Chapter 13, Problem 13.142QA

Interpretation Introduction

To find:

a) Write the rate law for the given reaction and the value of the rate constant at ${298}^{\mathrm{o}}\mathrm{K}$.

b) Calculate the activation energy of the reaction from given data.

Answer to Problem 13.142QA

Solution:

a) The rate law for the given reaction is $rate= k \left[O_3\right]\left[C_2{H}_4\right]$ and value of rate constant is $k=1.02\times {10}^3{M}^{-1}{s}^{-1}$

b) The activation energy of the reaction is$22kJ/mol$.

Explanation of Solution

1) Concept:

We can write the rate law for the reaction as $rate=k{\left[O_3\right]}^x{\left[C_2{H}_4\right]}^y$, where x and y are the order of reaction with respect to $O_3$ and $C_2{H}_4$ respectively.

We need to find the values of x, y to write complete rate law and then find the rate constant k of the reaction.

2) Given:

We are given the table for the values of initial concentrations and initial rates of ${\mathrm{N}\mathrm{H}}_2$ and $\mathrm{N}\mathrm{O}$ in the four experiments.

Experiment	${{[\mathrm{O}}_3]}_0\left(\mathrm{M}\right)$	${\left[{\mathrm{C}}_2{\mathrm{H}}_4\right]}_0\left(\mathrm{M}\right)$	Initial Rate $(\mathrm{M}/\mathrm{s})$
$1$	${0.86\times 10}^{-2}$	${1.00\times 10}^{-2}$	$0.0877$
$2$	${0.43\times 10}^{-2}$	${1.00\times 10}^{-2}$	$0.0439$
$3$	${0.22\times 10}^{-2}$	${0.50\times 10}^{-2}$	$0.0110$

3) Calculations:

a) We need to first find the order of reaction with respect to ${\mathrm{O}}_3$ and ${\mathrm{C}}_2{\mathrm{H}}_4$ .

Using the general equation of rate law,

$rate=k{\left[O_3\right]}^x{\left[C_2{H}_4\right]}^y$

 We can write rate equations for all the experiments as

$0.0877=k{\left[{0.86\times 10}^{-2}\right]}^x{\left[{1.00\times 10}^{-2}\right]}^y$------------ (1) Experiment 1

$0.0439=k{\left[{0.43\times 10}^{-2}\right]}^x{\left[{1.00\times 10}^{-2}\right]}^y$------------ (2) Experiment 2

$0.0110=k{\left[{0.22\times 10}^{-2}\right]}^x{\left[{0.50\times 10}^{-2}\right]}^y$------------- (3) Experiment 3

Now by taking the ratio of equation (1) and (2), we can find value of x.

$\frac{0.0877}{0.0439}=\frac{k{\left[{0.86\times 10}^{-2}\right]}^x{\left[{1.00\times 10}^{-2}\right]}^y}{k{\left[{0.43\times 10}^{-2}\right]}^x{\left[{1.00\times 10}^{-2}\right]}^y}$

$2={\left[2\right]}^x$

$x=1$

Now, to find value of y, we have to take ratio of equation (2) and equation (3) and put $\mathrm{x}=1$.

$\frac{0.0439}{0.0110}=\frac{k{\left[{0.43\times 10}^{-2}\right]}^1{\left[{1.00\times 10}^{-2}\right]}^y}{k{\left[{0.22\times 10}^{-2}\right]}^1{\left[{0.50\times 10}^{-2}\right]}^y}$

$4=2\times [{2]}^y$

$2={\left[2\right]}^y$

$y=1$

Thus, we can write the rate law equation as

$\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{k}\left[{\mathrm{O}}_3\right]\left[{\mathrm{C}}_2{\mathrm{H}}_4\right]$

1. Using this rate law and the values of concentrations and initial rates, we can find the rate constant,

i) So from experiment 1, rate law is

$0.0877=k\left[{0.86\times 10}^{-2}\right]\left[{1.00\times 10}^{-2}\right]$

$k=\frac{0.0877}{\left[{0.86\times 10}^{-2}\right]\left[{1.00\times 10}^{-2}\right]}$

$k={1.02\times 10}^3{M}^{-1}{s}^{-1}$

ii) From experiment 2, rate law is

$0.0439=k\left[{0.43\times 10}^{-2}\right]\left[{1.00\times 10}^{-2}\right]$

$k=\frac{0.0439}{\left[{0.43\times 10}^{-2}\right]\left[{1.00\times 10}^{-2}\right]}$

$k={1.02\times 10}^3{M}^{-1}{s}^{-1}$

iii) From experiment 3, rate law is

$0.0110=k\left[{0.22\times 10}^{-2}\right]\left[{0.50\times 10}^{-2}\right]$

$k=\frac{0.0110}{\left[{0.22\times 10}^{-2}\right]\left[{0.50\times 10}^{-2}\right]}$

$k={1.00\times 10}^3{M}^{-1}{s}^{-1}$

The average value of rate constant

$k={1.02\times 10}^3{M}^{-1}{s}^{-1}$

b) To find the activation energy, we have to plot the graph of $\mathrm{l}\mathrm{n}\mathrm{k}$ versus $1/\mathrm{T}$. From the graph, we can find the slope and then the activation energy usingthe formula

${\mathrm{E}}_{\mathrm{a}}=-\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}\times \mathrm{R}$

We have given table for the values of rate constants at different temperatures.

$\mathrm{T}\mathrm{ }\left(\mathrm{K}\right)$	$1/\mathrm{T}\left({\mathrm{K}}^{-1}\right)$	$\mathrm{k}\left({\mathrm{M}}^{-1}{\mathrm{s}}^{-1}\right)$	$\mathrm{l}\mathrm{n}\mathrm{k}$
$263$	$0.003802$	${3.28\times 10}^2$	$5.793$
$273$	$0.003663$	${4.73\times 10}^2$	$6.159$
$283$	$0.003534$	${6.65\times 10}^2$	$6.4997$
$293$	$0.003413$	${9.13\times 10}^2$	$6.8167$

The plot of $\mathrm{l}\mathrm{n}\mathrm{k}$ versus $1/\mathrm{T}$ is as follows, and the value of $\mathrm{s}\mathrm{l}\mathrm{o}\mathrm{p}\mathrm{e}=-2632.21$

The activation energy would be $E_a=-slope\times R$

$E_a=-2632.21 K\times (8.314J/K.mol$

$E_a=21885J/mol$

$E_a=22kJ/mol$

Conclusion:

a) The rate law for the given reaction is $\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{e}=\mathrm{ }\mathrm{k}\mathrm{ }\left[{\mathrm{O}}_3\right]\left[{\mathrm{C}}_2{\mathrm{H}}_4\right],$ and value of rate constant is $\mathrm{k}=1.02\times {10}^3{\mathrm{M}}^{-1}{\mathrm{s}}^{-1}$

b) The activation energy of the reaction is $22\mathrm{k}\mathrm{J}/\mathrm{m}\mathrm{o}\mathrm{l}$.