Question

Match the following (take reactant to be in stoichiometric proportion in case of two reactants)

Reaction(Homogeneous gaseous phase)	Degree of dissociation of reactant in terms of equilibrium constant
$1$ $A + B\,\,\overset {} \leftrightarrows \,\,\,2C$	$\,(\sqrt K )/(1 + \sqrt K )$
$2$ $2A\,\,\,\overset {} \leftrightarrows \,\,B + C$	$\,(\sqrt K )/(2 + \sqrt K )$
$3$ $A + B\,\,\,\overset {} \leftrightarrows \,\,\,C + D$	$\,2K/(1 + 2K)$
$4$ $AB\,\,\overset {} \leftrightarrows \,\,\dfrac{1}{2}{A_2} + \dfrac{1}{2}{B_2}$	$\dfrac{{2\sqrt K }}{{1 + 2\sqrt K }}$

A.$1$–d, $2$-c,$3$-b, $4$-a
B.$1$-a,$2$-c,$3$-b,$4$-d
C.$1$-b,$2$-d,$3$-a,$4$-c
D.$1$-b,$2$-a,$3$-d,$4$-c

Answer 1

In the above question, there are four reactions given which are homogenous in the gaseous phase. The degree of dissociation is defined as the fraction of molecules dissociated at a given time.

Complete answer:
Dissociation is the general process in which molecules or ionic compounds like salt and complexes dissociate into smaller particles like ions and free radicals. The degree of dissociation is defined as the fraction of solute molecules dissociated at a given time. It indicates the extent to which dissociation occurs. It is denoted by the symbol $\alpha$.
Degree of dissociation depends on several factors such as temperature, concentration of the solution and the nature of the solvent. When the temperature increases, the degree of dissociation also increases. Similarly when dilution of solution increases, the degree of dissociation also increases, but degree of dissociation decreases as the concentration of solution increases. The polar and non-polar nature of solvent also affects the degree of dissociation.
Under normal dilution, the value of $\alpha$ for strong electrolytes is equal to $1$ and for weak electrolytes is less than$1$.
In the above question, four homogeneous gaseous reactions are given:
$(1)$ $A\,\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,B\,\,\,\,\,\,\,\overset {} \leftrightarrows \,\,\,\,\,2C$
Initial concentration $1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$
At equilibrium $(1 - \alpha )\,\,\,\,\,\,\,\,\,\,\,\,(1 - \alpha )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha$
$K = \dfrac{{{{[C]}^2}}}{{[A][B]}}$
$K = \dfrac{{4{\alpha ^2}}}{{(1 - \alpha )(1 - \alpha )}}$
$K = \dfrac{{4{\alpha ^2}}}{{{{(1 - \alpha )}^2}}}$
Then, find the value of $\alpha$
$\sqrt K = \dfrac{{2\alpha }}{{1 - \alpha }}$
Therefore, $\alpha = \dfrac{{\sqrt K }}{{2 + \sqrt K }}$
This the relation between degree of dissociation and equilibrium constant for $(1)$ reaction
Similarly, for $(2)$ reaction
$(2)$ $2A\,\,\,\,\,\,\,\,\,\,\,\overset {} \leftrightarrows \,\,\,\,\,B\,\,\,\,\,\, + \,\,\,\,C$
Initial concentration $\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0$
At equilibrium $(2 - 2\alpha )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\alpha$
$K = \dfrac{{[B][C]}}{{{{[A]}^2}}}$
$K = \dfrac{{{\alpha ^2}}}{{{{(2 - 2\alpha )}^2}}}$
Take square roots on both sides.
$2\sqrt K = \dfrac{\alpha }{{1 - \alpha }}$
Then, solve for $\,\alpha$
$\alpha = \dfrac{{2\sqrt K }}{{1 + 2\sqrt K }}$
Similarly, for $(3)$reactions
$(3)$ $A\,\,\,\,\,\,\,\,\, + \,\,\,\,\,\,\,\,\,B\,\,\,\,\,\,\,\,\,\overset {} \leftrightarrows \,\,\,\,\,C\,\,\,\,\,\, + \,\,\,\,D$
Initial concentration $\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,0$
At equilibrium $(1 - \alpha )\,\,\,\,\,\,\,\,\,\,(1 - \alpha \,)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\alpha \,\,\,\,\,\,\,\,\,\,\,\,\alpha$
$K = \dfrac{{[C][D]}}{{[A][B]}}$
$K = \dfrac{{{\alpha ^2}}}{{{{(1 - \alpha )}^2}}}$
Taking square root on both sides,
$\sqrt K = \dfrac{\alpha }{{(1 - \alpha )}}$
Then, simplify it and solve for $\,\alpha$
$\alpha = \dfrac{{\sqrt K }}{{1 + \sqrt K }}$
Similarly, we can calculate for $(4)$ reaction
$(4)$ $AB\,\,\,\,\,\,\,\,\,\,\,\,\overset {} \leftrightarrows \,\,\,\,\dfrac{1}{2}{A_2}\,\, + \,\,\,\,\,\dfrac{1}{2}{B_2}$
Initial concentration $\,1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0$
At equilibrium $(1 - \alpha )\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{\alpha }{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\dfrac{\alpha }{2}\,\,\,$
$K = \dfrac{{{{[{A_2}]}^{1/2}}{{[{B_2}]}^{1/2}}}}{{[AB]}}$
Substituting the values, we get
$K = \dfrac{{\dfrac{\alpha }{2}}}{{1 - \alpha }}$
$K - \alpha K = \dfrac{\alpha }{2}$
$\alpha = \dfrac{{2K}}{{1 + 2K}}$
Hence, we got the relation between $\alpha$ and $K$ for all the four reactions.
Therefore, the correct answer is option (C).

Note:
Don’t get confused with the term dissociation constant and degree of dissociation. There is a difference between them. Degree of dissociation is the extent of dissociation that occurs at a particular time whereas dissociation constant is the ratio of dissociated ions to the original molecules present in the solutions. Strong acids and strong bases have degree of dissociation close to $1$, but weak acids and weak bases have low degree of dissociation.