Chem 152 Final
Section:______________
Name:_________________
You will have 1 hour and 50 minutes. Do not begin the exam until you are
instructed to start. Best of luck.
Question 1________________/80
Question 2________________/20
Question 3________________/20
Question 4________________/20
Question 5________________/20
Question 6________________/20
Question 7________________/20
Total____________________/200
Page 1
1. Multiple Choice (4 pts. each):
a. What is the correct order of the following bonds in terms of decreasing
polarity (highest to lowest)?
A: N-Cl, P-Cl, As-Cl
B: P-Cl, N-Cl, As-Cl
C: As-Cl, P-Cl, N-Cl
.
D: As-Cl, N-Cl, P-Cl
b. Of energy, work, enthalpy, entropy, Gibbs free energy, and heat, how many
are not state functions?
A: 1
B: 2
C: 3
D: 4
c. Which of the following molecules has a dipole moment?
A: CH4
B: CO2
C: BH3
D: OF2
d. In which of the following molecules is the bonding expected to be the most
ionic?
A: CO2
B: CaF2
C: CF4
D: BaF2
e. For the following reaction at 298 K and 1 atm, the change in enthalpy is
more positive than the change in energy by 2.5 kJ/mol:
H 2O(l) → H 2O(g)
The difference between the enthalpy and energy change is due to:
A:
B:
C:
D:
The heat flow required to maintain a constant temperature.
The work done by the system on the surroundings.
The difference in O-H bond strength in H2O(l) versus H2O(g)
None of the above.
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f. Given the following information:
C2 H 4 (g) + 3O2 (g) → 2CO2 (g) + 2H 2O(l) ∆H° = −1411kJ
1
C(graphite) + 3H 2 (g) + O2 (g) → C2 H 5OH(l) ∆H° = −278kJ
2
C2 H 4 (g) + H 2O(l) → C2 H 5OH(l) ∆H° = −44kJ
What is ∆H° for the following reaction:
C2 H 5OH(l) + 3O2 (g) → 3H 2O(l) + 2CO2 (g)
A: 44 kJ
B: -1367 kJ
C: -1742 kJ
D: 632 kJ
g. Given the following information:
1
Cu2O(s) + O2 (g) → 2CuO(s) ∆H° = −144kJ
2
Cu2O(s) → Cu(s) + CuO(s) ∆H° = +11kJ
What is the standard enthalpy of formation for CuO(s)
A: -166 kJ/mol
B: +155 kJ/mol
C: -155 kJ/mol
D: -299 kJ/mol
h. One mole of an ideal gas at 25° C is expanded isothermally and reversibly
from 125.0 liters to 250.0 liters. Which of the following statements is correct?
A: ∆Ssys=0
B: ∆Suniv=0
C: ∆Ssurr=0 D: ∆Ssys=Cpln(2)
j. The ∆G°f for NO(g) is 78 kJ/mol. What is K for the following reaction that
occurs at 1000 K?
→
N 2 (g) + O2 (g) ←
  2NO(g)
A: 8.4 x 10-5
B: 7.1 x 10-9
C: 1
Page 3
D: 1.8 x 10-19
k. Place the elements S, Cl, and F in order of increasing ionization energy
(lowest to highest):
A: S, Cl, F
B: F, Cl, S
C: Cl, F, S
D: F, S, Cl
l. How many of the following electron configurations are correct descriptions
of the lowest-energy configuration for a given species:
I. Ca: 1s22s22p63s23p64s2
II. Mg: 1s22s22p63s1
III. V: [Ar]3s23d3
IV: As: [Ar]4s23d104p3
A: 1
B: 2
C: 3
D: 4
m. An element with the electron configuration 1s22s22p63s23p4 would belong to
which of the following groups (or columns of the periodic table)?
A: Group 1
B: Group 4
C: Group 6
D: Group 8
n. Working with the following two 1/2 cells, you are asked to make a galvanic
electrochemical cell:
Co +2 + 2e− → Co E1/0 2 = −0.28V
Pb +2 + 2e− → Pb
E1/0 2 = −0.13V
What will be the cell potential of your galvanic cell when constructed?
