Question

In some reactions, there is a competition between kinetic control and thermodynamic control over product yields. Suppose compound A can undergo two elementary reactions to stable products:

$\mathrm{A}\underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\leftrightharpoons }}\mathrm{B}\mathrm{or}\mathrm{A}\underset{{\mathrm{k}}_{-2}}{\overset{{\mathrm{k}}_2}{\leftrightharpoons }}\mathrm{C}$

For simplicity we assume first-order kinetics for both forward and reverse reactions. We take the numerical values$1\times 108s^{-1}{k}^{-1}=1\times 102s^{-1}$

$k2=1\times 109s^{-1}{k}^{-2}=1\times 109s^{-1}$

a) Calculate the equilibrium constant for the equilibrium.

$\mathrm{B}\underset{}{\overset{}{\leftrightharpoons }}\mathrm{C}$

From this value, give the ratio of the concentration of B to that of C at equilibrium. This is an example of thermodynamic control.

b) In the case of kinetic control, the products are isolated (or undergo additional reaction) before the back reactions can take place. Suppose the back reactions in the preceding example (k-₁ and k_-2) can be ignored. Calculate the concentration ratio of B to C reached in this case.

Short answer

a) The equilbrium ratio of concendration of B and C is$1x{10}^1$

b)The concendration of B and C is found to be$1x{10}^{-1}$

Step-by-step solution

(a) Overall reaction

The given reactions in the question

$\mathrm{A}\underset{{\mathrm{k}}_{-1}}{\overset{{\mathrm{k}}_1}{\leftrightharpoons }}\mathrm{B}\mathrm{or}\mathrm{A}\underset{{\mathrm{k}}_{-2}}{\overset{{\mathrm{k}}_2}{\leftrightharpoons }}\mathrm{C}$

Where k₁ k₂k_-1k_-2 are the rate constants of forward and backward reactions.

On combining above two reactions, the reaction obtained is

$\mathrm{B}\underset{}{\overset{}{\leftrightharpoons }}\mathrm{C}$is the overall reaction.

Principle of detailed balance:

Principle of detailed balance explains the relation between equilbrium constant and rate constant is stated as below

$$\begin{gathered} {\mathrm{K}}_1=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_{-1}} \\ {\mathrm{K}}_2=\frac{{\mathrm{k}}_2}{{\mathrm{k}}_{-2}} \end{gathered}$$

The equilbrium constant of overall reaction can be obtained by multiplying the equilbrium constant of their elementary reaction steps.

localid="1661431373699" $$\begin{gathered} {\mathrm{K}}_{\left(\mathrm{overall}\right)}={\mathrm{K}}_1{\mathrm{K}}_2.......\left(1\right) \\ \mathrm{Substituting}\mathrm{both}{\mathrm{K}}_1\mathrm{and}{\mathrm{K}}_2\mathrm{values} \\ \mathrm{K}=\frac{{\mathrm{k}}_1}{{\mathrm{k}}_{-1}}\frac{{\mathrm{k}}_2}{{\mathrm{k}}_{-2}} \end{gathered}$$

Equilbrium constant of Overall reaction:

Substituting all the k values given in the equation

$$\begin{gathered} {\mathrm{k}}_1=1\times {10}^8{\mathrm{s}}^{-1} \\ {\mathrm{k}}_2=1\times {10}^9{\mathrm{s}}^{-1} \\ {\mathrm{k}}_{-2}=1\times {10}^9{\mathrm{s}}^{-1} \\ {\mathrm{k}}_{-1}=1\times {10}^2{\mathrm{s}}^{-1} \\ {\mathrm{K}}_{\mathrm{overall}}=\frac{1\mathrm{x}{10}^2\mathrm{x}1\mathrm{x}{10}^9}{1\mathrm{x}{10}^8\mathrm{x}1\mathrm{x}{10}^4} \\ {\mathrm{K}}_{\mathrm{overall}}=1\mathrm{x}{10}^1 \end{gathered}$$

 Ratio of concendrations:

$\mathrm{B}\underset{}{\overset{}{\leftrightarrows }}\mathrm{C}$

The equilbrium constant is given as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{overall}}=\frac{\left[\mathrm{C}\right]}{\left[\mathrm{B}\right]} \\ \frac{\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]}=\frac{1}{{\mathrm{K}}_{\mathrm{overall}}} \\ \frac{\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]}=\frac{1}{1\mathrm{x}{10}^{-1}} \\ \frac{\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]}=1\mathrm{x}{10}^1 \end{gathered}$$

Therefore at equilbrium ratio of concendration of B and C is $1x{10}^1$

(b) Ratio of concentrations when backward reaction is ignored:

When backward reactions are ignored , then the equation can be written as

$$\begin{gathered} {\mathrm{K}}_{\mathrm{overall}}=\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1} \\ \mathrm{Substituting}{\mathrm{k}}_1\mathrm{and}{\mathrm{k}}_2\mathrm{values}in\mathrm{the}\mathrm{equation} \\ {\mathrm{K}}_{\mathrm{overall}}=\frac{1\mathrm{x}{10}^9}{1\mathrm{x}{10}^8} \\ {\mathrm{K}}_{\mathrm{overall}}=1\mathrm{x}{10}^1 \end{gathered}$$

For the overall reaction

$\mathrm{B}\underset{}{\overset{}{\leftrightarrows }}\mathrm{C}$

$$\begin{gathered} {\mathrm{K}}_{\mathrm{overall}}= \frac{\left[\mathrm{C}\right]}{\left[\mathrm{B}\right]} \\ \frac{\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]}=\frac{1}{{\mathrm{K}}_{\mathrm{overall}}} \\ =\frac{1}{1\mathrm{x}{10}^1} \\ \frac{\left[\mathrm{B}\right]}{\left[\mathrm{C}\right]}=1\mathrm{x}{10}^{-1} \end{gathered}$$

Therefore, the ratio is$1x{10}^{-1}$