Chapter 6, Problem 6.3P

(a)

Interpretation Introduction

Interpretation:

Rate law has to be determined for the given reaction.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

$\text{aA}\text{+}\text{bB}+cC\stackrel{}{\to }\text{Products}$

Where,

A and B are reactants

a and b are stoichiometric coefficients

$\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

Where,

k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Explanation of Solution

Given information,

The reaction is ${\text{2 ICl}}_{\left(\text{g}\right)}{\text{ + H}}_{\text{2}\left(g\right)}\underset{}{\overset{}{\to }}{\text{ I}}_{\text{2}}{}_{\left(g\right)}{\text{ + 2 HCl}}_{\left(g\right)}$

Rate law of the equation is $k={\left[ICl\right]}^x{\left[H_2\right]}^y$

  $\begin{array}{cccc}\text{Expt}& {\text{[ICl]}}_{\text{0}}& {{\text{[H}}_{\text{2}}\text{]}}_{\text{0}}& {\text{v}}_{\text{0}}\\ 1& 1.5& 1.5& 3.7\times {10}^{-7}\\ 2& 3.0& 1.5& 7.4\times {10}^{-7}\\ 3& 3.0& 4.5& 22\times {10}^{-7}\\ 4& 4.5& 2.7& ?\end{array}$

Calculate the value of x from experiment 2 to 1

  $\begin{array}{l}x\text{=}\frac{\text{log}\left(\frac{{\text{Rate}}_2}{{\text{Rate}}_1}\right)}{\text{log}\left(\frac{{\left[ICl\right]}_2}{{\left[ICl\right]}_1}\right)}\text{=}\frac{\text{log}\left(\frac{7.4\text{×}{\text{10}}^{-\text{7}}}{3.7\text{×}{\text{10}}^{-\text{7}}}\right)}{\text{log}\left(\frac{3.0}{\text{1}\text{.5}}\right)}\\ \\ \text{=}\frac{\text{log}\left(2\right)}{\text{log}\left(2\right)}\text{=}\text{1}\end{array}$

Calculate the value of y from experiment 3 to 2

  $\begin{array}{l}y\text{=}\frac{\text{log}\left(\frac{{\text{Rate}}_3}{{\text{Rate}}_2}\right)}{\text{log}\left(\frac{{\left[{\text{H}}_2\right]}_3}{{\left[{\text{H}}_2\right]}_2}\right)}\text{=}\frac{\text{log}\left(\frac{22\text{×}{\text{10}}^{-\text{7}}}{7.4\text{×}{\text{10}}^{-\text{7}}}\right)}{\text{log}\left(\frac{\text{4}\text{.5}}{\text{1}\text{.5}}\right)}\\ \\ \text{=}\frac{\text{log}\left(3\right)}{\text{log}\left(\text{3}\right)}\text{=}\text{1}\end{array}$

Therefore, the rate law of the given reaction is $\text{Rate}\text{=}\text{k}\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]$.

(b)

Interpretation Introduction

Interpretation:

Value of rate with its unit has to be determined for the given reaction.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

$\text{aA}\text{+}\text{bB}+cC\stackrel{}{\to }\text{Products}$

Where,

A and B are reactants

a and b are stoichiometric coefficients

  $\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

Where,

k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Explanation of Solution

Given information,

The reaction is ${\text{2 ICl}}_{\left(\text{g}\right)}{\text{ + H}}_{\text{2}\left(g\right)}\underset{}{\overset{}{\to }}{\text{ I}}_{\text{2}}{}_{\left(g\right)}{\text{ + 2 HCl}}_{\left(g\right)}$

  $\begin{array}{cccc}\text{Expt}& {\text{[ICl]}}_{\text{0}}& {{\text{[H}}_{\text{2}}\text{]}}_{\text{0}}& {\text{v}}_{\text{0}}\\ 1& 1.5& 1.5& 3.7\times {10}^{-7}\\ 2& 3.0& 1.5& 7.4\times {10}^{-7}\\ 3& 3.0& 4.5& 22\times {10}^{-7}\\ 4& 4.5& 2.7& ?\end{array}$

The rate law of the given reaction is $\text{Rate}\text{=}\text{k}\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]$.

Calculate the value of rate constant from experiment 1

  $\begin{array}{c}\text{Rate}\text{=}\text{k}\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]\\ \\ \text{Rate}\text{constant, k}\text{=}\frac{\text{Rate}}{\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]}\text{=}\frac{\left(7.4\times {10}^{-7}\text{mol}{\text{.dm}}^{-3}.s^{-1}\right)}{\left(3.0\times {10}^{-3}\text{ mol}{\text{.dm}}^{-3}\right)\left(1.5\times {10}^{-3}\text{ mol}{\text{.dm}}^{-3}\right)}\\ \\ \text{=}0.164mo{l}^{-1}d{m}^3{s}^{-1}\\ \\ \text{=}0.164M^{-1}{s}^{-1}\end{array}$

Therefore, the value of rate constant is $0.164M^{-1}{s}^{-1}$.

(c)

Interpretation Introduction

Interpretation:

Rate reaction for the experiment 4 has to be predicted.

Concept introduction:

Rate law: It is an equation that related to the dependence of the reaction rate on the concentration of each substrate (reactants).

For a reaction,

$\text{aA}\text{+}\text{bB}+cC\stackrel{}{\to }\text{Products}$

Where,

A and B are reactants

a and b are stoichiometric coefficients

  $\text{Rate}\text{=}\text{-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{k}{\left[\text{A}\right]}^{\text{x}}{\left[\text{B}\right]}^{\text{y}}{\left[\text{C}\right]}^z$

Where,

k is the rate constant

Rate constant: Rate constant is an expression used to relate the rate of a reaction to the concentration of reactants participating in the reaction.

Explanation of Solution

Given information,

The reaction is ${\text{2 ICl}}_{\left(\text{g}\right)}{\text{ + H}}_{\text{2}\left(g\right)}\underset{}{\overset{}{\to }}{\text{ I}}_{\text{2}}{}_{\left(g\right)}{\text{ + 2 HCl}}_{\left(g\right)}$

  $\begin{array}{cccc}\text{Expt}& {\text{[ICl]}}_{\text{0}}& {{\text{[H}}_{\text{2}}\text{]}}_{\text{0}}& {\text{v}}_{\text{0}}\\ 1& 1.5& 1.5& 3.7\times {10}^{-7}\\ 2& 3.0& 1.5& 7.4\times {10}^{-7}\\ 3& 3.0& 4.5& 22\times {10}^{-7}\\ 4& 4.5& 2.7& ?\end{array}$

The rate law of the given reaction is $\text{Rate}\text{=}\text{k}\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]$.

Value of rate constant is $0.164M^{-1}{s}^{-1}$.

Calculate the value of rate constant from experiment 1

  $\begin{array}{c}\text{Rate}\text{=}\text{k}\left[\text{ICl}\right]\left[{\text{H}}_{\text{2}}\right]\\ \\ =\left(0.164M^{-1}{s}^{-1}\right)\left(4.7\times {10}^{-3}\text{ mol}{\text{.dm}}^{-3}\right)\left(2.7\times {10}^{-3}\text{ mol}{\text{.dm}}^{-3}\right)\\ \\ =\text{ 2}\text{.1}\times {\text{10}}^{-6}\text{ mol}{\text{.dm}}^{-3}.s^{-1}\end{array}$

Rate reaction for the experiment 4 is $\text{2}\text{.1}\times {\text{10}}^{-6}\text{ mol}{\text{.dm}}^{-3}.s^{-1}$