Question

${K_C}$ for the synthesis of HI (g) from ${H_2}$ (g) and ${I_2}$ (g) is 50. The degree of dissociation of HI is:
(A) 0.10
(B) 0.14
(C) 0.18
(D) 0.22

Answer 1

Hint: The value of ${K_C}$ for dissociation will be equal to reciprocal of the ${K_C}$ for synthesis of HI (g). ${K_C}$ is the ratio of molar concentration of products to reactants at equilibrium concentrations raised to the power of their stoichiometric coefficients.

Complete step by step answer:
-First of all we will write the balanced reaction for dissociation of HI. And we will take the initial concentration of HI to be = 1.
Let the degree of dissociation for the reaction be α.
The reaction and concentration of the reactants and products before and after the reaction will be written as:
$\begin{array}{*{20}{c}}
{}&{2HI(g)}& \rightleftharpoons &{{H_2}(g)}& + &{{I_2}(g)} \\
{Initially:}&1&{}&0&{}&0 \\
{Equilibrium:}&{1 - \alpha }&{}&{\dfrac{\alpha }{2}}&{}&{\dfrac{\alpha }{2}}
\end{array}$
-We will now see what ${K_C}$ is.
${K_C}$ is basically an equilibrium constant which represents the ratio of the equilibrium concentrations of products to the equilibrium concentrations of reactants, all raised to the power of their stoichiometric coefficients.
For a reaction: $aA \rightleftharpoons bB$
The value of ${K_C}$ for the above reaction will be:
${K_C} = \dfrac{{{{\left[ B \right]}^b}}}{{{{\left[ A \right]}^a}}}$
-Similarly we will write the expression of ${K_C}$ for the dissociation reaction of HI.
The question gives us the value of ${K_C}$ for the synthesis of HI to be 50. So, the value of ${K_C}$ for the dissociation of HI will be = $\dfrac{1}{{50}}$
-For it the expression of ${K_C}$ will be:
${K_C} = \dfrac{{\left( {\dfrac{\alpha }{2}} \right)\left( {\dfrac{\alpha }{2}} \right)}}{{{{\left( {1 - \alpha } \right)}^2}}} = \dfrac{1}{{50}}$
$50 \times {\left( {\dfrac{\alpha }{2}} \right)^2} = {\left( {1 - \alpha } \right)^2}$
$\dfrac{\alpha }{{2(1 - \alpha )}} = \dfrac{1}{{5\sqrt 2 }}$
 $5\sqrt 2 \alpha = 2 - 2\alpha $
 $\left( {2 + 5\sqrt 2 } \right)\alpha = 2$
 $\alpha = \dfrac{2}{{2 + 5\sqrt 2 }}$
  = 0.22
Hence the degree of dissociation of HI will be: 0.22

So, the correct option will be: (D) 0.22

Note: Equilibrium constant for any reaction depends on temperature and is independent of the actual quantities of reactants and products, the presence of catalyst and the presence of any inert material. It also does not depend on the concentrations, pressures and volumes of the reactants and products.