Chem 152 Midterm I
Section:______________
Name:_________________
You will have 50 minutes. Useful equations and other information is provided on the last
page of this exam...feel free to remove this page if it is convenient. You do not need to turn it in
with the exam.
Useful Constants:
R = 8.314 J/mol.K = 0.0821 liter.atm/mol.K
Avogadro’s Number: Na = 6.022 x 1023 molecules/mol
Boltzmann’s Constant: k = 1.38 x 10-23 J/K
Multiple Choice (4 pts. each). Please CIRCLE the correct answer.
1a. If ∆H° for the following reaction is –72 kJ:
H 2 (g) + Br2 (g) 
→ 2HBr(g)
Then ∆H°f for HBr(g) in kJ/mol is:
A. -72
B. 36
C. -36
D. -144
1b. ∆H°f for an element in its standard state is:
A. zero only at T = 0K
B. defined as zero.
C. defined by the second law.
D. Both A and B.
1c. Consider the following generic reaction:
1. A 
→ 2B
2. B 
→ C + D
3. E 
→ 2D
∆H for the following reaction?

A
→ 2C + E
A. ∆Η1 + ∆Η2 + ∆Η3
B. ∆Η1 + ∆Η2
C. ∆Η1 + ∆Η2 − ∆Η3
D. ∆Η1 + 2∆Η2 − ∆Η3
1d. For an ideal atomic crystal at 0K, the entropy is:
A.
 
B. –qrev
C. ∆Hrxn/T
D.
zero
1e. One mol of an ideal gas initially at standard state undergoes isothermal expansion until Vf =
10Vi, w is equal to:
A -R(298K)ln(10)
B. –(1atm)(10 l)
C. -R(298K)ln(0.1)
D. Not enough information given.
1f. Of energy, work, enthalpy, Gibbs free energy, entropy, and heat, how many are not state
functions?
A: 0
B: 1
C: 2
D: 4
1g. In lab you learned that the Ka for acetic acid was 1.8 x 10-5 at 25° C. What is ∆G° for the
dissociation of acetic acid?
A: -27,000 J
B: 27,000 J
C: 2,300 J
D: -270 J
1h. The condensation of water occurs spontaneously at 1 atm and 90 ° C. The change in Gibbs
free energy for this process is:
A: positive
B: negative
C: zero
D: undefined
1h. Two moles of an ideal monatomic gas undergo isochoric cooling from 225° C to 25° C . ∆S
for this process in units of J/K is:
A: -12.8
B: -24.7
C: -54.8
D: -4.3
1i. One mol of an ideal monatomic gas at standard conditions are expanded isothermally until
Pfinal = 2Pinitial. ∆H for this process in kJ is:
A: -w
GRADING: All or nothing.
B: q
C: ∆E
D: 0
Section II: Long-Answer/Numerical Questions
2) (20 pts. total) 1 mol of an ideal monatomic gas initial at standard temperature and pressure
undergoes isothermal expansion until Vfinal = 2Vinitial. In a subsequent process, the gas undergoes
isochoric heating until T = 600 K. Both steps are performed irreversibly. Calculate ∆E, q, w,
∆H, and ∆S total.
Step 1 (isothermal ∆T = 0):
∆E1 = ∆H1 = 0
101.3J
q1 = −w1 = Pext ∆V = (0.5atm)(2(24.5l) − 24.5l) = (0.5atm)(24.5l)
= 1.24kJ
l.atm
V 
∆S1 = nRln f  = (1mol) 8.314 J molK ln(2) = 5.76 J K
V 
(
i
)
Step 2 (isochoric, ∆V = 0):
w2 = 0
3

