Chapter 17, Problem 17B.17P

(a)

Interpretation Introduction

Interpretation:

The integrated form of a third-order rate law is derived and the reactant present initially in their stoichiometric proportions in terms of rate law has to be expresses.

Explanation of Solution

Given,

  $\text{2A+B}\to \text{P}$

Rate equation is shown below,

  $\begin{array}{l}\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{2}}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\\ \text{=}\frac{\text{d}\left[\text{B}\right]}{\text{dt}}\\ \text{rate=}\text{k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\\ \text{-}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{2k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\\ \frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{-2k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\end{array}$

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 $\left[\text{B}\right]\text{ =}{\text{B}}_{\text{0}}\text{- x}$

  $\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{-2k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]$

The reaction is rearranged,

  $\begin{array}{l}\frac{\text{d}\left[\text{A}\right]}{\text{dt}}\text{=}\text{-2k}{\left[\text{A}\right]}^{\text{2}}\left[\text{B}\right]\\ \text{-2}\frac{\text{dx}}{\text{dt}}\text{=}\text{-2k}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}\left({\left[\text{B}\right]}_{\text{0}}\text{-}\text{x}\right)\\ \frac{\text{dx}}{\text{dt}}\text{=}\text{k}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}\left(\frac{\text{1}}{\text{2}}{\left[\text{A}\right]}_{\text{0}}\text{-}\text{x}\right)\\ \frac{\text{dx}}{\text{dt}}\text{=}\frac{\text{1}}{\text{2}}\text{k}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{3}}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The rate law in terms of the reactant presents initially in twice that amount has to be expressed.

Explanation of Solution

The reaction is integrated,

  $\begin{array}{l}\frac{1}{2}k{\displaystyle \underset{\text{0}}{\overset{\text{t}}{\int }}\text{dt}}\text{ =}{\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\frac{\text{dx}}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^3}}\\ \text{kt =}\frac{1}{4}\left[{\left(\frac{1}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}\right)}^2-{\left(\frac{1}{{\left[\text{A}\right]}_{\text{0}}}\right)}^2\right]\\ \text{kt =}\frac{2x\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}{{\left[\text{A}\right]}_0^2{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2}\end{array}$

Where, ${\left[\text{A}\right]}_{\text{0}}\text{ =}{\left[\text{B}\right]}_{\text{0}}$, therefore,

  $\begin{array}{l}\frac{dx}{\text{dt}}\text{ =}k{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2\left({\left[\text{B}\right]}_{\text{0}}\text{- x}\right)\\ \frac{dx}{\text{dt}}\text{ =}k{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)\\ \text{kt =}{\displaystyle \underset{0}{\overset{x}{\int }}\frac{dx}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}}\\ \frac{1}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}=\frac{\alpha }{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^2}+\frac{\beta }{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}+\frac{\lambda }{\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}\end{array}$

By using partial fraction,

  $\begin{array}{l}\frac{\text{1}}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}\text{=}\frac{\text{α}}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}}\text{+}\frac{\text{β}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}\text{+}\frac{\text{γ}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}\\ \text{α}\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)\text{+β}\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)\text{+γ}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}\text{=1}\\ \left({\left[\text{A}\right]}_{\text{0}}\text{α}\text{+}{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\text{β+}{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\text{γ}\right)\text{-}\left(\text{α}\text{+3β}{\left[\text{A}\right]}_{\text{0}}^{}\text{+4γ}{\left[\text{A}\right]}_{\text{0}}^{}\right)\text{x+}\left(\text{2β+4γ}\right){\text{x}}^{\text{2}}\text{=1}\\ {\left[\text{A}\right]}_{\text{0}}\text{α}\text{+}{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\text{β+}{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\text{γ}\text{=}\text{1}\\ \text{α}\text{+3}{\left[\text{A}\right]}_{\text{0}}^{}\text{β+}{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\text{γ}\text{=}\text{0}\\ \text{3β+4γ}\text{=}\text{0}\end{array}$

Therefore,

  $\text{α}\text{=}\frac{2}{{\left[\text{A}\right]}_{\text{0}}},\text{β}=-\frac{2}{{\left[\text{A}\right]}_0^2}\text{ and}\text{γ}=\frac{1}{{\left[\text{A}\right]}_0^2}$

Now,

  $\begin{array}{l}\text{kt}\text{=}{\displaystyle \underset{\text{0}}{\overset{\text{x}}{\int }}\frac{\frac{\text{2}}{{\left[\text{A}\right]}_{\text{0}}}}{{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}^{\text{2}}}\text{-}\frac{\frac{\text{2}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}\text{+}\frac{\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}}\\ \text{kt =}{\left(\frac{\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}}}{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}\text{+}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}}\text{ln}\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)\text{-}\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}}\text{ln}\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)\right)}_{\text{0}}^{\text{x}}\\ \text{kt =}\left(\frac{\text{2x}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}\right)\text{+}\left(\frac{\text{1}}{{\left[\text{A}\right]}_{\text{0}}^{\text{2}}}\right)\text{ln}\left(\frac{\left({\left[\text{A}\right]}_{\text{0}}\text{- 2x}\right)}{\left({\left[\text{A}\right]}_{\text{0}}\text{- x}\right)}\right)\end{array}$