Chapter 21, Problem 13STP

Interpretation Introduction

Interpretation:

The values of m and n should be calculated for the given rate law expression.

$\text{Rate = k}{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$

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 rate of chemical reaction at specific temperature to the concentration of the reaction is known as rate constant.

Answer to Problem 13STP

The values of m and nare 1 and 2.

Rate law for the reaction is:

$\text{Rate = k}{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$

Explanation of Solution

Given information:

Data is given as:

The general expression of rate law is expressed as:

$\text{Rate = k}{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$

Where, m and n are the experimentally determined values.

Experiment 1:

Rate law expression is written as:

$7.93\text{ = k}{\left[\text{0}\text{.10}\right]}^{\text{m}}{\left[0.10\right]}^{\text{n}}\text{ (1)}$

Experiment 2: Rate law expression is written as:

$23.79\text{ = k}{\left[\text{0}\text{.30}\right]}^{\text{m}}{\left[0.10\right]}^{\text{n}}\text{ (2)}$

Experiment 3: Rate law expression is written as:

$95.16\text{ = k}{\left[\text{0}\text{.30}\right]}^{\text{m}}{\left[0.20\right]}^{\text{n}}\text{ (3)}$

Now, divide equation (2) by (1):

$\frac{23.79\text{ = k}{\left[\text{0}\text{.30}\right]}^{\text{m}}{\left[0.10\right]}^{\text{n}}}{7.93\text{ = k}{\left[\text{0}\text{.10}\right]}^{\text{m}}{\left[0.10\right]}^{\text{n}}}$

$\text{3=}{\left(\text{3}\right)}^{\text{m}}$

$\text{m =1}$

Now, divide equation (3) by (2):

$\frac{95.16\text{ = k}{\left[\text{0}\text{.30}\right]}^{\text{m}}{\left[0.20\right]}^{\text{n}}}{\text{23}\text{.79 = k}{\left[\text{0}\text{.30}\right]}^{\text{m}}{\left[0.10\right]}^{\text{n}}}$

$\text{4 =}{\left(2\right)}^{\text{n}}$

${\text{2}}^2\text{ =}{\left(2\right)}^{\text{n}}$

$\text{n =2}$

Thus, rate law expression is written as:

$\text{Rate = k}{\left[\text{A}\right]}^1{\left[\text{B}\right]}^2$