Download - Catalyst

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. Which of the following statements about quantum theory is incorrect:
A: The momentum and position of an electron can be determined
simultaneously.
B: Lower energy orbitals are filled with electrons before higher energy
orbitals.
C: Hund’s rule dictates that electrons will occupy unfilled orbitals before
occupied orbitals.
D: No two electrons can have the same four quantum numbers.
b. Of energy, enthalpy, heat, entropy, and the Gibbs energy how many are state
functions?
A: 2
B: 3
C: 4
D: 5
c. Using the following 1/2 cell reactions:
2H + + 2e- ⎯⎯
→ H2
E 10 = (you should know this)
2
→ 2H 2O
H 2O2 + 2H + + 2e- ⎯⎯
E 10 = 1.78 V
2
What is ∆G° for the following reaction (in kJ):
2H 2O ⎯⎯
→ H 2O2 + H 2
A: -343
B: 343
C: 687
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D: -687
d. For the following reaction under standard thermodynamic conditions:
H 2O (l ) ⎯⎯
→ H 2O (g )
∆H is more positive than ∆E by 2.5 kJ mol-1. This difference can be assigned
to:
A.
B.
C.
D.
The heat flow required to maintain a constant temperature
The work done by the system
The difference in the H-O bond strength between liquid and gaseous water.
That’s just how ∆H is calculated.
e. Consider the following reaction:
CH 4 (g ) + 4Cl2 (g ) ⎯⎯
→ CCl4 (g ) + 4HCl (g )
∆H = −434 kJ
If 19.2 g of CH4 undergoes this reaction at constant pressure, what amount of
heat is transferred between system and surroundings?
A: 520 kJ of heat are released to the surroundings
B: 520 kJ of heat are absorbed by the system
C: 360 kJ of heat are released to the surroundings
D: 360 kJ of heat are absorbed by the system
f. For the following reaction occurring in basic solution, in the balanced
equation how many electrons are transferred?
Fe(s) + NO3− (aq) → Fe+ (aq) + NH 3 (g )
A: 3
B: 4
C: 5
D: 8
g. Which of the following electron configurations is correct?
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A.
B.
C.
D.
Ga: [Kr]4s23d104p1
Bi: [Xe]6s24f145d106p3
Ca: [Ar]4s13d10
Mo: [Kr]5s24d5
Consider the following half reactions for questions h, j, k, and l:
Cr +3 + 3e− ⎯⎯
→ Cr
E 0 = −0.73 V
Br2 + 2e− ⎯⎯
→ 2Br −
E 0 = +1.09 V
h. What is E°cell for a galvanic cell based on these 1/2 cells?
A: +1.82 V
B: -1.82 V
C: 0.36 V
D: -0.36 V
C. 142
D.
j. What is ln(K) for the galvanic cell?
A. −425
B. 425
213
k. If [Cr+3] = 0.2 M, [Br-] = 0.1 M, and [Br2] = 0.5 M, what is the value of Ecell
at 298 K?
A: 2.21 V
B: 1.76 V
C: 2.12 V
D: 1.88 V
l. Which of the following statements is true about this cell?
A. Br - is reduced
B. Cr+3 is reduced
C. Cr is reduced
D. Br2 is reduced
m. Which of the following compounds has the bond with the most ionic
character?
Page 4
A: LiCl
B: KF
C: KCl
D: NaCl
n. Which of the following molecules has no dipole moment?
A: NO2-
C: PCl5
B: O3
D: SO2
o. Which of the following molecules does not contain a double or triple bond?
A: N2
B: H2CO
C: C2H6
D: SCN-
p. Which of the following molecules has a Lewis structure most like CO32-?
A: CO2
q.
C: NO3-
B: SO32-
D: O3
Consider the following reactions:
2ClF + O2 ⎯⎯
→ Cl2O + F2O
∆H = 167.4 kJ mol-1
2ClF3 + 2O2 ⎯⎯
→ Cl2O + 3F2O
∆H = 341.4 kJ mol-1
2F2 + O2 ⎯⎯
→ 2F2O
∆H = −43.4 kJ mol-1
At the same temperature, what is ∆H for the following reaction?
