Chapter 20, Problem 30E

Interpretation Introduction

Interpretation:

The approximate initial rate for the second order reaction is to be determined.

Set I		Set II		Set III
Time, s	[A], M	Time, s	[A], M	Time, s	[A], M
0	1.00	0	1.00	0	1.00
25	0.78	25	0.75	25	0.80
50	0.61	50	0.50	50	0.67
75	0.47	75	0.25	75	0.57
100	0.37	100	0.00	100	0.50
150	0.22			150	0.40
200	0.14			200	0.33
250	0.08			250	0.29

Concept introduction:

• Zero-order reaction- for this, rate of reaction is independent of the concentration.

$\text{Rate = k}{\text{.[A]}}^0=\text{ k}$

And the graph of zero order reaction is-

• The first order reaction is the one in which the rate is directly proportional to the concentration of one of the reactant.

$\begin{array}{l}\text{Rate}\propto \text{[reactant]}\\ \text{Rate = k[A]}\\ \text{where k - rate constant}\end{array}$

And half-life is

$t_{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=\raisebox{1ex}{$0.693$}\!\left/ \!\raisebox{-1ex}{$k$}\right.$

Integral rate law for first order reaction is −

$\ln [\text{A}]=\ln [{\text{A}}_0]-\text{kt}$

and

• The second order reaction is the one in which the rate is directly proportional to the square of the concentration of one of the reactant.

$\begin{array}{l}\text{Rate}\propto {\text{[reactant]}}^2\\ {\text{Rate = k[A]}}^2\end{array}$

Integral rate law for second order reaction is −

$\frac{1}{[\text{A]}}=\frac{1}{[{\text{A}}_{\text{0}}]}\text{+kt}$

And the graph of second order reaction is-