Chapter 13, Problem 13.124AP

Interpretation Introduction

Interpretation: The information regarding the use of Sodium chlorate in weed control and its ban in EU countries and the reaction involved in the testing of Sodium chlorate is given. The rate law and the rate constant for the reaction is to be calculated.

Concept introduction: According to the rate law the rate of reaction depends on the concentration of reactants that are involved in the reaction. Therefore, for a reaction like $\text{A}\to \text{B}$, the rate law equation is given as,

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

Where,

• $k$ is the rate constant.
• $m$ is the order of the reaction.

To determine: The rate law and the rate constant for the given reaction.

Answer to Problem 13.124AP

Solution

The rate law equation is,

$\text{Rate}=k{\left[{\text{MnO}}_4{}^{-}\right]}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^1{\left[{\text{H}}^{+}\right]}^{0.5}$

The value of rate constant is $\underset{\_}{16.44M^{-5/2}{s}^{-1}}$.

Explanation of Solution

Explanation

The rate law expression for the given reaction is,

$\text{Rate}=k{\left[{\text{MnO}}_4{}^{-}\right]}^a{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^b{\left[{\text{H}}^{+}\right]}^c$ (1)

The table for the given reaction is given as,

Experiment	${\left[{\text{MnO}}_4{}^{-}\right]}_0\left(\text{M}\right)$	${\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}_0\left(\text{M}\right)$	${\left[{\text{H}}^{+}\right]}_0\left(\text{M}\right)$	Initial Rate $\left(\text{M}/\text{s}\right)$
$1$	$0.10$	$0.10$	$0.10$	$5.2\times {10}^{-3}$
$2$	$0.25$	$0.10$	$0.10$	$3.3\times {10}^{-2}$
$3$	$0.10$	$0.30$	$0.10$	$1.6\times {10}^{-2}$
$4$	$0.10$	$0.10$	$0.20$	$7.4\times {10}^{-3}$

Table 1

The rate law equation for the four experiments is given as,

$5.2\times {10}^{-3}=k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c$ (2)

$3.3\times {10}^{-2}=k{\left[0.25\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c$ (3)

$1.6\times {10}^{-2}=k{\left[0.10\right]}^a{\left[0.30\right]}^b{\left[0.10\right]}^c$ (4)

$7.4\times {10}^{-3}=k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.20\right]}^c$ (5)

Divide equation (2) by equation (3),

$\begin{array}{c}\frac{5.2\times {10}^{-3}}{3.3\times {10}^{-2}}=\frac{k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c}{k{\left[0.25\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c}\\ 0.1575={\left(0.4\right)}^a\end{array}$

Take log on both the sides of the above equation,

$\begin{array}{c}\log 0.1575=a\log 0.4\\ -0.803=a\times \left(-0.398\right)\\ a=2.0\end{array}$

Divide equation (2) by equation (4),

$\begin{array}{c}\frac{5.2\times {10}^{-3}}{1.6\times {10}^{-2}}=\frac{k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c}{k{\left[0.10\right]}^a{\left[0.30\right]}^b{\left[0.10\right]}^c}\\ 0.325={\left(0.33\right)}^b\end{array}$

Take log on both the sides of the above equation,

$\begin{array}{c}\log 0.325=b\log 0.33\\ -0.488=b\times \left(-0.481\right)\\ b=1.0\end{array}$

Divide equation (2) by equation (5),

$\begin{array}{c}\frac{5.2\times {10}^{-3}}{7.4\times {10}^{-3}}=\frac{k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.10\right]}^c}{k{\left[0.10\right]}^a{\left[0.10\right]}^b{\left[0.20\right]}^c}\\ 0.703={\left(0.5\right)}^c\end{array}$

Take log on both the sides of the above equation,

$\begin{array}{c}\log 0.703=c\log 0.5\\ -0.153=c\times \left(-0.3010\right)\\ b=0.5\end{array}$

Substitute the values of $a$, $b$ and $c$ in equation (1),

$\text{Rate}=k{\left[{\text{MnO}}_4{}^{-}\right]}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^1{\left[{\text{H}}^{+}\right]}^{0.5}$

Substitute the value of rate and concentration of all species from experiment (1) in above equation,

$\begin{array}{c}\text{Rate}=k{\left[{\text{MnO}}_4{}^{-}\right]}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^1{\left[{\text{H}}^{+}\right]}^{0.5}\\ 5.2\times {10}^{-3}\text{M}/\text{s}=k{\left[0.10\text{M}\right]}^2{\left[0.10\text{M}\right]}^1{\left[0.10\text{M}\right]}^{0.5}\\ 5.2\times {10}^{-3}\text{M}/\text{s}=k\times 3.1\times {10}^{-4}{\text{M}}^{7/2}\\ k=\underset{\_}{16.44M^{-5/2}{s}^{-1}}\end{array}$

Conclusion

The rate law equation is,

$\text{Rate}=k{\left[{\text{MnO}}_4{}^{-}\right]}^2{\left[{\text{ClO}}_{\text{3}}{}^{-}\right]}^1{\left[{\text{H}}^{+}\right]}^{0.5}$

The value of rate constant is $\underset{\_}{16.44M^{-5/2}{s}^{-1}}$.