part I of the lab.
Known specific heat values—Iron: 0.444 J/g°C; Zinc: 0.390 J/g°C; Copper: 0.385 J/g°C, Aluminum: 0.900 J/g°C
0.8605/.900 x100 =95.6 %
So the percent error is 4.4% 1. Using the experimental specific heat capacity value that you determined in Part II of the lab, what is the most probable identity of the metal that you examined?
= 0.428 J / (g × °C)
It most likely a Nickel because the value is closest to the specific heat capacity~ 2. Assuming that is the identity of the metal, determine the percent error of your calculated specific heat capacity value. Metal | Specific Heat Capacity | Nickel | 0.440 | Tin | 0.210 | Silver | 0.237 | Magnesium | 0.140 | Calcium | 0.650 | Mercury | 0.140 | * Metal you examined: Metal B * Experimental specific heat capacity: 0.428 J / (g × °C) * Metal identity: Nickel * Known specific heat capacity value: 0.440 J / (g × °C)
0.428/0.0.440= 97.3%
The percent error is 2.7%. 3. In complete sentences, describe three sources of experimental error that could occur with this type of calorimetry lab. Explain, in detail, the effect that each specific error would have on the calculated specific heat capacity values.
The experimental error could occur when I round some of the values to the nearest hundredth or tenth. By this way, it is not exact.
Some questions from 6.02 and 6.03:
When ammonium nitrate is dissolved in water, energy is absorbed from the surroundings, causing a decrease in the resulting solution's temperature. Which of the following best describes this type of reaction? – endothermic
Which of the following best describes the relationship between products and reactants in an exothermic reaction?
Potential energy of the products is less than the potential energy of the reactants.
Which of the following accurately characterizes the process of freezing? Because energy is released, freezing is an exothermic process.
6.04 Enthalpy, Entropy and Free energy
I. Intro
A spontaneous change is a change in a system that proceeds without a net input of energy from an outside source. Spontaneous reaction still needs some activation energy to get it started, and the reaction itself may be fast or slow. II. Enthalpy * You would not be surprised to see a ball roll down a hill unassisted, but you might be surprised if it turned around and rolled back up to the top of the hill without anyone giving it a push. * From most of our experiences with the world around us, we have learned that processes that lead to a lower energy state tend to occur. * The ball rolls on its own from the top of the hill to the bottom because it has a more stable, lower energy state at the bottom of the hill. Some of the potential energy that the ball had at the top of the hill was released as kinetic energy as it rolled.
Many of the chemical reactions that occur naturally are exothermic, meaning that energy is released and the final products are a t a lower energy state than the original reactants.
Although the enthalpy change of a process can help us predict spontaneity, enthalpy change is not the only factor that affects spontaneity. Ex: A solid only melts spontaneously at temperatures above its melting point. Whether a process will proceed spontaneously can depend on temperature or pressure, and also on a property called entropy.
III. Entropy
Entropy is a measure of the disorder or molecular randomness, of a system.
Entropy is represented(ΔS) that occurs over the course of a chemical reaction or process.
Phase changes – an example of entropy change. (As a solid, the particles are packed closely with limited movement. As a liquid, it moves randomly but they are still condensed and packed relatively close. As a gas, the particles are spread far apart and moving randomly at greater speeds. These gas particles have more positions and arrangements available to them than the particles in the solid state, that is why entropy of a gas is greater than the entropy of a solid.
The table below shows a relative comparison of entropy between the common states of matter. State of substance | Relative Entropy (S) | Gas | highest S | Aqueous
(dissolved particles in solution) | high S | Liquid | medium S | Solid | lowest S | 1. Entropy increases as a substance changes to a less condensed state (solid to liquid or liquid to gas) or when the moles of gaseous products are greater than the moles of gaseous reactants in a process. This is because an increase in the motion of particles means an increase in disorder. 2. Entropy increases when substances mix or something is dissolved because mixing particles together increases the disorder. 3. Entropy increases when the total number of moles of products is greater than the number of moles of reactants because a greater number of particles lead to a greater chance of disorder. 4. The entropy of any sample increases when temperature increases because of the increase in kinetic energy and movement. 5. Entropy is greater for particles with a greater mass. For example, one mole of Cl2 has a greater mass than one mole of F2, so the entropy of the chlorine sample is greater. 6. Entropy is greater for weakly bonded compounds than for strongly bonded compounds. For example, the carbon atoms in graphite have weaker bonds than the carbon atoms in diamonds, so the entropy is greater for the sample of graphite. 7. Entropy is greater for compounds with greater complexity (greater number of atoms, greater mass, etc.).
