Chapter 11, Problem 11.30PAE

The rate of the decomposition of hydrogen peroxide, H₂O₂, depends on the concentration of iodide ion present. The rate of decomposition was measured at constant temperature and pressure for various concentrations of H₂O₂and of KI. The data appear below. Determine the order of reaction for each substance, write the rate law, and evaluate the rate constant.

Rate	[H2OJ	[Kll
(mL min-’)	(mol L ’)	(mol L ’)
0.090	0.15	0.033
0.178	0.30	0.033
0.184	0.15	0.066

Interpretation Introduction

To determine:

Order of reaction of each substance, rate law and rate constant

Explanation of Solution

Let following chemical reaction

$aA+bB\to cC+dD$

Rate law for this reaction can be written as

$R=K{\left[A\right]}^x{\left[B\right]}^y.$

$A=H_2{O}_2$

$B=KI.$

The rate law for the given reaction can be written as

$R=K{\left[H_2{O}_2\right]}^x{\left[KI\right]}^y$

You have given three set of data. Put each in the above equation

$R=0.090mL{\min }^{-1}$

$1mL={10}^{-3}L$

So,

$R=0.090\times {10}^{-3}mol{L}^{-1}{\min }^{-1}$

$\begin{array}{l}\left[H_2{O}_2\right]=0.15mol{L}^{-1}.\\ \left[KI\right]=0.033mol{L}^{-1}\end{array}$

Putting these values on equation, you get

$\left(0.09\times {10}^{-3}mol{L}^{-1}{\min }^{-1}\right)=K{\left(0.15mol{L}^{-1}\right)}^x{\left(0.033mol{L}^{-1}\right)}^y\cdots \cdots \left(1\right)$

$\begin{array}{c}R=0.178mol{L}^{-1}{\min }^{-1}\\ =0.178\times {10}^{-3}molL{\min }^{-1}\end{array}$

$\begin{array}{l}\left[H_2{O}_2\right]=0.30mol{L}^{-1}\\ \left[KI\right]=0.033mol{L}^{-1}\end{array}$

Putting these value in equation you, get

$\left(0.178\times {10}^{-3}mol{L}^{-1}{\min }^{-1}\right)=K{\left(0.30mol{L}^{-1}\right)}^x.{\left(0.033mol{L}^{-1}\right)}^y\cdots \cdots \left(2\right)$

$\begin{array}{c}R=0.184mol{L}^{-1}{\min }^{-1}\\ =0.184\times {10}^{-3}mol{L}^{-1}{\min }^{-1}\end{array}$

$\begin{array}{l}\left[H_2{O}_2\right]=0.15mol{L}^{-1}\\ \left[KI\right]=0.066mol{L}^{-1}\end{array}$

Putting these value in equation you, get

$0.184\times {10}^{-1}mol{L}^{-1}{\min }^{-1}=K{\left(0.15mol{L}^{-1}\right)}^x{\left(0.66mol{L}^{-1}\right)}^y\cdots \cdots \left(3\right)$

$\left(1\right)÷\left(2\right)$

$\frac{\left(0.09\times {10}^{-3}\right)mol{L}^{-1}{\min }^{-1}}{0.178\times {10}^{-3}mol{L}^{-1}{\min }^{-1}}=\frac{K{\left(0.15mol{L}^{-1}\right)}^x{\left(0.033mol{L}^{-1}\right)}^y}{K{\left(0.30mol{L}^{-1}\right)}^x{\left(0.033mol{L}^{-1}\right)}^y}$

$\begin{array}{c}\frac{1}{1.98}=\frac{1}{{\left(2\right)}^x}\\ 2^x=1.98\\ 2^x={\left(2\right)}^1\\ x=1\end{array}$

$\left(1\right)÷\left(3\right)$

$\frac{\left(0.09\times {10}^{-3}\right)L{\min }^{-1}}{\left(0.178\times {10}^{-3}\right)L{\min }^{-1}}=\frac{K{\left(0.15mol{L}^{-1}\right)}^x{\left(0.033mol{L}^{-1}\right)}^y}{K{\left(0.15mol{L}^{-1}\right)}^x{\left(0.066mol{L}^{-1}\right)}^y}$

$\begin{array}{c}\frac{1}{1.98}=\frac{1}{{\left(2\right)}^y}\\ {\left(2\right)}^y=1.98\\ {\left(2\right)}^y={\left(2\right)}^1\\ y=1\end{array}$

Hence

$x=1,y=1.$

Hence overall order of reaction $=1+1=2$

Rate law can be written as

$R=K\left[H_2{O}_2\right]\left[KI\right]$

Putting the value $R,$

$\left[H_2{O}_2\right],\left[KI\right],$ to find the value of $K.$

$\left(0.09\times {10}^{-3}\right)L^{-}{\min }^{-1}=K\left(0.15mol{L}^{-1}\right)\left(0.033mol{L}^{-1}\right).$

$\begin{array}{l}K=0.01818mo{l}^{-1}L{\min }^{-1}\\ K=18.18mo{l}^{-1}mL{\min }^{-1}.\end{array}$

Hence rate law can be written as

$R=18.18mo{l}^{-1}mL{\min }^{-1}\left[H_2{O}_2\right]\left[KI\right]$

Conclusion

It is the initial rates experiment method.