AP Chemistry: Kinetics Practice Problems
Directions: Write your answers to the following questions in the space provided. For problem solving, show all of
your work. Make sure that your answers show proper units, notation, and significant digits. Do not use a calculator
on the multiple choice questions.
1.
Three major methods used to increase the rate of a reaction are adding a catalyst, increasing the
temperature, and increasing the concentration of a reactant. From the perspective of collision theory,
explain how each of these methods increases the reaction rate.
a.
adding a catalyst
Adding a suitable catalyst lowers the energy demands of the reaction (the activation energy) by
providing a less demanding pathway. It does this by forming a different activated complex with a
lower activation energy.
b.
increasing the temperature
Increasing the temperature increases the average kinetic energy of the molecules. However, what
is most important here is that the fraction of molecules that have the energy necessary to react (the
activation energy) is increased.
c.
increasing the concentration of the reactants
Increasing the concentration of a reactant increases the probability of collisions, and therefore
increases the possibility of reaction, for the particles must collide to react. Note that not all
collisions result in successful reactions.
2.
Why do large crystals of sugar burn more slowly than finely ground sugar?
The rate of combustion is proportional to the surface area of sugar exposed to oxygen. Smaller crystals
have more surface area and burn faster.
3.
How do homogeneous catalysts and heterogeneous catalyst differ?
A homogeneous catalyst is a catalyst that exists in the same phase (liquid or gas) as the reactants. A
heterogeneous catalyst, on the other hand, is a catalyst that exists in a different phase form the reactants.
4.
Express the rate of reaction in terms of the rate of change of each reactant and each product in the
following.
a.
3ClO-(aq) → ClO3-(aq) + 2Cl-(aq)
rate of reaction = b.
3SO2(g) + O2(g) → 2SO3(g)
rate of reaction = c.
∆[SO 2 ]
∆[O 2 ] ∆[SO 3 ]
=−
=
∆t
3∆t
2 ∆t
C2H4(g) + Br2(g) → C2H4Br2(g)
rate of reaction = 5.
∆[ClO - ] ∆[ClO 3- ] ∆[Cl - ]
=
=
3∆t
2 ∆t
∆t
∆[C 2 H 4 ]
∆[Br2 ] ∆[C 2 H 4 Br2 ]
=−
=
∆t
∆t
∆t
In the Haber process for the production of ammonia,
N2(g) + 3H2(g) → 2NH3(g)
What is the relationship between the rate of production of ammonia and the rate of consumption of
hydrogen?
∆[H 2 ] ∆[NH 3 ]
rate of reaction = −
=
3∆t
2 ∆t
6.
The rate constant
A.
always shows an exponential increase with the Kelvin or absolute temperature.
B.
increases with increasing concentration.
C.
usually increases with increased pressure for gases.
D.
never changes (it is a constant).
E.
is the same for a given reaction at the same Kelvin or absolute temperature.
Answer: (E) In this question be sure to note that it is the rate constant that is involved. Refer to
the Arrhenius equation for the specific relationship.
1
7.
8.
What are the units for each of the following if the concentrations are expressed in moles per liter and the
time in seconds?
a.
rate of a chemical reaction
The units for rate are always mol/L•s.
b.
rate constant for a zero-order rate law
Rate=k; k must have units of mol/L•s
c.
rate constant for a first-order rate law
Rate=k[A], k must have units of s-1
d.
rate constant for a second-order rate law
Rate=k[A]2, k must have units of L/mol•s
The reaction
2I-(aq) + S2O82-(aq) → I2(aq) + 2SO42-(aq)
was studied at 25°C. The following results were obtained where
Rate = [S2O82-]0 (mol/L)
0.040
0.040
0.020
0.040
0.030
[I-]0 (mol/L)
0.080
0.040
0.080
0.032
0.060
a.
∆[S 2 O 8 2 − ]
∆t
Initial Rate (mol/L • s)
12.5×10-6
6.25×10-6
6.25×10-6
5.00×10-6
7.00×10-6
Determine the rate law.
