Question

The equilibrium constant ( ${K_p}$) for the decomposition of gaseous ${H_2}O$ . ${H_2}O(g) \rightleftharpoons {H_2}(g) + \dfrac{1}{2}{O_2}(g)$ is related to degree of dissociation ( $\alpha $ ) at a total pressure p is given by,
A.${K_p} = \dfrac{{{\alpha ^3}{p^{\dfrac{1}{2}}}}}{{\left( {1 + \alpha } \right){{\left( {2 + \alpha } \right)}^{\dfrac{1}{2}}}}}$
B.${K_p} = \dfrac{{{\alpha ^3}{p^{\dfrac{3}{2}}}}}{{\left( {1 + \alpha } \right){{\left( {2 + \alpha } \right)}^{\dfrac{1}{2}}}}}$
C.${K_p} = \dfrac{{{\alpha ^{\dfrac{3}{2}}}{p^2}}}{{\left( {1 + \alpha } \right){{\left( {2 + \alpha } \right)}^{\dfrac{1}{2}}}}}$
D.${K_p} = \dfrac{{{\alpha ^{\dfrac{3}{2}}}{p^{\dfrac{1}{2}}}}}{{\left( {1 - \alpha } \right){{\left( {2 + \alpha } \right)}^{\dfrac{1}{2}}}}}$

Answer 1

Hint: ${K_p}$ is the equilibrium constant written in terms of partial pressure of reactants and products. So in order to solve this question, we need to calculate partial pressure of each component in terms of degree of dissociation.

Complete step by step answer:
The given reaction is,
${H_2}O(g) \rightleftharpoons {H_2}(g) + \dfrac{1}{2}{O_2}(g)$
${K_p}$ for this reaction can be written as,
\[{K_p} = \dfrac{{\left( {p{H_2}} \right){{\left( {p{O_2}} \right)}^{\dfrac{1}{2}}}}}{{\left( {p{H_2}O} \right)}}\]
Where $p{H_2}$ is partial pressure of hydrogen gas, $p{O_2}$ is partial pressure of oxygen gas and $p{H_2}O$ is partial pressure of water vapour.
Let X be the initial pressure of water vapour. At this time, pressure of hydrogen gas and oxygen gas are zero. At equilibrium, pressure of water vapour, hydrogen gas and oxygen gas are $X(1 - \alpha )$ , $X\alpha $ and $\dfrac{{X\alpha }}{2}$ respectively.
$
  {H_2}O(g) \rightleftharpoons {H_2}(g) + \dfrac{1}{2}{O_2}(g) \\
  {\text{X 0 0 (at t = 0)}} \\
  {\text{X(1 - }}\alpha {\text{) X}}\alpha {\text{ }}\dfrac{{X\alpha }}{2}{\text{ (at equilibrium)}} \\
$
Given that total pressure is p. Hence we can write,
$p = X(1 - \alpha ) + X\alpha + \dfrac{{X\alpha }}{2} = X + \dfrac{{X\alpha }}{2} = \dfrac{{X(2 + \alpha )}}{2}$
From this we can write,
$X = \dfrac{{2p}}{{2 + \alpha }}$
Now let us substitute the values of partial pressures on the equation of ${K_p}$ .
\[{K_p} = \dfrac{{\left( {X\alpha } \right){{\left( {\dfrac{{X\alpha }}{2}} \right)}^{\dfrac{1}{2}}}}}{{X(1 - \alpha )}} = \left( {\dfrac{{{X^{\dfrac{3}{2}}}{\alpha ^{\dfrac{3}{2}}}}}{{{2^{\dfrac{1}{2}}}X(1 - \alpha )}}} \right) = \left( {\dfrac{{{X^{\dfrac{1}{2}}}{\alpha ^{\dfrac{3}{2}}}}}{{{2^{\dfrac{1}{2}}}(1 - \alpha )}}} \right)\]
Now substitute the value of X, $X = \dfrac{{2p}}{{2 + \alpha }}$
\[{K_p} = \left( {\dfrac{{{{\left( {\dfrac{{p\alpha }}{{(2 + \alpha )}}} \right)}^{\dfrac{1}{2}}}{\alpha ^{\dfrac{3}{2}}}}}{{{2^{\dfrac{1}{2}}}(1 - \alpha )}}} \right)\]
Simplifying the equation we get,
\[{K_p} = \dfrac{{{\alpha ^{\dfrac{3}{2}}}{p^{\dfrac{1}{2}}}}}{{\left( {1 - \alpha } \right){{\left( {2 + \alpha } \right)}^{\dfrac{1}{2}}}}}\]
Therefore, the correct option is D.

Note:
We can also do the calculation by first substituting the value of X on partial pressure of each component and then substituting the values on the equation of ${K_p}$ .
Partial pressure of water vapour in terms of degree of dissociation can be written as,
$p{H_2}O = X(1 - \alpha ) = \dfrac{{2p}}{{2 + \alpha }}(1 - \alpha )$
Partial pressure of hydrogen gas can be written as,
$p{H_2} = X\alpha = \dfrac{{2p\alpha }}{{2 + \alpha }}$
Partial pressure of oxygen gas can be written as,
$p{O_2} = \dfrac{{X\alpha }}{2} = \dfrac{{2p\alpha }}{{2\left( {2 + \alpha } \right)}} = \dfrac{{p\alpha }}{{\left( {2 + \alpha } \right)}}$
When we substitute these values on the equation of ${K_p}$ , we will get the same answer.