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Chapter 9
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3. Thermodynamics - ch 9
1. For a particular process q = – 10 kJ and w = 25 kJ. Which of the following statements is true?
a. Heat flows from the surroundings to the system.
b. The system does work on the surroundings.
c. ∆E = – 15 kJ
d. All of these are true.
e. None of these is true.

4. Thermodynamics - ch 9 Sign Notation
q > 0 ⇒ endothermic ⇒ heat is flowing from the surroundings into the system
q < 0 ⇒ exothermic ⇒ heat is flowing out of the system to the surroundings
w > 0 ⇒ work is being done on the system by the surroundings when a gas is getting compressed by an external pressure
w < 0 ⇒ the system is doing work on the surroundings when a gas is expanding against an external pressure
∆E > 0 ⇒ internal energy of a gas increases when the temperature of the gas increases
∆E < 0 ⇒ internal energy of a gas decreases when the temperature of the gas decreases
∆E = 0 ⇒ internal energy of a gas remains constant if isothermal

5. Thermodynamics - ch 9
2. A gas absorbs 4.8 J of heat and then performs 13.0 J of work. What is the change in internal energy of the gas?

6. Thermodynamics - ch 9 3. Which of the following statements is correct?
a. The internal energy of a system increases when more work is done by the system than heat is flowing into the system.
b. The internal energy of a system decreases when work is done on the system and heat is flowing into the system.
c. The system does work on the surroundings when an ideal gas expands against a constant external pressure.
d. All the statements are true.
e. All the statements are false.

7. Thermodynamics - ch 9
4. Which of the following are always endothermic?
a. Melting
b. Combustion
c. Condensation
d. All of the above

8. Thermodynamics - ch 9 Breaking bonds requires energy ⇒ q > 0
Making bonds releases energy ⇒ q < 0

9. Thermodynamics - ch 9
5. The following reaction occurs at constant temperature and pressure.
4 NH3 (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (g)
a. The system does work on the surroundings and w < 0
b. The surroundings do work on the system and w < 0
c. The system does work on the surroundings and w > 0
d. The surroundings do work on the system and w > 0
e. No work is done, w = 0

10. Thermodynamics - ch 9 3 Equations for Work 1) w = – Pext∆V
2) w = – ∆nRT
3) w = – nR∆T
equations 2 & 3
are derived from PV = nRT
choose the equation based on the
variables that are given in the problem

11. Thermodynamics - ch 9
6. There are two containers, one has 1 mole of CO2 (g) and the other has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC at constant pressure. For each of the following indicate true or false.
a. No work is done in heating either gas
b. More work is done in heating CO2 (g) than for Ne (g)
c. The change in internal energy of CO2 (g) is greater than it is for Ne (g)

12. Thermodynamics - ch 9
7. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1 J K-1 mol-1. When a sample of nitrogen gas is heated at constant pressure, what fraction of the energy is converted to heat?
a. 1.00 b c d e

13. Thermodynamics - ch 9 At constant volume the fraction
At constant pressure the fraction of energy converted to heat
⇒ Cv /Cp
At constant volume the fraction
of energy converted to heat
⇒ Cv /Cv or 1

14. Thermodynamics - ch 9 8. Which of the following are state functions?
a. work, heat
b. work, heat, enthalpy, energy
c. enthalpy, energy
d. work, heat, enthalpy
e. heat, enthalpy, energy

15. Thermodynamics - ch 9 State Functions
are pathway independent ⇒ since ∆ is defined as (final state – initial state) then ∆ anything can not depend on the states in between the initial and final states a.k.a. the “path” ⇒ if dealing with a state function you can use any path to solve for it ⇒ the easiest paths for ∆E and ∆H are…
∆E = qv
∆H = qp

16. Thermodynamics - ch 9 State Functions Think of all the
possible ways you
could go from peak A
to peak B – no matter which
path you choose the value of
Δh will be the same
Peak B
Peak A Δh

17. Thermodynamics - ch 9
9. Which one of the following statements is false?
a. The change in internal energy, ∆E, for a process is equal to the amount of heat absorbed at constant volume, qv.
b. The change in enthalpy, ∆H, for a process is equal to the amount of heat absorbed at constant pressure, qp.
c. A bomb calorimeter measures ∆H directly.
d. If qp for a process is negative, the process is exothermic.
e. The freezing of water is an example of an exothermic reaction.