Rate = k[I-]x[S2O82-]y
Compare experiments 1 and 2.
rate 2 k [I - ]2x [S 2 O 8 2 − ]2y
=
rate1 k [I - ] x1 [S 2 O 8 2 − ]1y
6.25 × 10-6
12.5 × 10-6
k ( 0.040) x ( 0.040) y
=
k ( 0.080) x ( 0.040) y
, 0.5 = 0.5 x , x = 1
rate 3 k [ I - ] x3 [S 2 O 8 2 − ]3y
=
rate1 k [ I - ] x1 [S 2 O 8 2 − ]1y
6.25 × 10-6
12.5 × 10-6
k ( 0.080) x ( 0.020) y
=
k ( 0.080) x ( 0.040) y
, 0.5 = 0.5y , y = 1
Rate = k[I-][S2O82-]
b.
Calculate a value for the rate constant for each experiment and an average value for the rate constant.
FG
IJ FG 0.040 mol IJ , k = 3.9 × 10
H
KH L K
mol
F 0.040 mol IJ FG 0.040 mol IJ , k = 3.9 × 10
= kG
H L KH L K
mol
F 0.080 mol IJ FG 0.020 mol IJ , k = 3.9 × 10
= kG
H L KH L K
mol
F 0.032 mol IJ FG 0.040 mol IJ , k = 3.9 × 10
= kG
H L KH L K
mol
F 0.060 mol IJ FG 0.030 mol IJ , k = 3.9 × 10
= kG
H L KH L K
Experiment 1 =
12.5 × 10-6 mol
0.080 mol
=k
L•s
L
Experiment 2 =
6.25 × 10-6
L•s
Experiment 3 =
6.25 × 10-6
L•s
Experiment 4 =
5.00 × 10-6
L•s
7.00 × 10-6
L•s
-3
Kmean = 3.9×10 L/mol•s
Experiment 5 =
2
-3
-3
-3
-3
-3
L
mol • s
L
mol • s
L
mol • s
L
mol • s
L
mol • s
9.
The following rate data were obtained at 25°C for the following reaction. What is the rate law expression?
What is the overall order of the reaction?
2A + B → 3C
Experiment
1
2
3
4
[A]0 (mol/L)
0.10
0.30
0.10
0.20
[B]0 (mol/L)
0.10
0.30
0.30
0.40
Initial Rate (mol/L • s)
2.0 × 10-4
6.0 × 10-4
2.0 × 10-4
6.0 × 10-4
Rate = k[A]x[B]y
Experiments 1 and 3:
[A] is constant = 0.10 M
[B] increases by a factor of 3 but the rate of reaction does NOT change
Therefore [B] does not affect the rate
Therefore the order of reaction with respect to B = y = 0
Experiments 2 and 3:
[B] is constant = 0.30 M
[A] decreases by a factor of 3 and the rate of the reaction decreases by a
factor of 3
Therefore the reaction is first order with respect to A = x = 1
Rate law expression: Rate = k[A]1[B]0 = k[A]
The overall order of reaction is (1+0) = 1
10.
What is the rate law expression for the following reaction, given the data below? What is the overall order
of the reaction?
2A + B + 2C → 3D + 2E
Experiment
1
2
3
4
[A]0 (mol/L)
0.20
0.20
0.20
0.10
[B]0 (mol/L)
0.10
0.30
0.10
0.60
[C]0 (mol/L)
0.10
0.20
0.30
0.40
Rate = k[A]x[B]y[C]z
Experiments 1 and 3:
Initial Rate (mol/L • min)
2.0 × 10-4
18.0 × 10-4
2.0 × 10-4
3.6 × 10-4
[A], [B] are constant, [C] increases by a factor of 3 but the rate of the
reaction does NOT change.
Therefore the reaction [C] does not affect the rate
Therefore the order of reaction with respect to C = z = 0 and we can
ignore this data.
