Chapter 21, Problem 14STP

Interpretation Introduction

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

Answer to Problem 14STP

The value of rate constant is equal to $7.93\times {10}^3{\text{ M}}^{-2}\cdot {\text{s}}^{-1}$.

Explanation of Solution

Given information:

Data is given as:

$\left[\text{A}\right]$ Initial	$\left[\text{B}\right]$ Initial	Initial rate ( $\text{mol/L}\cdot \text{s}$ )
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:

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

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:

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

Now,

For experiment 1:

$\left[\text{A}\right]=0.10\text{ M}$

Rate = 7.93

Put the values, in rate law expression:

Thus, value of rate constant is equal to