Thermodynamics
Thermodynamics Lab
Introduction:
Thermodynamics is the study of energy which can exist in many forms, such as heat, light, chemical energy, and electrical energy. The variables that thermodynamics can be used to define include temperature, internal energy, entropy, and pressure. Temperature, relating to thermodynamics, is the measure of kinetic energy in the particles of a substance. Light is usually linked to absorbance and emission in thermodynamics while pressure, linked with volume, can do work on an entire system. The entropy is the measure of the flow of heat through a system whose equation is for a thermodynamically reversible process as
Regarding thermodynamics, there are three laws the first of which is appropriately named the First Law of Thermodynamics. The First Law of Thermodynamics states that energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe must remains constant at all times, however it can change from one form to another. In other words energy is always conserved with the amount remaining constant. The Second Law of Thermodynamics states that in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state also referred to as entropy. The third law of thermodynamics is that the entropy of a perfect crystal at absolute zero is exactly equal to zero. When a system is at zero kelvin, the system will be in a state with the least possible energy. The third law is true if and only if the perfect crystal has only one minimum energy state. The entropy is zero because it only has one microstate. In simple terms the entropy of a system approaches a constant value as the temperature approaches zero. Enthalpy is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure, which enthalpy is greatly affected by. The unit of measurement for enthalpy is the joule since enthalpy is the measure of energy. With simplicity, enthalpy is energy transfer. Enthalpy accounts for energy transferred to the environment through the expansion of the system. The total enthalpy of a system, written as H, of a system cannot be measured directly so the change in enthalpy, written as ΔH, is more commonly used. The change ΔH is positive in endothermic reactions, and negative in exothermic processes which is a process that releases heat. The ΔH of a system is equal to the sum of work done on it and the heat supplied to it. For processes under constant pressure, ΔH is equal to the change in the internal energy of the system, in addition to the work that the system has done on its surroundings so the change in enthalpy would be the heat absorbed or released by a chemical reaction. The equation for enthalpy is as H=U+pV Equation 1 where H is the enthalpy of the system, U is the internal energy of the system, p is the pressure of the system, and V is the volume of the system. An equation for the change in enthalpy is ∆H=Hf-HI Equation 2 where delta H is the change in enthalpy. Hess’s Law states that if several reactions add up to produce an overall reaction, then the heat transfers of the reaction will add up to the value of the heat transfer of the total reaction. With relation to this lab, the enthalpy changes for the reaction of ammonia and hydrochloric acid can be determined using Hess’s law. First the enthalpy change for the reaction between sodium hydroxide and hydrochloric acid and the reaction between sodium hydroxide and ammonium chloride need to be determined. Then the enthalpy change for the reaction between ammonia and hydrochloric acid can be calculated. Then the second equation is reversed and added to Equation 1, the result is Equation 3. The equations for trial two three, and four are, respectively, as follows and coordinate to equation 1,2 and 3. NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) Equation 3 NH4Cl (aq) + NaOH (aq) NH3 (aq) +NaCl (aq) + H20 (l) Equation 4 NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) Equation 5
The enthalpy of the products and the reactants and is independent of the steps in going from reactants to products. Since acid-base neutralizations are exothermic processes, the enthalpy for this lab should be negative.
Materials:
The materials needed include a calorimeter, a hot plate, deionized water, magnetic stirrer, magnetic bar, a thermometer, 2.0M HCl (reagent), 2.0M NaOH (reagent), 2.0M NH4Cl (reagent), 2.0M NH3 (reagent), graduated cylinders, 50 mL beakers, and 100 mL beakers.
Procedure:
Part 1. Determine the Heat Capacity of the Calorimeter
1. Set-up a calorimeter with a cover having a hole in it to accept a thermometer.
2. Measure 50.0 mL of distilled water in a 50-mL graduated cylinder and transfer the water into the calorimeter.
3. Place the calorimeter assembly on a magnetic stirrer, add a magnetic stirring bar, and set the bar spinning slowly.
4. Measure and record the temperature of the water.
5. Heat approximately 75 mL of distilled water to about 70 degrees C in a 250-mL beaker.
6. Measure 50.0 mL of the 70 degrees C distilled water in a 50-mL graduated cylinder.
7. Measure and record the temperature of the hot water.
8. Immediately pour the hot water into the room temperature water in the calorimeter.
9. Cover the calorimeter, insert the thermometer, and stir the water.
10. Record the temperature every 20 seconds for a total of 3 minutes.
11. Empty the calorimeter and dry the inside of calorimeter when finished.
Part 2. Determine the Heats of Reaction
1. Measure 50.0 mL of a 2.0 M HCl solution in a 50-mL graduated cylinder and transfer to the calorimeter. 2. Record the temperature of the HCl solution.
