Chapter 7, Problem 7A.15E

(a)

Interpretation Introduction

Interpretation:

The order with respect to each reactant and the overall order of the reaction have to be determined.

Concept Introduction:

The rate of the reaction is referred to the change in the molar concentration in the distinct interval of time.  According to the rate law, the rate of the reaction is directly proportional to the initial concentration of the reactant of the reaction.

Answer to Problem 7A.15E

The order with respect to $\text{A}$, $\text{B}$ and $\text{C}$ is $1$, $2$ and $0.009$ respectively and the overall order of the reaction is $3.009$.

Explanation of Solution

The given chemical reaction is shown below.

  $\text{2A}\left(\text{g}\right)+2\text{B}\left(\text{g}\right)+\text{C}\left(\text{g}\right)\to 3\text{G}\left(\text{g}\right)+4\text{F}\left(\text{g}\right)$

Suppose the order of the reaction with respect to $\text{A}$, $\text{B}$ and $\text{C}$ is $a$, $b$ and $c$ respectively.

Therefore, the generic rate law expression for the given chemical reaction is shown below.

  $\text{Rate}=k_r{\left[\text{A}\right]}^a{\left[\text{B}\right]}^b{\left[\text{C}\right]}^c$        (1)

Substitute the values of $\left[\text{A}\right]$, $\left[\text{B}\right]$ and $\left[\text{C}\right]$ from the given data for experiment 1 in equation (1).

  $\text{2}\text{.0}=k_r{\left(10\right)}^a{\left(100\right)}^b{\left(700\right)}^c$        (2)

Substitute the values of $\left[\text{A}\right]$, $\left[\text{B}\right]$ and $\left[\text{C}\right]$ from the given data for experiment 2 in equation (1).

  $\text{4}\text{.0}=k_r{\left(20\right)}^a{\left(100\right)}^b{\left(300\right)}^c$        (3)

Substitute the values of $\left[\text{A}\right]$, $\left[\text{B}\right]$ and $\left[\text{C}\right]$ from the given data for experiment 3 in equation (1).

  $\text{16}=k_r{\left(20\right)}^a{\left(200\right)}^b{\left(200\right)}^c$        (4)

Substitute the values of $\left[\text{A}\right]$, $\left[\text{B}\right]$ and $\left[\text{C}\right]$ from the given data for experiment 4 in equation (1).

  $2.0=k_r{\left(10\right)}^a{\left(100\right)}^b{\left(400\right)}^c$        (5)

Divide equation (2) and equation (5) to calculate the value of $c$.

    $\begin{array}{c}\frac{2.0}{2.0}=\frac{k_r{\left(10\right)}^a{\left(100\right)}^b{\left(700\right)}^c}{k_r{\left(10\right)}^a{\left(100\right)}^b{\left(400\right)}^c}\\ 1={\left(1.75\right)}^c\\ c=0\end{array}$

Therefore, the order with respect to $\text{C}$ is $0$.

Divide equation (2) and equation (3) to calculate the value of $a$.

  $\begin{array}{c}\frac{2.0}{4.0}=\frac{k_r{\left(10\right)}^a{\left(100\right)}^b{\left(700\right)}^c}{k_r{\left(20\right)}^a{\left(100\right)}^b{\left(300\right)}^c}\\ \frac{1}{2}=\frac{k_r{\left(10\right)}^a{\left(700\right)}^{0.009}}{k_r{\left(20\right)}^a{\left(300\right)}^{0.009}}\\ \frac{1}{2}={\left(\frac{1}{2}\right)}^a\\ a=1\end{array}$

Therefore, the order with respect to $\text{A}$ is $1$.

Divide equation (2) and equation (4) to calculate the value of $a$.

  $\begin{array}{c}\frac{4.0}{16}=\frac{k_r{\left(20\right)}^a{\left(100\right)}^b{\left(300\right)}^c}{k_r{\left(20\right)}^a{\left(200\right)}^b{\left(200\right)}^c}\\ \frac{1}{4}={\left(\frac{1}{2}\right)}^b{\left(\frac{3}{2}\right)}^{0.009}\\ \frac{1}{4}={\left(\frac{1}{2}\right)}^b\\ b=2\end{array}$

Therefore, the order with respect to $\text{B}$ is $2$.

