Question

Write the overall reaction and rate laws that correspond to the following reaction mechanisms. Be sure to eliminate intermediates from the answers.

(a) $$\begin{gathered} \mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\underset{{\mathbf{k}}_{\mathbf{2}}}{\overset{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{\to }}}\mathbf{}\mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{D}\mathbf{}\left(Fastequilbrium\right) \\ \mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{E}\mathbf{}\stackrel{{\mathbf{k}}_{\mathbf{2}}}{\mathbf{\to }}\mathbf{}\mathit{F}\mathbf{}\left(Slow\right) \end{gathered}$$

(b) $$\begin{gathered} \mathit{A}\mathbf{}\underset{{\mathbf{k}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{⟷}}}\mathbf{}\mathit{B}\mathbf{}\mathbf{+}\mathbf{}\mathit{C}\left(Fastequilbrium\right) \\ \mathit{C}\mathbf{}\mathbf{+}\mathbf{}\mathit{D}\underset{{\mathbf{k}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{⟷}}}\mathbf{}\mathit{E}\mathbf{}\left(Fastequilbrium\right) \\ \mathit{E}\mathbf{}\stackrel{{\mathbf{k}}_{\mathbf{3}}}{\mathbf{\to }}\mathbf{}\mathit{F}\mathbf{}\left(Slow\right) \end{gathered}$$

Short answer

The rate of the reaction is$Rate=\frac{k_1{k}_2{k}_3\left[A\right]\left[D\right]}{k_{-2}{k}_{-1}\left[B\right]}$

Step-by-step solution

Step-1: Overall equation:

(a)

If the reaction contains many steps the overall reaction can be obtained summing up all the reactents and products on the either side.

$$\begin{gathered} A+B\underset{k_2}{\overset{k_1}{\to }}C+D\left(Fastequilbrium\right) \\ C+E\stackrel{k_2}{\to }F\left(Slow\right) \end{gathered}$$

Add all the steps to obtain the entire reaction

$A+B+C+E\to C+D+F.....\left(1\right)$

As C is common both sides ,one would cancel this term,so the overallreactionobtained is

$A+B+E\to D+F.....\left(2\right)$

Step-2: Rate determining step  

The rate-determining step is the slowest step in the reaction, even though the reaction contains many steps.

$rate=k\left[C\right]\left[E\right]..........\left(3\right)$

Step-3: Determining the equilbrium constant

The equilibrium constant for a reaction can be obtained by dividing the product concendration by reactent concendration.One would consider

The fast equilbrium step to calculate K as it is reversable reaction.

$A+B\underset{k_2}{\overset{k_1}{\to }}C+D\left(Fastequilbrium\right)$

$K=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}..............\left(4\right)$

According to principle of detailed balance

$K=\frac{k_1}{k_{-1}}..........\left(5\right)$

From equation 4 and 5 the following eqaution is obtained

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[C\right]\left[D\right]}{\left[A\right]\left[B\right]}............\left(6\right) \\ or \\ \left[C\right]=\frac{k_1\left[A\right]\left[B\right]}{k_{-1}\left[D\right]}...........\left(7\right) \end{gathered}$$

Substituting the value of $\left[C\right]$ in the 6 equation.

$Rate=\frac{k_1{k}_2\left[A\right]\left[B\right]\left[E\right]}{k_{-1}\left[D\right]}......\left(8\right)$

This is the rate law for the above reaction.

Step-4:  overall equation:

(a)

If the reaction contains many steps the overall reaction can be obtained by summing up all the reactents and products on the either side.

$$\begin{gathered} A\underset{k_{-1}}{\overset{k_1}{⟷}}B+C\left(Fastequilbrium\right) \\ C+D\underset{k_{-1}}{\overset{k_1}{⟷}}E\left(Fastequilbrium\right) \\ E\stackrel{k_3}{\to }F\left(Slow\right) \end{gathered}$$

Add all the steps to obtain the reaction

$A+D+C+E\to +B+C+E+F.....\left(9\right)$

Cancel common terms

$A+D+E\to +B+E+F.....\left(10\right)$

Step-5:  Rate determining step:

Rate determining step is the slowest step of several steps in the reaction path way

$$\begin{gathered} E\stackrel{k_3}{\to }F\left(Slow\right).......\left(11\right) \\ Rate=k_3\left[E\right]...........\left(12\right) \end{gathered}$$

Step-6:   Determining equilbrium constant:

The equilibrium constant for a reaction can be obtained by dividing the product concendration by reactent concendration.One would consider

The fast equilbrium step to calculate K as it is reversable reaction.

E is said to be equilbrium with C,D and F so the rate constant is calculated in the following way

$K_2=\frac{\left[E\right]}{\left[C\right]\left[D\right]}...........\left(13\right)$

The rate constant K₂is given as

$K_2=\frac{k_2}{k_{-2}}.........\left(14\right)$

From equation 13 and 14

$\frac{k_2}{k_{-2}}=\frac{\left[E\right]}{\left[C\right]\left[D\right]}..........\left(15\right)$

The equilbrium concendration of $\left[E\right]$ is found to be

$\left[E\right]=\frac{k_2\left[C\right]\left[D\right]}{k_{-2}}.........\left(16\right)$

Substituting $\left[E\right]$ value in 12 th eqaution the following rate is obtained

$Rate=\frac{k_2{k}_3\left[C\right]\left[D\right]}{k_{-2}}......\left(17\right)$

Again C is in equilbroum with first and seacond step of the reaction

$k_1=\frac{\left[B\right]\left[C\right]}{\left[A\right]}.......\left(18\right)$

The principle of detailed balance is the following

$K=\frac{k_1}{k_{-1}}........\left(20\right)$

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[B\right]\left[C\right]}{\left[A\right]} \\ \left[C\right]=\frac{\left[A\right]k_1}{\left[B\right]k_{-1}}........\left(21\right) \end{gathered}$$

Substituting C value in equation 17 to obtain the rate of the eqaution

$Rate=\frac{k_1{k}_2{k}_3\left[A\right]\left[D\right]}{k_{-2}{k}_{-1}\left[B\right]}......\left(22\right)$

Therfore the rate of the reaction is mentioned as above.