Question

Match List I (equations) with List II (process) and select the correct option:

	List I		List II
	Equation		Types of process
(a)	${{\text{K}}_{\text{P}}}{\text{ > Q}}$	(i)	Non-spontaneous
(b)	$\Delta {{\text{G}}^0}\, < \,{\text{RT}}\,{\text{lnQ}}$	(ii)	Equilibrium
(c)	${{\text{K}}_{\text{P}}}{\text{ = Q}}$	(iii)	Spontaneous and endothermic
(d)	\[{\text{T}}\,{\text{ > }}\,\dfrac{{\Delta {\text{H}}\,}}{{\Delta {\text{S}}}}\]	(iv)	spontaneous

A. (a)-(i), (b)-(ii), (c)-(iii), (d)-(iv)
B. (a)-(iii), (b)-(iv), (c)-(ii), (d)-(i)
C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii)
D. (a)-(ii), (b)-(i), (c)-(iv), (d)-(iii)

Answer 1

Hint:To answer we should know the condition required for the reaction to be spontaneous and non-spontaneous. We will write the equilibrium constant, reaction quotient values. We will also write the Gibbs free energy and equilibrium constant relationship and Gibbs free energy and enthalpy relation. Then by applying the given condition we will determine the correct options.

Complete step-by-step answer:The equilibrium constant is defined as the product of the pressure of products having a stoichiometric coefficient as power divided by the product of the concentration of reactants having a stoichiometric coefficient as power.
${\text{A}}\, + \,{\text{B}}\,\, \rightleftharpoons {\text{C}}\, + \,{\text{D}}$
The equilibrium constant expression for the above reaction is as follows:
${{\text{k}}_{\text{P}}}{\text{ = }}\,\dfrac{{\left[ {\text{C}} \right]\,\,\left[ {\text{D}} \right]}}{{\left[ {\text{A}} \right]\left[ {\text{B}} \right]}}\,$
When the reaction is not equilibrium means after or before the equilibrium the concentration of reactant and product are different. The ratio of product and reactant is shown by reaction quotient which is shown as follows:
${\text{Q = }}\,\dfrac{{\left[ {\text{C}} \right]\,\,\left[ {\text{D}} \right]}}{{\left[ {\text{A}} \right]\left[ {\text{B}} \right]}}\,$
${{\text{K}}_{\text{P}}}{\text{ > Q}}$ means Q will increase to approach the ${{\text{K}}_{\text{P}}}$. From the reaction quotient expression, we can say the reaction quotient Q is directly proportional to the concentration of product. So, to increase the value of the reaction quotient, the reaction will go in a forward direction.
${{\text{K}}_{\text{P}}}{\text{ = Q}}$ means the reaction is in equilibrium.
The Gibbs free energy is related to the enthalpy and entropy as follows:
$\Delta {\text{G}}\,{\text{ = }}\,\Delta {\text{H}}\, - {\text{T}}\Delta {\text{S}}$….$(1)$
It is given, ${\text{T}}\,{\text{ > }}\,\dfrac{{\Delta {\text{H}}\,}}{{\Delta {\text{S}}}}$
On rearranging,
${\text{T}}\Delta {\text{S}}\,{\text{ > }}\,\Delta {\text{H}}$
The value of ${\text{T}}\Delta {\text{S}}$ is negative so, the value of $\Delta {\text{G}}\,{\text{ = }}\, - \,{\text{ve}}$. So, the reaction will be spontaneous.
According to the equation $(1)$, the $\,{{\Delta H}}$ is positive, so the heat is required and hence the reaction is endothermic.
The relationship between free energy change in a reaction and the corresponding equilibrium constant, is as follows:
 $ - \Delta {{\text{G}}^0}\, = \,{\text{RT}}\,{\text{ln}}{{\text{K}}_{\text{C}}}$
It is given that, $\Delta {{\text{G}}^0}\, < \,{\text{RT}}\,{\text{lnQ}}$
So, on applying the given condition,
\[\Delta {{\text{G}}^0}\, = \, + \,{\text{ve}}\]
So, the reaction will be non-spontaneous.
So, the correct match is:

	List I		List II
	Equation		Types of process
(a)	${{\text{K}}_{\text{P}}}{\text{ > Q}}$	(iv)	Spontaneous
(b)	$\Delta {{\text{G}}^0}\, < \,{\text{RT}}\,{\text{lnQ}}$	(i)	Non-spontaneous
(c)	${{\text{K}}_{\text{P}}}{\text{ = Q}}$	(ii)	Equilibrium
(d)	${\text{T}}\,{\text{ > }}\,\dfrac{{\Delta {\text{H}}\,}}{{\Delta {\text{S}}}}$	(iii)	spontaneous and endothermic

Therefore, option (C) C. (a)-(iv), (b)-(i), (c)-(ii), (d)-(iii), is correct.

Note:At ${{\text{K}}_{\text{P}}}{\text{ < Q}}$ condition reaction goes in backward direction. The Gibbs free energy is the energy which is used in work. At equilibrium, the rate of forward direction and backward direction reaction becomes equal, so the concentration on both sides of the reaction becomes equal. So, entropy and enthalpy change tends to zero so the free energy change also tends to zero. The negative value of Gibbs free energy tells the reaction is spontaneous. The positive value of Gibbs free energy tells the reaction is nonspontaneous. The reaction release heat is known as exothermic reaction. The value of $\Delta {\text{H}}$ is negative for that reaction.