18. Thermodynamics - ch 9
10. When a gas is cooled at constant volume from an initial temperature of 50 oC to 25 oC it loses 60 J of heat. When the same gas is cooled from 50 oC to 25 oC at constant pressure, it loses 100 J of heat. What is the value of the enthalpy change, ∆H, when the gas is heated from 25 oC to 50 oC at constant volume?
a J
b. – 60 J
c. – 100 J
d J
e. – 40 J

19. Thermodynamics - ch 9
11. Suppose you add 45 J of heat to a system, let it do 10. J of expansion work, and then return the system to its initial state by cooling and compression. Which statement is true for this process?
a. ∆H < ∆E
b. The work done in compressing the system must exactly equal the work done by the system in the expansion step.
c. ∆H = 70. J
d. The change in the internal energy for this process is zero.

20. Thermodynamics - ch 9
12. Consider a gas in a 1.0-L bulb at STP that is connected via a valve to another bulb that is initially evacuated. After the valve is opened…
a. q > 0 q < 0 q = 0
b. w > 0 w < 0 w = 0
c. ∆H > 0 ∆H < 0 ∆H = 0
d. ∆E > 0 ∆E < 0 ∆E = 0

21. Thermodynamics - ch 9
13. For the following reactions at constant pressure, predict if ∆H > ∆E, ∆H < ∆E, or ∆H = ∆E. Why is there a difference? Is work being done by the system or the surroundings?
a. H2 (g) + Cl2 (g) → 2 HCl (g)
b. N2 (g) F2 (g) → 2 NF3 (g)

22. Thermodynamics - ch 9
14. Consider a sample containing 2 moles of xenon that undergoes the following changes:
Pa = 10.0 atm
Va = 10.0 L
Pb = 10.0 atm
Vb = 5.0 L
Pc = 20.0 atm
Vc = 25.0 L
Step 1
Step 2
For each step, assume that the external pressure is constant
and equals the final pressure of the gas for that step. Calculate q, w, ΔE and ΔH (in kJ) for each step and calculate the total values. Which values would change if you went directly from state a to state c?

23. Thermodynamics - ch 9 Definition Easiest Path Substitution ΔE = q + w
Constant Volume
ΔE = qv = nCvΔT
ΔE = 3 2 Δ(PV) monoatomic gases
ΔE = Cv R Δ(PV) polyatomic gases
ΔH = ΔE + Δ(PV)
Constant Pressure
ΔH = qp = nCpΔT
ΔH = 5 2 Δ(PV) monoatomic gases
ΔE = Cp R Δ(PV) polyatomic gases

24. Thermodynamics - ch 9
15. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1 atm and its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at 73°C.

25. Thermodynamics - ch 9 16. Consider the combustion of ethanol:
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = –1.37×103 kJ
When a 15.6-g sample of ethanol (molar mass = 46.1 g/mol) is burned, how much energy is released as heat?
A) 87.8 kJ
B) 2.14 × 104 kJ
C) 4.64 × 102 kJ
D) 4.05 × 103 kJ
E) 4.75 kJ

26. Thermodynamics - ch 9 17. Consider the following reaction:
CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = – 434 kJ
How many grams of methane must be react to raise the temperature of 1.2 L of water 50°C (CH2O = J/g°C)

27. Thermodynamics - ch 9
18. A 25.0 g piece of unknown metal was transferred from an oven at 115 °C into a coffee cup calorimeter containing 150. mL of water at 24 °C and allowed to come to equilibrium where the temperature was measured to be 28 °C. Calculate the specific heat capacity of the metal. (Ccal = 25 J/K and Cwater = 4.18 J/g°C)