Experiments 1 and 2:
[A] is constant = 0.20 M
[B] increases by a factor of 3
The rate increases by a factor of 9
Therefore the reaction is second order with respect to B = y = 2
There are no experiments where [A] varies and [B] remains constant, so we will have to evaluate the order
of the reaction with respect to A mathematically.
rate 4 k [A ]4x [ B]4y
=
rate 2 k [A ]2x [ B]2y
3.6 × 10-3
1.8 × 10-3
=
. ) x ( 0.6) 2
k ( 01
k (0.2) x ( 0.3) 2
2 = (0.5) x (2) 2
FG IJ
HK
x
2
1
=
4
2
x=1, therefore the reaction is first order with respect to A.
The rate law expression is Rate = k[A][B]2
The overall order is 1+2 = 3
3
11.
Use the rate law determined in question 9 to answer the following question. What would happen to the rate
of the reaction if the concentration of A was halved and the concentration of B was tripled during a
reaction?
Assume the initial concentrations of A and B were both 1 M.
Therefore Rate = k[A][B]2 = k(1)(1)2=k
After the change in the reaction, Rate = k(1/2)(3)2 = 9/2 k
Therefore the rate of the reaction would increase by a factor of 9/2.
12.
Rate data were collected for the following reaction at a particular temperature.
2ClO2(aq) + 2OH-(aq) → ClO3-(aq) + ClO2-(aq) + H2O(l)
Experiment
1
2
3
4
13.
[ClO2]0 (mol/L)
0.012
0.024
0.012
0.024
[OH-]0 (mol/L)
0.012
0.012
0.024
0.024
Initial Rate (M/s)
2.07 × 10-4
8.28 × 10-4
4.14 × 10-4
1.66 × 10-3
a.
What is the rate-law expression for this reaction?
Rate = k[ClO2]x[OH-]y
Experiments 1 and 3:
[ClO2] is constant.
[OH-] doubles
The rate doubles.
Therefore the reaction is first order with respect to B = y = 1
Experiments 3 and 4:
[ClO2] doubles
[OH-] is constant.
The rate quadruples.
Therefore the reaction is second order with respect to A= x = 2
Rate = k[ClO2]2[OH-]
b.
Calculate a value for k.
Using the data from Experiment 1 to calculate k, we have
2.07 × 10-4 M/s = k(0.012 M)2(0.012 M)
k = 1.2 × 102 M-2• s-1
c.
Describe the order of the reaction with respect to each reactant and to the overall order.
The reaction is second order with respect to ClO2, first order with respect to OH- and third order
overall.
Consider a chemical reaction between compounds A and B that is first order in A and first order in B.
From the information shown here, fill in the blanks.
Experiment
[A]
[B]
Rate (M•s-1)
1
0.20 M
0.050 M
0.24
0.030 M
0.20
2
3
0.40 M
0.80
Rate=k[A][B]
Use experiment 1 to find the value of k:
Rate
0.24 M • s-1
=
= 24 M -1 • s −1
[A][B] (0.20 M )(0.050 M )
Then in experiment 2, we solve for [A] and in experiment 3, we solve for [B]
k=
[A] =
Rate
0.20 M • s −1
=
= 0.28 M
k [ B] (24 M -1 • s −1 )(0.030 M )
[ B] =
4
Rate
0.80 M • s −1
=
= 0.083 M
k [ A] (24 M -1 • s −1 )(0.40 M )
17.
14.
The rate of a chemical reaction
A.
is always dependent of the concentration of all reactants.
B.
is always increased with increasing temperatures.
C.
is directly proportional to the value of ∆E.
D.
is greater with higher activation energies.
E.
is independent of surface area.
Answer: (B) The collision model indicates that molecular collision is necessary for reaction.
Because an increase in temperature raises molecular velocities and the percentage of effective
collisions, the reaction rate increases.
15.
For the first-order reactions of different substances A and X
A → B t1/2 = 30.0 min
X → Y t1/2 = 60.0 min
This means that
A.
doubling the concentration of A will have ½ the effect on half-life that doubling the
concentration of B will have on its half-life.
