Name_____[KEY]_________________________
Date:______________
Period:______
[KEY] Review Unit 7 Test (Chp 14): Chemical Kinetics (Free Response) [KEY]
Section II Free Response
Calculator Allowed
CLEARLY SHOW THE METHODS USED AND STEPS INVOLVED IN YOUR ANSWERS.
It is to your advantage to do this, because you may earn partial credit if you do and little or no credit if
you do not. Attention should be paid to significant figures.
Your responses to these questions will be graded on the basis of the accuracy and relevance of the
information cited. Explanations should be clear and well organized. Examples and equations may be
included in your responses where appropriate. Specific answers are preferable to broad, diffuse responses.
X → 2Y + Z
1.
The decomposition of gas X to produce gasses Y and Z is represented by the equation above. In a
certain experiment, the reaction took place in a 5.00 L flask at 428 K. Data from this experiment
were used to produce the information in the table below, which is plotted in the graphs that follow.
Time
(minutes)
0
10
20
30
50
70
100
[X]
(mol∙L–1)
0.00633
0.00520
0.00427
0.00349
0.00236
0.00160
0.000900
ln [X]
-5.062
-5.259
-5.456
-5.658
-6.049
-6.438
-7.013
[X]–1
(L∙mol–1)
158
192
234
287
424
625
1,110
1
(a) How many moles of X were initially in the flask?
[X] at 0 minutes is 0.00633 M, so…
5.00 L x 0.00633 mol X = 0.0317 mol X
1L
(b) How many molecules of Y were produced in the first 20. minutes of the reaction?
[X] at 20 minutes is 0.00427 M, so after 20. minutes of reaction, the number of moles of X remaining in
the flask is
5.00 L x 0.00427 mol X = 0.0214 mol X (left or remaining)
1L
So the number of moles of X that reacted in the first 20. minutes is
(0.0317 mol X) – (0.0214 mol X) = 0.0103 mol X (reacted or consumed)
(initial @ t0)
(final @ t20)
The number of molecules of Y produced in the first 20. minutes is
0.0103 mol X reacted x 2 mol Y produced x 6.02 x 1023 molecules Y = 1.24 x 1022 molecules Y
1 mol X reacted
1 mol Y
(c) What is the order of this reaction with respect to X? Justify your answer.
The reaction is first order with respect to X because a plot of ln X versus time produces a
straight line with a negative slope.
(d) Write the rate law for this reaction.
rate = k [X]1
(e) Calculate the specific rate constant for this reaction. Specify units.
ln [X]t – ln [X]o = –kt
From the first two data points in the plot of [X] vs time:
ln [0.00520] – ln [0.00633] = –k(10.0)
k = 0.0197 min–1
(f) Calculate the concentration of X in the flask after a total of 150. minutes of reaction.
ln [X]t – ln [X]o = –kt
ln [X]150 – ln [0.00633] = –(0.0197 min–1)(150)
ln [X]150 = –8.017
[X]150 = e–8.017
[X]150 = 3.30 x 10–4 M
2
2 NO(g) +2 H2(g)  N2(g) + 2 H2O(g)
2.
Experiments were conducted to study the rate of the reaction represented by the equation above.
Initial concentrations and rates of reaction are given in the table below.
Experiment
1
2
3
4
Initial Concentration
(mol/L)
[NO]
[H2]
0.0060
0.0060
0.0010
0.0020
0.0010
0.0020
0.0060
0.0060
Initial Rate of
Formation of N2
(mol/L∙ min)
1.8 10–4
3.6 10–4
0.30 10–4
1.2 10–4
(a)
(i) Determine the order for each of the reactants, NO and H2, from the data given and
show your reasoning.
expt. 1 & 2 held [NO] constant while [H2] doubled and the rate doubled, therefore the rate is
1st order with respect to [H2].
expt. 3 & 4 held [H2] constant while [NO] doubled and the rate quadrupled, therefore the rate is
2nd order with respect to [NO].
OR
rate = k [NO]x[H2]y
Exp2
Exp1
3.6 x 10–4 = k(0.0060)x(0.0020)y
1.8 x 10–4
k(0.0060)x(0.0010)y
2 = (2)y
Exp4
Exp3
y=1
1st order w.r.t. [H2]
1.2 x 10–4 = k(0.0020)x(0.0060)y
0.30 x 10–4 k(0.0010)x(0.0060)y
4 = (2)x
x=2
2nd order w.r.t. [NO]
(ii) Write the overall rate law for the reaction.
rate = k [NO]2[H2]
(b) Calculate the value of the rate constant, k, for the reaction. Include units.
using expt. 1,
rate = k [NO]2[H2]
1.810–4 mol/L.min = k (6.010–3 mol/L)2 (1.010–3 mol/L)
k = 5.0103 L2.mol–2.min–1 or 5.0x103 M–2min–1
3
(c) For experiment 2, calculate the concentration of NO remaining when exactly one-half of
the original amount of H2 had been consumed.
[H2] = ½ (0.0020 M) = 0.0010 M H2 consumed or reacted
2 NO(g) +2 H2(g)  N2(g) + 2 H2O(g)
0.0010 M H2 x 2 mol NO = 0.0010 M NO consumed or reacted
2 mol H2
[NO] = (0.0060 M) – (0.0010 M) = 0.0050 M
(initial)
(reacted)
(remaining)
(d) The following sequence of elementary steps is a proposed mechanism for the reaction.
