Chem 112
Question 1
6 Points
Spring 2008
Exam I Key
Whelan
Which of the following types of intermolecular forces are responsible for holding CH2Cl2
in the liquid phase? Check ALL those that apply.
Question 2
4 Points
Question 3
Which of the following compounds would have the highest boiling point?
3 Points
Question 4
6 Points
Question 5
6 Points
The vapor pressure of liquid octane, C8H18, is 40 mm Hg at 318K.
A sample of C8H18 is placed in a closed, evacuated container of constant volume
at a temperature of 460K. It is found that all of the C8H18 is in the vapor phase
and that the pressure is 54.0 mm Hg. If the temperature in the container is
reduced to 318K, which of the following statements are correct?
At 251K the vapor pressure of CS2 is 40 mm Hg, at 268K its vapor pressures is 100 mm
Hg. Determine the heat of vaporization for CS2 in kJ/mol.
Show Work
T1 = 252K
P1 = 40
T2 = 268
P2 = 100
Ln
100
40
=
∆Hvap
8.314
1
1
251
268
0.91629 =
∆Hvap
8.314
2.5272x10-4
KJ/mol
Question 6
6 Points
Question 7
6 Points
An aqueous solution is 18.0% by mass ammonia, NH3 (MM = 17.04 g.mol-1), and has a
density of 0.929 g/mL. Calculate the molarity of this solution.
Show Work
18g NH3 in 82g H2O making 100g Solution
100g Sol
moles NH3
V(L) Solution
M=
18.0g NH3
1 mol
17.04 g
1 mL
0.929 g
= 107.6 mL
= 1.06 mol NH3
9.81 M
Question 8
6 Points
An aqueous solution is 18.0% by mass ammonia, NH3 (MM = 17.04 g.mol-1), and has a
density of 0.929 g/mL. Calculate the molality of this solution.
Show Work
18g NH3 in 82g H2O making 100g Solution
moles of NH3
kg H2O
m=
18.0g NH3
1 mol
17.04 g
= 1.06 mol NH3
12.9 m
Question 9
4 Points
Match the following aqueous solutions with the appropriate letter from the column on
the right.
C
0.16m CH3CO2Na
– A. Lowest freezing point
B
0.09m Na3PO4
– B. Second lowest freezing point
A
0.19m KCl
– C. Third lowest freezing point
D
0.30m Nonelectrolyte
– D. The highest freezing point
Question 10
5 Points
What is the freezing point of a solution containing 13.4g of HOCH2CH2OH, a
nonelectrolyte (MM = 62.08 g.mol-1), dissolved in 802g of water? Kfp = -1.84 0Cm-1
∆TfP = i x KfP x m
13.4g
1 mol
62.08 g
m=
= 0.216 mol
0.216
0.802
= 0.269
∆TfP = 1 x (-1.84) x 0.269
-0.50
Question 11
4 Points
0
C
For the gas phase decomposition of phosphine at 120oC
4PH3(g) Æ P4(g) + 6H2(g)
the average rate of disappearance of PH3 over the time period from t = 0 s to t = 30 s is
found to be 1.33x10-3 Ms-1. The average rate of formation of H2 over the same time
period is
1.99x10-3 M.s-1
2 NO + O2 Æ 2 NO2
Question 12
6 Points
Exp
1
2
3
4
Question 13
6 Points
[NO]0, M
0.31
0.31
0.62
0.70
[O2]0, M I.R., Ms-1
1.45
1200
2.90
2400
2.90
9600
1.35
?
1. The overall order of the reaction is:
3
2. The rate constant k:
8612
3. The Initial Rate for Exp 4:
5697
The gas phase decomposition of N2O5 at 335K is first order in N2O5 with a rate constant
of 4.71x10-3 s-1. If the initial concentration of N2O5 is 0.208M, how long (in seconds)
will it take for the N2O5 concentration to reach 0.0291M?
Show Work
ln[R]t = -kt + ln[R]0
[R]0 = 0.208; [R]t = 0.0291; k = 4.71x10-3
ln(0.0291) = -4.71x10-3t + ln(0.208)
-3.537 = -4.71x10-3t –1.570
-4.71x10-3t = -2.037
418 Seconds
Question 14
6 Points
In a study of the decomposition of ammonia on a tungsten surface at 1100oC
2 NH3 Æ N2 + 3 H2
The concentration of NH3 was followed as a function of time. It was found that a graph
of [NH3] versus time in seconds gave a straight line with a slope of -4.96x10-6 Ms-1 and a
y-intercept of 7.68x10-3 M.
Based on this plot, the reaction is
a. The reaction is 0 order in NH3.
b. The rate constant for the reaction is 4.96x10-6 Ms-1.
c. The initial concentration of NH3 was 7.68x10-3 mol.L-1
Question 15
A reaction profile A Æ C is depicted below. Check the statement(s) that are correct.
6 Points
Question 16
A plot to determine the activation energy by
measuring the rate constant at different
temperatures is depicted on the left.
6 Points
a. Check the correct label on the plot for the
X and Y axis.
Question 17
6 Points
b. Activation Energy:
101
-1
(in kJ.mol )
At 400K the rate constant for a reaction whose activation energy is 250.0kJ.mol-1 is
found to be 0.243s-1. At what temperature would the rate constant equal 1.52s-1?
Show Work
T1 = 400K
k1 = 0.243
T2 = ?
K2 = 1.52
Ln
1.52
=
0.243
1.8334
250,000
8.314
= 30,070
1
400
0.0025 -
1
T2
1
T2
1.8334
=
-73.342 = -
75.175 -
30,070
T2
30,070
T2
K
Question 18
8 Points
The decomposition of hydrogen peroxide in the presence of potassium iodide is believed
to occur by the following mechanism:
Step 1
Slow
H2O2 + IÆ
H2O + OIStep 2
Fast
H2O2 + OI- Æ
H2O + O2 + I1.
Give the equation for the overall reaction?
Use the smallest integer coefficients possible.
2 H2O2 Æ 2 H2O + O2
2. Which species acts as a catalyst?
I-
3. Which species acts as an intermediate?
OI-
4. The rate law that is consistent with this mechanism:
Do Not Write Below This
Exam I Score
Rate = k[H2O2][I-]