Determine if each of the following changes would result in an increase or decrease in entropy (disorder). 1. Freezing water into ice.
Check your answer:
Decrease in entropy. The change in state of a liquid to a solid is a decrease in entropy (disorder) because the particles in a solid are arranged in a more orderly way than when they are randomly moving around as a liquid. 1. Dissolving salt in water.
Check your answer:
Increase in entropy. The dissolving process involves mixing two substances together, which increases the disorder of the system. 1. Evaporating rubbing alcohol.
Check your answer:
Increase in entropy. The change in state of a liquid to a gas is an increase in entropy (disorder) because the particles of a gas are moving faster and forming more random arrangements than they are in the more condensed liquid state. 1. The reaction of carbon dioxide gas with solid calcium oxide to produce solid calcium carbonate.
CO2 (g) + CaO (s) → CaCO3 (s)
Check your answer:
Decrease in entropy. Over the course of this reaction, one mole of gas and one mole of solid are converted to one mole of solid. As a general rule, the side of the equation with more moles of gas has a greater amount of entropy (disorder). Also, the total number of moles decreases over the course of the reaction, another indication of a decrease in entropy. 1. The decomposition of solid potassium chlorate to produce solid potassium chloride and oxygen gas.
2 KClO3 (s) → 2 KCl (s) + 3 O2 (g)
Check your answer:
Increase in entropy. Over the course of this reaction, two moles of solid decomposes into two moles of solid and three moles of gas. The presence of gas particles in the products indicates greater entropy (disorder). The fact that there are more moles of products than moles of reactants is a second indicator that entropy increased over the course of this reaction. V. Free energy
Entropy, enthalpy and temperature have an effect on the spontaneity of a chemical process. These properties can be combined in an equation that helps determine if a physical or chemical change is spontaneous at a given temperature.
ΔG = ΔH − (T × ΔS)
The temperature used here is Kelvin. (Positive value at all times)
The free energy change (ΔG) of a process is a function of temperature, enthalpy change, and entropy change
At a constant temperature and pressure, a process is spontaneous in the direction that gives a negative change in free energy (−ΔG)
Chemists can determine values for entropy change, enthalpy change, and temperature and plug them into the equation above to determine if a reaction or process is spontaneous under those conditions. In addition, the relationship given by the equation above is useful to make general predictions of the spontaneity of a process.
Refer to the table below to see if reactions with various combinations of entropy, enthalpy, and temperature will be spontaneous. Enthalpy Change | Entropy Change | Spontaneous Reaction? | Exothermic (-ΔH) | Increase (+ΔS) | Yes, -ΔG at any temperature | Exothermic (-ΔH) | Decrease (-ΔS) | Only at low temps, if |T ΔS| < |ΔH| | Endothermic (+ΔH) | Increase (+ΔS) | Only at high temps, if |T ΔS| > |ΔH| | Endothermic (+ΔH) | Decrease (-ΔS) | No, +ΔG at any temperature | 1. The melting of ice has an enthalpy value of ΔH = +333 kJ/kg. What is true about the spontaneity of this process?
Spontaneous at high temperatures
The melting of ice has a positive entropy change (melting from a solid to a liquid is always a positive entropy change) and a positive enthalpy change (ΔH = +333 kJ/kg). For this combination of entropy and enthalpy, the reaction will be spontaneous (-ΔG) at high temperatures. 1. The combustion of ethane gas has a negative enthalpy change and a positive entropy change.
