Name: ________KEY________________
Section: _______________
CHM 152 Exam 1: Chapters 14 (Kinetics) and 15 (Equilibrium)
Show all work and clearly mark your final answers!
1. (8 pts) Write the correct letter for or determine the value of each
part below using the graph shown on the right:
a. Enthalpy of reaction (H): __F___
b. Activation energy (Ea): __D___
c. Value of Ea: __40 kJ__
d. Transition state: ___B__
2. (8 pts) Consider the hypothetical reaction:
3X2(g) + 2Y(g) → 2X3Y(g) . Write the rate expression
for this equation and calculate the rate of X2 if the rate of X3Y = 8.0 X 10 3 M/s.
−
rate  
1 [X 2 ]
1 [X 3 Y]

3 t
2 t
∆[𝑋3 ]
∆[𝑌]
∆[𝑋3 𝑌]
= +
=+
3∆𝑡
2∆𝑡
2∆𝑡
Rate of X2 = -(3/2) * (8.0 x 103 M/s) = - 1.2 x 104 M/s
3. (6 pts) The rate law for the reaction 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) is rate = k [NO] 2 [H2] .
If the rate constant for this reaction is k = 0.739 M-2 ∙ s-1, what is the reaction rate when [NO] = 0.148 M and
[H2] = 0.356 M ?
Rate = (0.739 M-2 ∙ s-1) [NO] 2 [H2] = (0.739 M-2 ∙ s-1) (0.148 M) 2 (0.356 M) = 5.77 x 10 –3 M ∙ s -1
4. (5 pts) Which one of the graphs below shows the correct function of concentration versus time to yield a
linear plot for a first order reactant? Circle your answer.
CHM 152, Spring 2013
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Name: ________KEY________________
Section: _______________
5. (8 pts) A second-order reaction has a half-life of 1.20 x 102 seconds. If the initial concentration of the
reactant is 1.25 M, what is the reactant concentration after 47.0 seconds?
t1 
2
1
k [ A]0
where t½ = 1.20x102 s; [A]0 = 1.25 M.
k = 1 / (120 s *1.25 M) = 6.66667 x 10 −3 M -1 ∙ s -1
Now that the value for the 2nd-order rate constant, k, is calculated, plug this and the given time, 47 seconds,
into the integrated rate law for a 2nd-order reaction:
1
1
= kt +
[A]t
[A]0
= (6.6667 ´10-3 )(47) +
1
= 1.1149
And finally, [A]t = 1 / 1.1149 = 0.898M
1.25
__A_ 6. (5 pts) All of the following statements are true EXCEPT
a. In a series of stepwise reactions, the rate-determining step is the fast one.
b. The rate constant for a reaction can be changed by changing the temperature.
c. The rates of most chemical reactions change with time.
d. The rate constant is independent of the reactant concentrations.
7. (6 pts) A student gathered data for the reaction H2(g) + I2(g) ---> 2HI(g) at four different temperatures.
From a known rate law, the student was able to calculate the rate constant at those four different temperatures.
The student plotted (1/T) on the x-axis and (ln k) on the y-axis. The computer-generate best-fit line returned a
slope value of -1.95 x 104 K and a y-intercept value of +2.739. Calculate the activation energy (in kJ/mol) for
this reaction.
y = mx + b
m = slope
m = -(Ea/R)
Ea = m * -R
Ea = - (- 1.95 x 104 K) (8.314 J/K mol)
Ea= 1.62123 x 105 J/mol = 162 kJ/mol
8. (6 pts) Consider the two-step mechanism shown below for a given reaction.
Step 1: NO (g) + O2 (g)  NO3 (g)
Step 2: NO3 (g) + NO (g)  2NO2 (g)
What is the overall/net equation? ____ 2 NO(g) + O2 (g)  2 NO2 (g)________
If the overall rate law is: rate = k[NO]2[O2], which step is the rate-determining step? ___2_______
9. (10 pts) Write equilibrium expressions, Kc and Kp, for the following balanced equations:
N2 (g) + 2 H2 (g)  N2H4 (g)
Kc = [N2H4] / [N2][H2]2
Kp = (PN2H4) / (PN2)(PH2)2
Hg22+(aq) + 2 Cl-(g)  Hg2Cl2(s)
Kc = 1 / [Hg22+][Cl-]2
Kp = 1 / (PCl-)2
CHM 152, Spring 2013
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Name: ________KEY________________
Section: _______________
__B_ 10. (5 pts) Consider the following equation and equilibrium constant at a given temperature:
N2O4(g)  2 NO2(g), Kc = 0.37. What is the value of Kc for 2 NO2(g)  N2O4(g) at the same temperature?
a. 1.6
b. 2.7
c. 9.0x102
d. 18
e. 3.2x102
__A_ 11. (5 pts) For the following reaction at 25oC, 2 SO2 (g) + O2 (g) ⇋ 2 SO3 (g), Kc = 756. What is Kp?
a. 30.9
b. 179
c. 756
d. 1.85x104
e. Not enough information
n;
-1

Kp = Kc(RT) Kp = (756)(0.08206*298)
12. (10 pts) Calculate all equilibrium concentrations of a 0.250 M hydrazoic acid (HN3) solution?
