Question

For reaction $HI \rightleftharpoons \dfrac{1}{2}{H_2} + \dfrac{1}{2}{I_2}$ value of ${K_c}$ is $\dfrac{1}{8}$ then value of ${K_c}$ for ${H_2} + {I_2} \rightleftharpoons 2HI$
A. $\dfrac{1}{{64}}$
B. $64$
C. $\dfrac{1}{8}$
D. $8$

Answer 1

Hint: We can say in a chemical reaction chemical equilibrium is the condition in which the rates of forward reaction and the reverse reaction are almost identical. The result is that concentration of the reactants and the products remains fixed. We have to reciprocate the equilibrium constant of the original reaction (forward reaction) to get the equilibrium constant of the reverse reaction.

Formula used: The formula used to calculate the equilibrium constant for the second reaction is,
${{\text{K}}^{\text{'}}}{\text{ = }}\dfrac{{\text{1}}}{{\text{K}}}$
The equilibrium concentration of the second reaction (reverse reaction) is indicated by ${{\text{K}}^{\text{'}}}.$
The equilibrium concentration of the first reaction (forward reaction) is indicated by ${\text{K}}.$

Complete step by step answer:
The given data contains,
We know the value of equilibrium constant for the first reaction is $\dfrac{1}{8}$.
The given first reaction,
$HI \rightleftharpoons \dfrac{1}{2}{H_2} + \dfrac{1}{2}{I_2}$
We can write the rate expression for this reaction as,
${K_1} = \dfrac{{{{\left[ {{H_2}} \right]}^{1/2}}{{\left[ {{I_2}} \right]}^{1/2}}}}{{\left[ {HI} \right]}} \to \left( 1 \right)$
The given second equation is,
${H_2} + {I_2} \rightleftharpoons 2HI$
We can write the rate expression for this reaction as,
${K_2} = \dfrac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\left[ {{I_2}} \right]}} \to \left( 2 \right)$
From equations (2) and (1)
${K_1}^2 = \dfrac{1}{{{K_2}}}$
We know the value of the equilibrium constant for the first reaction is $\dfrac{1}{8}$.
${K_2} = \dfrac{1}{{{K_1}^2}}$
Let substitute the values in above formula we get,
$ \Rightarrow {K_2} = \dfrac{{\dfrac{1}{1}}}{{{{\left( {\dfrac{1}{8}} \right)}^2}}}$
On simplifying we get,
$ \Rightarrow {K_2} = 64$
The value for equilibrium constant for the reaction ${H_2} + {I_2} \rightleftharpoons 2HI$ is $64$.

So, the correct answer is Option B.

Note: We have remembered that the value of equilibrium constant is not affected by the rate of catalyst in forward and backward reaction. We can predict the extent of the reaction with the help of the equilibrium constant. The magnitude of the equilibrium constant tells about the relative amounts of the reactants and the products. We have to remember that if the value of equilibrium constant is huge, forward reaction is preferred and a reverse reaction is preferred when the value of equilibrium constant is very less.