B.
a certain number of grams of A will react twice as fast as the same number of grams of X.
C.
a certain number of grams of X will react twice as fast as the same number of grams of A.
D.
the rate constant for A → B is lower than the rate constant of X → Y.
E.
3 moles of A will react more rapidly than 3 moles of X.
Answer: (E) It takes half as much time for A to form B as for X to form Y, as seen by the smaller
half-life. Note that option “B” would be incorrect as the grams of A and the grams of X are not
the same number of moles.
16.
A reaction is first order with respect to [X] and second order with respect to [Y]. When [X] is
0.20 M and [Y] = 0.20 M the rate is 8.00 × 10-3 M/min. The value of the rate constant, including
correct units, is
A.
1.00 M min-1
B.
1.00 M-2 min-1
C.
2.00 M-1 min-1
D.
2.00 M-2 min-1
E.
8.00 × 10-3 min-3
Answer: (B) From these data, it follows that the rate law is Rate = k[X]Y]2. Solving for the rate
constant and substituting data for this reaction:
K = Rate/[X][Y]2
= 8.00 × 10-3 M/min/(0.200 M)(0.200 M)2
= 0.008 M/min/0.008 M3
= 1.00 M-2min-1
The rate constant for the decomposition of nitrogen dioxide
2NO2 → 2NO + O2
with a laser beam is 1.70 M-1· min-1. Find the time, in seconds, needed to decrease 2.00 mol/L of NO2 to
1.25 mol/L.
Since the units of the rate constant are M-1· min-1 , we know that the reaction is second order overall. Since
the reaction has only one reactant, NO2, the reaction is second order with respect to NO2. The integrated
rate equation for a reaction that is second order with respect to NO2 as the only reactant is
1
1
−
= akt where a = stoichiometric coefficient of NO 2
[ NO 2 ]t [ NO 2 ]0
1
1
Substituting,
−
= 2(1.70 L • mol -1 • min −1 )t
1.25 M 2.00 M
0.800 M −1 − 0.500 M −1
t=
= 0.0882 min or 5.29 s
(2)(1.70 L • mol -1 • min −1 )
18.
What is meant by the half-life of a reactant?
The half-life of a reactant is the time required for half of that reactant to be converted into products. For a
first order reaction, the half-life is independent of concentration so that the same time is required to
consume half of any starting amount or concentration of the reactant. On the other hand, the half-life of a
second-order reaction does depend on the starting amount of the reactant.
5
19.
The decomposition of hydrogen peroxide was studied, and the following data were obtained at a particular
temperature:
Time (s)
0
120.
300.
600.
1200.
1800.
2400.
3000.
3600.
assuming that Rate =
[H2O2] (mol/L)
1.00
0.91
0.78
0.59
0.37
0.22
0.13
0.082
0.050
∆H 2 O 2
determine
∆t
a.
the rate law
Use your graphing calculator to determine the rate law.
The plot of ln [H2O2] vs. time is linear. Thus, the reaction is first order.
The rate law is Rate=k[H2O2]
b.
the integrated rate law
ln[H2O2]t = -kt + - ln[H2O2]0
c.
the value of the rate constant
We determine the rate constant k by determining the slope of the ln[H2O2] vs time plot (slope = k). This can be done using your graphing calculator or by using the old fashioned method of using
two points on the curve.
By calculator: 8.35 × 10-4 s-1
Using two points on the curve gives:
∆y 0 − ( −3.00)
slope = − k =
=
= −8.3 × 10 −4 s-1
∆x
0 − 3600
d.
Calculate the [H2O2] at 4000. s after the start of the reaction.
To determine [H2O2] at 4000. s, use the integrated rate law where at t=0, [H2O2]0 = 1.00 M.
ln[ H 2 O 2 ]t − ln[ H 2 O 2 ]0 = − kt
ln[ H 2 O 2 ]t = ln[ H 2 O 2 ]0 − kt
ln[ H 2 O 2 ]t = ln[100
. ]0 − ( −8.3 × 10 −4 s-1 )4000. s = - 3.3
[ H 2 O 2 ] = e −3.3 = 0.037 M
20.
