Chapter 20, Problem 60E

Interpretation Introduction

Interpretation:

The order for an enzyme catalyzed reaction with respect to substrate concentration is to be determined.

Time, min	[S], M
0	1.00
20	0.90
60	0.70
100	0.50
160	0.20

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 has direct relationship with the concentration of one of the reactant.

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

Integral rate law for first order reaction is −

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

And the graph of first order reaction is-

and

• The second order reaction is the one in which the rate has direct relationship with 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-