Chapter 15, Problem 135CP

(a)

Interpretation Introduction

Interpretation: The order of the reaction with respect to ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ and ${\text{I}}^{-}$ needs to be determined.

Concept Introduction: The rate law for a reaction is represented as follows:

  $r=k{\left[A\right]}^x{\left[B\right]}^y$

According to above expression, the reaction is x order with respect to A and y order with respect to B and k is rate constant.

Answer to Problem 135CP

The order of the reaction with respect to both ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ and ${\text{I}}^{-}$ is 1.

Explanation of Solution

The given reaction is as follows:

  ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\left(aq\right)+3{\text{I}}^{-}\left(aq\right)+2{\text{H}}^{+}\left(aq\right)\to {\text{I}}_3^{-}\left(aq\right)+2{\text{H}}_{\text{2}}\text{O}\left(l\right)$

The initial concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is $8.0\times {10}^{-4}$ .

The rate law for the above reaction is represented as follows:

  $Rate=\frac{-d\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{dt}\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m{\left[{\mathrm{H}}_2{O}_2\right]}^n$

The given concentration of ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ is very less as compared to $\left[{\text{H}}^{+}\right]$ and $\left[{\text{I}}^{-}\right]$ .

Also, the plot between ln $\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$ vs t is linear thus, the reaction is first order with respect to ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ .

The rate law can be rewritten as follows:

  $\frac{d\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{dt}=K{\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]}^n$

Here,

  $K=\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m$

This K is the slope value for plot between ln $\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$ vs t .

Taking the ratio of slope from experiment 2 and 3:

  $\begin{array}{l}\frac{-0.480}{-0.360}=\frac{\left(k_1+k_2\left(0.04\right)\right){\left(0.4\right)}^m}{\left(k_1+k_2\left(0.04\right)\right){\left(0.3\right)}^m}\\ \frac{0.48}{0.36}={\left(\frac{0.4}{0.3}\right)}^m\\ 1.33={1.33}^m\end{array}$

Thus, value of m is 1.

Therefore, the order of reaction is 1 with respect to both $\left[{\text{H}}_{\text{2}}{\text{O}}_{\text{2}}\right]$ and $\left[{\text{I}}^{-}\right]$ .

(b)

Interpretation Introduction

Interpretation: The order of the reaction with respect to ${\text{H}}_{\text{2}}{\text{O}}_{\text{2}}$ and ${\text{I}}^{-}$ needs to be determined.

Concept Introduction: The rate law for a reaction is represented as follows:

  $r=k{\left[A\right]}^x{\left[B\right]}^y$

According to above expression, the reaction is x order with respect to A and y order with respect to B and k is rate constant.

Answer to Problem 135CP

The value of $k_1$ and $k_2$ is $\text{0}\text{.827 L/mol min}$ and $9.33{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{min}$ respectively.

Explanation of Solution

The slope is represented as follows:

  $K=\left(k_1+k_2\left[{\text{H}}^{+}\right]\right){\left[{\text{I}}^{-}\right]}^m$

Here, value of m is equal to 1 thus,

  $K=\left(k_1+k_2\left[{\text{H}}^{+}\right]\right)\left[{\text{I}}^{-}\right]$

Now, from experiment 1 and 4:

  $0.120=\left(k_1+k_2\left(0.04\right)\right)\left(0.1\right)$

Or,

  $0.120=0.1k_1+0.004k_2$

Or,

  $k_1=\frac{0.120-0.004k_2}{0.1}$ ....... (1)

Also,

  $0.0760=\left(k_1+k_2\left(0.02\right)\right)\left(0.075\right)$

Putting the value from equation (1),

  $\begin{array}{l}0.0760=\left(\frac{0.120-0.004k_2}{0.1}+k_2\left(0.02\right)\right)\left(0.075\right)\\ \frac{0.00760}{0.075}=0.120-0.004k_2+0.002k_2\\ 0.10133-0.120=-0.002k_2\\ -0.01867=-0.002k_2\\ k_2=\frac{0.01867}{0.002}=9.33{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{min}\end{array}$

Putting the calculated value in equation (1),

  $\begin{array}{l}{k}_1=\frac{0.120-0.004\left(9.33\right)}{0.1}\text{ L/mol min}\\ \text{=0}\text{.827 L/mol min}\end{array}$

Thus, the value of $k_1$ and $k_2$ is $\text{0}\text{.827 L/mol min}$ and $9.33{\text{ L}}^{\text{2}}{\text{/mol}}^{\text{2}}\text{min}$ respectively.

(c)

Interpretation Introduction

Interpretation: The reason for the two-term dependence of the rate on hydrogen ion concentration needs to be explained.

Concept Introduction: The rate law for a reaction is represented as follows:

  $r=k{\left[A\right]}^x{\left[B\right]}^y$

According to above expression, the reaction is x order with respect to A and y order with respect to B and k is rate constant.

Explanation of Solution

There are two possible pathways, one involving hydrogen ion and the rate law is represented as follows:

  $rate=k_2\left[H^{+}\right]\left[I^{-}\right]\left[H_2{O}_2\right]$

The other pathway is which do not involve hydrogen ion.

  $rate=k_2\left[I^{-}\right]\left[H_2{O}_2\right]$

The overall rate depends on the pathway that dominates and this depends on the concentration of hydrogen ion in the solution.