Chapter 11, Problem 25QRT

(a)

Interpretation Introduction

Interpretation:

The order of the reaction with respect to substance $I$ and substance $II$ has to be determined.

Explanation of Solution

The rate law can be represented as given below.

  $Rate=k{\left[I\right]}^i{\left[II\right]}^j$

Where, $k,iandj$ are unknown.  The values of $iandj$ can be calculated by taking given below assumptions.

Carefully examining the table, it can be said that $\left[I\right]$ is constant.  Then, the rate law can be simplified as follows,

  $Rate=k^{\text{'}}{\left[II\right]}^{j}Where,k\text{'}=k{\left[I\right]}^i$

By taking logarithm on both the sides, it can be written as follows,

  $\begin{array}{l}\log \left(Rate\right)=\log \left(k^{\text{'}}\right)+\log \left({\left[II\right]}^j\right)\\ \\ \log \left(Rate\right)=\log \left(k^{\text{'}}\right)+j\log \left(\left[II\right]\right)\\ \\ \log \left(Rate\right)=j\log \left(\left[II\right]\right)+\log \left(k^{\text{'}}\right)\end{array}$

The above equation is in the form of $y=mx+b.$  Thus, the slope of the plot between $\log \left(rate\right)and\log \left(\left[II\right]\right)$ will give the order of the reaction, j.

Similarly, by keeping $\left[II\right]$ constant, the rate law can be simplified as,

  $Rate=k"{\left[I\right]}^{i}Where,k"=k{\left[II\right]}^j$

By taking logarithm on both the sides, it can be written as follows,

  $\begin{array}{l}\log \left(Rate\right)=\log \left(k"\right)+\log \left({\left[I\right]}^i\right)\\ \\ \log \left(Rate\right)=\log \left(k"\right)+i\log \left(\left[I\right]\right)\\ \\ \log \left(Rate\right)=i\log \left(\left[I\right]\right)+\log \left(k"\right)\end{array}$

The above equation is in the form of $y=mx+b.$  Thus, the slope of the plot between $\log \left(rate\right)and\log \left(\left[I\right]\right)$ will give the order of the reaction, i.

Four data sets have constant $\left[I\right]$ and four data sets have constant $\left[II\right]$ as given below.

Figure 1

The graph of $\log \left(rate\right)and\log \left(\left[II\right]\right)$ is given below.

Figure 2

Therefore, the reaction order of reactant $\left[II\right]$ is one.

The graph of $\log \left(rate\right)and\log \left(\left[I\right]\right)$ is given below.

Figure 3

Therefore, the reaction order of reactant $\left[I\right]$ is one.

(b)

Interpretation Introduction

Interpretation:

The rate law for the reaction has to be derived.

Explanation of Solution

The order of the reaction with respect to substance $I$ and substance $II$ is one and one respectively.  So the rate law can be written as given below.

  $Rate=k\left[I\right]\left[II\right]$

(c)

Interpretation Introduction

Interpretation:

The rate constant $k$ has for the reaction has to be calculated and its appropriate units also have to be expressed.

Answer to Problem 25QRT

The average of the rate constant is $1.04Lmo{l}^{-1}{s}^{-1}.$

Explanation of Solution

The rate constant can be expressed as shown below.

  $\begin{array}{c}Rate=k\left[I\right]\left[II\right]\\ \\ k=\frac{Rate}{\left[I\right]\left[II\right]}\end{array}$

For first experiment the rate constant can be calculated by plugging all the data given in the table.

In first experiment, the initial concentration of $I$ and $II$ are $1.65\times {10}^{5}mol/Land10.6\times {10}^{5}mol/L$ respectively.  The initial rate of the reaction in first experiment is $1.50\times {10}^{9}molL{}^{-1}{s}^{-1}.$

Now, the rate constant for first experiment is given below.

  $k_1=\frac{Rate}{\left[I\right]\left[II\right]}=\frac{1.50\times {10}^{9}molL{}^{-1}{s}^{-1}}{\left(1.65\times {10}^{5}mol/L\right)\left(10.6\times {10}^{5}mol/L\right)}=0.853Lmo{l}^{-1}{s}^{-1}.$

Similarly, the rate constant for rest of the experiments can be calculated.  The calculated rate constant values are given in the table below.

$\begin{array}{c}\left[I\right]\times {10}^5\\ \left(mol/L\right)\end{array}$	$\begin{array}{c}\left[II\right]\times {10}^5\\ \left(mol/L\right)\end{array}$	Initial rate $\times {10}^9\left(molL{}^{-1}{s}^{-1}\right)$	Rate constant $\left({\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}}\right)$
$1.65$	$10.6$	$1.50$	$0.853$
$14.9$	$10.6$	$17.7$	$1.12$
$14.9$	$7.10$	$11.2$	$1.06$
$14.9$	$3.52$	$6.30$	$1.20$
$14.9$	$1.76$	$3.10$	$1.18$
$4.97$	$10.6$	$4.52$	$0.853$
$2.48$	$10.6$	$2.70$	$1.03$

The average of the rate constant is $1.04Lmo{l}^{-1}{s}^{-1}.$

(d)

Interpretation Introduction

Interpretation:

The initial rate of the reaction has to be calculated.

Answer to Problem 25QRT

The initial rate of reaction is $5.9\times 10^{-9}mol{L}^{-1}{s}^{-1}.$

Explanation of Solution

The rate law can be expressed as shown below.

  $Rate=k\left[I\right]\left[II\right]$

Given that, $\left[I\right]$ is $8.3\times {10}^{-5}mol/L$ and that of $\left[II\right]$ is $6.78\times {10}^{-5}mol/L.$  The value of rate constant is $\text{1}\text{.04}{\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}}\text{.}$  Then, by plugging all these values in the rate law, the initial rate of reaction can be calculated as follows,

  $\begin{array}{c}Rate=k\left[I\right]\left[II\right]\\ \\ =\left(\text{1}\text{.04}{\text{Lmol}}^{\text{-1}}{\text{s}}^{\text{-1}}\right)\left(8.3\times {10}^{-5}mol/L\right)\left(6.78\times {10}^{-5}mol/L\right)\\ \\ =5.9\times 10^{-9}mol{L}^{-1}{s}^{-1}.\end{array}$

Therefore, the initial rate of reaction is $5.9\times 10^{-9}mol{L}^{-1}{s}^{-1}.$