28. Thermodynamics - ch 9
19. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g of iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C. The mixture of hot aluminum and iron is then dropped into 97.3 mL of water at 22.0°C. Calculate the equilibrium temperature assuming no heat is lost to the surroundings. CH2O = 4.18 Jg-1°C-1

29. Thermodynamics - ch 9
20. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and 20.0 mL of 0.30 M HCl are mixed. The temperatures before and after the reaction were measured to be 23.0°C and 31.0°C. Calculate the molar enthalpy change for the neutralization reaction. Assume no heat is absorbed by the calorimeter. The density of each solution is 1.0 g/mL and the specific heat capacity for each solution is 4.18 J/g°C.

30. Thermodynamics - ch 9
21. Calculate the final temperature when 38 mL of 0.15 M Pb(NO3)2 is mixed with 42 mL of 0.23 M KI both initially at 23 °C. Assume no heat is lost to the surroundings. The specific heat capacity for the solution is 4.18 J/g°C and the density is 1.0 g/mL.
Pb I- → PbI ∆H = – 297 kJ/mol

31. Thermodynamics - ch 9
22. Given the following data calculate the ∆H° for the reaction: NO O → NO2
Rxn 1: 2 O3 → 3 O2 ∆H° = – 427 kJ
Rxn 2: O2 → 2 O ∆H° = +495 kJ
Rxn 3: NO + O3 → NO O2 ∆H° = – 195 kJ

32. Thermodynamics - ch 9 Hess’ 3 Laws 1. If a reaction is flipped ⇒
change the sign of ∆H
2. If a reaction is multiplied by a number ⇒
multiply ∆H by the same number
3. If two or more reactions are added ⇒
take the sum of the ∆Hs

33. Thermodynamics - ch 9 - Feldwinn
23. Calculate ΔH for the reaction
N2H4 (l) + O2 (g) → N2 (g) H2O (l)
given the following data:
2 NH3 (g) + 3 N2O (g) → 4 N2 (g) + 3 H2O (l) ΔH = – kJ
N2O (g) + 3 H2 (g) → N2H4 (l) + H2O (l) ΔH = – 317 kJ
2 NH3 (g) + ½ O2 (g) → N2H4 (l) + H2O (l) ΔH = – 143 kJ
H2 (g) + ½ O2 (g) → H2O (l) ΔH = – 286 kJ

34. Thermodynamics - ch 9
24. Choose the correct equation for the standard enthalpy of formation of CO (g), where ∆H°f for CO = –110.5 kJ/mol.
a. 2 Cgraphite (s) + O2 (g) → 2 CO (g) ∆H° = –110.5 kJ
b. Cgraphite (s) + O (g) → CO (g) ∆H° = –110.5 kJ
c. Cgraphite (s) + ½ O2 (g) → CO (g) ∆H° = –110.5 kJ
d. Cgraphite (s) + CO2 (g) → 2 CO (g) ∆H° = –110.5 kJ
e. CO (g) → Cgraphite (s) + O (g) ∆H° = –110.5 kJ

35. Thermodynamics - ch 9
25. The combustion of methanol takes place according to the reaction
2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)
Calculate the molar ∆H for the combustion under standard conditions. Use the following standard enthalpies of formation:
∆H°f for CH3OH (l) = – kJ/mol
∆H°f for CO2 (g) = – kJ/mol
∆H°f for H2O (l) = – kJ/mol

36. ∆H°rxn = Σ ∆H°f (products) – Σ ∆H°f (reactants)
Thermodynamics - ch 9
Hess’ Equation:
∆H°rxn = Σ ∆H°f (products) – Σ ∆H°f (reactants)

