Q16.102 CP - Chemistry: Molecular Nature Of Matter And Change

Question

Even when a mechanism is consistent with the rate law, later work may show it to be incorrect. For example, the reaction between hydrogen and iodine has this rate law: $\mathit{r}\mathit{a}\mathit{t}\mathit{e}\mathbf{=}\mathit{k}\left[H_2\right]\left[I_2\right]$ . The long-accepted mechanism had a single bimolecular step; that is, the overall reaction was thought to be elementary: 

${\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{I}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\to }\mathbf{2}\mathbf{H}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}$

In the 1960s, however, spectroscopic evidence showed the presence of free I atoms during the reaction. Kineticists have since proposed a three-step mechanism:

$$\begin{gathered} \left(1\right){\mathbf{I}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\left[fast\right] \\ \left(2\right){\mathbf{H}}_{\mathbf{2}\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }{\mathbf{H}}_{\mathbf{2}}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\left[fast\right] \\ \left(3\right){\mathbf{H}}_{\mathbf{2}}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{\rightleftharpoons }\mathbf{2}\mathbf{H}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\left[slow\right] \end{gathered}$$

Show that this mechanism is consistent with the rate law.

Step-by-Step Solution

Answer

 $k=\left(\frac{K_3{K}_2{K}_1}{K_{-2}{K}_{-1}}\right)$ thus, the predicted mechanism is consistent with the rate law.

Step 1: Rate law

The rate law is defined as the product of the reactants' concentrations raised to the power of their respective stoichiometric coefficients. For each mechanism, create a rate law.

Because it is a slow step, the total rate law will be represented using the third step.

$$\begin{gathered} {\mathit{H}}_{\mathbf{2}}{\mathit{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\mathbf{+}{\mathbf{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}}\stackrel{{\mathbf{k}}_{\mathbf{3}}}{\mathbf{\to }}\mathbf{2}\mathit{H}{\mathit{I}}_{\mathbf{\left(}\mathbf{g}\mathbf{\right)}} \\ \mathit{r}\mathit{a}\mathit{t}{\mathit{e}}_{\mathbf{3}}\mathbf{=}{\mathit{k}}_{\mathbf{3}}\left[H_{2}I\right]\left[I\right] \end{gathered}$$

Where ${\mathit{k}}_{\mathbf{3}}$ is the step is to take rate constant. We must replace the concentration of ${\mathit{H}}_{\mathbf{2}}\mathit{I}$ and I since they are intermediates.

Step 2: The second equilibrium stage

The second step is an equilibrium stage that includes both forward and backward reactions. 

Write the rate law for step 1:

$$\begin{gathered} H_{2\left(g\right)}+I_{\left(g\right)}\frac{\stackrel{K_2}{\to }}{\underset{K_{-2}}{\leftarrow }}{H}_2{I}_g \\ rat{e}_{forward}=K_2\left[H_2\right]\left[I\right] \\ rat{e}_{reverse}=K_{-2}\left[H_{2}I\right] \end{gathered}$$

The rate constants for forward and reverse steps are $K_2$ and $K_{-2}$ , respectively.

Forward and reverse reaction rates are equivalent at equilibrium. To get $\left[H_{2}I\right]$, solve the formula as follows:

$$\begin{gathered} rat{e}_{forward}=rat{e}_{reverse} \\ {k}_2\left[H_2\right]\left[I\right]=k_{-2}\left[H_{2}I\right] \\ \left[H_{2}I\right]=\left(\frac{k_2}{k_{-2}}\right)\left[H_2\right]\left[I\right] \end{gathered}$$

Substitute the value in rate law expression:

$$\begin{gathered} rate3=k3\left(\frac{k_2}{k_{-2}}\right)\left[H_2\right]\left[I\right]\left[I\right] \\ =k3\left(\frac{k_2}{k_{-2}}\right)\left[H_2\right]{\left[I\right]}^2 \end{gathered}$$

Step 3: The first equilibrium stage

The first step is an equilibrium phase that includes both forward and backward reactions.

Write the rate law for step 1:

$$\begin{gathered} I_{2\left(g\right)}\frac{\stackrel{K_1}{\to }}{\underset{K_{-1}}{\leftarrow }}2I_g \\ rat{e}_{forward}=K_1\left[I_2\right] \\ rat{e}_{reverse}=K_{-1}{\left[I\right]}^2 \end{gathered}$$

The rate constants for forward and reverse steps are $k_1$ and $k_{-1}$ , respectively.

Forward and reverse reaction rates are equivalent at equilibrium. To get [I], solve the equation.

$$\begin{gathered} rat{e}_{forward}=rat{e}_{reverse} \\ {k}_1\left[I_2\right]=k_{-1}{\left[I\right]}^2 \\ \left[I\right]={\left(\frac{k_1}{k_{-1}}\right)}^{1/2}{\left[I_2\right]}^{1/2} \end{gathered}$$

Substitute the value in rate law expression:

$$\begin{gathered} rate3=\left(\frac{k_3{k}_2}{k_{-2}}\right)\left[H_2\right]{\left(\frac{k_1}{k_{-1}}\right)}^{1/2}{\left[I_2\right]}^{1/2} \\ =\left(\frac{K_3{K}_2{K}_1}{K_{-2}{K}_{-1}}\right)\left[H_2\right]\left[I_2\right] \\ =k\left[H_2\right]\left[I_2\right] \end{gathered}$$

Where, $k=\left(\frac{K_3{K}_2{K}_1}{K_{-2}{K}_{-1}}\right)$

Thus, the predicted mechanism is consistent with the rate law.