Question

At a certain temperature, only $ 50\% $ $ {\text{HI}} $ is dissociated into $ {{\text{H}}_2} $ and $ {{\text{I}}_2} $ at equilibrium. The equilibrium constant is:
(A) 1.0
(B) 3.0
(C) 0.5
(D) 0.25

Answer 1

Hint: To answer this question, you must recall the formula for equilibrium constant of a reaction, in terms of the concentration of the constituents of the reaction mixture. In the reaction given in the question, 2 moles of $ {\text{HI}} $ dissociate to form one mole each of $ {{\text{H}}_2} $ and $ {{\text{I}}_2} $ .
Formula used: For the reaction given in the question, the equilibrium constant can be written as,
 $ {{\text{K}}_{\text{c}}} = \dfrac{{\left[ {{{\text{H}}_2}} \right]\left[ {{{\text{I}}_2}} \right]}}{{{{\left[ {{\text{HI}}} \right]}^2}}} $
Where, $ {{\text{K}}_{\text{c}}} $ represents the equilibrium constant in terms of concentrations.
 $ \left[ {{{\text{H}}_{\text{2}}}} \right] $ represents the concentration of hydrogen gas at equilibrium
 $ \left[ {{{\text{I}}_{\text{2}}}} \right] $ represents the concentration of iodine gas at equilibrium
And, $ \left[ {{\text{HI}}} \right] $ represents the concentration of hydrogen iodide at equilibrium.

Complete step by step solution

At t=0	2	0	0
At equilibrium	$ 2 - 2{\text{X}} $	X	X

The dissociation of $ {\text{HI}} $ is given as $ {\text{2HI}} \rightleftharpoons {{\text{H}}_{\text{2}}} + {{\text{I}}_{\text{2}}} $
We are given the initial amount of $ {\text{HI}} $ as 2 moles. We assume the amount of $ {\text{HI}} $ dissociating as $ {\text{X}} $ .
We are given that $ 50\% $ $ {\text{HI}} $ is dissociated. So, $ {\text{X}} = \dfrac{{50}}{{100}} $ . Thus, we can now find the concentrations of all the substances and write the equilibrium constant for the reaction as,
 $ {{\text{K}}_{\text{c}}} = \dfrac{{\left[ {{{\text{H}}_2}} \right]\left[ {{{\text{I}}_2}} \right]}}{{{{\left[ {{\text{HI}}} \right]}^2}}} $
Substituting:
 $ \Rightarrow {{\text{K}}_{\text{c}}} = \dfrac{{\left( {0.5} \right)\left( {0.5} \right)}}{{{{\left( 1 \right)}^2}}} $
 $ \Rightarrow {{\text{K}}_{\text{c}}} = 0.25 $
Thus, the correct answer is D.

Note
The numerical calculation for equilibrium constant is done by allowing the chemical reaction to reach the equilibrium and then measuring the concentrations of each constituent at that point. Since the concentrations of these constituents are measured at equilibrium point, thus equilibrium constant will always have the same value for a given reaction no matter what is the initial concentration of the reactants.
Using this information about the equilibrium constant, we can derive a standard expression that serves as a standard model for all reactions. This form of the equilibrium constant can be modified as per the requirement of the chemical reaction and is given as
 $ {K_{eq}} = \dfrac{{{{\left( {{a_C}} \right)}^c}}}{{{{\left( {{a_A}} \right)}^a}{{\left( {{a_B}} \right)}^b}}} $ for a reaction: $ aA + bB \rightleftharpoons cC $