Chapter 18, Problem 18.58P

(a)

Interpretation Introduction

Interpretation:

It has to be shown that the observed rate law for the overall reaction is consistent with the proposed mechanism.

  $Rateofreaction=k\left[H_2\right]{\left[B{r}_2\right]}^{1/2}$

Concept Introduction:

Rate law or rate equation: The relationship between the reactant concentrations and reaction rate is expressed by an equation.

  $\begin{array}{l}\text{aA + bB}\to x\text{X}\\ \\ {\text{Rate of reaction = k [A]}}^{\text{m}}{\text{[B]}}^{\text{n}}\\ \\ \text{Total}\text{order}\text{of reaction = }\left(\text{m + n}\right).\end{array}$

Order of a reaction: The order of a reaction with respect to a particular reactant is the exponent of its concentration term in the rate law expression, and the overall reaction order is the sum of the exponents on all concentration terms.

Rate constant, k: It is a proportionality constant that relates rate and concentration at a given temperature.

Steady – state approximation:

Steady – state approximation is applicable on an intermediate.  At steady – state approximation, the rate of formation is equal to the rate of decomposition.

Explanation of Solution

The given rate of reaction is,

  $Rateofreaction=k\left[H_2\right]{\left[B{r}_2\right]}^{1/2}$

As known, the overall reaction rate is same as the rate of the slowest reaction step.

The slowest step of the given mechanism is,

  $Br\left(\text{g}\right)\text{+}{H}_{\text{2}}\left(\text{g}\right)\stackrel{{\text{k}}_2}{\to }HBr\left(\text{g}\right)\text{+}H\left(\text{g}\right)$

The rate law for the above slowest reaction is, $\text{Rate}{\text{ = k}}_2\left[\text{Br}\right]\left[{\text{H}}_{\text{2}}\right]\mathrm{.......}\left(1\right)$

The concentration of the intermediate $\left(Br\right)$ cannot be involved in the overall rate law. Hence, substitution is needed for the intermediate.

For equilibrium reaction $\left(1\right)$, ${\text{Br}}_{\text{2}}\left(\text{g}\right)\underset{{\text{k}}_{\text{-1}}}{\overset{{\text{k}}_{\text{1}}}{\rightleftharpoons }}\text{2Br}\left(\text{g}\right)$

The rate of forward reaction, $\text{Rate}{\text{ = k}}_{\text{1}}\left[{\text{Br}}_{\text{2}}\right]\mathrm{........}\left(2\right)$

The rate of reverse reaction, $\text{Rate}{\text{ = k}}_{\text{-1}}{\left[\text{Br}\right]}^{\text{2}}\mathrm{........}\left(\text{3}\right)$

Equating both $\left(2and3\right)$, the value of $\left[\text{Br}\right]$ can be obtained as given below.

  $\begin{array}{l}{\text{k}}_{\text{1}}\left[{\text{Br}}_{\text{2}}\right]{\text{= k}}_{\text{-1}}{\left[\text{Br}\right]}^{\text{2}}\\ \\ \left[\text{Br}\right]\text{=}{\left(\frac{{\text{k}}_{\text{1}}\left[{\text{Br}}_{\text{2}}\right]}{{\text{k}}_{\text{-1}}}\right)}^{\text{1/2}}\mathrm{.....}\left(\text{4}\right)\end{array}$

Now, substituting equation $\left(4\right)$ into the equation $\left(1\right)$ results as,

  $\begin{array}{l}\text{Rate}{\text{ = k}}_2\left[\text{Br}\right]\left[{\text{H}}_{\text{2}}\right]\\ \\ \text{Rate}\text{ =}\text{k}\left[{\text{H}}_{\text{2}}\right]{\left(\frac{{\text{k}}_{\text{1}}\left[{\text{Br}}_{\text{2}}\right]}{{\text{k}}_{\text{-1}}}\right)}^{\text{1/2}}\\ \\ \text{Rate}\text{ =}\text{k }{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{1/2}\left[{\text{H}}_{\text{2}}\right]{\left[{\text{Br}}_{\text{2}}\right]}^{1/2}\\ \\ \text{Rate}\text{ =k}\left[{\text{H}}_{\text{2}}\right]{\left[{\text{Br}}_{\text{2}}\right]}^{1/2}\text{where,}\text{k}\text{=}\text{k }{\left(\frac{{\text{k}}_{\text{1}}}{{\text{k}}_{\text{-1}}}\right)}^{\text{1/2}}\\ \end{array}$

Thus, the rate law of the overall reaction becomes,

  $Rateofreaction=k\left[H_2\right]{\left[B{r}_2\right]}^{1/2}$

Therefore, the rate law of the given reaction mechanism is obtained as $Rateofreaction=k\left[H_2\right]{\left[B{r}_2\right]}^{1/2}$ is consistent with the given rate law.

(b)

Interpretation Introduction

Interpretation:

The units of the rate constant for the reaction have to be written.

Concept Introduction:

Refer to part (a).

Explanation of Solution

The rate of reaction is,

  $Rateofreaction=k\left[H_2\right]{\left[B{r}_2\right]}^{1/2}$

The order of the reaction is $3/2.$  Then the unit of the rate constant can be written as given below.

  $\begin{array}{l}Rate\text{constant}\left(k\right)=\frac{Rate}{\left[H_2\right]{\left[Br\right]}^{1/2}}\\ \\ Rate\text{constant}\left(k\right)=\frac{Concentration}{time}\times \frac{1}{{\left(Concentration\right)}^{3/2}}\\ \\ Rate\text{constant}\left(k\right)=\frac{{\left(Concentration\right)}^{-1/2}}{time}\\ \\ Rate\text{constant}\left(k\right)=\frac{{\left(mol.L^{-1}\right)}^{-1/2}}{s}=L^{1/2}.mo{l}^{-1/2}.s^{-1}\end{array}$

Therefore, the units of rate constant is $L^{1/2}.mo{l}^{-1/2}.s^{-1}.$

(c)

Interpretation Introduction

Interpretation:

The reaction occurs in a single step mechanism or not has to be decided.

Explanation of Solution

The reaction is given below.

  $H_2\left(g\right)+B{r}_2\left(g\right)\to 2HBr\left(g\right)$

No, it is not possible that this reaction occurs in a single step.