Question

Match the List I (Equations) with the List II (Type of Processes).

List I (Equations)	List II (Type of Processes)
a) ${K_P} > Q$	i. Non-spontaneous
b) $\Delta {G_o} < RT{\log _c}Q$	Ii. Equilibrium
c) ${K_P} = Q$	Iii. Spontaneous and endothermic
d) $T > \dfrac{{\Delta H}}{{\Delta S}}$	Iv. Spontaneous

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

Answer 1

$\Delta G$, which is the Gibbs free energy is related to the equilibrium constant, k as if k < 1, then the reaction favours the reactants and hence it is non-spontaneous which makes $\Delta G > 0$. And if k > 1, then the reaction favours the products and hence it is spontaneous which makes $\Delta G < 0$. $\Delta G$ is also related to the enthalpy and the entropy of the system. It is equal to the change in enthalpy ($\Delta H$) minus the temperature multiplied with the change in entropy ($\Delta S$).

Complete Solution:
Let us takes an example, at equilibrium
$A(g) + B(g) \to C(g)$
The equilibrium constant in terms of partial pressure is given as ratio of product of the partial pressures of the products raised to their respective stoichiometric coefficients to the product of the partial pressures of the Reactants raised to their respective stoichiometric coefficients. It is calculated at equilibrium and is written as follows.
${K_P} = \dfrac{{{P_C}}}{{{P_A} \times {P_B}}}$

Whereas, Q is the active mass of the reaction, calculated at any time of the reaction.
$A(g) + B(g) \to C(g)$
It is given as follows.
$Q = \dfrac{{{\rm{concentration~of~product}}}}{{{\rm{concentration~of~reactant}}}}$

In the (a) part, it is given that ${K_P} > Q$. And we know that,
${{\rm{K}}_{\rm{P}}}{\rm{ = }}\dfrac{{{\rm{rate~of~forward~reaction}}}}{{{\rm{rate~of~backward ~reaction}}}}$
And ${K_P} > Q$, therefore, when this happens, the reaction moves in the forward direction, or we can say that the rate of forward reaction is greater than the rate of the backward reaction because the concentration of the reaction is more than the concentration of the products in this case. This happens when the reaction is spontaneous. Hence, (a)-iv.

In the (b) part, it is given that $\Delta {G^o} < RT {\log _c}Q$. Here, $\Delta {G^o}$ is the Gibbs free energy and can be defined as the difference of the change in enthalpy and the product of temperature and entropy. When $\Delta {G^o} < 0$, the reaction is said to be spontaneous and when $\Delta {G^o} > 0$, then the reaction is said to be non-spontaneous. This is due to the reason that, when $\Delta G < 0$, the reactants have more free energy than the products and hence the reaction moves in forward direction and is spontaneous. And when $\Delta G > 0$, the reactants have more free energy than the products and more extra energy is required for the reaction to proceed and thus the reaction is nonspontaneous. In the given scenario, $\Delta {G^o} < RT{\log _c}Q$, but we know that $\Delta {G^o} = - RT {\log _c}Q$. Now, when we equate these two equations,
$- RT {\log _c}Q < RT {\log _c}Q$
we can see that $K$ < $Q$, which shows that the concentration of the reactants is less than that of the products. Therefore, $\Delta {G^o} > 0$. This shows that the reaction is nonspontaneous. Hence, (b)-i.

In the (c) part, it is given that ${K_P} = Q$. This means that the equilibrium constant and the active mass at any time are equal. This happens only in the case of equilibrium. Hence, (c)-ii.
In the (d) part, it is given that $T > \dfrac{{\Delta H}}{{\Delta S}}$. This means that $T\Delta S > \Delta H$. And the Gibbs free energy is given as $\Delta {G^o} = \Delta H - T\Delta S$, and if $T\Delta S > \Delta H$, then $\Delta {G^o} < 0$and the nature of the reaction will be spontaneous and also it is an endothermic reaction. Hence, (d)-iii.

So, the correct answer is “Option C”.

Note: Students tend to confuse between entropy and enthalpy. There is a difference between the two. Enthalpy is the total heat present in a thermodynamic system where the pressure is constant. Whereas, entropy is the degree of disorder in a thermodynamic system.