Chemistry 112
Exam 3C
Harwood/Fall 04
KEY
1. Which order is correct for increasing polarity (least polar first)?
C—C
C—Cl
C—F
C—I
A
B
C
D
(a)
A<D<B<C
(d) A < D < C < B
(b)
D<C<B<A
(e) A < B < C < D
(c)
C<B<A<D
–
2. How many valence electrons are shown in the Lewis structure of OH ?
(a) 8
(b) 7
(c) 6
(d) 9
(e) 10
3. Which of the following molecules is polar?
(a) CO2
(d) CO
(b) O2
(e) CH4
(c) C2H2 (H–C–C–H is skeleton structure)
4.
Which pair is mismatched?
(a) polarizability — dispersion forces
(b) N–H or O–H bonds — hydrogen bonding
(c) polarity — dipole moment
(d) dipole moment — dipole-dipole forces
(e) polar — low boiling point
5. Drawing (1) shows the equilibrium vapor pressure of a pure liquid. Which drawing
1
Chemistry 112
Exam 3C
Harwood/Fall 04
KEY
6. The following pictures represent the equilibrium state for four different reactions of the
type A2 + X2 ↔ 2 AX (X = B, C, D, E). A atoms are open circles. X atoms are closed
circles. Which reaction has the smallest equilibrium constant?
(a) A2 + B2 ↔ 2 AB
(b) A2 + C2 ↔ 2 AC
(c) A2 + D2 ↔ 2 AD
(d) A2 + E2 ↔ 2 AE
7.
A2 + B2 ↔ 2 AB
A2 + C2 ↔ 2 AC
A2 + D2 ↔ 2 AD
A2 + E2 ↔ 2 AE
A 6.0 m H2SO4 solution has d=1.34 g/mL. What is the mass percent H2SO4 in the
solution?
" 98.0794 g %
mass H2SO 4 = ( 6.0 mol H2SO 4 )$
' = 588 g
# 1 mol &
total mass of solution = 1000 g + 588 g = 1588 g
" 588 g %
$
' = 0.370
# 1588 g &
0.370 ( 100% = 37.0%
(a) 37.0
(b) 43.9
(c) 63.0
(d) 56.1
(e) 58.8
8. Which of the following solutions has the lowest freezing point?
Molality of particles
!
(a) 1.0 m glucose in water
1.0
lowest f.p. is sol’n with
the largest molality of
(b) 1.0 m CaCl2 in water
3.0
particles
(c) 1.0 m NaBr in water
2.0
(d) 1.0 m NaCl in water
2.0
(e) water
0.0
9. KP for the reaction H2(g) + I2(g) ↔ 2 HI(g) is 54.3 at 698°C. What is the value of KC at this
temperature?
(a) 85.6
–2
(b) 1.80 × 10
–4
(c) 2.30 × 10
KP = KC (RT)
Δn = 2-2 = 0
Δn
so KP = KC
3
(d) 4.33 × 10
(e) 54.3
2
Chemistry 112
Exam 3C
Harwood/Fall 04
KEY
10.
Which direction will the following reaction (in a 5.0 L flask) proceed if the pressure of
CO2(g) is 1.0 atm?
CaCO3(s) ↔ CaO(s) + CO2(g)
KP = 1.9 × 10
(a) to the left because Q < KP
–23
Q = 1.0
(b) to the right because Q < KP
Q>K
(c) to the left because Q > KP
(d) to the right because Q > KP
(e) The reaction is at equilibrium.
11.
For the system CaO(s) + CO2(g) ↔ CaCO3(s), the equilibrium constant expression is
(a) [CO2]
(d) [CaCO3] / [CaO] [CO2]
(b) 1 / [CO2]
(e) 1 / [CaO] [CO2]
(c) [CaO] [CO2] / [CaCO3]
12. Which change will increase the quantity of product in the following reaction?
2 SO2(g) + O2(g) ↔ 2 SO3(g)
ΔH° = -198 kJ
(a) an increase in temperature
(b) an increase in the volume of the container
(c) an increase in pressure
(d) a decrease in amount of SO2
Increasing T will cause shift to left
(exothermic).