Thus, the overall order of the reaction is $1+2+0.009=3.009$.

(b)

Interpretation Introduction

Interpretation:

The expression for the rate law has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7A.15E

The rate law for the given reaction, $\text{R}=k_r\left[\text{A}\right]{\left[\text{B}\right]}^2{\left[\text{C}\right]}^{0.009}$.

Explanation of Solution

The order with respect to $\text{A}$, $\text{B}$ and $\text{C}$ is $1$, $2$ and $0.009$ respectively.

Substitute the value of $a$ and $b$ in equation (1).

  $\text{Rate}=k_r\left[\text{A}\right]{\left[\text{B}\right]}^2{\left[\text{C}\right]}^{0.009}$

Thus, the expression for rate law for the given chemical reaction is shown below.

  $\text{Rate}=k_r\left[\text{A}\right]{\left[\text{B}\right]}^2{\left[\text{C}\right]}^{0.009}$

(c)

Interpretation Introduction

Interpretation:

The rate constant for the given reaction has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7A.15E

The rate constant of the reaction is $\underset{\_}{1.88\times 10^{-5}{L}^2\cdot mmo{l}^{-2}\cdot s^{-1}}$.

Explanation of Solution

The expression for the rate law for the given reaction is shown below.

    $\text{Rate}=k_r\left[\text{A}\right]{\left[\text{B}\right]}^2{\left[\text{C}\right]}^{0.009}$        (6)

Substitute the value of $\text{R}$, $\left[\text{A}\right]$, $\left[\text{B}\right]$ and $\left[\text{C}\right]$ from the given data for experiment 1 in equation (6).

  $\begin{array}{c}2.0\text{mmol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}=k_r\left(10\text{mmol}\cdot {\text{L}}^{-1}\right){\left(100\text{mmol}\cdot {\text{L}}^{-1}\right)}^2{\left(700\text{mmol}\cdot {\text{L}}^{-1}\right)}^{0.009}\\ k_r=\frac{\text{2}\text{.0}\text{mmol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}}{\left(10\text{mmol}\cdot {\text{L}}^{-1}\right){\left(100\text{mmol}\cdot {\text{L}}^{-1}\right)}^2{\left(700\text{mmol}\cdot {\text{L}}^{-1}\right)}^{0.009}}\\ =1.88\times {10}^{-5}{\text{L}}^2\cdot {\text{mmol}}^{-2}\cdot {\text{s}}^{-1}\end{array}$

Thus, the rate constant of the reaction is $\underset{\_}{1.88\times 10^{-5}{L}^2\cdot mmo{l}^{-2}\cdot s^{-1}}$.

(d)

Interpretation Introduction

Interpretation:

The reaction rate for the experiment 5 has to be determined.

Concept Introduction:

Same as part (a).

Answer to Problem 7A.15E

The reaction rate for the experiment 5 is $\underset{\_}{2.78\times 10^{-6}mmol\cdot L^{-1}\cdot s^{-1}}$.

Explanation of Solution

Substitute the value of $k_r$, $\left[\text{A}\right]$ and $\left[\text{B}\right]$ from the given data for experiment 5 in equation (6).

  $\begin{array}{c}R=\left(\begin{array}{l}\left(1.88\times {10}^{-5}{\text{L}}^2\cdot {\text{mmol}}^{-2}\cdot {\text{s}}^{-1}\right)\left(4.62\text{mmol}\cdot {\text{L}}^{-1}\right){\left(0.177\text{mmol}\cdot {\text{L}}^{-1}\right)}^2\\ {\left(12.4\text{mmol}\cdot {\text{L}}^{-1}\right)}^{0.009}\end{array}\right)\\ =2.78\times {10}^{-6}\text{mmol}\cdot {\text{L}}^{-1}\cdot {\text{s}}^{-1}\end{array}$

Thus, the reaction rate for the experiment 5 is $\underset{\_}{2.78\times 10^{-6}mmol\cdot L^{-1}\cdot s^{-1}}$.