Kinetics Review

Chem I | Chem II AP
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1)
Explain how collisions of reactant molecules determine the rate of a reaction.
A)
What factors can increase the number of effective collisions in a reaction?
B)
What other factors influence reaction rate?
2)
Explain activation energy.
A)
How does activation energy affect the rate of a reaction?
B)
If activation energy is high, will the reaction be fast or slow?
3)
Look at a potential energy diagram that shows the effect of a catalyst.
A)
Explain how to find activation energy (Ea)
B)
Explain the purpose of a catalyst and how it changes Ea, ∆H and rate of a reaction.
4)
How does doubling the concentration of a first order reaction affect the rate?
A)
A second order reaction?
B)
A zero order reaction?
5)
What do you graph on the x axis and y axis to check the order of a reaction (linear test plot)?
Zero order =
vs.
First order =
vs.
Second order =
vs.
6)
Write the integrated rate law and the half life equation for the following types of reactions:
Zero order:
[A]t = [A]o - kt
t½ = [A]o / 2k
First order:
ln ( [A]o / [A]t ) = kt
t½ = .693 / k
Second order:
1/[A]t = 1/[A]o + kt
t½ = 1 / k[A]o
7)
A certain first-order reaction is 50% complete in 3.5 hours. What period of time is required for the reaction to be 88% complete?
t½ =
0.693
k
= 3.5 hr
k =
0.693
t½
=
0.693
3.5 hr
= 0.198 hr-1

Since the reaction is 50% complete in 3.5 hours, that is the half life. For the reaction to be 88% complete, there has to be 12% of the orginal amount left. I set the [A]o as 1 and the [A]t as 0.12 to reflect this.

