Name:
Period:
AP Chemistry
Date:
Rate Laws (Differential Rate Laws) Practice 1
1. The reaction of t-butyl bromide is represented by the equation:
(CH3)3CBr + H2O  (CH3)3COH + HBr
The following data were obtained:
[(CH3)3CBr] (mol/L)
[H2O] (mol/L)
Rate (mol/L min)
Exp 1
0.05
0.02
2.0 x10-6
Exp 2
0.05
0.04
2.0 x10 -6
Exp 3
0.1
0.04
4.0 x 10-6
Exp 4
0.02
0.05
???
a. What happens to the rate when the [(CH3)3CBr] is doubled? What order is the reaction with respect to (CH3)3CBr?
b. What happens to the rate when the [H2O] is doubled? What order is the reaction with respect to H2O?
c. Write the rate law for this reaction:
d. Solve for the value of k, including units! Show work below.
e. Use your rate law and value of k to calculate the rate in experiment 4. Show work below.
2. Consider the reaction:
2NO + O2  2NO2
The following data was collected:
Exp 1
Exp 2
Exp 3
[NO] (mol/L)
0.010
0.020
0.010
[O2] (mol/L)
0.010
0.010
0.020
Rate of NO2 formed (mol/L min)
2.5 x10-5
1.0 x10 -4
5.0 x 10-5
a. Determine the rate order with respect both reactants.
b. Write the rate law equation for this reaction.
c. Solve for the value of k, including units! Show work below.
d. Looking at the balanced equation, how would the rate of disappearance of O2 compare to that of NO2?
2020-2021
3. For the reaction:
2NO2 + O3  N2O5 + O2
The following data was collected:
Exp 1
Exp 2
Exp 3
[NO2] (mol/L)
0.10
0.10
0.20
[O3] (mol/L)
0.20
0.10
0.40
Rate of O2 formed (mol/L min)
1300
650
5200
a. Determine the rate orders for each of the reactants.
b. Write out the rate law and determine the value of k.
4. For the reaction:
2NO + 2H2  N2 + 2H2O
The following data was collected:
Exp 1
Exp 2
Exp 3
Exp 4
Exp 5
[NO] (mol/L)
0.420
0.210
0.210
???
0.350
[H2] (mol/L)
0.122
0.122
0.244
0.488
0.205
Rate of N2 formed (mol/Lmin)
0.136
0.0339
0.0678
0.0339
???
a. Determine the rate orders with respect to each of the reactants.
b. Write out the rate law and determine the value of k.
c. Use your rate law & k to determine the initial [NO] in experiment 4.
d. Determine the rate of N2 formed in experiment 5.
Numerical answers in order:
k = 4E-5/min
k = 6.4E4/min
rate = 8E-7 M/min
k = 6.32/M2min
k = 25 M/min
[NO] = 0.105 M
rate O2 would be ½
rate N2 = 0.159 M/min