KINETICS Practice Problems and Solutions Name: Period:
AP Chemistry Dr. Mandes
Date:
The following questions represent potential types of quiz questions. Please answer each question completely and thoroughly. The solutions will be posted on-line on Monday. 5.
Please do #18 in chapter 12 of your text. a. This increases the [H2], which will increase the rate, but has no effect on k b. Due to Arrhenius equation, changing temperature changes the value of k. c. Catalyst function by lowering the activation energy, so due to Arrhenius equation, changing the activation energy changes the value of k. Consider the reaction: P4 + 6 H2 → 4 PH3. data that were obtained are shown in the table.
6.
a.
A rate study of this reaction was conducted at 298 K. The
[P4], mol/L
[H2], mol/L
Initial Rate, mol/(L . s)
0.0110
0.0075
3.20 x 10-4
0.0110
0.0150
6.40 x 10-4
0.0220
0.0150
6.39 x 10-4
What is the order with respect to:
P4 __0___. H2 __1___.
b.
Write the rate law for this reaction.
rate = k[H2]
c. Determine the value and units of the rate constant, k. plug and chug using the rate law & data from exp’t 1 and solving for k, we get k = 0.0427 s-1 Consider the reaction: SO2 + O3 → SO3 + O2. data that were obtained are shown in the table.
7.
a.
b.
A rate study of this reaction was conducted at 298 K. The
[SO2], mol/L
[O3], mol/L
Initial Rate, mol/(L . s)
0.25
0.40
0.118
0.25
0.20
0.118
0.75
0.20
1.062
What is the order with respect to:
Write the rate law for this reaction.
SO2
___2__.
O3
___0__.
rate = k[SO2]2[O3]0
c. Determine the value and units of the rate constant, k. plug and chug using the rate law & data from exp’t 1 and solving for k, we get k = 2.36 mol.L-1. s-1
8.
Consider the following mechanism.
A2 +
B2
→
R
A2 +
R
→
C +
+
C
(slow) (fast)
B2 →
a.
Write the overall balanced chemical equation. 2 A2
b.
Identify any intermediates within the mechanism. R
c.
What is the order with respect to each reactant? A2 1st; B2 1st
2C
KINETICS Practice Problems and Solutions d. 9.
Write the rate law for the overall reaction. rate = k [A2][B2]
Consider the following mechanism.
O3 O3 +
O
→
O2 +
→
2 O2
O
(fast) (slow)
→ 3 O2
a.
Write the overall balanced chemical equation. 2 O3
b.
Identify any intermediates within the mechanism. O
c.
What is the order with respect to each reactant? O3 2nd (once in rds, then once when sub for intermediate)
d.
Write the rate law for the overall reaction. rate = k [O3]2
10. Consider the reaction: 2B → C + 3D. In one experiment it was found that at 300 K the rate constant is 0.134 L/(mol.s). A second experiment showed that at 450 K, the rate constant was 0.569 L/(mol .s). Determine the activation energy for the reaction. Ea
at 300 K:
k300
Ae RT
at 450 K:
k450
Ae RT
ln
k450 A
Ea
Ea RT
ln(k450 ) ln( A)
Ea RT
where ln( A) ln(k300 )
Ea RT
so that Ea RT
ln(k450 ) [ln(k300 ) 450 ln( kk300 )
Ea R
1 ( T300
1 T450
]
Ea RT
)
plug and solve for Ea, Ea = 10.8 kJ
MORE PROBLEMS>>>> Determining rate law from mechanisms (use the rate-determining step to get the orders). 1.
One method for the destruction of ozone in the upper atmosphere is:
overall rxn
O3
+
NO
NO2 +
O2
(slow)
NO2
+
O
NO
O2
(fast)
O3
+
O
2O2
a.
Which species is an intermediate?
b.
Which species is a catalyst?
c.
Which is the rate-determining step (rds)?
d.
Number of times each reactant is used in the rds?
e.
Write the rate law for the reaction.
+
KINETICS Practice Problems and Solutions Determining rate law from Initial Rates. (Use the ratio of initial rates to get the orders). 2.
Consider the table of initial rates for the reaction: Experiment
[ClO2]o, mol/L
2ClO2 + 2OH1-
[OH1-] o, mol/L
ClO31- + ClO21- + H2O.
Initial Rate, mol/(L . s)
1
0.050
0.100
5.75 x 10-2
2
0.100
0.100
2.30 x 10-1
3
0.100
0. 050
1.15 x 10-1
a . Order with respect to ClO2: b.
Order with respect to OH1-:
c.
Rate law for this reaction:
d.
Value and units for the rate constant:
3.
Consider the table of initial rate for the reaction between hemoglobin (Hb) and carbon monoxide. Experiment
[HB]o, mol/L
[CO] o, mol/L
Initial Rate, mol/(L . s)
1
2.21
1.00
0.619
2
4.42
1.00
1.24
3
3.36
2.40
2.26
a . Order with respect to HB:
Part II 1.
b.