37. Thermodynamics - ch 9
26. The heat combustion of acetylene, C2H2 (g), at 25°C, is –1299 kJ/mol. At this temperature, ∆H°f values for CO2 (g) and H2O (l) are –393 and –286 kJ/mol, respectively. Calculate ∆H°f for acetylene.
a kJ/mol
b kJ/mol
c kJ/mol
d. –625 kJ/mol
e. none of these

38. Thermodynamics - ch 9
You have completed Ch.9

39. Answer key – ch. 9
1. For a particular process q = – 10 kJ and w = 25 kJ. Which of the following statements is true?
a. Heat flows from the surroundings to the system.
b. The system does work on the surroundings.
c. ∆E = – 15 kJ
d. All of these are true.
e. None of these is true. If q < 0 heat flows out of the system. If w > 0 work is done on the system by the surroundings. ∆E = q + w = -10 kJ + 25 kJ = 15 kJ

40. Answer key – ch. 9
2. A gas absorbs 4.8 J of heat and then performs 13.0 J of work. What is the change in internal energy of the gas?
∆E = q + w
∆E = (4.8 J) + (-13.0 J) = -8.2 J

41. Answer key – ch. 9 3. Which of the following statements is correct?
a. The internal energy of a system increases when more work is done by the system than heat is flowing into the system.
b. The internal energy of a system decreases when work is done on the system and heat is flowing into the system.
c. The system does work on the surroundings when an ideal gas expands against a constant external pressure.
w = -Pext∆V (“expand” => ∆V > 0 => w < 0)
d. All the statements are true.
e. All the statements are false.

42. Answer key – ch. 9 4. Which of the following are always endothermic?
a. Melting => heat is absorbed when bonds are broken
b. Combustion => always exothermic
c. Condensation => heat is released when bonds are formed
d. All of the above

43. Answer key – ch. 9
5. The following reaction occurs at constant temperature and pressure.
4 NH3 (g) + 5 O2 (g)  4 NO (g) + 6 H2O (g)
a. The system does work on the surroundings and w < 0
The reaction goes from 9 moles of gas to 10 moles of gas => ∆V > 0 => w < 0
b. The surroundings do work on the system and w < 0
c. The system does work on the surroundings and w > 0
d. The surroundings do work on the system and w > 0
e. No work is done, w = 0

44. Answer key – ch. 9
6. There are two containers; one has 1 mole of CO2 (g) and the other has 1 mole of Ne (g). Both gases are heated from 25 oC to 30 oC at constant pressure. For each of the following indicate true or false.
a. No work is done when heating either gas False => w = 0 only at constant volume or if expanding in a vacuum
b. More work is done in heating CO2 (g) than for Ne (g) False => work is dependent on the change in temperature which is the same for both gases
c. The change in internal energy of CO2 (g) is greater than it is for Ne (g) True => the heat capacity for CO2 is higher than Ne due to the bonds in the molecule

45. Cv/Cp = 20.8JK-1mol-1/29.1JK-1mol-1 = 0.715
Answer key – ch. 9
7. For nitrogen gas at 25 oC, Cv = 20.8 J K-1 mol-1 and Cp = 29.1 J K-1 mol-1. When a sample of nitrogen gas is heated at constant pressure, what fraction of the energy is converted to heat?
a. 1.00 b c d e
Cv/Cp = 20.8JK-1mol-1/29.1JK-1mol-1 = 0.715

46. Answer key – ch. 9 8. Which of the following are state functions?
a. work, heat
b. work, heat, enthalpy, energy
c. enthalpy, energy
d. work, heat, enthalpy
e. heat, enthalpy, energy

47. Answer key – ch. 9 9. Which one of the following statements is false?
a. The change in internal energy, ∆E, for a process is equal to the amount of heat absorbed at constant volume, qv.
b. The change in enthalpy, ∆H, for a process is equal to the amount of heat absorbed at constant pressure, qp.
c. A bomb calorimeter measures ∆H directly. In a bomb calorimeter the ∆H is measured indirectly by observing how much heat is absorbed by the calorimeter
d. If qp for a process is negative, the process is exothermic.
e. The freezing of water is an example of an exothermic reaction.

