Question

The equilibrium constant for a reaction \[A + 2B \rightleftharpoons 2C\] is 40. The equilibrium constant for reaction \[C \rightleftharpoons B + \dfrac{1}{2}A\] is:
A. \[\left( {\dfrac{1}{{40}}} \right)\]
B. \[{\left( {\dfrac{1}{{40}}} \right)^{\dfrac{1}{2}}}\]
C. \[{\left( {\dfrac{1}{{40}}} \right)^2}\]
D. 40

Answer 1

Hint: At equilibrium the forward and backward reaction rates become the same. As a result, equilibrium constant can be written as the ratio of product side concentration to reactant side concentration.

Formula used: \[{R_f} = {k_f}\left[ A \right]{\left[ B \right]^2}\] and \[{R_b} = {k_b}{\left[ C \right]^2}\]

Complete step- by- step answer:
For a reversible reaction at a situation when the amount of product is formed is equal to the amount of reactant is formed then it is called equilibrium. At equilibrium the amount of product and reactant become constant.
From the given reaction, \[A + 2B \rightleftharpoons 2C\] let, the rate constant of forward reaction is \[{K_f}\] and the rate constant for backward reaction is \[{k_b}\]. therefore, the rates of forward and backward reactions are,
 \[{R_f} = {k_f}\left[ A \right]{\left[ B \right]^2}\] and \[{R_b} = {k_b}{\left[ C \right]^2}\] respectively. Now, at equilibrium the forward and backward reaction rates become the same. As a result,
 \[
  {R_f} = {R_b} \\
  or,{k_f}\left[ A \right]{\left[ B \right]^2} = {k_b}{\left[ C \right]^2} \\
  or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
  {k_{eq}} = \dfrac{{{{\left[ C \right]}^2}}}{{\left[ A \right]{{\left[ B \right]}^2}}} \\
 \] ….(1)
Where \[{k_{eq}}\] is the equilibrium constant of the first reaction.
Now in the same process write the expression of equilibrium constant for the second reaction as follows.
 \[C \rightleftharpoons B + \dfrac{1}{2}A\]
 \[{R_f} = {k_f}\left[ C \right]\] and \[{R_b} = {k_b}{\left[ A \right]^{\dfrac{1}{2}}}\left[ B \right]\]
And at equilibrium,
 \[
  {R_f} = {R_b} \\
  or,{k_f}\left[ C \right] = {k_b}{\left[ A \right]^{\dfrac{1}{2}}}\left[ B \right] \\
  or,\dfrac{{{k_f}}}{{{k_b}}} = \dfrac{{{{\left[ A \right]}^{\dfrac{1}{2}}}\left[ B \right]}}{{\left[ C \right]}} \\
  {k'}_{eq} = \dfrac{{{{\left[ A \right]}^{\dfrac{1}{2}}}\left[ B \right]}}{{\left[ C \right]}} \\
 \] ….(2)
Where \[{k'}_{eq}\] is equilibrium constant of second reaction
Now compare equation (1) and (2), and find out the relation as follows,
 \[{k'}_{eq} = \dfrac{1}{{\sqrt {{k_{eq}}} }}\]
The value of \[{k_{eq}}\] is 40. Therefore, the value of \[{k'}_{eq}\] is,
 \[
  {k'}_{eq} = \dfrac{1}{{\sqrt {{k_{eq}}} }} \\
  or,{k'}_{eq} = \dfrac{1}{{\sqrt {40} }} \\
 \]

So, the correct option is B.

Note: At equilibrium forward reaction rate and backward reaction rate become the same and the equilibrium constant is the ratio of product side concentration to reactant side concentration.