ClF + F2 ⎯⎯
→ ClF3
r.
A: -127.5 kJ mol-1
B: -108.7 kJ mol-1
C: -130.2 kJ mol-1
D: 217.5 kJ mol-1
For a certain reaction, ∆H° = 40 kJ and ∆S° = 50 J K-1. This reaction will
be:
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A:
B:
C:
D:
spontaneous at temperatures greater than 800 K
spontaneous at temperatures less than 10 K
spontaneous at all temperatures
no temperature at which the reaction is spontaneous
s. Based on Lewis dot structures, which of the following would you expect to
not be a stable molecule?
A: NH3
t.
B: N2H2
C: N2H4
D: N2H6
In nonpolar KrCl4, the Cl-Kr-Cl bond angle is:
A: 120°
B: 90°
C: 180°
D: 109°
u. For which of the following reactions is ∆S° expected to be most positive?
→ 2H 2O (g )
A: O2 (g ) + 2H 2 (g ) ⎯⎯
→ H 2O (g )
B: H 2O (l ) ⎯⎯
→ 2N 2 (g ) + O2 (g ) + 4H 2O (g )
C: 2NH 4 NO3 (s) ⎯⎯
→ 2NO2 (g )
D: N 2O4 (g ) ⎯⎯
_______/80
Page 6
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 if appropriate.
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 Cl O
Cl O Cl
Cl Cl O
geometry _bent_________________.
-
a) IO2
geometry __________________.
b) CCl4
geometry __________________.
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Problem 2 continued.
c) ONF
geometry __________________.
d) N2O
geometry __________________.
_______/20
Page 8
3 (20 pts.) One mol of an ideal monatomic gas at standard temperature and
pressure undergoes reversible isochoric heating until a final pressure of 5 atm is
reached. Next, the gas undergoes reversible isothermal expansion until the
original pressure is reached. Calculate ∆E, q, w, ∆H, and ∆S for this
thermodynamic process. Use Cv(Ne) = R if you do not remember the constantvolume heat capacity for an ideal monatomic gas.
Step 1: Isochoric Heating
⎛3 ⎞
∆E = nCV ∆T = (1 mol)⎜ R ⎟ (1490 K − 298 K ) = 14.9 kJ
⎝2 ⎠
⎛5 ⎞
∆H = nCP ∆T = (1 mol )⎜ R⎟ (1490 K − 298 K ) = 24.8 kJ
⎝2 ⎠
q = ∆E
w=0
⎛ Tf ⎞
⎛3 ⎞
∆S = nCV ln ⎜ ⎟ = (1 mol)⎜ R ⎟ ln (5 ) = 20.1 J K -1
⎝2 ⎠
⎝ Ti ⎠
Step 2: Isothermal Expansion
∆E = ∆H = 0
⎛ Vf ⎞
q = −w = nRT ln ⎜ ⎟ = (1 mol )(R )(1490 K )ln (5 ) = 19.9 kJ
⎝ Vi ⎠
⎛ Vf ⎞
∆S = nR ln ⎜ ⎟ = (1 mol )(R )ln (5 ) = 13.4 J K -1
⎝ Vi ⎠
q: 34.8 kJ
w: -19.9 kJ
∆E: 14.9 kJ
(more room on next page)
∆H: 24.8 kJ
∆S:33.5 J K-1
_______/20
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4) (more room for problem 3)
Page 10
4. (20 pts) Consider the following reaction:
⎯⎯
⎯
→ N 2O4 (g)
2NO2 (g) ←
⎯
One initially begins a reaction with only N2O4 present at a pressure of 10 atm.
What is the pressure of NO2 at equilibrium? The following thermodynamic data
will prove useful, and you can employ the standard assumptions of equilibrium
chemistry.
