Question

For the process ${{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right) \to {{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{1bar,373K}}} \right)$, the correct set of thermodynamic parameters is:
A.${{\Delta G = 0,\Delta S = + ve}}$
B.${{\Delta G = 0,\Delta S = - ve}}$
C.${{\Delta G = + ve,\Delta S = 0}}$
D.${{\Delta G = - ve,\Delta S = + ve}}$

Answer 1

Hint: Entropy is defined as a measure of the randomness or the disorderliness of a system. Greater the randomness of the system, greater is the entropy. The Gibbs free energy change of a system is zero if the reaction is at equilibrium.

Complete step by step answer:
Entropy is the property of a system which measures the randomness or the amount of disorder of a system quantitatively. It is a state function just like internal energy and enthalpy because the magnitude of its change also depends on the entropies of the system in the initial and final states.For the given process of ${{\text{H}}_{\text{2}}}{\text{O}}\left( {\text{l}} \right) \to {{\text{H}}_{\text{2}}}{\text{O}}\left( {{\text{1bar,373K}}} \right)$, we can see that the conversion of water takes place at the same temperature which is equal to 373K or 100 degree Celsius and also at the same pressure. Thus, water is converted from the liquid phase to the gas phase at its boiling point. Thus, the process is the vaporization of water. Since it involves absorption of heat energy, the enthalpy change of vaporization will be positive.
Now, the atoms in gases are more apart from one another than those in liquids. So, they are much more disordered than liquids. Thus, the entropy of the product which is in the gas phase is more than that of the reactant which is in the liquid phase. Therefore, entropy change will be positive for this reaction.
The Gibbs Helmholtz equation gives the relation between the free energy change and enthalpy and entropy change as:
$\Delta {\text{G}} = \Delta {\text{H - T}}\Delta {\text{S}}$
Since, water in liquid phase is in equilibrium with water in gas phase, $\Delta {\text{H = T}}\Delta {\text{S}}$.
Hence, ${{\Delta G = 0}}$.

Hence, the correct option is A.

Note:
The Gibbs free energy concept is useful in determining the spontaneity or the feasibility of a process. The process will be spontaneous if $\Delta {\text{G}}$ is negative and non-spontaneous when $\Delta {\text{G}}$ is positive.
For endothermic reactions, the enthalpy change is positive. If in this case, if the entropy change is negative, then from Gibbs Helmholtz equation, it can be deduced that free energy change is positive and the reaction will be non-spontaneous. But, if the entropy change is positive, then from Gibbs Helmholtz equation, it can be deduced that free energy change is negative and the reaction will be spontaneous.