HN3(aq) + H2O(l) ⇋ H3O+(aq) + N3-(aq), Kc = 1.9 x 10-5 .
Initial
Change
Equilibrium
Equilibrium
HN3(aq) +
0.250 M
-x
0.250 – x
0.248 M
H2O(l)
-
⇋ H3O+(aq) +
0
+x
x
2.2. x 10-3 M
Ka = x2 / 0.250 = 1.9 x 10-5
assume x <<0
check x: (2.18 x 10-3 M / 0.250 M) x 100% = 0.872%
N3-(aq)
0
+x
x
2.2. x 10-3 M
x2 = 4.75 x 10-6
Ka = x2 / (0.250 – x)
= 1.9 x 10-5
x = 2.18 x 10-3 M
[HN3] = 0.250 – 2.18 x 10-3 M = 0.248 M
Use the reaction 2NO (g) + O2 (g)  2 NO2 (g), H = -374 kJ and Kc = 6.9 x 105 at 500 K for # 13 – 16.
13. (2 pts) Is this system
reactant-favored
or
product-favored
? Circle your answer.
14. (6 pts) If a closed container is filled with 0.012 M NO, 0.20 M O2, and 0.16 M NO2, is the system at
equilibrium? Show your work (a simple yes or no answer is NOT sufficient)!
Qc = [NO2]2 / ([NO2][O2]) = (0.16 M)2 / ( (0.012 M)2 (0.20 M) = 889
Qc < Kc; these concentration values indicate that the system is NOT at equilibrium
15. (4 pts) If it is NOT at equilibrium (based on your answer to #14), in which direction will it need to shift to
reach equilibrium? Explain why!
This system needs to shift to the right () to reach equilibrium. The ratio of products to reactants will
increase until K is reached. The ratio increases when the product concentration increases.
16. (6 pts) which of the following stresses to the system will cause it to shift in the direction indicated in the
above question? Circle all that apply.
a. Add [NO]
b. Add a catalyst
c. Increase temperature
d. Increase volume
e.
f. Increase pressure
Remove [NO2]
CHM 152, Spring 2013
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Name: ________KEY________________
Section: _______________
_______________________________________________________________
Extra Questions:
2. For the reaction: 2 A + 3 B  C + 1/2 D + 2 E; the rate, with respect to A, was experimentally
determined to be 0.0185 Mmin-1. What would the rate be with respect to D?
A: 0.0185 Mmin-1
B: 0.00925 Mmin-1
C: 0.00463 Mmin-1
D: 0.037 Mmin-1
E: 0.074 Mmin-1
2. The rate law for the reaction 2 NO(g) + 2 H2(g) → N2(g) + 2 H2O(g) is rate = k [NO] 2 [H2] . If the
rate of this reaction is 5.63 x 10 –2 M ∙ s -1 when [NO] = 0.242 M and [H2] = 1.30 M, what is the reaction rate
when [NO] = 0.148 M and [H2] = 0.356 M ?
To solve this problem, plug in the values for rate, [NO] , and [H2], solving for the rate constant k. After
k is calculated, use this value and the new [NO] and [H2] values to determine the new rate.
rate = k [NO] 2 [H2] = 5.63 x 10 –2 M ∙ s -1 = k (0.242 M) 2 (1.30 M)…… k = 7.39 x 10 – 1
and the new rate is: 7.39 x 10 – 1[NO] 2 [H2] = 7.39 x 10 – 1 (0.148 M) 2 (0.356 M) = 5.77 x 10 –3 M ∙ s -1
3. When two compounds, A and B, are mixed together, they form compound C, by a reaction that’s not well
understood. Fortunately, the following rate information was experimentally determined, as shown
below:
[A] (mol/L)
0.050
0.10
0.050
[B] (mol/L)
0.050
0.050
0.10
Rate (mol/L.sec)
4.0 x 10-3
8.0 x 10-3
1.6 x 10-2
a)
Determine the rate law for this reaction.
b)
Determine the rate constant for this reaction.
The reaction A --> products is second order in A. Initially [A]0 = 1.00 M; and after 25 mins,
[A] = 0.25 M. What is the rate constant for this reaction?
