Chapter 13, Problem 13.51QA

Interpretation Introduction

To find:

The rate law for the reaction $2BrO\left(g\right)\to {Br}_2\left(g\right)+O_2\left(g\right)$ under each of the given reaction conditions.

Answer to Problem 13.51QA

Solution:

a) The rate doubles when $\left[BrO\right]$ doubles; $Rate=k\left[BrO\right]$

b) The rate quadruples when $\left[BrO\right]$ doubles; $Rate=k{\left[BrO\right]}^2$

c) The rate is halved when $\left[BrO\right]$ is halved; $Rate=k\left[BrO\right]$

d) The rate is unchanged when $\left[BrO\right]$ doubles; $Rate=k$

Explanation of Solution

1) Concept:

Rate law is the equation, which determines the relation between the rate of reaction and the concentration of reactant.

$2BrO\left(g\right)\to {Br}_2\left(g\right)+O_2\left(g\right)$

Consider the general rate law for this reaction, which is

$Rate=k{\left[BrO\right]}^n$

where $k$ is the rate constant and $n$ is the order in $BrO$.

2) Calculation:

a) The rate doubles when $\left[BrO\right]$ doubles.

Let the initial rate be $Rate 1,$ and when the concentration of $BrO$ doubles, the rate also doubles. Consider it as $Rate 2.$

i.e., $Rate 2 = 2 \left(Rate 1\right)$ and $\left[BrO\right] = 2 \left[BrO\right]$

So the rate law for both will be

$Rate 1=k{\left[BrO\right]}^n$

$Rate 2=k{\left(2\left[BrO\right]\right)}^n$

Take the ratio of $Rate 2$ to $Rate 1$,

$\frac{Rate 2}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

Plug the value of $Rate 2=2 \left(Rate 1\right)$,

Therefore,$\frac{2 Rate 1}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

$2=\frac{2^n {\left[BrO\right]}^n}{ {\left[BrO\right]}^n}$

${2=2}^n$

$n = 1$

That means reaction is first order in $BrO$. Hence the rate law can be written as

$Rate=k\left[BrO\right]$

b) The rate quadruples when $\left[BrO\right]$ doubles.

Let the initial rate be $Rate 1m,$ and when the concentration of $BrO$ doubles, the rate quadruples. Consider it as $Rate 2$.

i.e., $Rate 2 = 4 \left(Rate 1\right)$ when $\left[BrO\right] = 2 \left[BrO\right]$

So the rate law for both will be

$Rate 1=k{\left[BrO\right]}^n$

$Rate 2=k{\left(2\left[BrO\right]\right)}^n$

Take the ratio of $Rate 2$ to $Rate 1$.

$\frac{Rate 2}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

Plug the value of $Rate 2=4 \left(Rate 1\right)$.

Therefore,$\frac{4 Rate 1}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

$4=\frac{2^n {\left[BrO\right]}^n}{ {\left[BrO\right]}^n}$

${4=2}^n$

$n =2$

i.e., reaction is second order in $BrO,$ and the rate law will be

$Rate=k{\left[BrO\right]}^2$

c) The rate is halved when $\left[BrO\right]$ is halved.

Let the initial rate be $Rate 1,$ and when the concentration of $BrO$ halves, the rate also halves. Consider it as $Rate 2$.

i.e., $Rate 2 = 1/2 \left(Rate 1\right)$ when $\left[BrO\right] = 1/2 \left[BrO\right]$

so the rate law for both will be

$Rate 1=k{\left[BrO\right]}^n$

$Rate 2=k{\left(1/2\left[BrO\right]\right)}^n$

Take the ratio of $Rate 2$ to $Rate 1$.

$\frac{Rate 2}{Rate 1}=\frac{k{\left(1/2 \left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

Plug the value of $Rate 2=\frac{1}{2}\left(Rate 1\right).$

Therefore,

$\frac{1/2 Rate 1}{Rate 1}=\frac{k{\left(1/2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

$1/2=\frac{{\left(1/2\right)}^n {\left[BrO\right]}^n}{ {\left[BrO\right]}^n}$

$1/2= {\left(1/2\right)}^2$

$n =1$

Hence, reaction is first order in $BrO,$ and rate law is

$Rate=k\left[BrO\right]$

d) The rate is unchanged when $\left[BrO\right]$ doubles.

When rate the concentration of $BrO$ doubles, the rate remain unchanged,

i.e.,$\left[BrO\right] =2 \left[BrO\right]$ , $Rate 2 = Rate 1$

So the rate law for both will be

$Rate 1=k{\left[BrO\right]}^n$

$Rate 2=k{\left(2\left[BrO\right]\right)}^n$

Take the ratio of $Rate 2$ to $Rate 1$.

$\frac{Rate 2}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

Since, $Rate 2 = Rate 1,$ equation becomes

$\frac{Rate 1}{Rate 1}=\frac{k{\left(2\left[BrO\right]\right)}^n}{k {\left[BrO\right]}^n}$

$0=\frac{2^n {\left[BrO\right]}^n}{ {\left[BrO\right]}^n}$

${0=2}^n$

$n = 0$

Hence, the reaction is zero order in $BrO$.

The rate law will be

$Rate=k{\left[BrO\right]}^o$

$Rate=k$

Conclusion:

The effect of change in concentration of reactants on rate depends on the order of reaction with respect to that reactant.