1. Chapter 5 Thermochemistry
Chemistry, The Central Science, 11th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 5 Thermochemistry
John D. Bookstaver
St. Charles Community College
Edited by Debbie Amuso
Camden High School
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2. Thermochemistry
Thermodynamics = study of energy and its transformations Thermochemistry = study of chemical reactions involving changes in heat energy
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3. Energy Energy = the ability to do work or transfer heat energy.
Work = energy used to cause an object with mass to move (w = f x d)
Heat = energy used to cause the temperature of an object to increase
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4. Major Types of Energy
Potential energy = energy an object possesses by virtue of its position or chemical composition.
Kinetic energy = energy an object possesses by virtue of its motion.
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5. Kinetic Energy 1 2
KE =  m v 2
m = mass in kilograms (kg) v = velocity in meters per second (m/s) KE = kinetic energy in joules (J) 1 Joule = 1 kg-m2/s2 A mass of 2 kg moving at a speed of one meter per second possesses a kinetic energy of 1 Joule.
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6. Potential Energy PE = m g h
m = mass in kilograms (kg) g = acceleration due to gravity (9.8 m/s2) h = height (m) PE = potential energy in joules (J) 1 Joule = 1 kg-m2/s2
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7. Units of Energy The SI unit of energy is the joule (J).
An older, non-SI unit is still in widespread use: the calorie (cal).
1 cal = J
1000 calories = one nutritional Calorie
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8. First Law of Thermodynamics
Energy is neither created nor destroyed, but it can undergo a transformation from one type to another. (Law of Conservation of Energy)
The total energy of the universe is a constant.
The energy lost by a system must equal the energy gained by its surroundings, and vice versa.
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9. System and Surroundings
System = the molecules we want to study (here, the hydrogen and oxygen molecules).
Surroundings = everything else (here, the cylinder and piston).
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10. E = Efinal − Einitial (It’s a state function)
Internal Energy
The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E.
E = Efinal − Einitial (It’s a state function)
If E is positive, the system absorbed energy from the surroundings.
If E is negative, the system released energy to the surroundings.
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11. E = q + w
When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w).
That is, E = q + w.
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12. q, w, and their signs + q = system gains or takes in heat
q = system loses or gives off heat
+ w = work is done on the system by the surroundings (piston pushed in)
- w = work is done by the system on its surroundings (piston moves out)
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13. The system gave off 790 J of energy to its surroundings
Example
As hydrogen and oxygen gas are ignited in a cylinder, the system loses 550 J of heat to its surroundings. The expanding gases move a pistion to do 240 J of work on its surroundings. E for system = ?
Answer:
E = q + w
E = (-550 J) + (-240 J)
E = J
What does it mean?
The system gave off 790 J of energy to its surroundings
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14. Enthalpy & DH The symbol for enthalpy is H.
Enthalpy is the internal energy plus the product of pressure and volume:
At constant pressure:
H = E = q
So at constant pressure,
H = heat lost or gained by the system.
H = E + PV
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15. Recall: Endothermic
When heat is absorbed (taken in) by the system from the surroundings, the process is endothermic.
H = Hfinal − Hinitial
H = Hproducts − Hreactants
H = positive value
for endothermic
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16. Recall: Exothermic
When heat is released (given off) by the system into the surroundings, the process is exothermic.
H = Hfinal − Hinitial
H = Hproducts − Hreactants
H = negative value
for exothermic
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17. Enthalpy of Reaction
This quantity, H, is called the enthalpy of reaction, or the heat of reaction.
Enthalpy is an extensive property.
Reverse Rx  Negate H value.
Phase (state) matters.
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18. Enthalpies of Formation
An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms.
Example:
2 Al (s) O2 (g)  2 Al2O3 (s)
DHf = kJ
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19. Standard Enthalpies of Formation
Standard enthalpies of formation, Hf°, is defined as the enthalpy change for the reaction in which one mole of a compound is made from its constituent elements in their standard states at 25o C (298 K) and 1.00 atm pressure.
Example: 2 Al (s) + 3/2 O2 (g)  Al2O3 (s) Hf°= kJ/mol
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20. Calculation of H H = nHf°products – mHf° reactants
where n and m are the coefficients.
Hf°Values are listed in Appendix C of your Textbook.
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21. H = nHf products – mHf° reactants
Calculation of H
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
H = [3( kJ) + 4( kJ)] – [1( kJ) + 5(0 kJ)]
= [( kJ) + ( kJ)] – [( kJ) + (0 kJ)]
= ( kJ) – ( kJ) = kJ
H = nHf products – mHf° reactants
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22. Calorimetry is Another Way to Measure DH values
We measure H through calorimetry, which measures heat flow.
But first, some definitions.