CHM 152, Spring 2013
Leedy
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Name: ________KEY________________
Section: _______________
An elementary reaction, 2 A + C --> D, is second order in A and first order in C. The rate
of this reaction is 2.5 x 10-1 M/s when the concentrations of A, C, and D are all 1.00 mM. What is
the rate constant for the reaction?
concentration of A in a sample is 1.6 M, what will be the concentration of A after 80 min?
(A) 0.40 M (B) 0.20 M (C) 0.10 M (D) 0.050 M
3. The half-life of 214-Bi is 19.7 min. Starting with 1.00 x 10-3 grams of 214Bi, how many grams remain
after 59.1 min?
5.00 x 10-4
2.50 x 10-4
3.33 x 10-4
1.25 x 10-4
10. If the rate constant of a reaction increases by a factor of 5 when the temperature is increased from 22 to
35°C, what is the activation energy (Ea) of the reaction.
A: 93.5 kJ/mol
B: 923 J/mol
C: 792 J/mol
D: 53 kJ/mol
E: There is not enough information to answer the question.
8. The reaction mechanism for a reaction is as follows...
. Step1
. Step2
What is the overall reaction?
Identify the molecularity of each step.
Identify all that apply: reactants, products, , [reaction intermediates]
What is the rate law expression for this mechanism?
5: Researchers have proposed the following mechanism for a reaction:
Step 1: NO2 (g) + F2 (g)  NO2F (g) + F (g) (slow step)
Step 2: F (g) + NO2 (g)  NO2F (g) (fast step)
Which of the following statements is/are true?
i. F is an intermediate
ii. Step 1 is unimolecular
CHM 152, Spring 2013
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Name: ________KEY________________
Section: _______________
iii. The rate law for the overall reaction would be: Rate = k[NO2][F2]
iv. The overall reaction would be third order
A: i only
B: ii only
C: i and ii only
D: ii and iii
E: i and iii
For the following reaction, the rate law is found to be Rate = k[Ce4+][Mn2+].
One mechanism for this reaction, containing the following elementary steps, is shown below:
1. Ce4+ + Mn2+ ---> Ce3+ + Mn3+
2. Ce4+ + Mn3+ ---> Ce3+ + Mn4+
3.
Tl+ + Mn4+ ---> Tl3+ + Mn2+
What is/are the intermediates? _______________ What is the catalyst? _________
What is the overall equation? ________________________________
8. (4 pts) Which of the following equations represents the rate law for the following elementary process: A +
B ----> C + D?
a. Rate = k[C][D]
b. Rate = k[A]
c. Rate = k[A][B]2
d. Rate = k[A][B]
e. Rate = k[B]
6. The reaction X + Y 2M has Kc = 0.89 at 672 K. At equilibrium,
a) products predominate substantially
b) reactants predominate substantially
c) roughly equal molar amounts of products and reactants are present
d) only products exist
e) only reactants exist
What is the value of Kc for the reaction H2(g) + I2 (g) <===> 2 HI (g)
if the equilibrium concentrations for H2, I2, and HI are 0.15M, 0.033M, and 0.55M respectively?
A: 23
B: 111
C: 0.0090
D: 61.1
E: 0.016
37. A mixture of 2.0 mol of CO(g) and 2.0 mol of H2O(g) was allowed to come to equilibrium in a l L flask at
a high temperature. If Kc = 4.0, what is the molar concentration of H2(g) in the equilibrium mixture?
31. The value of the equilibrium constant K for a reaction at equilibrium is altered by
(A) changing the effective concentration of reactants.
(B) changing the effective concentration of products.
(C) changing the temperature.
(D) adding a catalyst.
(E) adding water.
CHM 152, Spring 2013
Leedy
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Name: ________KEY________________
Section: _______________
20. Consider the following reaction at equilibrium: 2NH3 (g)  N2 (g) + 3H2 (g), H = +92.4 kJ. Adding N2
(g) to the system at equilibrium will __________.
a) decrease the concentration of NH (g) at equilibrium
b) decrease the concentration of H (g) at equilibrium
c) increase the value of the equilibrium constant
d) cause the reaction to shift to the right
e) remove all of the H (g)
13. (4 pts) For which one of the following reactions are Kc and Kp the same?
a) H2(g) + Cl2(g)  2HCl(g)
b) 2SO3(g)  2SO2 (g) + O2 (g)
c) N2O4 (g)  2NO2 (g)
d) C(s) + CO2 (g)  2CO(g)
e) 2CO(g)  C(s) + CO2 (g)
15. For the vapor-phase reaction 2AZ  A2 + Z2 Kc = 16 at 523 K. If 0.030 mol AZ is introduced into an
evacuated 1.00 L vessel at 523 K, then at equilibrium [Z2] is _____________ M.
a) 0.003
b) 0.013
c) 0.017
d) 0.24
e) 0.0052
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