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The equilibrium constant of the reaction, ${H_2}(g) + {I_2}(g) \rightleftharpoons 2HI(g)$ is $50$. If the volume of the container is reduced to one half of its original value, the equilibrium constant will be:
A. $25$
B. $50$
C. $75$
D. $100$

Last updated date: 13th Jun 2024
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Hint: Equilibrium constant in concentration units i.e., ${K_c}$ is the ratio of product of molar concentration of products to the product of molar concentration of reactants with each concentration term raise to the power equal to stoichiometric coefficient in balanced reversible reaction.

Complete step-by-step answer:The reaction is given as below:
${H_2}(g) + {I_2}(g) \rightleftharpoons 2HI$
Equilibrium constant $({K_c})$ for this reaction is 50.
$\therefore $ ${K_c} = \dfrac{{\left[ {Product} \right]}}{{\left[ {\operatorname{Re} ac\tan t} \right]}}$
$ \Rightarrow $ \[{K_c} = \dfrac{{{{\left[ {HI} \right]}^2}}}{{\left[ {{H_2}} \right]\;\left[ {{I_2}} \right]}}\] $ - \left( 1 \right)$
We know that Concentration $ = \dfrac{{no.\;of\;moles}}{{volume}}$
$\therefore $ Concentration \[\propto \;\dfrac{1}{{volume}}\]
As given in the question, we have changed the volume and hence there will be a change in the value of concentration which will change the value of the equilibrium constant ${K_c}$.
If the volume of the container is reduced to half of its original value.
i.e., initial volume of Container $ = V$
On reducing the volume of Container$ = \dfrac{V}{2} = 0.5V$, the concentration of the container is increased two times.
New concentration of HI i.e., ${[HI]'}$ = $2[HI]$
Similarly, ${[H_2]'}$= $2[H_2]$
And, ${[I_2]'}$ = $2[I_2]$
Thus, the new equilibrium constant ${K_c}^1$ is given by
$K_c^1 = \dfrac{{{{(2[HI])}^2}}}{{2 \times [{H_2}] \times 2[{I_2}]}}$
$K_c^1 = 1 \times \dfrac{{{{[HI]}^2}}}{{[{H_2}] \times [{I_2}]}}$ $ - \left( 2 \right)$
Substituting $\left( 1 \right)\,in\,\left( 2 \right)$
$\therefore $ $K_c^1 = 1 \times {K_c}$
Given equilibrium constant ${K_c}$ $ = 50$
Then, $K_c^1 = 1 \times 50$
$K_c^1 = 50$
Thus, the value of the new equilibrium constant is 50.

Additional information: It can also be seen in another way that change in number of moles $(\Delta n)$ of the reaction is zero.
i.e.$1{H_2}(g) + 1{I_2}(g) \rightleftharpoons 2HI$
$\Delta ng = \Sigma {n_p} - {\rm E}{n_r} = 2 - 2$
$ = 0$
Hence, there would be no impact of change in concentration on the equilibrium constant ${K_c}$when $\Delta {n_g} = 0$. Hence it will remain unaffected for any change in concentration.

Hence, the correct option is (B).

Note: The value of equilibrium constant is only temperature dependent. It is not dependent on the values of concentration. The value of reaction quotient is concentration dependent. It is compared with the value of equilibrium constant to predict the extent of the reaction. For reactions involving $\Delta {n_g} = 0$ , the values of all the equilibrium constants ${K_p},{K_c},{K_n},{K_x}$ are same.