3. Rinse the 50-mL graduated cylinder with distilled water.
4. Measure 50.0 mL of a 2.0 M NaOH solution in a 50-mL graduated cylinder.
5. Record the temperature of the NaOH solution.
6. Put a magnetic stirring bar into the calorimeter and start the bar spinning slowly in the HCI solution. 7. Quickly add the 50.0 mL, of 2.0 M NaOH solution to the calorimeter, cover, and insert the thermometer. 8. Record the temperature after 20 seconds, and then every 20 seconds for a total of 3 minutes.
9. Thoroughly rinse and dry the calorimeter, thermometer, stirrer bar, and graduated cylinder used for Reaction 1.
10. Repeat steps 1-8 of Part 2 using 2.0 M NaCl solution and 2.0 M NaOH solution. Be sure to perform this procedure in the fume hood.
11. Thoroughly rinse and dry the calorimeter, thermometer, stirrer bar, and graduated cylinder used for reaction 2.
12. Repeat steps 1-8 of Part 2 using 2.0 M NH3 solution and 2.0 M HCl solution. Be sure to perform this procedure in the fume hood.
Data:
Part one data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 45 | 120 | 44 | 40 | 44 | 140 | 44 | 60 | 44 | 160 | 44 | 80 | 44 | 180 | 44 | 100 | 44 | | |
Part two data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 33 | 120 | 34 | 40 | 33.5 | 140 | 34 | 60 | 33.5 | 160 | 34 | 80 | 33.5 | 180 | 34 | 100 | 34 | | |
Part three data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 25 | 120 | 25 | 40 | 25 | 140 | 25 | 60 | 25 | 160 | 25 | 80 | 25 | 180 | 25 | 100 | 25 | | |
Part four data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 28 | 120 | 29 | 40 | 29 | 140 | 29 | 60 | 29 | 160 | 29 | 80 | 29 | 180 | 29 | 100 | 29 | | |
Analysis:
Part One Using Data Set One
Tmix=44.444 degrees Celsius
The mixing temperature can be found either by extending the best fit line back to the y-axis or by simply using the y intercept, b, given in the y=mx+b equation on the graph.
Average initial temperature=42.25 degrees Celsius
The average temperature can be found by adding the final to the initial and dividing the sum by two.
Heat lost by water=-2.194 Joules
The difference between the average and mixing temperature will give the heat loss from the water. qwater = (grams of water) x (specific heat of water) x (Tavg – Tmix) Equation 6 * qwater =50g x 4.18 J/(g ˚C) x (44.444-42.25) * * qwater = -458.546 J * * qwater = -qcal * * -458.546 J = -qcal * * 458.546 J = qcal * * qcal = q water /(T mix – T initial) qcal=-458.546 J/ (42.25C° - 48°C) qcal= -2,636.63395 J/ C°
The heat capacity of the calorimeter can be found by using the equation * qcal = q water /(T mix – T initial) Equation 7 * also written as * qcal=∆T(°C)xheat capacity (J/°C) Equation 8 * Part Two Using Data Sets Two, Three, and Four
Tmix=33.139 degrees Celsius
Tmix=25 degrees Celsius
Tmix=29 degrees Celsius
To calculate the amount of heat involved for each reaction the equation Qrxn=[heat absorbed by solution+heat absorbed by calorimeter] Equation 8 which can be written as * Qrxn = -[(grams of solution x specific heat x ∆Tsolution) + (Ccal x ∆Tsolution)] Equation 9 * Reaction 1: * Qrxn = -[ (103 g of solution) x (4.18 J/g ˚C) x (33.139-28)] + [(-2,636.63395) x (33.139-28)] * Qrxn= -15762.2 * Reaction 2: * Qrxn = -[(103 g of solution) x (4.18 J/g ˚C) x (25-22.5)] + [(-2,636.63395) x (25-22.5)] * Qrxn = -7667.93 * Reaction 3: * Qrxn=- [(103 g of solution) x (4.18 J/g ˚C) x (29-24)] + [(-2,636.63395) x (29-24)] * Qrxn = -15335.9 * Then the enthalpy change can be found using the equation * ∆H = (Q of reaction) / (moles in reaction)
Reaction 1:
∆H= / -15762.2∙(1kJ / 1000J)
∆H=-15.7622 KJ/mol
Reaction 2:
∆H= / -7667.93∙(1kJ / 1000J)
∆H=-7.66793 KJ/mol
Reaction 3:
∆H= / -15335.9∙(1kJ / 1000J)
∆H=-15.3359 KJ/mol
NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) ∆H =-15.7622 kJ/mol
-(NH4Cl (aq) + NaOH (aq) --> NH3 (aq) +NaCl (aq) + H20 (l)) ∆H = -(-7.66793)kJ/mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
-NH4Cl (aq) + HCl (aq) --> -NH3 (aq) ∆H = -8.09427 kJ/mol
The percent error can be found by using the equation percent error=(measured-actual)/actual Equation 10
Percent error=(-8.09427-(-15.3359))/ -8.09427
The percent error was calculated to be 89.4661 percent which could have been do to incomplete mixing, letting the solution cool down too much before putting it in the calorimeter, etc.
Conclusion:
In summarization, Hess’s law is used to calculate the enthalpy of a third reaction resulting from the two before it. Unfortunately, the data does not prove hess’s law because of the high percent error, however it does hold true. The high percent error was due to the probabilities of lack of mixing, and letting the solution cool before transfer to the thermometer.
