Chapter 15, Problem 136MP

(a)

Interpretation Introduction

Interpretation:

The rate law for the reaction should be determined by using concentration versus time.

  $\text{2A(g)+2B(g)}\to \text{C(g)+2D(g)}$

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

Answer to Problem 136MP

Rate law for the reaction is:

  $\text{Rate = k }\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

Explanation of Solution

Given information:

Data is given as:

Time (s)	Experiment 1 $\left[\text{A}\right]$ mole/L	Experiment 2 $\left[\text{A}\right]$ mole/L
0	$1.0\times {10}^{-2}$	$1.0\times {10}^{-2}$
10	$\text{8}\text{.4}\times {\text{10}}^{-3}$	$5.0\times {\text{10}}^{-3}$
20	$7.1\times {10}^{-3}$	$2.5\times {10}^{-3}$
30	?	$1.3\times {10}^{-3}$
40	$5.0\times {10}^{-3}$	$6.3\times {10}^{-4}$

In experiment 1, concentration of B is 10.0 M and in experiment 2, concentration of B is

20.0 M.

  $\text{Rate}=\frac{-\text{d}\left[\text{A}\right]}{\text{dt}}$

The order of reaction can be determined by the concentration and time data.

The general expression of rate law is expressed as:

  $\text{Rate = k}{\left[\text{A}\right]}^{\text{m}}{\left[\text{B}\right]}^{\text{n}}$

Where, m and n are the experimentally determined values.

In both experiments, the concentration of B is more than the concentration of A, thus $\left[\text{B}\right]$ is assuming constant over experiments. From this, one can obtained pseudo-order rate law equation.

In first experiment, 40 seconds are required to undergo one half-life of reactant A. In experiment 2, the half-life decreases by factor 4 as concentration of B doubles.

This observation implies that the reaction is second order with respect to B whereas the reaction is first order with respect to A as the $\text{ln}\left[\text{A}\right]$ vs time shows straight line graph.

Thus, rate law for the reaction is expressed as:

  $\text{Rate = k }\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

(b)

Interpretation Introduction

Interpretation:

The value of rate constant including units should be calculated.

  $\text{2A(g)+2B(g)}\to \text{C(g)+2D(g)}$

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

The proportionality coefficient which relates the rate of chemical reaction at specific temperature to the concentration of the reaction is known as rate constant.

Answer to Problem 136MP

Rate constant = $1.73\times {10}^{-4}{\text{ L}}^2/{\text{mol}}^{\text{2}}\cdot \text{s}$

Explanation of Solution

Given information:

Data is given as:

Time (s)	Experiment 1 $\left[\text{A}\right]$ mole/L	Experiment 2 $\left[\text{A}\right]$ mole/L
0	$1.0\times {10}^{-2}$	$1.0\times {10}^{-2}$
10	$\text{8}\text{.4}\times {\text{10}}^{-3}$	$5.0\times {\text{10}}^{-3}$
20	$7.1\times {10}^{-3}$	$2.5\times {10}^{-3}$
30	?	$1.3\times {10}^{-3}$
40	$5.0\times {10}^{-3}$	$6.3\times {10}^{-4}$

In experiment 1, concentration of B is 10.0 M and in experiment 2, concentration of B is

20.0 M.

  $\text{Rate}=\frac{-\text{d}\left[\text{A}\right]}{\text{dt}}$

In second experiment, the concentration decreases by half every ten seconds. As the reaction is first order with respect A, the below mathematical expression is used for determining value of rate of constant (k’).

  $k^{\prime }\text{ =}\frac{0.693}{t}$

Put the value of half-life,

  $k^{\prime }\text{ =}\frac{0.693}{10\text{ s}}$

  $k^{\prime }\text{ = 0}{\text{.0693 s}}^{-1}$

Rate law for the given reaction is expressed as:

  $\text{Rate = }k\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

Since, concentration of B is more in comparison to A, thus, rate law is written as:

  $\text{Rate}=k^{\prime }\left[\text{A}\right]$

Where, $k^{\text{'}}=k{\left[\text{B}\right]}^2$

  $k=\frac{k^{\text{'}}}{{\left[\text{B}\right]}^2}$

Put the values,

  $k=\frac{0.0693{\text{ s}}^{-1}}{{\left(20\text{ M}\right)}^2}$

  $k=1.73\times {10}^{-4}{\text{ L}}^2/{\text{mol}}^{\text{2}}\cdot \text{s}$

(c)

Interpretation Introduction

Interpretation:

The concentration of $\left[\text{A}\right]$ in experiment 1 at t = 30 s should be calculated.

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

The concentration of a reaction at any time is calculated by the values of concentration and time or by drawing the graph between concentration and time.

The rate constant expression for first order reaction is:

  $k\text{=}\frac{2.303}{t}\log \frac{C_o}{C}$

Answer to Problem 136MP

Concentration of A for first experiment at t = 30 s is $6.0\times {10}^{-3}\text{ M}$ .