Spontaneous at all temperatures
A negative enthalpy change and a positive entropy change are both “preferred” by nature, so this process is spontaneous at any temperature. 1. The process of freezing methane (CH4 (l) → CH4 (s)) has a negative enthalpy change (ΔH = -58.4 kJ/kg).
Spontaneous at low temperatures
The freezing of methane has a negative entropy change (freezing from a liquid to a solid is always a negative entropy change) and a negative enthalpy change (ΔH = -58.4 kJ/kg). For this combination of entropy and enthalpy, the reaction will be spontaneous (-ΔG) at low temperatures.
6.05: Rate and Collision theory
I. intro
The area of chemistry that studies reaction rate is Chemical Kinetics. The measure of the rate of a reaction involves measuring the change in something (amount, progress, motion etc) measured by ↑ or↓ concentration or amount of a reactant/ product.
II. Collision Theory
Chemists believe that for compounds, atoms, or ions to react successfully, the particles must collide together. This is the basis for a scientific theory known simply as the collision theory
This theory is based on the idea that the rate of a reaction is determined by the number of successful collisions that occur over a given amount of time. A successful collision is one that results in the making of the product(s).
The two factors are sufficient energy and correct orientation. These must be true for a given collsion to successfully result in the production of products. A. Sufficient Energy
The collision between reactants must have enough energy for the reactants’ valence energy levels to penetrate each other. This interaction between valence energy levels allows the electrons to rearrange to form new bonds.
A Maxwell-Boltzmann distribution graph.
The y-axis of the graph is labeled “number of particles,” increasing as you go up the axis, and the x-axis is labeled (energy), increasing left to right. The shape of the graph is like that of a bell or hill, showing that the majority of the particles have a kinetic energy near the average kinetic energy of the sample, while some particles have more energy (far right of the graph) and some particles have less energy (far left of the graph).
A Maxwell-Boltzmann distribution graph.
The y-axis of the graph is labeled “number of particles,” increasing as you go up the axis, and the x-axis is labeled (energy), increasing left to right. The shape of the graph is like that of a bell or hill, showing that the particles in this area of the curve do not have enough energy to react when they collide.
However, the particles to the right of the activation energy value are the only particles in the sample with enough energy to possibly react when they collide. B. Correct Orientation
When reactants come in contact with each other during their random movement, the orientation of their collision will determine if the collision is successful. The atoms from each reactant that will bond together to form the new products must come in contact with each other if a bond is going to form between them.
The more complex a reaction is, the slower the rate of reaction. For example, if more than two reactants need to collide together at the same time with correct orientation the reaction time would be slow.
For the reaction between ethene (CH2CH2) and hydrogen chloride (HCl) to occur, the double bond between the carbon atoms in ethene must be broken.
The double bond is converted to a single bond, allowing the carbon atoms to each bond with one of the atoms from the hydrogen chloride.
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The reaction can only happen if the hydrogen end of the hydrogen chloride compound collides with the double bond between the carbon atoms of the ethene.
Any other orientation will not be successful in producing the new product; the two reactants will just “bounce” off of each other.
06.05 Rate and Collision Theory: Factors Affecting Reaction Rate—Text Version
For most reactions, there is no practical way to have an effect on the orientation of particles in collisions. However, there are ways that chemists can increase the number of collisions that have sufficient energy to successfully produce the products.
Concentration and Rate
For many reactions involving liquids or gases, increasing the concentration of the reactants increases the rate of reaction. In a few cases, increasing the concentration of one of the reactants may have little noticeable effect of the rate.
In reactions that involve two or more reactants, increasing the concentration of the reactants will increase the rate of the reaction. According to the collision theory, a reaction between two different particles can only happen if those particles collide together. If the concentration of reactants is higher, meaning there are more moles of reactants in a given volume, the chances of the particles coming in contact increases. As the number of collisions between reactants increases, the number of successful collisions in a given amount of time will also increase.
The greater the concentration of solute particles in the solution, the greater the chance of the solute particles colliding with the surface of the solid reactant.
Be careful: We cannot assume that by doubling the concentration of one of the reactants you will double the rate of the reaction. The amount by which an increase in concentration increases the rate of a reaction depends on many different properties of the reaction.
