Question

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

a) $$\begin{gathered} \mathbf{2}\mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\underset{{\mathbf{k}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{⟷}}}\mathbf{}\mathit{D}\mathbf{}\left(Fastequilbrium\right) \\ \mathit{D}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\stackrel{{\mathbf{k}}_{\mathbf{2}}}{\mathbf{\to }}\mathbf{}\mathit{E}\mathbf{}\mathbf{+}\mathbf{}\mathit{F}\left(Slow\right) \\ \mathit{F}\mathbf{}\stackrel{{\mathbf{k}}_{\mathbf{3}}}{\mathbf{\to }}\mathbf{}\mathit{G}\left(Fast\right) \end{gathered}$$

b) $$\begin{gathered} \mathit{A}\mathbf{}\mathbf{+}\mathbf{}\mathit{B}\mathbf{}\underset{{\mathbf{k}}_{\mathbf{-}\mathbf{1}}}{\overset{{\mathbf{k}}_{\mathbf{1}}}{\mathbf{⟷}}}\mathbf{}\mathit{C}\mathbf{}\left(Fastequilbrium\right) \\ \mathit{C}\mathbf{+}\mathbf{}\mathit{D}\underset{{\mathbf{k}}_{\mathbf{-}\mathbf{2}}}{\overset{{\mathbf{k}}_{\mathbf{2}}}{\mathbf{⟷}}}\mathbf{}\mathbf{}\mathit{F}\left(Fastequilbrium\right) \\ \mathit{F}\mathbf{}\stackrel{{\mathbf{k}}_{\mathbf{3}}}{\mathbf{\to }}\mathbf{}\mathit{G}\mathbf{}\mathbf{}\left(Slow\right) \end{gathered}$$

Short answer

The rate law for the corresponding mecahnism is $Rate=\frac{k_3{k}_2{k}_1\left[D\right]\left[A\right]\left[B\right]}{k_{-2}{k}_{-1}}$

Step-by-step solution

Step-1: Over all reaction

If the reaction contains several steps overall reaction can be obtained by adding reactants and products on either side and canceling the common terms.

$2A+B+D+F\to D+F+E+G....\left(1\right)$

Canceling common terms on both sides

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

Step-2: Rate determining step:

If the expression contains several steps slowest step is the rate-determining step

$$\begin{gathered} D+B\stackrel{k_2}{\to }E+F\left(Slow\right) \\ Rate=k_2\left[D\right]\left[B\right]......\left(3\right) \end{gathered}$$

Step-3: Determination of rate law

D is found to be equilibrium with E,A ,F,B .the equilibrium constant for the first reaction is

$$\begin{gathered} K_1=\frac{\left[D\right]}{\left[B\right]{\left[A\right]}^2}......\left(4\right) \\ {K}_1=\frac{k_1}{k_{-1}}.......\left(5\right) \end{gathered}$$

From equation 4 and 5 following equation is derived

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[D\right]}{\left[B\right]{\left[A\right]}^2}......\left(6\right) \\ or\left[D\right]=\frac{k_1\left[B\right]{\left[A\right]}^2}{k_{-1}} \end{gathered}$$

Substituting D value in equation 3

$Rate=\frac{k_2{k}_1{\left[B\right]}^2{\left[A\right]}^2}{k_{-1}}.....\left(7\right)$

Therefore rate law corresponds to the above mechanism.

Step-4: Overall reaction

(b)

If the reaction contains several steps overall reaction can be obtained by adding reactants and products on either side and canceling the common terms.

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

Canceling common terms on both sides

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

Step-5: Rate determining step

If the expression contains several steps s lowest step is the rate-determining step

$$\begin{gathered} F\stackrel{k_3}{\to }G\left(Slow\right) \\ Rate=k_3\left[F\right]......\left(3\right) \end{gathered}$$

Step-6: Determination  of Rate law

But F is in equilibrium with C, D and G then equilibrium constant K₂

is defined as

$K_2=\frac{\left[F\right]}{\left[C\right]\left[D\right]}......\left(4\right)$

According to principle of detailed balance

$K_2=\frac{k_2}{k_{-2}}.......\left(5\right)$

From equations 4 and 5

$$\begin{gathered} \frac{k_2}{k_{-2}}=\frac{\left[F\right]}{\left[C\right]\left[D\right]}.......\left(6\right) \\ \left[F\right]=\frac{k_2\left[C\right]\left[D\right]}{k_{-2}}......\left(7\right) \end{gathered}$$

Substituting the value in equation-3

$$\begin{gathered} Rate=k_3\left[F\right]......\left(3\right) \\ Rate=\frac{k_3{k}_2\left[C\right]\left[D\right]}{k_{-2}}.....\left(8\right) \end{gathered}$$

Again C is in equilibrium with A and B in the first step, if the equilibrium constant of the first step is K₁ then

$K_1=\frac{\left[C\right]}{\left[A\right]\left[B\right]}........\left(9\right)$

According to principle of detailed balance

$K_1=\frac{k_1}{k_{-1}}......\left(10\right)$

$$\begin{gathered} \frac{k_1}{k_{-1}}=\frac{\left[C\right]}{\left[A\right]\left[B\right]}......\left(11\right) \\ or \\ \left[C\right]=\frac{k_1\left[A\right]\left[B\right]}{k_{-1}}.....\left(12\right) \end{gathered}$$

From equations 5 and 6 following equation is derived

Substituting the value of C in equation-8

$Rate=\frac{k_3{k}_2{k}_1\left[D\right]\left[A\right]\left[B\right]}{k_{-2}{k}_{-1}}..........\left(13\right)$

Therefore the rate law for the corresponding mechanism is$Rate=\frac{k_3{k}_2{k}_1\left[D\right]\left[A\right]\left[B\right]}{k_{-2}{k}_{-1}}$.