It took 143 s for 50.0% of a particular substance to decompose. If the initial concentration was 0.060 M
and the decomposition reaction follows second-order kinetics, what is the value of the rate constant?
1
1
For a second order reaction: t 1/2 =
or k =
k [A]0
t 1/2 [A]0
1
= 0.12 L / mol • s
k =
143 s(0.060 mol / L)
6
21.
The dimerization of butadiene
2C4H6(g) → C8H12(g)
was studied at 500. K, and the following data were obtained:
[C4H6] (mol/L)
1.6 × 10-2
1.5 × 10-2
1.3 × 10-2
1.1 × 10-2
0.68 × 10-2
Time (s)
195
604
1246
2180
6210
Assuming that
Rate = -
∆[C 4 H 6 ]
∆t
determine the
a.
form of the rate law
Use your graphing calculator to graph the data.
L1=time, L2=[C4H6], L3=ln[C4H6], L4=1/[C4H6]
Make linear regressions for L1, L2; L1, L3; and L1, L4 and identify the equation with the best “r”.
Since the second order plot of 1/[C4H6] vs. t is linear, then we can conclude that the reaction is
second order in butadiene. The rate law is:
Rate = k[C4H6]2
b.
the integrated rate law
1
1
−
= kt
[ C 4 H 6 ]t [ C 4 H 6 ] 0
which can be written as
1
1
= kt +
[ C 4 H 6 ]t
[ C 4 H 6 ]0
c.
the rate constant
The slope of the straight line equals the value of the rate constant. You can either use two points
on the line and determine the ∆y/ ∆x or obtain the value from your graphing calculator.
The slope equals 0.014 L/mol·s.
22.
A certain first-order reaction is 45.0% complete in 65 s. What are the rate constant and the half-life for this
process?
If [A]0 = 100.0, then after 65 s, 45.0: of A has reacted or [A] = 55.0. For first order reactions:
ln[A]t - ln[A]0 = -kt
ln[55.0] - ln[100.0] = -k(65 s)
k = 9.2 × 10-3 s-1
0.693
0.693
=
= 75 s
t1/ 2 =
k
9.2 × 10 −3 s-1
23.
The rate law for the decomposition of phosphine (PH3) is
Rate = -
∆[PH 3 ]
= k [ PH 3 ]
∆t
It takes 120. s for 1.00 M PH3 to decrease to 0.250 M. How much time is required for 2.00 M PH3 to
decrease to a concentration of 0.350 M?
ln[ PH 3 ]t − ln[ PH 3 ]0 = − kt
ln[0.250]t − ln[1.00]0 = − k (120 s)
k = 0.0116 s-1
ln[0.350]t − ln[2.00]0 = −0.0116 s-1 (t )
t = 150. s
7
24.
The rate of the reaction
NO2(g) + CO(g) → NO(g) + CO2(g)
depends only on the concentration of nitrogen dioxide below 225°C. At a temperature below 225°C, the
following data were collected:
Time (s)
0
1.20 × 103
3.00 × 103
4.50 × 103
9.00 × 103
1.80 × 104
[NO2] (mol/L)
0.500
0.444
0.381
0.340
0.250
0.174
Determine the
a.
rate law
Use your graphing calculator to graph the data.
L1=time, L2=[NO2], L3=ln[NO2], L4=1/[NO2]
Make linear regressions for L1, L2; L1, L3; and L1, L4 and identify the equation with the best “r”.
Since the second order plot of 1/[NO2] vs. t is linear, then we can conclude that the reaction is
second order in NO2. The rate law is:
Rate = k[NO2]2
b.
integrated rate law
1
1
−
= kt
[ NO 2 ]t [ NO 2 ]0
which can be written as
1
1
= kt +
[ NO 2 ]t
[ NO 2 ]0
c.
value of the rate constant.
The slope of the plot 1/[NO2] vs. t gives the value of k.
