Question

For the reaction: $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$
The degree of dissociation $(\alpha )$ of $HI(g)$ is related to equilibrium constant, ${K_p}$ by
A.$\dfrac{{1 + 2\sqrt {{K_p}} }}{2}$
B.$\sqrt {\dfrac{{1 + 2{K_p}}}{2}} $
C.$\sqrt {\dfrac{{2{K_p}}}{{1 + 2{K_p}}}} $
D.$\dfrac{{2\sqrt {{K_p}} }}{{1 + 2\sqrt {{K_p}} }}$

Answer 1

Hint:Degree of dissociation is the process of generating free ions carrying current. Equilibrium constant ${K_p}$ is a constant which is calculated by the partial pressures of the compounds in a chemical equation. For the given equation $2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$ we shall apply the formula for calculating Equilibrium constant ${K_p}$.
Formula used: ${K_p} = \dfrac{{{P_{{H_2}}} \times {P_{{I_2}}}}}{{{{\left( {{P_{HI}}} \right)}^2}}}$

Complete step by step answer:
The degree of dissociation refers to the generation of free ions which carry current and dissociate from the solute of given concentration. If the dissolution is exothermic in nature then, the dissociation constant will rise with increase in temperature and will be directly proportional to temperature whereas if the dissolution is endothermic in nature then, the dissociation constant will decrease with increase in temperature and will be inversely proportional to temperature. The percentage of dissociation is represented by $\alpha $ which gives the ratio of concentration of dissociated hydrogen ions to the concentration of undissociated substance.
Equilibrium constant ${K_p}$ is a constant which is calculated by the partial pressures of the compounds in a chemical equation. It gives us the relationship between the partial pressures of the reactants to the partial pressures of the products. It has no unit.
For the given question we consider the balanced equilibrium reaction of $HI$
$2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g)$
Now substituting the number of moles at equilibrium,
\[
  2HI(g) \rightleftharpoons {H_2}(g) + {I_2}(g) \\

  {{ }}1 - \alpha {{ }}\dfrac{\alpha }{2}{{ }}\dfrac{\alpha }{2} \\
 \]
We know that the formula for calculating the equilibrium constant ${K_p}$ for the above given equation is,
${K_p} = \dfrac{{{P_{{H_2}}} \times {P_{{I_2}}}}}{{{{\left( {{P_{HI}}} \right)}^2}}}$
Now substitute the moles in the equilibrium constant ${K_p}$ formula,
${K_p} = \dfrac{{{{\left( {\dfrac{\alpha }{2}{P_T}} \right)}^2}}}{{{{\left( {1 - \alpha } \right)}^2}P_T^2}}$
$ \Rightarrow {K_p} = \dfrac{\alpha }{{1 - \alpha }}$
$ \Rightarrow {K_p} = 2\sqrt {{K_p}} $
Hence we conclude that,
$\alpha = \dfrac{{2\sqrt {{K_p}} }}{{1 + 2\sqrt {{K_p}} }}$

So, the correct option for the above question is (D) $\dfrac{{2\sqrt {{K_p}} }}{{1 + 2\sqrt {{K_p}} }}$

Note:
It is important to understand the extent, feasibility and applications of a reaction to use it effectively but it is equally important to understand the rate of reaction and the factors affecting the rate of reaction. The rate of a reaction at a particular moment is known as the instantaneous rate.
 $Rate = k{\left[ A \right]^x}{\left[ B \right]^y}$