ln (
[A]o
[A]t
) = ktln(
1
0.12
)= (0.198 hr-1)(t)
8)
A reaction mechanism is proposed to consist of three steps:
2 A + C → A2C
fast
A2C + B → A2BC
slow
A2BC + B → A2B2 + C
fast
A)
Write the balanced equation for the overall reaction.
+
B)
Does this reaction use a catalyst?
If so, identify the catalyst(s):
C)
Does this reaction contain an intermediate?
If so, identify the intermediate(s):
&
D)
Which step is the rate determining step?
E)
Write the rate law.
Rate =
9)
Given: 2 NOBr → 2 NO + Br2     rate = k[NOBr]2     k = 0.80 M-1 s-1
A)
Determine half life when [NOBr]o = 0.650 M.
seconds
B)
Calculate [NOBr] at t = 5.80 X 10-3 s if [NOBr]o = 0.650 M.
M
C)
If [NOBr]o = 1.00 M, how long would it take for 80% NOBr to react?
seconds
10)
The reaction 2 N2O5 → 4 NO2 + 5 O2 obeys the rate law: rate = Δ[N2O5] / Δt = k[N2O5]
A)
If the initial concentration of N2O5 is 4.0 x 10-2 M at the same temperature as above, what is the initial rate of the reaction? (k = 3.4 x 10-4 sec-1)
M/sec
B)
What is the half-life for this reaction?
seconds
C)
What is the concentration after 10 minutes?
M
D)
How much time will have passed when the concentration of N2O5 reaches 0.015 M?
seconds
11)
Observe the data in the chart for the following reaction and answer the questions.
A + B + C → D
[A]
[B]
[C]
Rate, M/sec
0.20
0.60
0.30
0.010
0.20
0.60
0.60
0.020
0.40
0.60
0.30
0.010
0.40
0.20
0.30
0.00037
A)
Determine the order for each reactant.
[A]
[B]
[C]
[A]: Look at reactions 1 & 3 when B and C are constant. A doubles and Rate stays constant, so it is zero order in A.
[B]: Look at reactions 3 & 4 when A and C are constant. B triples and the rate goes up by a factor of 27 (which is 33), so it is third order in B.
[C]: Look at reactions 1 & 2 when A and B are constant. C doubles and the rate also doubles, so it is 1st order in C.
B)
Write the rate law for this reaction.
Rate = k [A]
 [B]
 [C]
C)
Find the value and units for "k."
Rate = k [A]o [B]3 [C]
0.01 M/sec = k (0.6 M)o (0.40 M)3 (0.20 M) = 0.0648 M4
k = 0.154 M-3sec-1
D)
What is the rate when [A] = 0.60 M, [B] = 0.40 M & [C] = 0.20 M?
M/sec
Rate = k [A]o [B]3 [C]
Rate = (0.154 M-3sec-1) (0.40 M)3 (0.20 M) = 0.00197 M/sec
12)
The decomposition of N2O obeys zero order kinetics.
Given a rate constant = 2.46 X 10-3 M/s and [N2O] at 120 seconds is 0.155 M, calculate [N2O]o
M
[A]t = [A]0 - kt
0.155 M = [A]o - (0.00246 M/sec)(120 sec)
[A]o = 0.450 M
13)
Determine rate law from the following mechanism:
Make sure that you base your rate law on the SLOW step.
A)
A + B → C (slow)
C + E → F (fast)
Rate =
B)
A + B → C (fast)
C + E → F (slow)
Rate =
14)
Consider the hypothetical reaction: A + B → P. For each of the possible rate laws listed below, indicate the reaction order with respect to A, with respect to B, and the overall reaction order.
A)
Rate = k[A][B]
[A] =
[B] =
Overall =
B)
Rate = k[A]2
[A] =
[B] =
Overall =
C)
Rate = k[A][B]2
[A] =
[B] =
Overall =
15)
What units would each of the rate constants in question #14 have if concentration is expressed as M and rate as M/sec?
A)
Rate = k[A][B]
Units for k =
B)
Rate = k[A]2
Units for k =
C)
Rate = k[A][B]2
Units for k =
16)
Given the data and the graphs below determine the order of this reaction.
t(s)
[C2H6]
0
.01000
1000
.00629
2000
.00459
3000
.00361
What is the order of the reaction?
How do you know?
What is the value for k in this reaction?
k =
1
[A]t
=
1
[A]o
+ kt
1
0.00629 M
=
1
0.01 M
+ (k)(1000 sec)159 M-1 = 100 M-1 + (k)(1000 sec)59 M-1 = (k)(1000 sec)
What is [C2H6] after 5000 seconds?
[A]t =
M
1
[A]t
=
1
0.01 M
+ (0.059 M-1 sec-1)(5000 sec)
1
[A]t
= 395 M-1
[A]t = 0.00253 M
17)
Consider the following reaction: A + 2 B → C for which the following data is given:
[A]
[B]
Rate
0.2
0.1
7.0 x 10-5
0.4
0.1
7.0 x 10-5
0.2
0.2
28 x 10-5
A)
Write the rate law expression for this reaction.
Rate =
This reaction is 2nd order with B because in reactions 1 & 3, when B doubles, the rate increases by a factor of 4. The reaction is zero order with A because in reactions 1 & 2 when A doubles, the rate doesn't change.
B)
What is the order with respect to A?
With respect to B?
What is the overall reaction order?
C)
What is the numerical value for the rate constant with units?
From reaction 1 and Rate = k [B]2:
7.0 x 10-5 M/sec = k (0.1 M)2
k = 0.007 M-1 sec-1
D)
What is the rate of appearance of C when [A] = 0.1 and [B] = 0.3?
M/sec
Rate = k [B]2Since the reaction is zero order with A, we don't need A in the equation.
Rate = (0.007 M-1 sec-1) (0.3 M)2 = 6.3 x 10-4 M/sec
Multiple Choice:
18)
For the following reaction: A → B + C that is second order, a linear plot will result when time is plotted against:
19)
Which of the following statements is TRUE for the catalyzed and uncatalyzed versions of the same reaction?
20)
Catalysts lower the activation energy of a reaction by:
21)
For a first order reaction with a rate constant k = 3.13 x 102 s-1, how long does it take for the concentration of the only reactant to become 40% of the original amount?
ln (
[A]o
[A]t
) = kt
ln (
1
0.4
) = (313 sec-1) t
0.916 = (313 sec-1) tt = 0.00293 sec
22)
Increasing the temperature at which a reaction occurs speeds up the reaction by:
23)
The decomposition of a substance is found to be first order. If it takes 2.5 x 103 sec for the concentration of that substance to fall to half of its original value, find the rate constant.
sec-1
t1/2 =
0.693
k
2.5 x 103 sec =
0.693
k
k = 2.77 x 10-4 sec-1
24)
A reaction that is second order in A is 50.0% complete after 350 min. If [A] = 1.35 M, what is the rate constant?
M-1min-1
For a 2nd order reaction:t1/2 =
1
k [A]o
350 min =
1
k (1.35 M)
472.5 k = 1k = 2.12 x 10-3 M-1min-1
25)
The decomposition of NOBr is second order: 2 NOBr → 2 NO + Br2 The rate constant for the reaction at 10°C is 0.80 M-1sec-1. If [NOBr]o = 0.052 M, what is [NOBr] after the reaction has run for 1.0 min?
M
1
[A]t
=
1
[A]o
+ kt
1
[A]t
=
1
0.052 M
+ (0.80 M-1sec-1)(60 sec)= 67.2 M-1[A]t = 0.015 M
26)
Given: SO2Cl2 → SO2 + Cl2 First order with a rate constant of 2.2 x 10-5 sec-1 at 320°C. What percentage of a sample of SO2Cl2 will remain if it is heated for 5.00 hr at 320°C?
%

Because it is looking for the percentage remaining, I set [A]o equal to 100, so that when I find [A]t, it will automatically be a percentage of [A]o.

ln (
[A]o
[A]t
) = kt = (2.2 x 10-5 sec-1)(1.8 x 104 sec) = 0.396
100
[A]t
= 1.48[A]t = 67.6 %
27)
The rate constant for the first order reaction of the conversion of cyclopropane to propene at 500°C is 5.5 x 10-4 sec-1.
a)
Find the half-life of cyclopropane at 500°C.
M
t1/2 =
0.693
k
=
0.693
5.5 x 10-4 sec-1
= 1260 sec
b)
Given an initial cyclopropane concentration of 1.0 x 10-3 M at 500°C, find the concentration of cyclopropane that remains after 2.0 hr.
sec
ln [A]t = ln [A]o - kt = ln (1.0 x 10-3 M) - (5.5 x 10-4 sec-1)(7200 sec)
ln [A]t = -6.9 - 3.96 = -10.9
[A]t = 1.91 x 10-5 M
28)
Calculate the time required for the concentration to decrease to 1/10 of its initial value for a first order reaction with a k = 10.0 sec-1.
sec
ln (
[A]o
[A]t
) = kt
ln (
10
1
) = 10 t
2.3 = 10 tt = 0.23 sec