Slope = 2.09 × 10-4 L/mol·s
d.
Calculate the [NO2] at 2.70×104 s after the start of the reaction.
1
2.08 × 10 −4 L
1
=
× 2.70 × 10 4 s +
[ NO 2 ]t
mol • s
[0.500]0
1
.
= 7.62, [NO 2 ]t = 0131
M
[ NO 2 ]t
25.
26.
A reaction mechanism
A.
is the sum of all steps in a reaction except the rate determining step.
B.
has a ∆H equal to the ∆H for the most demanding step.
C.
always has a rate determining step.
D.
may be absolutely proven from the rate law.
E.
is determined from the balanced expression only.
Answer: (C) In all series of steps (the mechanism for the overall reaction), there must always be a
slowest step. The rate determining step is the slowest step.
Write the rate laws for the following elementary reactions.
a.
CH3NC(g) → CH3CN(g)
Rate = k[CH3NC]
b.
O3(g) + NO(g) → O2(g) + NO2(g)
Rate = k[O3][NO]
8
27.
Most reactions occur by a series of steps. The energy profile for a certain reaction that proceeds by a twostop mechanism is
On the energy profile, indicate
a.
The positions of reactants (R) and products (P).
b.
The activation energy for the overall reaction (Ea).
c.
∆E for the reaction.
d.
Which point on the plot represents the energy of the intermediate in the two-step reaction?
e.
Which step in the mechanism for this reaction is rate determining, the first or the second step?
Explain. In a mechanism, the rate of the slowest step determines the rate of the reaction. The
activation energy for the slowest step will be the largest energy barrier that the reaction must
overcome. Since the second hump in the diagram is at the highest energy, then the second step
has the largest activation energy and will be the rate determining step (the slow step).
28.
A proposed mechanism for a reaction is
C4H9Br → C4H9+ + BrC4H9 + H2O → C4H9OH2+
C4H9OH2+ + H2O → C4H9OH + H3O+
29.
Slow
Fast
Fast
a.
What is the overall balanced equation for the reaction?
C4H9Br + 2H2O → C4H9OH + Br- + H3O+
b.
What are the intermediates in the proposed mechanism?
C4H9+ and C4H9OH2+
c.
Write the rate law expected for this mechanism.
Since step 1 is the rate determining step, the rate law for this mechanism is:
Rate = k[C4H9Br]
The mechanism for the reaction of nitrogen dioxide with carbon monoxide to form nitric oxide and carbon
dioxide is though to be
NO2 + NO2 → NO3 + NO
NO3 + CO → NO2 + CO2
Slow
Fast
a.
What is the overall balanced equation for the reaction?
NO2 + CO → NO + CO2
b.
What is the intermediate in the proposed mechanism?
NO3
c.
Write the rate law expected for this mechanism.
Since step 1 is the rate determining step, the rate law for this mechanism is:
Rate = k[NO2]2
9
30.
The ozone, O3, of the stratosphere can be decomposed by reaction with nitrogen oxide (commonly called
nitric oxide), NO from high-flying jet aircraft.
O3(g) + NO(g) → NO2(g) + O2(g)
The rate expression is rate = k[O3][NO]. Which of the following mechanisms are consistent with the
observed rate expression?
a.
NO + O3 → NO3 + O
slow
Consistent, Slow first step determines the order.
NO3 + O → NO2 + O2
fast
O3 + NO → NO2 + O2
overall
31.
b.
NO + O3 → NO2 + O2
slow, one step
Consistent, One step reaction mechanism
c.
O3 → O2 + O
O + NO → NO2
O3 + NO → NO2 + O2
slow
fast
overall
Not consistent, slow first step, no dependence on
[NO] is predicted by this mechanism.
d.
NO → N + O
O + O3 → 2O2
O2 + N → NO2
O3 + NO → NO2 + O2
slow
fast
fast
overall
Not consistent, slow first step, no dependence on
[O3] is predicted by this mechanism.
e.