q2 = ∆E 2 = nCv ∆T = (1mol) * 8.314 J molK (600K − 298K ) = 3.77kJ
2

5

∆H 2 = nC p ∆T = (1mol) * 8.314 J molK (600K − 298K ) = 6.28kJ
2

T 
3
  600K 
∆S2 = nCv ln f  = (1mol) * 8.314 J molK  ln
 = 8.73 J K
2
  298K 
T 
i
Total:
∆E = 3.77kJ
∆H = 6.28kJ
w = −1.24kJ
q = 5.01kJ
∆S = 14.5 J
Grading:
K
+2 pts for each thermodynamic quantity in a given step.
-3 pts for units (for example, l.atm instead of J)
-3 pts for math errors (including totaling quantities).
3a) (10 pts.) Ethanol is employed as a gasoline additive. The reaction describing the combustion
of ethanol is as follows:
7
C2 H 5OH(l) + O2 (g) 
→ 2CO2 (g) + 3H 2O(g)
2
(Note: I blew the balancing here. Full credit was given if you used the reaction as is, or
balanced it correctly).
Using the following thermodynamic data, determine ∆G° for this process.
C2H5OH (l)
O2 (g)
CO2 (g)
H2O (g)
∆H°f (kJ/mol)
-277.6
----393.5
-241.8
∆G°f (kJ/mol)
-------394.4
-228.6
S° (J/mol.K)
161.0
205.0
213.7
188.7
∆H°rxn = 2∆H° f (CO2 (g)) + 3∆H° f (H 2O(g)) − ∆H° f (C2 H 5OH(l))
= 2(−393.5kJ ) + 3(−241.8kJ ) − (−277.6kJ ) = −1,235kJ
7
∆S °rxn = 2 S°(CO2 ( g)) + 3S°(H 2O(g)) − S °(C2 H 5OH ( l)) − S°(O2 ( g))
2
7
= 2 213.7 J K + 3 188.7 J K − 161.0 J K − 205.0 J K = 115 J K
2
(
) (
)(
) (
)
∆G°rxn = ∆H°rxn − T∆S°rxn
(
= (−1,235kJ ) − (298K ) 115 J
Grading:
 1kJ 
= −1,269.3kJ
K  1000J 
)
+ 4 for calculation of ∆H
+4 for calculation of ∆S
+2 for calculation of ∆G.
-3 for units (for example, not matching J and kJ)
-3 for math errors.
+2 for trying to calculate ∆G using ∆Gf. (doesn’t work since
∆G°f for ethanol is not provided)
3b) (10 pts.) For the sake of comparison, ∆G°rxn = -5.3x103 kJ for the combustion of one mole of
octane (C8H18). The combustion reaction for octane is:
C 8 H18 (l) +
25
O2 (g) 
→ 8CO2 (g) + 9H 2O(g)
2
Determine ∆G°f for octane using the thermodynamic data provided in the first part of this
problem.
∆G°rxn = 8∆G° f (CO2 ( g)) + 9∆G° f (H 2O( g)) − ∆G° f (C8 H18 (l))
−5,300kJ = 8(−394.4 kJ ) + 9(−228.6 kJ ) − ∆G° f (C8 H18 ( l))
−87.4 kJ = −∆G° f (C8 H18 ( l))
∆G° f (C8 H18 ( l)) = 87.4 kJ mol
Grading:
+4 for setup of problem (i.e., using the right equation)
+4 for calculating ∆G°f of ethanol
+2 for units of ∆G°f in kJ/mol, or J/mol.
-3 math errors.
4) Consider the following reaction:
→
2SO2 (g) + O2 (g) ←
  2SO3 (g)
a. (10 pts.) Determine the equilibrium constant, K, at T = 298 K using the following data:
∆H°f (kJ/mol)
-296.0
----396.0
SO2 (g)
O2 (g)
SO3 (g)
S° (J/mol.K)
248.1
205.0
256.7
∆H°rxn = 2∆H° f (SO3 (g)) − 2∆H° f (SO2 (g))
= 2(−396.0kJ ) − 2(−296.0kJ ) = −200kJ
∆S °rxn = 2 S°(SO3 ( g)) − 2 S°(SO2 ( g)) − S °(O2 ( g))
(
) (
)(
)
= 2 256.7 J K − 2 248.1 J K − 205.0 J K = −187.8 J K
∆G°rxn = ∆H°rxn − T∆S°rxn
(
= (−200kJ ) − (298K ) −187.8 J
K =e
−∆G° rxn
−(−144,000)J
RT
Grading:
=e
 1kJ 
= −144.0kJ
K  1000J 
)
(8.314 J K )(298K ) = e 58.12 = 1.7x10 25
+ 3 for calculation of ∆H
+3 for calculation of ∆S
+2 for calculation of ∆G.
+2 for calculating K.
-2 for units (for example, not matching J and kJ)
-2 for math errors.
b. (10 pts.) If the initial pressure of SO2 is 0.01 atm, SO3 is 0.5 atm, and O2 is 0.1 atm, will the
reaction be spontaneous? You must support your answer numerically for credit.
Q=
2
PSO
3
2
PSO
P
2 O2
(0.5) = 2.5x10 4
=
2
(0.01) (0.1)
2
∆Grxn = ∆G°rxn + RT lnQ
= −144.0kJ + RT ln(2.5x10 4 )
 1kJ 
= −144.0kJ + 8.314 J K (1000K )ln(2.5x10 4 )

1000J 
(
)
= −59.8kJ
∆Grxn < 0; therefore, reaction is spontaneous.
Grading:
+ 4 for calculation of Q
+4 for calculation of ∆G
+2 for concluding that the reaction is spontaneous (or consistent
with your ∆G).
-2 for units (for example, not matching J and kJ)
-2 for math errors.
+2 if right equation is written for ∆Grxn, but all else is wrong.