Temperature and Rate
Because temperature is a measure of average kinetic energy, we know that the temperature of a system will affect the number of collisions that have sufficient energy to react. You have experienced the effects of temperature on rate if you have ever cooked something on the stove top or in the oven.
You know that increasing the temperature at which you cook your dinner increases the rate of the cooking process. This can be a great time saver, but you need to be careful that you don’t end up burning your food!
Increasing the temperature of a system increases the rate of a reaction. This direct relationship between temperature and rate is due to two reasons, both related to the collision theory. 1. Increasing the temperature of a system increases the average kinetic energy of the particles. This means more collisions between the reactants will have enough energy to react successfully. 2. Increasing the temperature of a system increases the kinetic energy of the particles, meaning that the particles are moving faster. This means that there are more collisions per minute at a higher temperature than there are in the same system at a lower temperature. Having more collisions in a given amount of time means there are more opportunities for collisions to meet both requirements for success.
As you can see on the Maxwell-Boltzmann distribution curve below, a slight increase of temperature changes the distribution of energy in a sample. Although a slight increase in temperature (T2) may only increase the average kinetic energy by a small amount, there is a significant increase to the number of particles that meet and exceed the activation energy requirement for a given reaction.
Catalysts and Rate
We have seen that an increase in temperature increases the rate of a reaction; if a reaction does not happen fast enough at normal temperatures, raising the temperature can help get the job done in less time. However, sometimes it is not possible to raise the temperature when we need a reaction to occur at a faster rate. In industrial settings, scientists need to keep safety and cost in mind, and so sometimes cannot exceed a certain temperature. In your body, living cells can only survive within a narrow temperature range.
Raising the temperature within the cells to speed up the necessary chemical reactions would result in damage to the very cells that the reactions support.
So, how do we get the complicated biochemical reactions in your body to occur at a sufficient rate to keep you alive without raising the temperature to a point that would kill the cells you are trying to support? Thankfully, your body contains a variety of enzymes that speed up the rate of these reactions without the need to raise the temperature. Almost every important biological reaction occurs in living organisms with the assistance of an enzyme.
Enzymes are biological catalysts. A catalyst is a substance that speeds up a reaction without being consumed by the reaction. Catalysts are important in many biological and industrial reactions where the speed of a reaction is important but temperature cannot exceed a certain range. So, how do catalysts manage to speed up a reaction without adding heat to increase the temperature?
If temperature cannot be increased to meet the activation energy requirement of a given reaction, another way to speed up the reaction is to provide an alternative reaction pathway with a lower activation energy requirement. As you can see in the potential energy diagram below, a catalyst provides a different pathway between reactants and products that requires a lower amount of activation energy.
Imagine you are hiking through the woods and there are two paths that will take you to your destination. The main path requires you to climb a steeper hill than the alternate path. Both paths take you to the same exact destination; the alternative path just requires less energy to get there. That is how a catalyst works: It provides an alternative path to the same final products but requires less activation energy to get there.
The lower activation energy requirement of this alternative reaction pathway results in a larger percentage of the given collisions between reactants that will be successful and produce products. Notice in the Maxwell-Boltzmann distribution curve below that the lower activation energy requirement of the catalyzed pathway allows many more particles to meet the energy requirement for a successful collision without any increase in temperature.
Did You Know?
Some substances, called inhibitors, actually decrease the rate of catalyzed chemical reactions. They do this by interfering with the catalysts’ ability to participate in the reaction. Many biological poisons are inhibitors that interfere with the enzymes in your body.
Speeding up a Reaction
If all the students in a math class are about to take a very tough exam, what are the possible ways to increase the percentage of the class that successfully passes the exam with a grade of C or higher? If all the students work hard studying and practicing for the test, they would probably see an increase in the number of students who pass the exam successfully.
Another option is for the teacher to lower the standard for a C on the grading scale. If your teacher changed the grading scale to make 55 percent the requirement for a C grade, it would be easier for more students to earn a C or higher with less effort on their part.
This story gives an analogy for the two main ways we can increase the rate of a chemical reaction.