NO N + O
O + O3 → 2O2
O2 + N → NO2
O3 + NO → NO2 + O2
fast, equilibrium
slow
fast
overall
Consistent, slow second step predicts the first order
dependence on O, But O is produced in a previous
fast step, which predicts the first order dependence on
NO.
A proposed mechanism for the following reaction, H2 + I2 → 2HI is
I2 2I
I + H 2 H 2I
H2I + I → 2HI
fast, equilibrium
fast, equilibrium
slow
a.
Identify any reaction intermediates in this proposed mechanism.
The intermediates are I, H2I
b.
Show that this mechanism predicts the correct rate law, rate = k[H2][I2].
From step 3 (the slow, rate-determining step):
Rate = k3[H2I][I]
From step 1, Rate1f = Rate1r or k1r[I2] = k1r[I]2
k1f [I 2 ]
rearranging, [I] =
k1r
From step 2, Rate2f = Rate2r or k2r[I][H2] = k1r[IH2]
k [I][H 2 ]
rearranging, [IH 2 ] = 2f
k 2r
Substituting the expression for [I] from step 1 into this latter expression,
k 2f
k1 f [ I 2 ]
k1r
[IH 2 ] =
[H 2 ]
k 2r
Substituting this expression for [I] and [IH2] into the rate law deduced from step 1, and the
combining terms,
k 2f
Rate = k 3 ×
k1 f [ I 2 ]
k1r
k 2r
[H 2 ]
×
k1 f [ I 2 ]
k1r
=
k 3 k 2 f k1f
[ H 2 ][I 2 ]
k 2r k1r
Rate = k[H2][I2]
The mechanism is consistent with the observed rate law.
10
32.
The rearrangement of cyclopropane to propene has been studied at various temperatures. The following
values for the specific rate constant have been determined experimentally.
T(K)
600.
650.
700.
750.
800.
850.
900.
k (s-1)
3.30×10-9
2.19×10-7
7.96×10-6
1.80×10-4
2.74×10-3
3.04×10-2
2.58×10-1
a.
From the appropriate plot of these data, determine the value of the activation energy for this
reaction.
Slope = -Ea/R = -3.27×104 K
Ea = (-3.27×104 K)(8.31 J/mol·K) = 2.72×105 J/mol or 272 kJ/mol
b.
Use the graph to estimate the value of k at 500. K.
ln k = -32713 (1/500 K) + 34.99
ln k = -30.44
k = 6.05 × 10-4 s-1
Use the graph to estimate the temperature at which the value of k would be equal to 5.00×10-5 s-1.
ln (5.00×10-5 s-1) = -32713 (1/x) + 34.99
-9.90=-32713 (1/x) + 34.99
-44.89 = -32713 (1/x)
x = 729 K
The activation energy for the reaction
c.
33.
NO2(g) + CO(g) → NO(g) + CO2(g)
Is 125 kJ/mol, and ∆E for the reaction is -216 kJ/mol. What is the activation energy for the reverse
reaction [NO(g) + CO2(g) → NO2(g) + CO(g)]?
The activation energy for the reverse reaction is
Ea, reverse = 216 kJ/mol + 125 kJ/mol = 341 kJ/mol
34.
The reaction
(CH3)3CBr + OH- → (CH3)3COH + Brin a certain solvent is first order with respect to (CH3)3CBr and zero order with respect to OH-. In several
experiments the rate constant k was determined at different temperatures. A plot of ln k versus 1/T was
constructed resulting in a straight line with a slope value of -1.10×104 K and y-intercept of 33.5. Assume k
has units of s-1.
a.
Determine the activation energy for this reaction.
8.31 J
slope = - E a / R, E a = 110
. × 104 K ×
= 9.14 × 104 J / mol = 91.4 kJ / mol
K mol
b.
Determine the value of the frequency factor A.
y intercept = ln A, A = e33.5 = 3.54×1014 s-1
c.
Calculate the value of k at 25°C.
ln k = -Ea/RT + ln A
−E
k = A exp( a ) = 3.54
RT
11
35.
