Chapter 11, Problem 21QAP

The reaction

$\text{ICl}\left(g\right)+\frac{1}{2}{\text{H}}_2\left(g\right)\to \frac{1}{2}{\text{I}}_2\left(g\right)+\text{HCl}\left(g\right)$is first-order in both reactants. At a certain temperature, the rate of the reaction is $2.86\times {10}^{-5}\text{mol}/\text{L}\cdot \text{s}$ when $\left[\text{ICl}\right]=0.0500M$ and $\left[{\text{H}}_2\right]=0.0150M$.

(a) What is the value of k at that temperature?

(b) At what concentration of hydrogen is the rate $2.66\times {10}^{-4}\text{mol}/\text{L}\cdot \text{s}$ and $\left[\text{ICl}\right]=0.100M$?

(c) What is $\left[\text{ICl}\right]$ when the rate is $0.0715\text{mol}/\text{L}\cdot \text{s}$ and ${\text{H}}_2=3\left[\text{ICl}\right]$?

Interpretation Introduction

(a)

Interpretation:

To determine the value of rate constant for the given reaction.

Concept introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 21QAP

Rate constant for the given reaction is 3.81×10-2 L/mol.s

Explanation of Solution

Here the chemical reaction is:

$\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}$

Since the order of reaction with respect to ICl and H₂ is first order. Thus, rate law equation will look like:

$\begin{array}{l}\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}\\ \text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}]\mathrm{........}(1)\end{array}$

Here we have:

[ICl] = 0.0500 M

[H₂ ] = 0.0150 M

Rate of reaction = 2.86×10-5 mol/L.s

Plugging value of rate of reaction in equation 1 to get the value of rate constant as:

$\begin{array}{l}\text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}]\\ 2.86\times {10}^{-5}\text{mol/L}\text{.s}\text{=}\text{k}\times \text{(0}\text{.0500)}\times (\text{0}\text{.0150)}\\ \Rightarrow \text{k}\text{=}3.81\times {10}^{-2}\text{L/mol}\text{.s}\\ \end{array}$

Hence, the rate constant for the given reaction is 3.81×10-2 L/mol.s

Interpretation Introduction

(b)

Interpretation:

To determine the concentration of H₂ gas when rate of reaction is 2.66×10-4 mol/L.s and concentration of ICl is 0.100 M.

Concept Introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 21QAP

The concentration of hydrogen gas is 0.0698 mol/L.

Explanation of Solution

Here the chemical reaction is:

$\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}$

Since the order of reaction with respect to ICl and H₂ is first order. Thus, rate law equation will look like:

$\begin{array}{l}\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}\\ \text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}]\mathrm{........}(1)\end{array}$

Here we have:

[ICl] = 0.100 M

[H₂ ] = let it ‘t’ M

Rate of reaction = 2.66×10-4 mol/L.s

Rate constant = 3.81×10-2 L/mol.s

Plugging value of rate of reaction in equation 1 to get the value of rate constant as:

$\begin{array}{l}\text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}]\\ 2.66\times {10}^{-4}(\text{mol/L}\text{.s)}\text{=}\left(3.81\times {10}^{-2}(\text{L/mol}\text{.,s})\right)\times \left(\text{0}\text{.100}\text{(mol/L)}\right)\times (\text{t)}\\ \Rightarrow \text{t}\text{=}\text{0}\text{.0698}\text{mol/L}\\ \end{array}$

Hence, the concentration of hydrogen gas is 0.0698 mol/L.

Interpretation Introduction

(c)

Interpretation:

To determine the concentration of ICl when rate of reaction is 0.0715 mol/L.s and concentration of H₂ is 3 times the concentration ICl i.e. [H₂ ] = 3×[ICl].

Concept Introduction:

Rate of a chemical reaction: It tells us about the speed at which the reactants are converted into products.

Mathematically, rate of reaction is directly proportional to the product of concentration of each reactant raised to the power equal to their respective stoichiometric coefficients.

Let’s say we have a reaction:

$\begin{array}{l}\text{aA+bB}\stackrel{}{\to }\text{cC+dD}\\ \text{then,}\\ \text{rate}\text{α}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \Rightarrow \text{rate}{\text{=K}}_{\text{f}}{\text{[A]}}^{\text{a}}{\text{[B]}}^{\text{b}}\\ \text{where}{\text{K}}_{\text{f}}\text{= rate}\text{constant}\\ \text{a and b are order of reaction with respect to A and B }\\ \end{array}$

Answer to Problem 21QAP

The concentration of ICl is 0.791 mol/L

Explanation of Solution

Here the chemical reaction is:

$\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}$

Since the order of reaction with respect to ICl and H₂ is first order. Thus, rate law equation will look like:

$\begin{array}{l}\text{ICl(g)}\text{+}\frac{\text{1}}{\text{2}}{\text{H}}_{\text{2}}\text{(g)}\stackrel{}{\to }\frac{\text{1}}{\text{2}}{\text{I}}_{\text{2}}\text{(g)}\text{+}\text{HCl}\text{(g)}\\ \text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}]\mathrm{........}(1)\end{array}$

Here we have:

[H₂ ] = 3×[ICl]

Rate of reaction = 0.0715 mol/L.s

Rate constant = 3.81×10-2 L/mol.s

Plugging value of rate of reaction in equation 1 to get the value of rate constant as:

$\begin{array}{l}\text{rate}\text{=}{\text{k[ICl][H}}_{\text{2}}\text{]}\\ \text{0}\text{.0715}\text{=}\left(\text{3}{\text{.81×10}}^{\text{-2}}\right)\text{×}\left(\text{[ICl]}\right)\text{×}\left(\text{3×[ICl]}\right)\\ \Rightarrow {\text{[ICl]}}^{\text{2}}\text{=}\frac{\text{0}\text{.0715}}{\text{3}{\text{.81×10}}^{\text{-2}}\text{×}\text{3}}\\ \Rightarrow \text{[ICl]}\text{=}\sqrt{\text{0}\text{.626}}\\ \Rightarrow \text{[ICl]}=0.791\text{mol/L}\end{array}$

Hence, the concentration of ICl is 0.791 mol/L