A: -0.15 V
B: 0.15 V
C: -0.41 V
Page 4
D: 0.41 V
o. Which of the following statements about ionization energies (IE) is true:
A: The IE for Al will be greater than that of Mg because loss of two
electrons by Mg will result in a noble-gas configuration.
B: The IE for Al will be greater than that of Mg because the larger atomic
number of Al.
C: The IE of Al will be lower than that of Mg because the electron is taken
from a p orbital and is thus easier to remove.
D: none of the above.
p. For a particular process, q = 20 kJ and w = 15 kJ. Which of the following
statements is true?
A.
B.
C.
D.
Heat flows from the system to the surroundings.
The system does work on the surroundings.
The change in system energy is 35 kJ.
None of the above.
q. On another planet there exists the element negordyh that is similar to
hydrogen on earth except that the lowest energy orbital corresponds to n = 4 of
the hydrogen atom. The ratio of ionization energies for negordyh relative to
hydrogen is:
A: 1:4
B: 4:1
C: 16:1
D: 1:16
r. A concentration cell is constructed using two Cu electrodes with Cu2+
concentrations of 1.0 M and 10-4 M in anode and cathode sides of the cell,
respectively. The cell potential will be:
A: -0.118 V
B –0.0592 V
C: +0.0592
Page 5
D: +0.118 V
s. Which of the following experiments demonstrates the particle nature of
light?
A:
B:
C:
D:
Emission spectrum of hydrogen
Diffraction X-rays by a crystal
The photoelectric effect
The ultraviolet catastrophe
t. Consider the following reaction for which ∆H° = -200 kJ and ∆S° = -187
J/K:
2SO2 (g) + O2 (g) → 2SO3 (g)
At what temperature will K = 1 for this reaction?
A: 970 K
B: 1070 K
C: 2070 K
D: 298 K
u. The standard molar Gibbs free energies for the formation of NO2(g) and
N2O4(g) are 51.84 and 98.28 kJ/mol, respectively. What is the value of K for
the following reaction:
→
2NO2 (g) ← N 2O4 (g)
A: 1.37 x 108
B: 1.14 x 104
Page 6
C: 8.84
D: 0.113
Section II: Long-Answer/Numerical Questions
2 (20 pts.) Lewis Dot Structures. Write the Lewis dot structures for the following
compounds. Include all structural isomers, and resonance structures. Underline
the one you believe is the most-reasonable structure. Finally, describe the 3D
geometry of your most-reasonable structure. An example is provided below:
Ex: Cl2O
Cl O Cl
Cl O
Cl Cl O
Cl
Cl Cl O
geometry _bent_________________.
a) NO2
geometry __________________.
b) CO32-
geometry __________________.
Page 7
Problem 2 continued.
c) N3-
geometry __________________.
d) SF4O
geometry __________________.
Page 8
3 (20 pts.) 0.5 mol of an ideal monatomic gas initially at standard temperature and
pressure undergoes reversible isobaric cooling until the volume is reduced by a
factor of two.. Calculate the total ∆E, q, w, ∆H, and ∆S for this process. Use Cv
= R if you do not remember the constant-volume heat capacity for an ideal
monatomic gas.
Answer: Process is isobaric such that pressure is constant. A two-fold
decrease in volume will lead to a two-fold decrease in temperature via the ideal
gas law such that Ti = 298 K and Tf = 149 K. With these temperatures, the
thermodynamic properties of interest are readily calculated:
3 
∆E = nCv ∆T = (0.5mol) R(−149K ) = −929J
2 
5 
∆H = nC p ∆T = (0.5mol) R(−149K ) = −1548J
2 
q = ∆H (isobaric)
w = ∆E − q = −929J + 1548J = 619J
T 
5 
∆S = nC p ln f  = (0.5mol) R ln(0.5) = −7.2 J K
2 
 Ti 
Page 9
4. (20 pts.) Given the following reaction:
1
3
N 2 (g) + F2 (g) → NF3 (g)
2
2
o
∆H rxn
= −103kJ /mol
and
N2 bond energy = 941 kJ/mol, F2 bond energy = 154 kJ/mol
Determine the bond energy for a NF bond.
