Chapter 11, Problem 30QAP

Interpretation Introduction

(a)

Interpretation:

To determine the order of reaction with respect to each reactant in following reaction:

${\text{2NO(g)}}_{}\text{+}{\text{O}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{2NO}}_{\text{2}}\text{(g)}$

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 stochiometric coefficient.

Let’s say we have a reaction:

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

Answer to Problem 30QAP

Order of given reaction:

With respect to NO =4

With respect to O₂ =1

Explanation of Solution

Given Information:

Here the chemical reaction is:

${\text{2NO(g)}}_{}\text{+}{\text{O}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{2NO}}_{\text{2}}\text{(g)}$

Experiment No.	Concentration of NO	Concentration of ${\text{O}}_{\text{2}}$	Initial rate (mol/L.min)
1	0.0244	0.0372	$2.20\times {10}^{-3}$
2	0.0244	0.0122	$7.23\times {10}^{-4}$
3	0.0244	0.0262	$1.55\times {10}^{-3}$
4	0.0732	0.0372	$1.98\times {10}^{-2}$

Let’s assume the reaction to be ‘t’ order with respect to NO and ‘y’ order with respect to O₂.

Then, rate law for experiment 1, 2, 3 and 4 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[NO]}}^{\text{t}}{{\text{[O}}_{\text{2}}}^{\text{]}}\\ \text{2}{\text{.20×10}}^{\text{-3}}\text{=}\text{k[0}{\text{.0244]}}^{\text{t}}{\text{[0}}^{\text{.0372]}}\mathrm{......}\text{(1)}\\ \text{7}{\text{.53×10}}^{\text{-4}}\text{=}\text{k[0}{\text{.0244]}}^{\text{t}}{\text{[0}}^{\text{.0122]}}\mathrm{......}\text{(2)}\\ \text{1}{\text{.55×10}}^{\text{-3}}\text{=}\text{k[0}{\text{.0244]}}^{\text{t}}{\text{[0}}^{\text{.0262]}}\text{.}\mathrm{.....}\text{(3)}\\ \text{1}{\text{.98×10}}^{\text{-2}}\text{=}\text{k[0}{\text{.0732]}}^{\text{t}}{\text{[0}}^{\text{.0372]}}\mathrm{......}\text{(4)}\end{array}$

Dividing (1) by (2) to get value of ‘y’.

$\begin{array}{l}\frac{\text{2}{\text{.20×10}}^{\text{-3}}}{\text{7}{\text{.53×10}}^{\text{-4}}}\text{=}{\left(\frac{\text{0}\text{.0372}}{\text{0}\text{.0122}}\right)}^{\text{y}}\\ \Rightarrow 3.01\text{=}{\left(3.01\right)}^{\text{y}}\\ \Rightarrow \text{y=1}\end{array}$

Thus, order with respect to O₂ is 1

Dividing (1) by (4) to get value of ‘t’.

$\begin{array}{l}\frac{\text{2}{\text{.30×10}}^{\text{-4}}}{\text{1}{\text{.98×10}}^{\text{-2}}}\text{=}{\left(\frac{\text{0}\text{.0244}}{\text{0}\text{.0732}}\right)}^{\text{t}}\\ \Rightarrow \text{0}\text{.012}\text{=}{\left(\text{0}\text{.333}\right)}^{\text{t}}\\ \text{taking}\text{natural}\text{log}\text{both}\text{sides:}\\ \Rightarrow \text{ln(0}\text{.012)}\text{=}\text{t×ln(0}\text{.333)}\\ \text{t}\text{=}\frac{\text{-4}\text{.42}}{\text{-1}\text{.10}}\text{=4}\text{.02}\text{»}\text{4}\end{array}$

Thus, order with respect to NO is 4.

Interpretation Introduction

(b)

Interpretation:

To write the rate expression 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 order of reaction.