48. Answer key – ch. 9
10. When a gas is cooled at constant volume from an initial temperature of 50 oC to 25 oC it loses 60 J of heat. When the same gas is cooled from 50 oC to 25 oC at constant pressure, it loses 100 J of heat. What is the value of the enthalpy change, ∆H, when the gas is heated from 25 oC to 50 oC at constant volume?
a J
b. – 60 J
c. – 100 J
d J
e. – 40 J
∆H is a state function
which means we can pick any path we
want to calculate ∆H => the easiest path
is constant pressure =>
∆H = qp = +100 J

49. Answer key – ch. 9
11. Suppose you add 45 J of heat to a system, let it do 10. J of expansion work, and then return the system to its initial state by cooling and compression. Which statement is true for this process?
a. ∆H < ∆E
b. The work done in compressing the system must exactly equal the work done by the system in the expansion step.
c. ∆H = 70. J
d. The change in the internal energy for this process is zero. Since it has returned to it’s initial state there has been no overall change to the system

50. No energy is required to
Answer key – ch. 9
12. Consider a gas in a 1.0-L bulb at STP that is connected via a valve to another bulb that is initially evacuated. After the valve is opened…
a. q > 0 q < 0 q = 0
b. w > 0 w < 0 w = 0
c. ∆H > 0 ∆H < 0 ∆H = 0
d. ∆E > 0 ∆E < 0 ∆E = 0
No energy is required to
expand into a vacuum

51. Answer key – ch. 9
13. For the following reactions at constant pressure, predict if ∆H > ∆E, ∆H< ∆E, or ∆H = ∆E. Why is there a difference? Is work being done by the system or the surroundings?
a. H2 (g) Cl2 (g)  HCl (g)
∆H = ∆E + ∆(PV) or ∆H = ∆E + ∆(nRT) since the moles of gas remain constant ∆H = ∆E
b. N2 (g) F2 (g)  NF3 (g)
since the moles of gas decrease => ∆H < ∆E

52. Answer key – ch. 9
14. Consider a sample containing 2 moles of xenon that undergoes the following changes:
Pa = 10.0 atm
Va = 10.0 L
Step 1
Pb = 10.0 atm
Vb = 5.0 L
Step 2
Pc = 20.0 atm
Vc = 25.0 L
For each step, assume that the external pressure is constant and equals the final pressure of the gas for that step. Calculate q, w, ΔE and ΔH (in kJ) for each step and calculate the total values. Which values would change if you went directly from state a to state c?
Step 1 => constant pressure => ΔH = q = nCp ΔT => if you use
PV=nRT to find the temperatures you get Ta = 609K, Tb = 305K
and Tc = 3047 K (for step 2) => continued on next slide…

53. Answer key – ch. 9 14. …continued
ΔH = q = (2mol)(5/2)(8.3145x10-3kJmol-1K-1)(305K-609K) = kJ
w = -PΔV => w = -(10atm)(5.0L-10L)(0.1013kJatm-1L-1) = 5.1 kJ
ΔE = q + w = (-12.6 kJ) + (5.1 kJ) = -7.5 kJ
Step 2 => neither constant pressure or constant volume =>
w = -(20atm)(25L-5L)(0.1013kJatm-1L-1) = kJ
ΔH = nCp∆T =(2mol)(5/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 114 kJ
ΔE = nCv ∆T =(2mol)(3/2)(8.3145x10-3kJmol-1K-1)(3047K-305K)= 68.4 kJ
q = ΔE – w = (68.4 kJ) – (-40.5 kJ) = 109 kJ

54. Answer key – ch. 9
15. Calculate ∆E (in kJ) when 1 mol of liquid is vaporized at 1 atm and its boiling point (73°C). ∆Hvap = 28.6 kJ/mol at 73°C.
∆E = ∆H - ∆(PV) => ∆E = ∆H - ∆(nRT)
since T and R are constants
∆E = ∆H - ∆nRT
∆E = (1 mol)(28.6 kJ/mol) –
(1mol-0mol )(8.3145x10-3kJmol-1K-1)( K)
= 25.7 kJ