Introduction:
Thermodynamics is the study of energy which can exist in many forms, such as heat, light, chemical energy, and electrical energy. The variables that thermodynamics can be used to define include temperature, internal energy, entropy, and pressure. Temperature, relating to thermodynamics, is the measure of kinetic energy in the particles of a substance. Light is usually linked to absorbance and emission in thermodynamics while pressure, linked with volume, can do work on an entire system. The entropy is the measure of the flow of heat through a system whose equation is for a thermodynamically reversible process as
Regarding thermodynamics, there are three laws the first of which is appropriately named the First Law of Thermodynamics. The First Law of Thermodynamics states that energy can be changed from one form to another, but it cannot be created or destroyed. The total amount of energy and matter in the Universe must remains constant at all times, however it can change from one form to another. In other words energy is always conserved with the amount remaining constant. The Second Law of Thermodynamics states that in all energy exchanges, if no energy enters or leaves the system, the potential energy of the state will always be less than that of the initial state also referred to as entropy. The third law of thermodynamics is that the entropy of a perfect crystal at absolute zero is exactly equal to zero. When a system is at zero kelvin, the system will be in a state with the least possible energy. The third law is true if and only if the perfect crystal has only one minimum energy state. The entropy is zero because it only has one microstate. In simple terms the entropy of a system approaches a constant value as the temperature approaches zero. Enthalpy is a measure of the total energy of a thermodynamic system. It includes the internal energy, which is the energy required to create a system, and the amount of energy required to make room for it by displacing its environment and establishing its volume and pressure, which enthalpy is greatly affected by. The unit of measurement for enthalpy is the joule since enthalpy is the measure of energy. With simplicity, enthalpy is energy transfer. Enthalpy accounts for energy transferred to the environment through the expansion of the system. The total enthalpy of a system, written as H, of a system cannot be measured directly so the change in enthalpy, written as ΔH, is more commonly used. The change ΔH is positive in endothermic reactions, and negative in exothermic processes which is a process that releases heat. The ΔH of a system is equal to the sum of work done on it and the heat supplied to it. For processes under constant pressure, ΔH is equal to the change in the internal energy of the system, in addition to the work that the system has done on its surroundings so the change in enthalpy would be the heat absorbed or released by a chemical reaction. The equation for enthalpy is as H=U+pV Equation 1 where H is the enthalpy of the system, U is the internal energy of the system, p is the pressure of the system, and V is the volume of the system. An equation for the change in enthalpy is ∆H=Hf-HI Equation 2 where delta H is the change in enthalpy. Hess’s Law states that if several reactions add up to produce an overall reaction, then the heat transfers of the reaction will add up to the value of the heat transfer of the total reaction. With relation to this lab, the enthalpy changes for the reaction of ammonia and hydrochloric acid can be determined using Hess’s law. First the enthalpy change for the reaction between sodium hydroxide and hydrochloric acid and the reaction between sodium hydroxide and ammonium chloride need to be determined. Then the enthalpy change for the reaction between ammonia and hydrochloric acid can be calculated. Then the second equation is reversed and added to Equation 1, the result is Equation 3. The equations for trial two three, and four are, respectively, as follows and coordinate to equation 1,2 and 3. NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) Equation 3 NH4Cl (aq) + NaOH (aq) NH3 (aq) +NaCl (aq) + H20 (l) Equation 4 NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) Equation 5
The enthalpy of the products and the reactants and is independent of the steps in going from reactants to products. Since acid-base neutralizations are exothermic processes, the enthalpy for this lab should be negative.