∆H rxn = ∑ Dbondsbroken − ∑ Dbondsformed
1
3
-103 kJ mol = DN 2 + DF2 − 3DNF
2
2
1
3
-103 kJ mol = 941 kJ mol + 154 kJ mol − 3DNF
2
2
268 kJ mol = DNF
(
) (
Page 10
)
5. (20 pts.) Consider the following reaction:
PCl3 (g) + Cl2 (g) → PCl5 (g)
What is ∆Grxn if the pressures of all gases are 5 atm? The following
thermodynamic data will prove useful.
∆H°f (kJ/mol)
-287.0
-374.9
----
PCl3(g)
PCl5(g)
Cl2(g)
S° (J/mol K)
311.8
364.6
223.1
∆H°rxn = ∆H° f (PCl5 ) − ∆H° f (PCl3 ) − ∆H° f (Cl2 )
= −374.9 kJ mol + 287.0 kJ mol + 0
= −87.9 kJ mol
∆S°rxn = S° (PCl5 ) − S° (PCl3 ) − S° (Cl2 )
= 364.6 J molK − 311.8 J molK − 223.1 J molK
= −170 J molK
 1kJ 
∆G°rxn = ∆H°rxn − T∆S°rxn = −87.9 kJ mol − (298K) −170 J molK 

 1000J 
= −37.2 kJ mol
(
)
Now, need to include non-standard conditions to get quantity of interest:
PPCl5
∆Grxn = ∆G°rxn + RT lnQ = −37.2 kJ mol + 8.314 J molK (298K )ln
PCl2 PPCl3
(
= −37.2 kJ
mol
(
+ 8.314 J
)


1kJ
298K )
 ln(0.2) = −41.2 kJ mol
(
)
molK
 1000J 
Page 11
6a. (10 pts.) Consider the following electrochemical cell:
+3
−
Cr(s)Cr (aq) NO3 (aq) NO(g) (with a Pt electrode)
Write a balanced reaction for the above reaction occurring in acidic solution
and determine the cell potential.
Cr → Cr +3 + 3e−
−
+
−
3
E 01 = +0.74V
2
3e + 4H + NO → NO + 2H 2O
E
Cr + 4H + + NO3− → NO + Cr +3 + 2H 2O
E 0cell = +1.70V
Page 12
0
1
2
= +0.96V
6b. (10 pts.) A precipitating reagent is added to the anode side of the
electrochemical cell and the cell potential increases to 1.9 V. Determine
the concentration of the original ionic anode species after the addition of
the precipitating reagent.
Answer: At the anode, the ionic species is Cr+3 such that the precipitating
reagent is reducing the concentration of Cr+3. The concentration can be
determined using the Nernst Eq.”
0.0591V
log(Q)
n
0.0591V
1.9V = 1.7V −
log [Cr +3 ]
3
0
E cell = E cell
−
(
0.2V = −0.02V log([Cr ])
−10 = log([Cr ])
+3
+3
10−10 M = [Cr +3 ]
Page 13
)
7. (20 pts total) “The Kitchen Sink”. On the first day of lecture we discussed the
autoinoization of water:
+
−
→
H 2O ←
  H + OH
In this problem, we will determine the autoionization constant (Kw) using
thermodynamics and electrochemistry.
a. (8 pts.). Using the 1/2 cells for the reduction of O2 in acidic and basic media
(see list of standard 1/2 cells), determine the cell potential for the following
“redox” reaction:
+
−
→
4H 2O ←
  4H + 4OH
2H 2O → 4e− + 4H + + O2
−
O2 + 2H 2O + 4e → 4OH
E 01 = −1.23V
2
−
4H 2O → 4OH − + 4H +
Page 14
E
0
1
2
= +0.40V
0
E cell
= −0.83V
b. (8 pts.) Using your answer to part a, determine the equilibrium constant for this
redox reaction.
0
0
∆Grxn
= −nFE cell
= −RT lnK ⇒ K = e
0
nFE cell
(
4 96,485C
RT
=e
mol
)(−0.83V )
(8.314 J molK )(298K )
K = 7.06x10−57
c. (4 pts). Using the equilibrium constant determined in part b, calculate the
autoionization constant of water (hint: compare the reaction in part a to the
autoionization reaction given at the beginning of this problem).
K w = (K ) 4 = (7.06x10−57 ) 4 = 9.2x10−15 ≈ 1x10−14
1
1
Page 15