Order with respect to CO:
c.
Rate law for this reaction:
d.
Value and units for the rate constant:
Select Response If the reaction: 2HI a. c.
2.
rate = k[X] rate = k[Y]
Y + Z. Which of the following is a possible rate law for the reaction? b. d.
rate = k[Y][Z] rate = k[X][Y]
e.
rate = k[Z]
Consider the rate law: rate = k[Y]m[Z]n. How are the exponents m and n determined? a. b. c. d. e.
4.
2H2 + I2 HI is first order, which of the following will yield a linear plot? log [HI] vs time b. 1/[HI] vs time e. [HI]2 vs time [HI] vs time d. ln[HI] vs time
Consider the reaction: X a. c.
3.
Select the best answer to each question.
by using the balanced chemical equation by using the subscripts of the chemical formulas by using the coefficients of the chemical formulas by educated guess by experiment
The following data were obtained for the reaction of NO with O 2. [NO]o [O2]o Initial Rate 1 x 1018 2 x 1018
1 x 1018 1 x 1018
2.0 x 1018 8.0 x 1018
KINETICS Practice Problems and Solutions 3 x 1018 1 x 1018 1 x 1018
1 x 1018 2 x 1018 3 x 1018
18.0 x 1018 4.0 x 1018 6..0 x 1018
Which of the following is the correct rate law?
5.
a.
rate = k[NO][O2]
b.
rate = k[NO][O2]2
c.
rate = k[NO]2[O2]
d.
rate = k[NO]2
log [HI] vs time [HI] vs time
b. d.
1/[HI] vs time ln[HI] vs time
e.
[HI]2 vs time
Which of the following statements is typically true for a catalyst? a. b. c. d. e.
7.
rate = k[NO]2[O2]2
If the reaction 2HI → H2 + I2 is second order, which of the following will yield a linear plot? a. c.
6.
e.
The concentration of the catalyst will go down as the reaction proceeds. The catalyst provides a new pathway in the reaction mechanism. The catalyst speeds up the reaction. Two of the above. None of the above.
The catalyzed reaction has a _____ activation energy and thus causes a _____ reaction rate. a. c.
higher, lower lower, higher
b. d.
higher, higher lower, steady
e.
higher, steady
KINETICS Practice Problems and Solutions Part II Constructed Response
8.
Thoroughly and completely answer each question on a separate piece of paper.
Consider the exothermic reaction between reactants A and B? A + B → E (fast) E + B → C + D (slow) a.
What is the order with respect to reactants A and B? 1, 2
b.
What is the rate law for the reaction?
c.
Sketch a potential energy diagram for this reaction. Identify the activation energy for the overall forward reaction. Identify the location of reactants, intermediate(s), activated complex(es), and products.
9.
A first-order reaction is 38.5% complete in 480 s. a.
Calculate the value of the rate constant.
b.
What is the value of the half-life.
c.
How long will it take for the reaction to reach 95% completion.
The rate of the reaction NO2 + CO → NO + CO2 depends only on the concentration of nitrogen dioxide. The following data were collected. time, (s) [NO2]o, (M) 0 0.500 1.20 x 103 0.444 3.00 x 103 0.381 4.50 x 103 0.340 9.00 x 103 0.250 1.80 x 104 0.174
10.
a.
Determine the rate law.
b.
Write the integrated rate law.
c.
Determine the value of the rate constant for the reaction.
d.
Calculate the [NO2] at 2.40 x 104 s after the start of the reaction.
KINETICS Practice Problems and Solutions Determining rate law from time and concentration data. (Use the integrated rate laws and graphing to get orders). 4.
The rate of this rxn depends only on NO2:
NO2 + CO
The following data were collected.
NO + CO2.
Time (s)
[NO2] (mol/L)
0
0.500
1200.
0.444
3000.
0.381
4500.
0.340
9000.
0.250
18000.
0.174
a. Order with respect to NO2: b. Rate law for this reaction: c. [NO2] at 2.7 x 104 s after the start of the rxn. 5.
The following data were obtained for the decomposition of N 2O5 in CCl4. The following data were collected.
Time (s)
[N2O5] (mol/L)
0
1.46
423
1.09
753
0.89
1116
0.72
1582
0.54
1986
0.43
2343
0.35
a. Order with respect to N2O5: b. Rate law for this reaction: c. [N2O5] at 3.5 x 103 s after the start of the rxn.
SOLUTIONS!!!!!! TO “MORE PROBLEMS”>>>>> 1.
a.
Which species is an intermediate?
NO2
b.
Which species is a catalyst?
NO
c.
Which is the rate-determining step (rds)?
slow step
d.
Number of times each reactant is used in the rds?
O3 is used once so order is 1 O is used zero times, so order is 0
e.