Materials:
The materials needed include a calorimeter, a hot plate, deionized water, magnetic stirrer, magnetic bar, a thermometer, 2.0M HCl (reagent), 2.0M NaOH (reagent), 2.0M NH4Cl (reagent), 2.0M NH3 (reagent), graduated cylinders, 50 mL beakers, and 100 mL beakers.
Procedure:
Part 1. Determine the Heat Capacity of the Calorimeter
1. Set-up a calorimeter with a cover having a hole in it to accept a thermometer.
2. Measure 50.0 mL of distilled water in a 50-mL graduated cylinder and transfer the water into the calorimeter.
3. Place the calorimeter assembly on a magnetic stirrer, add a magnetic stirring bar, and set the bar spinning slowly.
4. Measure and record the temperature of the water.
5. Heat approximately 75 mL of distilled water to about 70 degrees C in a 250-mL beaker.
6. Measure 50.0 mL of the 70 degrees C distilled water in a 50-mL graduated cylinder.
7. Measure and record the temperature of the hot water.
8. Immediately pour the hot water into the room temperature water in the calorimeter.
9. Cover the calorimeter, insert the thermometer, and stir the water.
10. Record the temperature every 20 seconds for a total of 3 minutes.
11. Empty the calorimeter and dry the inside of calorimeter when finished.
Part 2. Determine the Heats of Reaction
1. Measure 50.0 mL of a 2.0 M HCl solution in a 50-mL graduated cylinder and transfer to the calorimeter. 2. Record the temperature of the HCl solution.
3. Rinse the 50-mL graduated cylinder with distilled water.
4. Measure 50.0 mL of a 2.0 M NaOH solution in a 50-mL graduated cylinder.
5. Record the temperature of the NaOH solution.
6. Put a magnetic stirring bar into the calorimeter and start the bar spinning slowly in the HCI solution. 7. Quickly add the 50.0 mL, of 2.0 M NaOH solution to the calorimeter, cover, and insert the thermometer. 8. Record the temperature after 20 seconds, and then every 20 seconds for a total of 3 minutes.
9. Thoroughly rinse and dry the calorimeter, thermometer, stirrer bar, and graduated cylinder used for Reaction 1.
10. Repeat steps 1-8 of Part 2 using 2.0 M NaCl solution and 2.0 M NaOH solution. Be sure to perform this procedure in the fume hood.
11. Thoroughly rinse and dry the calorimeter, thermometer, stirrer bar, and graduated cylinder used for reaction 2.
12. Repeat steps 1-8 of Part 2 using 2.0 M NH3 solution and 2.0 M HCl solution. Be sure to perform this procedure in the fume hood.
