Chapter 11, Problem 22QAP

The hypothetical reaction

$\text{X}\left(g\right)+\frac{1}{2}\text{Y}\left(g\right)\to \text{products}$is first-order in X and second-order in Y. The rate of the reaction is 0.00389 mol/L $\cdot$

min when [X] is 0.150 M and [Y] is 0.0800 M.

(a) What is the value for k?

(b) At what concentration of [Y] is the rate 0.00948 mol/L $\cdot$

min and [X] is 0.0441 M?

(c) At what concentration of [X] is the rate 0.0124 mol/L $\cdot$

min and $\left[\text{Y}\right]=2\left[\text{X}\right]$?

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 22QAP

Rate constant for the given reaction is 4.05 L/mol.min

Explanation of Solution

Here the chemical reaction is:

$\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}$

Since the order of reaction with respect to X and Y is first order and second order respectively. Thus, rate law equation will look like:

$\begin{array}{l}\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}\\ \text{rate}\text{=}\text{k[X][Y}{]}^2\mathrm{........}(1)\end{array}$

Here we have:

[X] = 0.150 M

[Y] = 0.0800 M

Rate of reaction = 0.00389 mol/L.min

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

$\begin{array}{l}\text{rate}\text{=}\text{k[X][Y}{]}^2\\ 0.00389\text{=}\text{k}\times \text{(0}\text{.150)}\times {(}^{\text{0}}\\ \Rightarrow \text{k}\text{=}4.05\text{L/mol}\text{.min}\\ \end{array}$

Hence, the rate constant for the given reaction is 4.05 L/mol.min

Interpretation Introduction

(b)

Interpretation:

To determine the concentration of Y when rate of reaction is 0.00948 mol/L.min and concentration of X is 0.0441 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 22QAP

The concentration of Y is 0.230 mol/L.

Explanation of Solution

Here the chemical reaction is:

$\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}$

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

$\begin{array}{l}\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}\\ \text{rate}\text{=}\text{k[X][Y}{]}^2\mathrm{........}(1)\end{array}$

Here we have:

[X] = 0.0441 M

Rate of reaction = 0.00948 mol/L.min

Rate constant = 4.05 L/mol.s

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

$\begin{array}{l}\text{rate}\text{=}\text{k[X][Y}{]}^2\\ 0.00948\text{=}(4.05)\times \text{(0}\text{.0441)}\times {\text{[Y}}^{]}\\ \Rightarrow {\text{[Y}}^{]}\text{=}0.0531\\ \Rightarrow \text{[Y}]\text{=}\sqrt{\text{0}\text{.0531}}=0.230M\\ \end{array}$

Hence, the concentration of Y is 0.230 mol/L.

Interpretation Introduction

(c)

Interpretation:

To determine the concentration of X when rate of reaction is 0.0124 mol/L.min and concentration of Y is 2 times the concentration X i.e., [Y] = 2×[X].

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 22QAP

The concentration of X is $0.09145$ mol/L

Explanation of Solution

Here the chemical reaction is:

$\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}$

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

$\begin{array}{l}\text{X}\text{+}\frac{\text{1}}{\text{2}}\text{Y}\stackrel{}{\to }\text{products}\\ \text{rate}\text{=}\text{k[X][Y}{]}^2\mathrm{........}(1)\end{array}$

Here we have:

[Y] = 2[X]

Rate of reaction = 0.0124 mol/L.min

Rate constant = 4.05 L/mol.min

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

$\begin{array}{l}\text{rate}\text{=}{\text{k[X][Y]}}^{\text{2}}\\ \text{0}\text{.0124}\text{=}\left(4.05\right)\text{×}\left(\text{[X]}\right)\text{×}{\left(\text{2[X]}\right)}^2\\ \Rightarrow {\text{[X]}}^3\text{=}\frac{\text{0}\text{.0124}}{\text{4}\text{.05×}4}\\ \Rightarrow \text{[X]}\text{=}\sqrt[3]{7.65\times {10}^{-4}}=0.09145\text{M}\\ \end{array}$

Hence, the concentration of X is $0.09145$ mol/L