I.
II.
III.
NO + NO  N2O2
The rate law determined in part (a) is
rate = k [NO]2[H2]
N2O2 + H2  H2O + N2O
Step I is not the RDS b/c then the rate law would be
rate = k [NO]2
N2O + H2  N2 + H2O
Based on the data presented, which of the above is the rate-determining step? (circle it)
Show that the mechanism is consistent with
(i) the observed rate law for the reaction, and
Based on the rate-determining step II. N2O2 + H2  H2O + N2O (slow),
the rate law is
rate = k2 [N2O2] [H2]
[N2O2] (an internediate) is substituted for by the following process:
NO + NO  N2O2 (Step I. is a fast equilibrium step)
ratef = rater
kf[NO]2 = kr[N2O2]
therefore
therefore
kf/kr [NO]2 = [N2O2]
Substuting into the rate law
rate =

kf
kr
(at equilibrium rates are equal)

rate = k2 [N2O2] [H2]
[NO]2 (k2 [H2])
for [N2O2], the rate law becomes:
kf/kr  k2 = k therefore,
rate = k [NO]2[H2]
(ii) the overall stoichiometry of the reaction.
I.
NO + NO  N2O2
II.
N2O2 + H2  H2O + N2O
III.
N2O + H2  N2 + H2O
------------------------------------------------2 NO + 2 H2  2 H2O + N2
Intermediates N2O2 and N2O cancel, but 2 moles of NO, H2 , H2O, and 1 mole of N2 remain in the overall rxn.
4
3.
The graphs below show results from a study of the decomposition of H2O2.
(a) Write the rate law for the reaction. Justify your answer.
rate = k[H2O2]1
A plot of ln[H2O2] versus time is a straight line, so the reaction follows simple first-order kinetics
(b) Determine the half-life of the reaction.
Using the graph showing [H2O2] versus time, the half-life is estimated to be about 650 minutes.
OR
Determine k from the slope of the graph of ln [H2O2] vs. time by picking 2 points:
point 1 (0 , 0.00)
point 2 (2800 , –3.00)
–k = y2-y2
x2-x1
Then, calculate from
t1/2 = 0.693
k
0.693
0.0011
–k = (–3.00 – 0.00)
(2800 – 0)
k = 0.0011
(which comes from
ln 0.5[H2O2]o – ln [H2O2]o = –kt )
= 630 min
(c) Calculate the value of the rate constant, k . Include appropriate units in your answer.
t1/2 = 0.693
k
650 = 0.693
k
0.693 = 0.0011 min–1
650
OR
k can be determined from the slope of the line in the ln [H2O2] vs. time as shown in part (b)
(d) Determine [H2O2] after 2,000 minutes elapse from the time the reaction began.
ln [H2O2]t – ln [H2O2]o = –kt
ln [H2O2]t – 0.00 = –(0.0011)(2000)
ln [H2O2]t = –2.2
[H2O2] = e−2.2
[H2O2] = 0.11 M
OR
From the graph of ln[H2O2] versus time,
ln[H2O2] = –2.2
[H2O2] = e−2.2
[H2O2] = 0.11 M
OR
From the graph of [H2O2] versus time, at 2000 min [H2O2] is approximately 0.12 M
5
4.
The reaction between NO and H2 is believed to occur in the following three-step process.
NO + NO  N2O2
(fast)
N2O2+ H2  N2O + H2O (slow)
N2O + H2  N2 + H2O
(fast)
(a) Write a balanced equation for the overall reaction.
2 NO + 2 H2  N2 + 2 H2O
(b) Identify the intermediates in the reaction. Explain your reasoning.
N2O2 and N2O; they are part of the mechanism but are neither original reactants nor products in
the overall reaction because they are produced in one step, but consumed in another step in the process.
(c) From the mechanism represented above, a student correctly deduces that the rate law for
the reaction is rate = k[NO]2[H2]. The student then concludes that
(1) the reaction is third-order and
(2) the mechanism involves the simultaneous collision of two NO molecules and one H2 molecule.
Are conclusions (1) and (2) correct? Explain.
conclusion (1) is correct; the sum of the exponents in the rate law (2 + 1) = 3, the overall order.
conclusion (2) is incorrect; the three steps in the mechanism are all bimolecular collisions.
(d) Explain why an increase in temperature increases the rate constant, k, given the rate law in (c).
An increase in temperature increases the rate and since there is no increase in concentrations,
then considering the rate law, rate = k[NO]2[H2] , the rate constant, k, must increase.
OR
Increasing temperature will increase the average kinetic energy of the reactant particles which will
increase the frequency of collisions and the energy of collisions.
This will increase the frequency of successful collisions with sufficient activation energy and proper
orientation required to react.
6
Answer Key
ln [X]t – ln [X]o = –kt
1.
2005B #3(c)-(f)
rate law from 3 graphs,
2.
1994 B (KINETICS)
rate law from data,
3.
2004B #3(b)
rate law from graph,
½ life,
4.
1996 D (KINETICS)
overall rxn,
mech
interm,
k & units,
mech
k
7