One mechanism for the destruction of ozone in the upper atmosphere is
O3(g) + NO(g) → NO2(g) + O2(g)
Slow
NO2(g) + O(g) → NO(g) + O2(g)
Fast
a.
What is the overall reaction?
O3(g) + O → 2O2(g)
b.
Which species is a catalyst?
NO
c.
Which species is an intermediate?
NO2
d.
Ea for the uncatalyzed reaction is 14.0 kJ. Ea for the same reaction when catalyzed is 11.9 kJ.
What is the ratio of the rate constant for the catalyzed reaction to that for the uncatalyzed reaction
at 25°C? Assume that the frequency factor A is the same for each reaction.
E (un) - E a (cat)
k cat A exp [-E a (cat) / RT]
k = A exp( − E a / RT );
=
= exp a
RT
k un
A exp [-E a (un) / RT]
FG
H
IJ
K
FG
H
IJ
K
k cat
2100 J / mol
= exp
= e 0.85 = 2.3
k un
8.31 J / mol • K × 298 K
The catalyzed reaction is 2.3 times faster than the uncatalyzed reaction at 25°C.
36.
The activation energy for this reaction, X + 2Y → 3Z, shown in the
potential energy diagram, could be
A.
increased by increasing [X].
B.
increased by increasing [X] and [Y].
C.
increased by increasing the temperature.
D.
decreased by removing Z from the system as it forms.
E.
decreased by adding a suitable catalyst.
Adding a catalyst suitable for this reaction will lower the energy barrier
(activation energy) by forming a different activated complex which has a lower potential energy.
37.
For all zero-order reactions
A.
a plot of time vs. concentration squared is linear.
B.
Ea is very low.
C.
the rate is independent of time.
D.
the rate constant is zero.
Answer: (D) For zero-order reactions, Rate = k[X]°. Because anything raised to the zero power
is equal to one, Rate = k. This is another way of saying that the rates of zero-order reactions do
not change; they do not speed up and they do not slow down, they either take place or they do not.
38.
If both ∆H and Ea for the forward reaction are known, the reverse reaction would have an Ea
A.
of (-∆H → ) + Ea→
B.
of ∆H → + Ea→
C.
equal Ea for the forward reaction
D.
equal (-Ea) for the forward reaction.
E.
but none of the above describes the value of Ea.
Answer: (A) This activation energy for the reverse reaction is the difference between the
potential energy of the products and the energy of the activated complex, hence the sum of the
reverse of the enthalpy change of the forward reaction plus the activation energy of the forward
reaction.
39.
The values for the change in enthalpy, ∆H, and the activation energy, Ea, for a given reaction are
known. The value of Ea for the reverse reaction equals
A.
Ea for the forward reaction
B.
-(Ea) for the forward reaction.
C.
the sum of – ∆H and Ea.
D.
the sum of ∆H and Ea
E.
the difference of ∆H and Ea
Answer: (C) Look at a reaction progress diagram.
12
40.
Write balanced net ionic equations for each of the following.
a.
Solid calcium carbonate is strongly heated.
CaCO3 → CaO + CO2
b.
Solid barium oxide is added to distilled water.
BaO + H2O → Ba2+ + 2OH-
c.
Solutions of manganese(II) sulfate and ammonium sulfide are mixed.
Mn2+ + S2- → MnS
d.
Carbon disulfide vapor is burned in excess oxygen.
CS2 + 3O2 → CO2 + 2SO2
e.
A solution of potassium dichromate is added to an acidified solution of iron(II) chloride
This is a redox reaction.
Cr2O72- → Cr3+
Fe2+ → Fe3+
2+
3+
+
Fe → Fe
14H + Cr2O72- → 2Cr3+ + 7H2O
2+
3+
6e + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
Fe → Fe + 1e
2+
3+
6Fe → 6Fe + 6e
6e- + 14H+ + Cr2O72- → 2Cr3+ + 7H2O
22+
+
6Fe + 14H + Cr2O7 → 6Fe3+ + 2Cr3+ + 7H2O
13