Explanation of Solution

Given information:

Data is given as:

Time (s)	Experiment 1 $\left[\text{A}\right]$ mole/L	Experiment 2 $\left[\text{A}\right]$ mole/L
0	$1.0\times {10}^{-2}$	$1.0\times {10}^{-2}$
10	$\text{8}\text{.4}\times {\text{10}}^{-3}$	$5.0\times {\text{10}}^{-3}$
20	$7.1\times {10}^{-3}$	$2.5\times {10}^{-3}$
30	?	$1.3\times {10}^{-3}$
40	$5.0\times {10}^{-3}$	$6.3\times {10}^{-4}$

In experiment 1, concentration of B is 10.0 M and in experiment 2, concentration of B is

20.0 M.

  $\text{Rate}=\frac{-\text{d}\left[\text{A}\right]}{\text{dt}}$

Since, the reaction is first order with respect to A, thus, the rate constant expression is:

  $k\text{=}\frac{2.303}{t}\log \frac{C_o}{C}$

Where, k = rate constant

t = time

C_o = Initial concentration

C = Concentration at given time.

Put the values from experiment 1,

  $k\text{=}\frac{2.303}{t}\log \frac{C_o}{C}$

  $\text{=}\frac{2.303}{10\text{ s}}\log \frac{1.0\times {10}^{-2}}{8.4\times {10}^{-3}}$

  $\text{=}0.2303\log \left(1.1904\right)$

  $\text{=}1.73\times {10}^{-2}{\text{ s}}^{-1}$

Now, at t = 30 s

  $k\text{=}\frac{2.303}{t}\log \frac{C_o}{C}$

Put the values,

  $1.73\times {10}^{-2}{\text{ s}}^{-1}=\frac{2.303}{30\text{ s}}\log \frac{1.0\times {10}^{-2}}{C}$

  $0.2253=\log \frac{1.0\times {10}^{-2}}{C}$

  $C=6.0\times {10}^{-3}\text{ M}$

Thus, concentration of A for first experiment at t = 30 s is $6.0\times {10}^{-3}\text{ M}$ .

(d)

Interpretation Introduction

Interpretation:

Among the given three mechanisms, the best mechanism for the given reaction should be determined. Also, reason should be explained for excluding any mechanism along with if all the three mechanism are equally good, reason should be explained.

  $\begin{array}{l}i.\text{ 2B}\rightleftharpoons {\text{B}}_2\text{ (fast equilibrium)}\\ {\text{ B}}_2+\text{A}\to \text{E+D (slow)}\\ \text{ E+A}\to \text{C+D (fast)}\end{array}$

  $\begin{array}{l}ii\text{ A+B}\to \text{D+F (slow)}\\ \text{ F+B}\to \text{C+G (fast)}\\ \text{ G+A}\to \text{D (fast)}\end{array}$

  $\begin{array}{l}iii.\text{ A+2B}\to \text{E+D (slow)}\\ \text{ E+A}\to \text{C+D (fast)}\end{array}$

Concept Introduction:

Rate Law can be expressed as an integrated rate law and a differential rate law.

Differential Rate Law: This describes the change in the concentrations of reactant as a function of time.

Integrated Rate Law: This describes the initial concentrations and the measured concentration of one or more reactants as a function of time.

Answer to Problem 136MP

The best mechanism is first mechanism.

  $\begin{array}{l}i.\text{ 2B}\rightleftharpoons {\text{B}}_2\text{ (fast equilibrium)}\\ {\text{ B}}_2+\text{A}\to \text{E+D (slow)}\\ \text{ E+A}\to \text{C+D (fast)}\end{array}$

Explanation of Solution

Given information:

  $\begin{array}{l}i.\text{ 2B}\rightleftharpoons {\text{B}}_2\text{ (fast equilibrium)}\\ {\text{ B}}_2+\text{A}\to \text{E+D (slow)}\\ \text{ E+A}\to \text{C+D (fast)}\end{array}$

  $\begin{array}{l}ii\text{ A+B}\to \text{D+F (slow)}\\ \text{ F+B}\to \text{C+G (fast)}\\ \text{ G+A}\to \text{D (fast)}\end{array}$

  $\begin{array}{l}iii.\text{ A+2B}\to \text{E+D (slow)}\\ \text{ E+A}\to \text{C+D (fast)}\end{array}$

The given reaction is:

  $\text{2A(g)+2B(g)}\to \text{C(g)+2D(g)}$

For the given reaction, rate law is expressed as:

  $\text{Rate = }k\left[\text{A}\right]{\left[\text{B}\right]}^{\text{2}}$

Among the given mechanisms, first and third mechanisms give same rate law to the rate law of the given reaction. Thus, both first and third mechanism is possible. In case of second mechanism, rate law is not same to the rate law of given reaction.

Also, third mechanism is very rare to takes place. Thus, best mechanism is first mechanism.