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23. DT = change in temperature
Heat Capacity
The amount of energy (joules) required to raise the temperature of a substance by 1 K (1C) is its heat capacity. Units are J/K or J/ C
q = C * DT or C = q / DT
q = heat in Joules
C = heat capacity in J/K
DT = change in temperature
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24. Specific Heat
Specific heat capacity (or simply specific heat) is the amount of energy (joules) required to raise the temperature of 1 g of a substance by 1 K. Units are J/g-K
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25. Molar Heat Capacity
Molar heat capacity is the amount of energy (joules) required to raise the temperature of 1 mole of a substance by 1 K. Units are J/mol-K
Also helpful: # mol = mass/gfm
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26. Use Units
What is the molar heat capacity of CH4 (g) if its specific heat capacity is 2.20 J/g-K?
Answer:
2.20 J x g =
g – K mol
35.2 J/mol-K
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27. Constant Pressure Calorimetry
By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the DH for the system by measuring the heat change for the water in the calorimeter.
q = m  sh  T
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28. Calorimetry q = m * sh * T q = Joules of heat
m = total mass in calorimeter
(If a solution, use m = D * V to obtain the mass)
Water has a density of 1.00 g/mL
Most water solutions have a density of 1.02 g/mL
sh = specific heat (for water, 4.18 J/g-K)
DT = Tf - Ti
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29. Calorimetry Σ q = 0 qcalorimeter + qreaction+ qresulting solution = 0
In other words, the heat released (or absorbed) by the reaction of interest = the sum of the heat gained (or released) by the resulting solution & calorimeter.
qcalorimeter + qreaction+ qresulting solution = 0
If we assume qcalorimeter is negligible the equation reduces to the following:
qreaction = - qsolution = -m*sh*ΔT
where DT = Tf - Ti and m = D * V
DH = qreaction /# moles
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30. Specific Heat of a Metal
How to determine specific heat of a metal:
1) Mass metal
2) Put metal in HOT water & measure initial temp of hot metal
3) Measure temp of mL (100.0 g) of COLD water
4) Put hot metal in cold water
5) Record temp of water with metal in it (that temp is the final temp for both the metal & water)
6) Calculate the sh of the metal
(m * sh * T)metal = (m * 4.18 * T)water
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31. Bomb Calorimetry
Reactions can be carried out in a sealed “bomb” such as this one.
The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction.
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32. Bomb Calorimetry
Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H.
For most reactions, the difference is very small.
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33. DH via Heat Capacity
q = C * DT q = heat in Joules C = heat capacity in J/K DT = change in temperature where DT = Tf – Ti DH = q / # moles
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34. Stoichiometric Determination of DH
Given the DH for a reaction, you can use mole ratios to determine DH values of different sample sizes.
See examples.
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35. H2 (g) + I2 (g)  2 HI (g) DH = +53.0 kJ
Sample Problem #1
H2 (g) + I2 (g)  2 HI (g) DH = kJ
How many joules of heat are required to form 5 moles of HI (g)?
5 mol HI __x___
2 mol HI = kJ
x = +133 kJ
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36. H2 (g) + I2 (g)  2 HI (g) DH = +53.0 kJ
Sample Problem #2
H2 (g) + I2 (g)  2 HI (g) DH = kJ
How many moles of HI (g) are formed by the expenditure of 235 kJ of heat energy?
__x___ kJ
2 mol HI = kJ
x = 8.87 mol HI
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37. H2 (g) + I2 (g)  2 HI (g) DH = +53.0 kJ
Sample Problem #3
H2 (g) + I2 (g)  2 HI (g) DH = kJ
How many joules of heat are required to form 109 grams of HI?
# mol HI = mass/gfm
# mol HI = 109 g / 128 g/mol = mol HI
0.852 mol HI __x___
2 mol HI = kJ
x = kJ
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38. Hess’s Law
H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested.
However, we can estimate H using published H values and the properties of enthalpy.
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39. Hess’s Law
Hess’s law states that “If a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.”
DH = DH1 + DH2 + DH3
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40. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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41. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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42. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
Imagine this as occurring
in three steps:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
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43. Calculation of H C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
The sum of these equations is:
C3H8 (g)  3 C (graphite) + 4 H2 (g)
3 C (graphite) + 3 O2 (g)  3 CO2 (g)
4 H2 (g) + 2 O2 (g)  4 H2O (l)
C3H8 (g) + 5 O2 (g)  3 CO2 (g) + 4 H2O (l)
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44. Energy in Foods
Most of the fuel in the food we eat comes from carbohydrates and fats.
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45. Energy in Fuels
The vast majority of the energy consumed in this country comes from fossil fuels.
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46. Summary: Ways to Calculate DH
H = nHf°products – mHf° reactants
Using DHfo values from Appendix C Or
By Using a Calorimeter
qreaction = - qsolution = - m*sh*ΔT
where DT = Tf - Ti and m = D * V
DH = qreaction /# moles
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47. Summary: Ways to Calculate DH
Using Heat Capacity
q = C * DT
q = heat in Joules
C = heat capacity in J/K
DT = change in temperature
where DT = Tf – Ti
DH = q / # moles
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48. Summary: Ways to Calculate DH
Hess’s Law
H = DH1 + DH2 + DH3 Or
By Stoichiometric Determination
Given the DH for a reaction, you can use mole ratios to determine DH values of different sample sizes.
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