Interpretation:
The values of k should be calculated for the reaction.
Concept introduction:
Rate Law can be expressed as an integrated rate law and a differential rate law.
Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.
Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.
The proportionality coefficient which relates the

Answer to Problem 14STP
The value of rate constant is equal to 7.93×103 M−2⋅s−1.
Explanation of Solution
Given information:
Data is given as:
[A] Initial | [B] Initial | Initial rate ( mol/L⋅s ) |
0.10 M | 0.10 M | 7.93 |
0.30 M | 0.10 M | 23.79 |
0.30 M | 0.20 M | 95.16 |
The general expression of rate law is expressed as:
Rate = k[A]m[B]n
Where, m and n are the experimentally determined values.
Experiment 1:
Rate law expression is written as:
7.93 = k[0.10]m[0.10]n (1)
Experiment 2:
Rate law expression is written as:
23.79 = k[0.30]m[0.10]n (2)
Experiment 3:
Rate law expression is written as:
95.16 = k[0.30]m[0.20]n (3)
Now, divide equation (2) by (1):
23.79 = k[0.30]m[0.10]n7.93 = k[0.10]m[0.10]n
3=(3)m
m =1
Now, divide equation (3) by (2):
95.16 = k[0.30]m[0.20]n23.79 = k[0.30]m[0.10]n
4 =(2)n
22 =(2)n
n =2
Thus, rate law expression is written as:
Rate = k[A]1[B]2
Now,
For experiment 1:
[A]=0.10 M
Rate = 7.93
Put the values, in rate law expression:
Thus, value of rate constant is equal to
Chapter 21 Solutions
Chemistry: Matter and Change
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