Chapter 17, Problem 17B.15P

(a)

Interpretation Introduction

Interpretation:

An ${\mathrm{n}}^{th}$ order in a reaction A is given by an equation ${\text{t}}_{\text{1/2}}$ has to be shown

Concept introduction:

Rate law for the nth order is given below,

  $\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}-{\text{k}}_{\text{r}}{\left[\text{A}\right]}^{\text{n}}$

Explanation of Solution

Rate law for the nth order is given below,

  $\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}-{\text{k}}_{\text{r}}{\left[\text{A}\right]}^{\text{n}}$

Here,

Initial concentration of reactant is $\left[\text{A}\right]$.

Rate constant of the reaction is ${\text{k}}_{\text{r}}$.

The reaction is rearranged,

  $\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}-{\text{k}}_{\text{r}}{\left[\text{A}\right]}^{\text{n}}$

The reaction is integrated,

  $\begin{array}{l}\text{-}{\displaystyle \underset{\left[{\text{A}}_{\text{0}}\right]}{\overset{\left[\text{A}\right]}{\int }}\frac{\text{d}\left[\text{A}\right]}{{\left[\text{A}\right]}^{\text{n}}}}\text{ =}{\text{k}}_{\text{r}}{\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}\text{dt}}\\ \frac{\text{-1}}{\text{1-n}}{\left(\frac{\text{1}}{{\left[\text{A}\right]}^{\text{n-1}}}\right)}_{\left[{\text{A}}_{\text{0}}\right]}^{\left[\text{A}\right]}\text{ =}{\text{k}}_{\text{r}}\text{t}\\ \frac{\text{-1}}{\text{1-n}}{\left(\frac{\text{1}}{{\left[\text{A}\right]}^{\text{n-1}}}\text{-}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{n-1}}}}\right)}_{\left[{\text{A}}_{\text{0}}\right]}^{\left[\text{A}\right]}\text{ =}{\text{k}}_{\text{r}}\text{t}\\ {\left[\text{A}\right]}^{\text{1-n}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}\text{t}\left(\text{n-1}\right)\end{array}$

An ${\mathrm{n}}^{th}$ order in a reaction A is given by an equation ${\text{t}}_{\text{1/2}}$ is given.

(b)

Interpretation Introduction

Interpretation:

Time for the concentration of a substance to fall to one·third the initial value in an n^th·order reaction has to be derived.

Concept introduction:

Balancing reaction:

Refer to part (a)

Explanation of Solution

Half-life of the reaction is t_1/2.

Half-life of the nth order reaction is given below,

Lets,

  $\left[\text{A}\right]=\text{0}\text{.5}{\left[\text{A}\right]}_0$

Therefore,

    $\begin{array}{l}{\left[\text{A}\right]}^{\text{1-n}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}\text{t}\left(\text{n-1}\right)\\ {\left[\text{0}\text{.5A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}{\text{t}}_{\text{1/2}}\left(\text{n-1}\right)\\ {\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\left(\text{0}{\text{.5}}^{\text{1-n}}\text{-1}\right)\text{=}{\text{k}}_{\text{r}}{\text{t}}_{\text{1/2}}\left(\text{n-1}\right)\\ {\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\left({\left({\text{2}}^{\text{-1}}\right)}^{\text{1-n}}\text{-1}\right)\text{=}{\text{k}}_{\text{r}}{\text{t}}_{\text{1/2}}\left(\text{n-1}\right)\end{array}$

  $\begin{array}{l}{\text{t}}_{\text{1/2}}\text{=}\frac{{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\left({\left({\text{2}}^{\text{-1}}\right)}^{\text{1-n}}\text{-1}\right)}{{\text{k}}_{\text{r}}\left(\text{n-1}\right)}\\ {\text{t}}_{\text{1/2}}\text{=}\frac{{\text{2}}_{}^{{}^{\text{n-1}}}\text{-1}{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}}{{\text{k}}_{\text{r}}\left(\text{n-1}\right)}\\ {\text{t}}_{\text{1/2}}\text{=}\frac{{\text{2}}_{}^{{}^{\text{n-1}}}\text{-1}}{\left(\text{n-1}\right){\text{k}}_{\text{r}}{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{n-1}}}}\end{array}$

Time is calculated when the concentration of a substance to fall to one third the initial value in an n^th·order reaction.

  $\begin{array}{l}{\left[\text{A}\right]}^{\text{1-n}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}\text{t}\left(\text{n-1}\right)\\ {\left[\left(\frac{1}{3}\right)\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}\text{t}\left(\text{n-1}\right)\\ 3^{n-1}{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{- }{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\text{=}{\text{k}}_{\text{r}}\text{t}\left(\text{n-1}\right)\\ {\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\left(3^{n-1}\text{-1}\right)\text{=}{\text{k}}_{\text{r}}{\text{t}}_{\text{1/2}}\left(\text{n-1}\right)\\ \text{t}\text{=}\frac{{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{1-n}}}\left(3^{n-1}\text{-1}\right)}{{\text{k}}_{\text{r}}\left(\text{n-1}\right)}\\ {\text{t}}_{\text{1/2}}\text{=}\frac{3_{}^{{}^{\text{n-1}}}\text{-1}}{\left(\text{n-1}\right){\text{k}}_{\text{r}}{\left[\text{A}\right]}_{\text{0}}^{{}^{\text{n-1}}}}\end{array}$

Time for the concentration of a substance to fall to one·third the initial value in an n^th·order reaction is shown.