55. Answer key – ch. 9 16. Consider the reaction:
C2H5OH (l) + 3O2 (g) → 2CO2 (g) + 3H2O (l) ∆H = –1.37×103 kJ
When a 15.6-g sample of ethyl alcohol (molar mass = 46.1 g/mol) is burned, how much energy is released as heat?
A) 87.8 kJ
B) 2.14 × 104 kJ
C) 4.64 × 102 kJ
D) 4.05 × 103 kJ
E) 4.75 kJ
1 mol C2H5OH
1370 kJ
15.6 g C2H5OH x x
= 464 kJ
46.1 g
1 mol C2H5OH

56. Answer key – ch. 9 qH2O = mC ∆T = (1.2 kg)(4.184 J/g°C)(50°C) = 251 kJ
17. Consider the following reaction:
CH4 + 4 Cl2 (g) → CCl4 (g) + 4 HCl (g), ∆H = –434 kJ
How many grams of methane must be burned to raise the temperature of 1.2 L of water 50°C (CH2O = J/g°C)
qH2O = mC ∆T = (1.2 kg)(4.184 J/g°C)(50°C) = 251 kJ
1 mol CH4
16 g CH4
251 kJ x x
= 9.3 g CH4
434 kJ
1 mol CH4

57. Answer key – ch. 9
18. A 25.0 g piece of unknown metal was transferred from an oven at 115 °C into a coffee cup calorimeter containing 150. mL of water at 24 °C and allowed to come to equilibrium where the temperature was measured to be 28 °C. Calculate the specific heat capacity of the metal. (Ccal = 25 J/K and Cwater = 4.18 J/g°C)
heat lost by metal = heat absorbed by water and the calorimeter
-qmetal = qwater + qcal
-mC∆T = mC∆T + C∆T
-(25g)(C)(28°C-115°C) = (150g)(4.18 J/g°C)(28°C- 24°C) + ( J/K)(28°C- 24°C)
Cmetal = 1.2 J/g°C

58. Answer key – ch. 9
19. A 5.00 g of aluminum pellets (CAl = 0.89 J°C-1g-1) and 10.0 g of iron pellets (CFe = 0.45 J°C-1g-1) are heated to 100.0°C. The mixture of hot aluminum and iron is then dropped into 97.3 mL of water at 22.0°C. Calculate the equilibrium temperature assuming no heat is lost to the surroundings. CH2O = 4.18 Jg-1°C-1
-(qAl +qFe) = qwater
-(mC∆T + mC∆T) = mC∆T
-((5g)(0.89 J°C-1g-1)(Tf-100°C) + (10g)(0.45 J°C-1g-1)(Tf-100°C)) = (97.3g)(4.18 Jg-1°C-1)(Tf-22°C)
Tf = 23.7°C

59. Answer key – ch. 9
20. In a coffee cup calorimeter 50.0 mL of 0.10 M NaOH and 20.0 mL of 0.30 M HCl are mixed. The temperatures before and after the reaction were measured to be 23.0°C and 31.0°C. Calculate the molar enthalpy change for the neutralization reaction. Assume no heat is absorbed by the calorimeter. The density of each solution is 1.0 g/mL and the specific heat capacity for each solution is 4.18 J/g°C.
-qrxn = +qsoln
-qrxn = mC∆T = (50g+20g)(4.18 J/g°C)(31°C-23°C) = 2341J
qrxn = kJ
continued on next slide…

60. Answer key – ch. 9 24. …continued HCl = > (20 mL)(0.3M) = 6 mmol
NaOH => (50 mL)(0.1M) = 5 mmol => Limiting reagent
since HCl + NaOH  NaOH + H2O
qrxn = kJ/0.005mol = -468 kJ/mol