Increasing V (decreasing P) will cause shift
toward most number of moles of gas (left).
Decreasing amount of reactant will cause
shift to left.
Increasing P will cause shift toward least
number of moles of gas (right).
13. Consider the reaction CO(g) + H2O(g) ↔ CO2(g) + H2(g) ΔH° = -41 kJ
All of the following changes would shift the equilibrium to the right except one. Which
one would not cause the equilibrium to shift to the right?
(a) Add some CO.
(b) Remove some CO2.
(c) Decrease the container volume.
(d) Decrease the temperature.
(e) Increase the partial pressure of H2O.
Decreasing T will cause shift to right
(exothermic).
Decreasing V (increasing P) will cause
shift toward least number of moles of
gas (no effect).
Increasing amount of reactant will
cause shift to right.
Decreasing amount of product will
cause shift to right.
END OF SCANTRON PORTION OF EXAM
3
Chemistry 112
Exam 3C
Harwood/Fall 04
KEY
14.
(a) What is the shape (molecular geometry) of CHCl3? (Circle one and show work.) 4 pts
tetrahedral
linear
bent
Cl
|
Cl – C – Cl
|
H
(b) Is PCl3 polar or nonpolar (C is the central atom)? (Show work)
!
Polar
••
Cl – P – Cl
|
4 pts
Cl
(c) List all intermolecular forces present in a sample of PCl3?
dipole-dipole
dispersion
4 pts
!
15. Consider the following endothermic reaction:
NH4HS(s) ↔ NH3(g) + H2S(g)
(a) NH4HS(s) is placed in a vessel and decomposes at 22°C. After equilibrium is
reached, 0.32 atm of both NH3 and H2S are present along with solid NH4HS.
What is the equilibrium constant, KP, for this reaction?
4 pts
KP = (0.32)(0.32)
KP = 0.10
(b) Suppose that 0.128 atm of pure NH3(g) and 0.128 atm of pure H2S(g) are placed in a
5.00 L flask at 50°C (KP @ 50°C = 0.52). What are the partial pressures of NH3 and
H2S at equilibrium?
5 pts
NH4HS(s)
NH3(g)
↔
+
H2S(g)
Initial
----
0.128
0.128
Change
---
–x
–x
Equilibrium
---
0.128 – x
0.128 – x
2
0.52 = (0.128 – x) = 0.016384 – 0.256x + x
2
2
x – 0.256x – 0.503616 = 0
x = 0.849, –0.593
PNH3 = PH2S = 0.128 – (–0.593) = 0.721 atm
(c) Will a decrease in temperature favor the decomposition of NH4HS?
4 pts
No
(d) Will an increase in volume favor the decomposition of NH4HS?
4 pts
Yes
4
Chemistry 112
Exam 3C
Harwood/Fall 04
KEY
BONUS QUESTION
16. Lysine, one of the amino acid building blocks found in proteins, contains 49.29% C,
9.65% H, 19.16% N, and 21.89% O by elemental analysis. A solution prepared by
dissolving 30.0 mg of lysine in 1.200 g of the organic solvent biphenyl, gives a freezingpoint depression of 1.37°C. What is molecular formula of lysine? (Lysine does not
dissociate.) (Kf for biphenyl is 8.00°C/m)
10 pts
C:
H:
N:
O:
(49.29 g)(1 mol/12.011 g) = 4.104 mol ==> 3
(9.65 g)(1 mol/1.0079 g) = 9.5744 mol ==> 7
(19.16 g)(1 mol/14.0067 g) = 1.36792 mol ==> 1
(21.89 g)(1 mol/15.9994 g) = 1.36818 mol ==> 1
Empirical formula: C3H7NO
Molar mass = 73 g/mol
ΔTf = Kfm
1.37°C = (8.00°C/m) m
m = 0.17125 mol / kg
(1.200 × 10
–3
kg)(0.17125 mol/ kg) = 2.055 × 10
Molar mass = (30.0 × 10
–3
g) / (2.055 × 10
–4
–4
mol
mol) = 146 g/mol
(146 g/mol)/(73 g/mol) = 2
Molecular formula = C6H14N2O2
5