Write the rate law for the reaction.
rate = k[O3]1
________________________________________________________________________________________________
KINETICS Practice Problems and Solutions Note that for a free-response question you must show the work (ratio of rate laws), but not for multiple choice 2.
rate2 rate1 0.230 0.0575 4
k[ClO2 ]2m [OH ]2n k[ClO2 ]1m [OH ]1n
rate2 rate3
k[ClO2 ]2m [OH ]2n k[ClO2 ]3m [OH ]3n
0.100m 0.0500m
0.230 0.115
0.100m 0.0500m
2m
2
2 = m, so order is 2
2n
1 = n, so order is 1
a. Order with respect to ClO2:
2
b. Order with respect to OH1-:
1
c. Rate law for this reaction:
so, rate = k[ClO2]2[OH1-]1
d. Value and units for the rate constant:
k = 230
L2 mol 2 s
get the value by subbing the data for exp’t 1 into the rate law and solving for k
________________________________________________________________________________________________
3.
rate2 rate1 1.24 0.619 2
k[ HB]2m [CO]2n k[ HB]1m [CO]1n 4.42m s 2.21m
rate3 rate1 2.26 0.619
2m
2.4
1 = m, so the order is 1
k[ HB]3m [CO]3n k[ HB]1m [CO]1n 3.36 2.21
1 1
2.41 1.00
n n
2.4 n
1 = n, so the order is 1
a. Order with respect to HB:
1
b. Order with respect to CO:
1
c. Rate law for this reaction:
so, rate = k[HB]1[CO]1
d. Value and units for the rate constant:
k = 0.28
L mol s
get the value by subbing the data for exp’t 1 into the rate law and solving for k
________________________________________________________________________________________________
KINETICS Practice Problems and Solutions 4. Graph for zeroeth order: [NO2] vs. time [y vs. x; y = ax +b] slope = -1.72 x 10-5
y-intercept =
General integrated rate law:
[A] = -kt + [A]o
This reaction's integrated rate law:
[H2O2] = (-1.72 x 10-5)t + 0.451
Graph for first order:
r2 =
0.451
0.901
r2 = 0.901
n[NO2] vs. time [y vs. x; y = ax +b]
slope = -5.78 x 10-5
r2 =
y-intercept = -0.770
General integrated rate law:
n[A] = -kt +
This reaction's integrated rate law:
n [NO2] = (-5.78 x 10-5)t + (-0.770)
0.971
n[A]o r2 = 0.971
Graph for second order: [NO2]-1 vs. time [y vs. x; y = ax +b] slope = 2.10 x 10-4 order is 2
+ 2.01”
r2 = 0.999 - best so
y-intercept = 2.01 1
[ A]o1
General integrated rate law:
[ A]
This reaction's integrated rate law:
[NO2] -1 = 2.10 x 10-4t + 2.01 r2 = 0.999
Graph with the greatest r2 value:
[NO2]-1
= kt +
vs. time, so the order is second order
a. Order with respect to NO2:
2
b. Rate law for this reaction:
rate = k[NO2]2
c. [NO2] at 2.7 x 104 s after the start of the rxn.
Subbing 2.7 x 104 s for time in “[NO2] -1 = 2.10 x 10-4t [NO2] = 0.130 mol/L
_____________________________________________________________________________________________ 5. Graph for zeroeth order: [N2O5] vs. time [y vs. x; y = ax +b] slope = -4.54 x 10-4
y-intercept =
General integrated rate law:
[A] = -kt + [A]o
This reaction's integrated rate law:
[N2O5] = (-4.54 x 10-4)t + 1.31 r2 = 0.947
Graph for first order: slope = -6.05 x 10
1.31
r2 =
0.947
n[N2O5] vs. time [y vs. x; y = ax +b] -4
General integrated rate law: This reaction's integrated rate law: order is 1
y-intercept = 0..353 n[A] = -kt +
r2 =
0.999
n[A]o
n[ N2O5] = (-6.05 x 10-4)t + 0.353
r2 = 0.999 - best so
KINETICS Practice Problems and Solutions Graph for second order: [N2O5]-1 vs. time [y vs. x; y = ax +b] slope = 9.18 x 10-4
y-intercept = 0.517
General integrated rate law:
[ A]
This reaction's integrated rate law:
[N2O5] -1 = 9.18 x 10-4t + 0.517 r2 = 0.971
Graph with the greatest r2 value:
1
= kt - +
n [N2O5]
r2 = 0.971s
[ A]o1
vs. time, so the order is first order
Order with respect to N2O5: Rate law for this reaction:
4)
a. Order with respect to N2O5:
1
b. Rate law for this reaction:
rate = k[N2O5]1
c. [N2O5] at 3.5 x 103 s after the start of the rxn. t + 1.31”
Subbing 3.5 x 103 s for time in “ n[ N2O5] = (-6.05 x 10[N2O5] = 0.171 mol/L
