Question

At what temperature liquid water will be in equilibrium with water vapor?
$\Delta {{\text{H}}_{{\text{vap}}}} = 40.73{\text{kJ}}.{\text{mo}}{{\text{l}}^{ - 1}},\Delta {{\text{S}}_{{\text{vap}}}} = 0.109{\text{kJ}}.{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}$
A. $282.4{\text{K}}$
B. ${\text{373}}{\text{.6K}}$
C. ${\text{100K}}$
D. ${\text{400K}}$

Answer 1

Equilibrium is a state in which there are no observable changes as time goes by. It is of two types-physical and chemical equilibrium. Conversion of water to vapor comes under physical equilibrium. Gibbs free energy is the energy available to do work. It is related to enthalpy, temperature and entropy.

Complete step by step answer:
A reaction is said to be in dynamic equilibrium when the rate of forward reaction is the same as that of backward reaction.
The corresponding reaction of water to vapor conversion is given below:
${{\text{H}}_2}{{\text{O}}_{\left( {\text{l}} \right)}} \rightleftharpoons {{\text{H}}_2}{{\text{O}}_{\left( {\text{g}} \right)}}$
At equilibrium, there is no change in liquid water and water vapor. Therefore gibbs free energy change is zero, i.e. $\Delta {\text{G}} = 0$.
Gibbs free energy change can be calculated from the values of enthalpy, temperature and entropy.
i.e. $\Delta {\text{G}} = \Delta {\text{H}} - {\text{T}}\Delta {\text{S}}$, where $\Delta {\text{G}}$ is the Gibbs free energy, $\Delta {\text{H}}$ is the enthalpy change, ${\text{T}}$ is the temperature and $\Delta {\text{S}}$ is the change in entropy.
Substituting the values of free energy as zero, we get
$0 = \Delta {\text{H}} - {\text{T}}\Delta {\text{S}}$
$\therefore \Delta {\text{H}} = {\text{T}}\Delta {\text{S}}$
From the above equation, we can calculate the temperature.
i.e. ${\text{T}} = \dfrac{{\Delta {\text{H}}}}{{\Delta {\text{S}}}}$
It is given that $\Delta {{\text{H}}_{{\text{vap}}}} = 40.73{\text{kJ}}.{\text{mo}}{{\text{l}}^{ - 1}},\Delta {{\text{S}}_{{\text{vap}}}} = 0.109{\text{kJ}}.{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}$
Substituting the values of enthalpy change and entropy change, we get
${\text{T = }}\dfrac{{40.73{\text{kJ}}.{\text{mo}}{{\text{l}}^{ - 1}}}}{{0.109{\text{kJ}}.{{\text{K}}^{ - 1}}.{\text{mo}}{{\text{l}}^{ - 1}}}}$
On simplification,
${\text{T}} = 373.6{\text{K}}$

Note:
A spontaneous reaction occurs naturally and favors the formation of products at the specified conditions. Spontaneous reactions produce substantial amounts of products at equilibrium and release free energy. Gibbs free energy is the maximum amount of energy that can be coupled to another process to do useful work. $\Delta {\text{G}}$ is negative in spontaneous reactions since it loses energy.