Just as increasing the effort made by the students will result in more students successfully meeting the grade requirement of a C, increasing the temperature of a system allows more particle collisions to meet the energy requirement for a successful collision.
On the other hand, lowering the standard for a passing grade can also increase the number of students who successfully make the grade. This is how a catalyst works. Lowering the activation energy requirement of a reaction allows more collisions to be successful without changing any properties of the collisions themselves.
Although using a catalyst to lower the energy standard in a chemical reaction is a common practice in chemistry, do not expect your teachers to lower their standards for your success anytime soon.
Some practice: he reaction shown below has a positive enthalpy change and a negative entropy change.
2C (s) + 2H2 (g) C2H4 (g) The reaction will not be spontaneous at any temperature.
Which of the following conditions will result in a reaction that is spontaneous only at low temperatures? negative enthalpy change and negative entropy change .
Compare the following words. Which word best describes a scientific model? Analogy
The collision theory states that the rate of a chemical reaction is determined by collisions occurring at the molecular level. Which of the following will best help us visualize collision theory? A scientific model
Scientific models help predict experimental results. The model of an ideal gas is useful because it approximates the behavior of most gas molecules.
Catalytic converters on automobiles use the metals rhodium and platinum as catalysts to convert harmful gases to carbon dioxide, nitrogen, and water. Which answer best explains why rhodium and platinum do not have to be continuously replaced in the system?
The metals speed up the reactions but are not used up in the reaction.
Quiz: Which of the following changes will always be true for a spontaneous reaction? - G
Which of the following changes has a decrease in entropy? 3Fe (s) + 2O2 (g) Fe3O4 (g)
Which of the following processes involves an increase in entropy? melting gold bars
Which of the following combinations will result in a reaction that is never spontaneous? positive enthalpy change and negative entropy change
Describe in detail what you expect for the changes in enthalpy, entropy, and free energy when a sample of liquid evaporates. How does temperature affect these changes?
For a sample of liquid to evaporate, heat is needed to allow the liquid to become gas.
As the temperature rises, the kinetic energy affects the liquid to have higher entropy.
Also, when the sample changes to a less condense state such as liquid to gas, the entropy is high.
The enthalpy of the sample would be endothermic (+ triangle H) because it first absorbed the heat energy. So the free energy would be only present at high temperatures.
+ΔH (enthalpy change) +ΔS (entropy change) + or - ΔG (free energy change) depending on temperature This process will be spontaneous (have a negative ΔG) at high temperatures. 1. What factors did you investigate in your procedure, and why did you choose to compare these two factors?
I checked if the amount of water and temperature would affect the rate of reaction. I learned that increased temperature means higher kinetic energy, which would affect the movement of the particles in the mixture. I was just really curious if the amount of water would have anything with dissolving the tablet. 2. What other factors did you need to control during your investigation? Explain how you controlled each one in your procedure.
The water need to be controlled during the investigation. So I used filtered water for each procedure (except the ones that I had to heat up).
The tablet had to be the same size for each experiment. 3. What is your prediction about the results of each trial in your lab? Explain your predictions based on your knowledge of the dissolving process, collision theory, and reaction rates.
I hypothesize that the hotter water would have a faster rate of reaction than the one with the room temperature water. This will occur because the kinetic energy will cause the particles to move faster, causing them to collide more easily. I also hypothesize that the large amount of water would be cause the tablet to dissolve faster because the collision theory states that some factors such as the change in amount, progress, motion etc, can affect the rate of reaction. 4. Explain the collision theory, in your own words, and what is necessary for a collision to be successful.
According to the theory, if the molecules in the reaction are allowed to collide more, the faster the reaction rate would be. In order for a collision to be successful, sufficient energy and orientation must be true for the collision to be successful. 5. A specific catalyst was not provided for this reaction, but catalysts are useful for increasing the rate of many slow reactions. In your own words, give a detailed explanation of how catalysts can increase the rate of a reaction or process.
A catalyst can speed up the reaction rate, by allowing the reactants and products to collide to only require a lower amount of activation energy.
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