-1
-1
NO2(g)
N2O4(g)
-1
∆H 0f (kJ mol )
S 0 (J mol K )
34
10
240
304
∆H 0 = ∆H 0f (N 2O4 ) − 2∆H 0f (NO2 ) = −58 kJ
∆S 0 = S 0 (N 2O4 ) − 2S 0 (NO2 ) = −176 J K -1
⎛ 1 kJ ⎞
= −6 kJ
∆G 0 = ∆H 0 − T ∆S 0 = −58 kJ - (298 K ) −176 J K -1 ⎜
⎝ 1000 J ⎟⎠
(
K=e
− ∆G 0
RT
K = 11.3 =
)
= e+2.4 = 11.3
PN2 O4
2
NO2
P
=
(10 − x ) ≈ 10
(2x )2 (2x )2
4 x 2 = 0.885 ⇒ x = 0.47
PNO2 = 2x = 0.94 atm
_______/20
Page 11
5. (20 pts.) Using the bond-enthalpy method, determine ∆Hrxn for the
following:
C2 H 4 + 3O2 ⎯⎯
→ 2CO2 + 2H 2O
Bond enthalpies (in kJ mol-1)
C C
347
C C 614
C C 839
O O
146
O O 495
C O 799
C O
1072
O H 467
C H 413
∆H = (CCdouble ) + 4 (CH ) + 3(OOdouble ) − 4 (COdouble ) − 4 (OH )
(
) (
) (
) (
) (
= 614 kJ mol-1 + 4 413 kJ mol-1 + 3 495 kJ mol-1 − 4 799 kJ mol-1 − 4 467 kJ mol-1
= −1313 kJ mol-1
_______/20
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)
6. (20 pts) For a number of years it was unclear whether mercury cation existed in
solution as Hg+ or Hg2+2. A electrochemical cell was constructed using mercury
electrodes at both the anode and cathode, A solution of 0.263 g of mercury nitrate
(HgNO3) in 100 ml of water was placed at the anode, and a corresponding solution
of 0.790 g of mercury nitrate in 50 ml of water was placed at the cathode. A
voltage of 0.023 V was measured for this cell. How does mercury cation exist in
solution, as Hg+ or Hg2+2? You must show all work to receive any credit on this
problem.
0.059 V
log (Q )
n
0.059 V
0.023 V = −
log (Q )
n
0
Ecell = Ecell
−
0.263 g
[ ]anode =
(0.1 L )(MW ) = 0.166
Q=
[ ]cathode 0.790 g
(0.05 L )(MW )
(note: MW cancels out in Q; therefore, it is irrelevant)
0.059 V
0.059 V
log (0.167 ) = −
(−0.777 ) =
n
n
0.046 V
0.023 V=
n
n=2
0.023 V = −
Need cation that corresonds to a two electron reduction:
Hg2+2 + 2e− ⎯⎯
→ Hg
_______/20
Page 13
7. (20 pts.) a) Consider the following reaction:
⎯⎯
⎯
→ 2NO(g)
N 2 (g) + O2 (g) ←
⎯
Given that ∆G 0f (NO(g)) = 86.7 kJ mol-1 , determine the equilibrium constant for this reaction.
0
∆Grxn
= 2∆G 0f (NO ) = 173.4 kJ
K=e
0
− ∆Grxn
RT
= 4.0 × 10 −31
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b. (12 pts) Given that S 0 (N 2 (g)) = 192 J mol-1 K -1 , S 0 (O2 (g)) = 205 J mol-1 K -1 , and
S 0 (NO(g)) = 211 J mol-1 K -1 , determine ∆H 0f (NO( g )).
∆S 0 = 2S 0 (NO) − S 0 (O2 ) − S 0 (N 2 ) = 25 J K -1
∆G 0 = ∆H 0 − T ∆S 0
⎛ 1 kJ ⎞
= 180.8 kJ
∆H 0 = ∆G 0 + T ∆S 0 = 173.4 kJ + (298 K ) 25 J K -1 ⎜
⎝ 1000 J ⎟⎠
(
)
∆H 0 = 2∆H 0f (NO)
180.8 kJ = 2∆H 0f (NO)
90.4 kJ mol-1 = ∆H 0f (NO)
_______/20
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