Chapter 13, Problem 13.64QA

Interpretation Introduction

To calculate:

The rate law and the rate constant and rate of appearance of ${CO}_2$ from initial rate and concentration of the ${\mathrm{N}\mathrm{O}}_2$ and $\mathrm{C}\mathrm{O}$ at $225 ℃.$

Answer to Problem 13.64QA

Solution:

a) $Rate=k {\left[{NO}_2\right]}^2$

b) $k=2.08 \times {10}^{-4}{M}^{-1}{s}^{-1}$ at $225 ℃.$

c) $Rate of appearance of {CO}_2=5.20 \times {10}^{-5} M{s}^{-1}$ at $225 ℃.$

Explanation of Solution

1) Concept:

There are two methods to determine the order of the reaction.

1. Graphical method - Plot the graph, and match it with the reference plot of order of reaction

2. Initial rate method - Determine the order of individual from ratio of two rate laws by keeping one constant.

In this problem, we are given the initial rates of the experiments. So we can use initial rate method to determine the order of the ${\mathrm{N}\mathrm{O}}_2$ and $CO$ from the general rate law equation.

We can calculate the rate of appearance from the rate of disappearance from the balanced reaction.

2) Given:

Experiment	${\left[{\mathrm{N}\mathrm{O}}_2\right]}_0$(M)	${\left[\mathrm{C}\mathrm{O}\right]}_0$(M)	Initial rate (M/s)
1	0.263	0.826	1.44 x 10-5
2	0.263	0.413	1.44 x 10-5
3	0.526	0.413	5.76 x 10-5

3) Formulae:

The general rate law equation for given reaction is

$Rate=k {\left[{NO}_2\right]}^m{\left[CO\right]}^n$

m and n are the order of the ${\mathrm{N}\mathrm{O}}_2$ and $\mathrm{C}\mathrm{O}$ respectively

4) Calculation

We are asked to determine the order of the reaction and rate constant from the initial rate and initial concentration of ${NO}_2$ and $\mathrm{C}\mathrm{O}$

The general rate law equation for given reaction is

$Rate=k {\left[{NO}_2\right]}^m{\left[CO\right]}^n$

We can calculate the m, n, and k from the initial rate values and initial concentrations.

a) To determine the value of m, we will select the pair of experiments with the same concentration of CO, so they can cancel out.

Experiment 3 and experiment 2 has the same value of CO. So, we can take the ratio of rate3 and rate 2 to calculate the order of NO₂

$\frac{{Rate}_3}{{Rate}_2}=\frac{k {{\left[{NO}_2\right]}_3}^m{{\left[CO\right]}_3}^n}{k {{\left[{NO}_2\right]}_2}^m{{\left[CO\right]}_2}^n}$

$\frac{5.75 \times {10}^{-5} M/s}{1.44 \times {10}^{-5} M/s }=\frac{ k {\left[0.526 M\right]}^m{\left[0.413 M\right]}^n}{k {\left[0.263 M\right]}^m{\left[0.413 M\right]}^n}$

$4=\frac{{\left[0.526\right]}^m}{{\left[0.263\right]}^m}$

$4= {\left(2\right)}^m$

$m=2$

So the order of reaction with respect toNO₂is2.

Experiment 1 and experiment 2 have the same value of NO₂. So, we can take the ratio of rate 1 and rate 2 to calculate the order of CO

$\frac{{Rate}_1}{{Rate}_2}=\frac{k {{\left[{NO}_2\right]}_1}^m{{\left[CO\right]}_1}^n}{k {{\left[{NO}_2\right]}_2}^m{{\left[CO\right]}_2}^n}$

$\frac{1.44 \times {10}^{-5} M/s}{1.44 \times {10}^{-5} M/s }=\frac{ k {\left[0.263 M\right]}^m{\left[0.826 M\right]}^n}{k {\left[0.263 M\right]}^m{\left[0.413 M\right]}^n}$

$1=\frac{{\left[0.826\right]}^n}{{\left[0.413\right]}^n}$

$1= {\left(2\right)}^n$

$n=0$

So the order of reaction with respect to COis 0.

Therefore, the rate law for the given reaction is

$Rate=k {\left[{NO}_2\right]}^2{\left[CO\right]}^0$

$Rate=k {\left[{NO}_2\right]}^2$

b) Now we will calculate the k value at $225 ℃.$ from the experiment 1.

$1.44 \times {10}^{-5} M/s=k {\left[0.263 M\right]}^2$

$k= \frac{1.44 \times {10}^{-5} M/s}{{\left( 0.263 M\right)}^2}$

$k= 2.08 \times {10}^{-4}{M}^{-1}{s}^{-1}$

Therefore, the rate constant at $225 ℃.$ is $2.08 \times {10}^{-4}{M}^{-1}{s}^{-1}$

c) To calculate the rate of appearance of CO₂ , at $225 ℃.$ we are given [NO₂] = [CO] = 0.500 M.

$Rate of disappearance of {NO}_2 =\frac{d\left[N{O}_2\right]}{dt}=2.08 \times {10}^{-4}{M}^{-1}{s}^{-1} \times {\left(0.500 M\right)}^2$

$Rate of disappearance {NO}_2=\frac{d\left[N{O}_2\right]}{dt}=5.20 \times {10}^{-5} M{s}^{-1}$

From the balanced reaction, rate of disappearance of NO₂ = rate of appearance of CO₂.

$Rate of appearance of {CO}_2=5.20 \times {10}^{-5} M{s}^{-1}$ at $225 ℃.$

Conclusion:

We have calculated the rate law from the initial rate method and initial concentration given in the problem. Wehave also calculated the rate constant at $225 ℃$ from the rate law and initial concentration at $225 ℃$ from experiments. We calculated the rate of appearance at $225 ℃$ from the rate and concentration of reactant at $225 ℃.$