61. Answer key – ch. 9
21. Calculate the final temperature when 38 mL of 0.15 M Pb(NO3)2 is mixed with 42 mL of 0.23 M KI both initially at 23 °C. Assume no heat is lost to the surroundings. The specific heat capacity for the solution is 4.18 J/g°C and the density is 1.0 g/mL.
Pb I-  PbI ∆H = -297 kJ/mol
-qrxn = qsoln
-qrxn = mC∆T
to find qrxn we need to identify the limiting reagent
Continued on next slide…

62. Answer key – ch. 9 21. …continued
Pb2+ => (38 mL)(0.15 M) = 5.7 mmol
I- => (42 mL)(0.23 M) = 9.7 mmol => Limiting Reagent
qrxn = ( mol I-)(-297kJ/2 mol I-) = 1440J
now using qrxn = mC∆T
-(-1440J) = (38g + 42g)(4.18 J/g°C)(Tf - 23°C)
Tf = 27.3 °C

63. Answer key – ch. 9
22. Given the following data calculate the ∆H° for the reaction: NO O  NO2
2 O3  3 O2 ∆H° = -427 kJ
O2  2 O ∆H° = +495 kJ
NO + O3  NO O2 ∆H° = -195 kJ
Rxn 1 => flip and divide by 2
Rxn 2 => flip and divide by 2
Rxn 3 => keep as is
∆H°total = (-427 kJ)(-0.5) + (495 kJ)(-0.5) + (-195 kJ) = -49 kJ

64. Answer key – ch. 9 continued on next slide…
23. Calculate ΔH for the reaction
N2H4 (l) + O2 (g)  N2 (g) H2O (g)
given the following data:
2 NH3 (g) + 3 N2O (g)  4 N2 (g) + 3 H2O (l) ΔH = kJ
N2O (g) + 3 H2 (g)  N2H4 (l) + H2O (l) ΔH = -317 kJ
2 NH3 (g) + ½ O2 (g)  N2H4 (l) + H2O (l) ΔH = -143 kJ
H2 (g) + ½ O2 (g)  H2O (l) ΔH = -286 kJ
continued on next slide…

65. Answer key – ch. 9 23. …continued Rxn 1 => multiply by 1/4
Rxn 2 => flip and multiply by 3/4
Rxn 3 => flip and multiply by 1/4
Rxn 4 => multiply by 9/4
∆Htotal = (1/4)( kJ) + (-3/4)(-317 kJ) + (-1/4)(-143 kJ) + (9/4)(-286 kJ) = kJ

66. Answer key – ch. 9
24. Choose the correct equation for the standard enthalpy of formation of CO (g), where ∆H°f for CO = –110.5 kJ/mol.
a. 2 Cgraphite (s) + O2 (g) → 2 CO (g) ∆H° = –110.5 kJ
b. Cgraphite (s) + O (g) → CO (g) ∆H° = –110.5 kJ
c. Cgraphite (s) + ½ O2 (g) → CO (g) ∆H° = –110.5 kJ
d. Cgraphite (s) + CO2 (g) → 2 CO (g) ∆H° = –110.5 kJ
e. CO (g) → Cgraphite (s) + O (g) ∆H° = –110.5 kJ

67. Answer key – ch. 9
25. The combustion of methanol takes place according to the reaction
2 CH3OH (l) + 3 O2 (g) → 2 CO2 (g) + 4 H2O (l)
Calculate ∆H° for the combustion of 1 mol of methanol under standard conditions. Use the following standard enthalpies of formation:
∆H°f for CH3OH (l) = kJ/mol
∆H°f for CO2 (g) = kJ/mol
∆H°f for H2O (l) = kJ/mol
∆H°rxn = Σ∆H°(products) - Σ∆H°(reactants)
∆H°rxn = (2 mol CO2)( kJ/mol) + (4 mol H2O)( kJ/mol) – (2 mol CH3OH)( kJ/mol) = kJ
if only 1 mol of CH3OH => (0.5)(-1452 kJ) = -726 kJ