Data:
Part one data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 45 | 120 | 44 | 40 | 44 | 140 | 44 | 60 | 44 | 160 | 44 | 80 | 44 | 180 | 44 | 100 | 44 | | |
Part two data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 33 | 120 | 34 | 40 | 33.5 | 140 | 34 | 60 | 33.5 | 160 | 34 | 80 | 33.5 | 180 | 34 | 100 | 34 | | |
Part three data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 25 | 120 | 25 | 40 | 25 | 140 | 25 | 60 | 25 | 160 | 25 | 80 | 25 | 180 | 25 | 100 | 25 | | |
Part four data set Time (sec) | Temperature (Celsius) | Time (sec) | Temperature (Celsius) | 20 | 28 | 120 | 29 | 40 | 29 | 140 | 29 | 60 | 29 | 160 | 29 | 80 | 29 | 180 | 29 | 100 | 29 | | |
Analysis:
Part One Using Data Set One
Tmix=44.444 degrees Celsius
The mixing temperature can be found either by extending the best fit line back to the y-axis or by simply using the y intercept, b, given in the y=mx+b equation on the graph.
Average initial temperature=42.25 degrees Celsius
The average temperature can be found by adding the final to the initial and dividing the sum by two.
Heat lost by water=-2.194 Joules
The difference between the average and mixing temperature will give the heat loss from the water. qwater = (grams of water) x (specific heat of water) x (Tavg – Tmix) Equation 6 * qwater =50g x 4.18 J/(g ˚C) x (44.444-42.25) * * qwater = -458.546 J * * qwater = -qcal * * -458.546 J = -qcal * * 458.546 J = qcal * * qcal = q water /(T mix – T initial) qcal=-458.546 J/ (42.25C° - 48°C) qcal= -2,636.63395 J/ C°
The heat capacity of the calorimeter can be found by using the equation * qcal = q water /(T mix – T initial) Equation 7 * also written as * qcal=∆T(°C)xheat capacity (J/°C) Equation 8 * Part Two Using Data Sets Two, Three, and Four
Tmix=33.139 degrees Celsius
Tmix=25 degrees Celsius
Tmix=29 degrees Celsius
To calculate the amount of heat involved for each reaction the equation Qrxn=[heat absorbed by solution+heat absorbed by calorimeter] Equation 8 which can be written as * Qrxn = -[(grams of solution x specific heat x ∆Tsolution) + (Ccal x ∆Tsolution)] Equation 9 * Reaction 1: * Qrxn = -[ (103 g of solution) x (4.18 J/g ˚C) x (33.139-28)] + [(-2,636.63395) x (33.139-28)] * Qrxn= -15762.2 * Reaction 2: * Qrxn = -[(103 g of solution) x (4.18 J/g ˚C) x (25-22.5)] + [(-2,636.63395) x (25-22.5)] * Qrxn = -7667.93 * Reaction 3: * Qrxn=- [(103 g of solution) x (4.18 J/g ˚C) x (29-24)] + [(-2,636.63395) x (29-24)] * Qrxn = -15335.9 * Then the enthalpy change can be found using the equation * ∆H = (Q of reaction) / (moles in reaction)
Reaction 1:
∆H= / -15762.2∙(1kJ / 1000J)
∆H=-15.7622 KJ/mol
Reaction 2:
∆H= / -7667.93∙(1kJ / 1000J)
∆H=-7.66793 KJ/mol
Reaction 3:
∆H= / -15335.9∙(1kJ / 1000J)
∆H=-15.3359 KJ/mol
NaOH (aq) + HCl (aq) NaCl (aq) + H20 (l) ∆H =-15.7622 kJ/mol
-(NH4Cl (aq) + NaOH (aq) --> NH3 (aq) +NaCl (aq) + H20 (l)) ∆H = -(-7.66793)kJ/mol
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
-NH4Cl (aq) + HCl (aq) --> -NH3 (aq) ∆H = -8.09427 kJ/mol
The percent error can be found by using the equation percent error=(measured-actual)/actual Equation 10
Percent error=(-8.09427-(-15.3359))/ -8.09427
The percent error was calculated to be 89.4661 percent which could have been do to incomplete mixing, letting the solution cool down too much before putting it in the calorimeter, etc.
Conclusion:
In summarization, Hess’s law is used to calculate the enthalpy of a third reaction resulting from the two before it. Unfortunately, the data does not prove hess’s law because of the high percent error, however it does hold true. The high percent error was due to the probabilities of lack of mixing, and letting the solution cool before transfer to the thermometer.