Chapter 17, Problem 17B.16P

Interpretation Introduction

Interpretation:

Integrated expression for a stoichiometry reaction $\text{2A+3B}\to \text{P}$ for a second-order rate law has to be derived and rate law has to be expressed.

Explanation of Solution

Given,

  $\text{2A+3B}\to \text{P}$

Rate equation is shown below,

  $\begin{array}{l}\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{2}}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\\ \text{=}\text{-}\frac{\text{1}}{\text{3}}\frac{\text{d}\left[\text{B}\right]}{\text{dt}}\\ \text{=}\text{+}\frac{\text{d}\left[\text{P}\right]}{\text{dt}}\end{array}$

Rate of change of product:

$\frac{\text{d}\left[\text{P}\right]}{\text{dt}}\text{=}{\text{k}}_{\text{r}}\left[\text{A}\right]\left[\text{B}\right]$

Here,

Initial concentration of reactant: $\text{A =}{\text{A}}_{\text{0}}$

Initial concentration of reactant: $\text{B =}{\text{B}}_{\text{0}}$

Concentration of A at time t $\left[\text{A}\right]\text{ =}{\text{A}}_{\text{0}}\text{- 2x}$

Concentration of A at time t $\left[\text{B}\right]\text{ =}{\text{B}}_{\text{0}}\text{- x}$

  $\begin{array}{l}\frac{\text{d}\left[\text{P}\right]}{\text{dt}}\text{=}{\text{k}}_{\text{r}}\left[\text{A}\right]\left[\text{B}\right]\\ \frac{\text{dx}}{\text{dt}}\text{=}{\text{k}}_{\text{r}}\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{3x}\right)\end{array}$

The reaction is rearranged,

  $\frac{\text{dx}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{3x}\right)}\text{=}{\text{k}}_{\text{r}}\text{dt}$

The reaction is integrated,

  $\frac{\text{dx}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{3x}\right)}\text{=}{\text{k}}_{\text{r}}\text{dt}$

  $\begin{array}{l}{\text{k}}_{\text{r}}\text{dt =}\frac{\text{dx}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{3x}\right)}\\ {\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}{\text{k}}_{\text{r}}\text{dt}}\text{ =}{\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\frac{\text{dx}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{3x}\right)}}\\ {\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}{\text{k}}_{\text{r}}\text{dt}}\text{ =}{\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\left(\frac{\text{6}}{\left(\text{2}{\left[\text{B}\right]}_{\text{0}}\text{- 3}{\left[\text{A}\right]}_{\text{0}}\right)}\right)\left(\frac{\text{1}}{\text{3}\left({\left[\text{A}\right]}_{\text{0}}\text{-2x}\right)}\text{-}\frac{\text{1}}{\text{2}\left({\left[\text{B}\right]}_{\text{0}}\text{-3x}\right)}\right)\text{dx}}\\ {\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}{\text{k}}_{\text{r}}\text{dt}}\text{ =}\left(\frac{\text{-1}}{\left(\text{2}{\left[\text{B}\right]}_{\text{0}}\text{- 3}{\left[\text{A}\right]}_{\text{0}}\right)}\right)\left({\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\frac{\text{dx}}{\text{x-}\left(\frac{\text{1}}{\text{2}}\right){\left[\text{A}\right]}_{\text{0}}}}\text{-}{\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\frac{\text{dx}}{\text{x-}\left(\frac{\text{1}}{\text{2}}\right){\left[\text{B}\right]}_{\text{0}}}}\right)\end{array}$

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

Therefore, the integrated expression for a second-order rate law is,

  ${\text{k}}_{\text{r}}\text{t =}\left(\frac{\text{1}}{\left(\text{3}{\left[\text{A}\right]}_{\text{0}}\text{-}\text{2}{\left[\text{B}\right]}_{\text{0}}\right)}\right)\text{ln}\left(\frac{\left(\text{2x-}{\left[\text{A}\right]}_{\text{0}}\right){\left[\text{B}\right]}_{\text{0}}}{{\left[\text{A}\right]}_{\text{0}}\left(\text{3x}\text{-}{\left[\text{B}\right]}_{\text{0}}\right)}\right)$