Question

For the reaction at $ 298K $
 $ A(g) + B(g) \rightleftharpoons C(g) + D(g) $
 $ \Delta {H^\circ } = 29.8kcal;\Delta {S^\circ } = 0.1kcal{K^{ - 1}} $
Calculate $ \Delta {G^\circ }andK. $
A) $ \Delta {G^\circ } = 0;K = 1 $
B) $ \Delta {G^\circ } = 1;K = e $
C) $ \Delta {G^\circ } = 2;K = {e^2} $
D) None of these

Answer 1

Hint: A reaction is said to be spontaneous if and only if the Gibbs free energy of the reaction is zero. The equilibrium constant can be said as the ratio of concentration of products to that of concentration of reactants.
Formula used: $ \Delta {G^\circ } = \Delta {H^\circ } - T\Delta {S^\circ } $ and $ \Delta {G^\circ } = - RT\ln K $

Complete step by step solution
Gibbs Free Energy (G) can be defined as the energy related with a chemical reaction that has to be used to do work. It can be represented as the Sum of total enthalpy change minus the product of temperature and the entropy change for that reaction. Mathematically it can be represented as
 $ \Delta {G^\circ } = \Delta {H^\circ } - T\Delta {S^\circ } $ Here $ \Delta {H^\circ } $ will be the total enthalpy change of the reaction, $ T $ is the temperature of the reaction and $ \Delta {S^\circ } $ is the total entropy change for the reaction.
Enthalpy of a system can be termed as the thermodynamic quantity which is the sum of internal energy of the system and the product of pressure of the system and the change in volume of the system.
 $ \Delta {H^\circ } = \Delta {U^\circ } + P\Delta V $ Here $ \Delta {H^\circ } $ will be the total enthalpy change of the reaction, $ \Delta {U^\circ } $ is the internal energy of the reaction and $ P $ is the pressure for the reaction and $ \Delta V $ is the volume change during reaction.
The given values from the question for given reaction: $ A(g) + B(g) \rightleftharpoons C(g) + D(g) $ are
 $ T = 298K $ ; $ \Delta {H^\circ } = 29.8kcal;\Delta {S^\circ } = 0.1kcal{K^{ - 1}} $ .
On substituting in the equation we get,
$ \Delta {G^\circ } = \Delta {H^\circ } - T\Delta {S^\circ } \\
   \Rightarrow \Delta {G^\circ } = 29.8kcal - 298K \times 0.1kcal{K^{ - 1}} \\ $
Simplifying the equation,
$ \Rightarrow \Delta {G^\circ } = 29.8kcal - 29.8kcal \\
   \Rightarrow \Delta {G^\circ } = 0 \\ $
Hence Gibbs energy is zero.
Now the formula in terms of equilibrium constant will be $ \Delta {G^\circ } = - RT\ln K $
Substituting the values,
 $ \Delta {G^\circ } = - RT\ln K \\
   \Rightarrow 0 = 2kcal{K^{ - 1}}mo{l^{ - 1}} \times 298K \times \ln K \\ $
On solving we get,
$ \Rightarrow \ln K = 0 \\
   \Rightarrow K = {e^0} \\
   \Rightarrow K = 1 \\ $
Hence equilibrium constant is equal to one.
 $ \therefore \Delta {G^\circ } = 0;K = 1 $
The correct option is (A).

Additional information:
When the value of entropy is low and decreasing the reaction becomes enthalpy driven as only if the enthalpy is negative the spontaneity of the reaction will be continued and like that in some cases the reaction is entropy driven.

Note:
When a system is endothermic, the enthalpy is positive hence the value of Gibbs free energy and the spontaneity of the reaction depends upon the increasing entropy of the reaction. As if the entropy of reaction will be increasing the value of $ - T\Delta S $ will increase. $ \Delta G $ Will become negative and spontaneous.