Question

Compare and contrast the mechanisms for the two gas phase reactions.

$$\begin{gathered} {\text{H}}_2+\text{B}{r}_2\to 2\text{HBr} \\ {\text{H}}_2+{\text{I}}_2\to 2\text{HI} \end{gathered}$$

Short answer

The order of formation of HBr reaction is found to be $\frac{3}{2}$

The order of formation of HI reaction is found to be 2 same as given in the question.

Step-by-step solution

Reactions

Consider the reactions which occur in gas phase.

$$\begin{gathered} {\text{H}}_2+{\text{Br}}_2\to 2\text{HBr} \\ {\text{H}}_2+{\text{I}}_2\to 2\text{HI} \end{gathered}$$

Theoritically it looks that these both reactions follow the same order,but experimentally it looks likethat they follow different mechanism and reaction is of different order.

HBr formation mechanism

$\begin{array}{rcl}& & {\text{Br}}_2+\text{M}\underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\leftrightharpoons }}\text{Br+M}\left(\text{Fastequilbrium}\right)\\ & & \text{Br}+{\text{H}}_2\stackrel{{\mathrm{k}}_2}{\to }\text{HBr}+\text{H}\left(\text{Slow}\right)\\ & & \text{H}+\text{B}{\mathrm{r}}_2\stackrel{{\mathrm{K}}_3}{\to }\mathrm{HBr}+\text{Br}\left(\text{Fast}\right)\\ & & \end{array}$

Where k₁,k_-1, are the rate constants of forward and backward reactions.

k₂ is the rate constant of the second reaction.

k₃ is the rate constant of the third reaction.

The Rate always depends on the slow step of the reaction, thus it is considered as rate -determining step.

${\text{Rate\hspace{0.05em}=k}}_2\left[\mathrm{Br}\right]\left[{\text{H}}_2\right]......\left(1\right)$

But $\left[\text{Br}\right]$ exist in equilibrium in first reaction.

$$\begin{gathered} {\text{K}}_1=\frac{{\left[\mathrm{Br}\right]}^2}{\left[{\mathrm{Br}}_2\right]}........\left(2\right) \\ {\mathrm{K}}_1=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}........\left(3\right) \end{gathered}$$

K₁ is the equilibrium constant of the reaction.

$\left[\mathrm{Br}\right]={{\mathrm{K}}_1}^{1/2}{\left[\mathrm{Br}\right]}^{1/2}......\left(4\right)$

Substitute bromine concentration value in eqaution-1

$$\begin{gathered} \mathrm{Rate} ={\mathrm{k}}_2{{\mathrm{K}}_1}^{1/2}{\left[\mathrm{Br}\right]}^{1/2}\left[{\mathrm{H}}_2\right]......\left(1\right) \\ \mathrm{Rate}={\mathrm{k}}_{\mathrm{obs}}{\left[\mathrm{Br}\right]}^{1/2}\left[{\mathrm{H}}_2\right] \end{gathered}$$

Therefore the order of the reaction is found to be $\frac{3}{2}$

Hydrogen-iodide formation mechanism

Reaction mechanism of HI formation

${\mathrm{H}}_2+{\mathrm{I}}_2\to 2\mathrm{HI}$

This reaction follows the following rection mechanism

$$\begin{gathered} {\mathrm{I}}_2+\mathrm{M}\underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\leftrightharpoons }}\mathrm{I}+\mathrm{I}+\mathrm{M}\left(\mathrm{Fast}\mathrm{equilbrium}\right) \\ {\mathrm{I}}_2+{\mathrm{H}}_2\stackrel{{\mathrm{k}}_2}{\to }2\mathrm{HI}\left(\mathrm{Slow}\right) \end{gathered}$$

Where k₁ and k_-1 are rate constant of forward and backward reaction, and k2 is the rate constant of second reaction,M denotes molecule.The rate always depends on the slow step which is called the rate determining step.

$\mathrm{Rate}={\mathrm{k}}_2\left[{\mathrm{I}}_2\right]\left[{\mathrm{H}}_2\right].....\left(1\right)$

But iodine exists in equilibrium in first reaction.

$$\begin{gathered} {\mathrm{K}}_1=\frac{{\left[\mathrm{I}\right]}^2}{\left[{\mathrm{I}}_2\right]}......\left(2\right) \\ {\mathrm{K}}_1=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}.......\left(3\right) \\ \mathrm{Concentration}\mathrm{of}\mathrm{iodine}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as} \\ {\left[\mathrm{I}\right]}^2={\mathrm{K}}_1\left[{\mathrm{I}}_2\right]......\left(4\right) \end{gathered}$$

Substituting the value of 4^th equation in 1 st equation

$$\begin{gathered} \mathrm{Rate}={\mathrm{k}}_2{\mathrm{K}}_1\left[{\mathrm{I}}_2\right]\left[{\mathrm{H}}_2\right].....\left(1\right) \\ {\mathrm{k}}_2{\mathrm{k}}_1={\mathrm{k}}_{\mathrm{obs}} \\ \mathrm{Rate}={\mathrm{k}}_{\mathrm{obs}}\left[{\mathrm{I}}_2\right]\left[{\mathrm{H}}_2\right].....\left(5\right) \end{gathered}$$

This represents that observed order and expected order are one and the same for the above reaction mechanism.

Determination of rate of reaction by rate law is not a correct procedure, but reaction mechanism explains the order accurately.