Chapter 6, Problem 6.2P

(a)

Interpretation Introduction

Interpretation:

The order of the given reaction with respect to the given complex and the given reactant has to be found.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area…

General rate reaction is,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Explanation of Solution

Record the given data’s

$\begin{array}{cccc}{\text{[Complex]/(mmol dm}}^{\text{-3}}\text{)}& 8.01& 9.22& 12.11\\ \begin{array}{l}{\text{v}}_{\text{0}}{\text{/(mol dm}}^{\text{-3}}{\text{s}}^{\text{-1}}\text{)}\left(a\right)\\ \left(b\right)\end{array}& \begin{array}{l}125\\ 640\end{array}& \begin{array}{l}144\\ 730\end{array}& \begin{array}{l}190\\ 960\end{array}\end{array}$

Given,

$\begin{array}{l}\left[Y\right]=2.7mmold{m}^{-3}\\ Rate=R{\left[X\right]}^a{\left[Y\right]}^b\\ 125=R{\left(8.01\right)}^a\times \left(2.7\right)mmold{m}^{-3}\\ 144=R{\left(9.22\right)}^a\times \left(2.7\right)mmold{m}^{-3}\end{array}$

Therefore, the rate is

$\begin{array}{r}\frac{125}{144}={\left(\frac{8.01}{9.22}\right)}^a\\ 0.868={\left(0.868\right)}^a\\ \underset{\_}{a=1}\end{array}$

Thus, the complete order is one. That means this is first order reaction.

(b)

Interpretation Introduction

Interpretation:

The rate constant for the given reaction has to be found.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and $\text{pH}$ of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area.

General rate reaction is,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

The above expression is called integrated rate law for first order reaction.

Explanation of Solution

Record the given data’s

$\begin{array}{cccc}{\text{[Complex]/(mmol dm}}^{\text{-3}}\text{)}& 8.01& 9.22& 12.11\\ \begin{array}{l}{\text{v}}_{\text{0}}{\text{/(mol dm}}^{\text{-3}}{\text{s}}^{\text{-1}}\text{)}\left(a\right)\\ \left(b\right)\end{array}& \begin{array}{l}125\\ 640\end{array}& \begin{array}{l}144\\ 730\end{array}& \begin{array}{l}190\\ 960\end{array}\end{array}$

Given,

$\begin{array}{l}\left[Y\right]=6.1mmold{m}^{-3}\\ Rate=R{\left[X\right]}^a{\left[Y\right]}^b\\ 125=R{\left(8.01\right)}^a\times {\left(2.7\right)}^b\\ 640=R{\left(8.01\right)}^b\times {\left(6.1\right)}^b\end{array}$

Therefore, the rate is

$\begin{array}{r}\frac{125}{640}={\left(\frac{2.7}{6.0}\right)}^b\\ 0.195={\left(0.443\right)}^b\\ \underset{\_}{b=2}\end{array}$

The reaction order is $Y=2$

Calculation for rate constant for given reaction:

$\begin{array}{l}125=R{\left(8.01\right)}^1\times {\left(2.7\right)}^2\\ R=\frac{125}{58.3929}\\ R=2.1\times {10}^{9}d{m}^{6}mo{l}^{-2}{s}^{-1}\end{array}$

Therefore, the rate constant is $\underset{\_}{R=2.1\times {10}^{9}d{m}^{6}mo{l}^{-2}{s}^{-1}}$