Let’s say we have a reaction:

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

Answer to Problem 30QAP

Rate law expression for above reaction will be;

$\text{rate}\text{=}{\text{k[NO]}}^4{{\text{[O}}_{\text{2}}\text{]}}^{}$

Explanation of Solution

Here the chemical reaction is:

${\text{2NO(g)}}_{}\text{+}{\text{O}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{2NO}}_{\text{2}}\text{(g)}$

Order of reaction with respect to NO = 4

Order of reaction with respect to O₂ = 1

Let the rate constant be ‘k’.

Then, rate law expression for above reaction will be:

$\text{rate}\text{=}{\text{k[NO]}}^4{{\text{[O}}_{\text{2}}\text{]}}^{}$

Interpretation Introduction

(c)

Interpretation:

To determine the rate constant and its unit 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 order of reaction.

Let’s say we have a reaction:

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

Answer to Problem 30QAP

Rate constant is $1.67\times {10}^5\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}$

And its unit is $\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}$

Explanation of Solution

Here the chemical reaction is:

${\text{2NO(g)}}_{}\text{+}{\text{O}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{2NO}}_{\text{2}}\text{(g)}$

Writing rate law for experiment 1 in above reaction will be;

$\begin{array}{l}\text{rate}\text{=}{\text{k[NO]}}^4{{\text{[O}}_{\text{2}}}^{\text{]}}\\ \text{2}\text{.20}\times {\text{10}}^{\text{-3}}\left(\frac{\text{mol}}{\text{L}\text{.min}}\right)\text{=}\text{k}\left(\text{U}\right)\text{×}{\left[\text{0}\text{.0244}\left(\frac{\text{mol}}{\text{L}}\right)\right]}^{\text{4}}\text{×0}\text{.0372}\left(\frac{\text{mol}}{\text{L}}\right)\\ \text{where}\text{U}\text{represent}\text{unit}\text{of}\text{rate}\text{constant}\text{.}\\ \Rightarrow \text{k}\left(\text{U}\right)\text{=}\frac{\text{2}\text{.20}\times {\text{10}}^{\text{-3}}\left(\frac{\text{mol}}{\text{L}\text{.min}}\right)}{{\left[\text{0}\text{.0244}\left(\frac{\text{mol}}{\text{L}}\right)\right]}^4\times \text{0}\text{.0372}{\left(\frac{\text{mol}}{\text{L}}\right)}^{}}=1.67\times {10}^5\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}\\ \end{array}$

Hence, the rate constant is $1.67\times {10}^5\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}$

And its unit is $\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}$

Interpretation Introduction

(d)

Interpretation:

To determine the rate of reaction at given concentration of reactants.

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 order of reaction.

Let’s say we have a reaction:

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

Answer to Problem 30QAP

Rate of reaction is $7.75\times {\text{10}}^{\text{-5}}\frac{\text{mol}}{\text{L}\text{.min}}$.

Explanation of Solution

Here the chemical reaction is:

${\text{2NO(g)}}_{}\text{+}{\text{O}}_{\text{2}}\text{(g)}\stackrel{}{\to }{\text{2NO}}_{\text{2}}\text{(g)}$

Rate law expression for above reaction:

$\text{rate}\text{=}{\text{k[NO]}}^4{{\text{[O}}_{\text{2}}\text{]}}^{}$

Here we have:

[NO] = 0.0100 M

[O₂ ] = 0.0462 M

Rate constant = $1.67\times {10}^5\frac{{\text{L}}^{\text{4}}}{{\text{mol}}^{\text{4}}\text{.min}}$

Plugging values in rate law as:

$\begin{array}{l}\text{rate}\text{=}1.67\times {10}^5\text{×(0}{\text{.0100)}}^{\text{4}}\text{×(0}\text{.0462)}\\ \Rightarrow \text{rate}\text{=}7.75\times {\text{10}}^{\text{-5}}\frac{\text{mol}}{\text{L}\text{.min}}\end{array}$

Hence, the rate of reaction is $7.75\times {\text{10}}^{\text{-5}}\frac{\text{mol}}{\text{L}\text{.min}}$.