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Question: At \( {25^{\text{o}}}{\text{C}} \) , for the process \( {H_2}O(l) \rightleftharpoons {H_2}O(g) \) ; ...

At 25oC{25^{\text{o}}}{\text{C}} , for the process H2O(l)⇌H2O(g){H_2}O(l) \rightleftharpoons {H_2}O(g) ; ΔGo\Delta {{\text{G}}^{\text{o}}} is 8.6kJ8.6{\text{kJ}} . The vapour pressure of water at this temperature is nearly:
(A) 29 torr29{\text{ torr}}
(B) 285 torr285{\text{ torr}}
(C) 23.17 torr23.17{\text{ torr}}
(D) 100 torr100{\text{ torr}}

Explanation

Solution

To answer this question, you must recall the relation between the equilibrium constant of a reaction and its free energy. Gibbs free energy or the Gibbs function is a thermodynamic function which is used for the calculation of the maximum work performed by a system at constant temperature and pressure. We shall calculate the equilibrium constant for the reaction and substitute the value in the equation given below.
Formulae used: ΔGθ=−RTln⁡Keq\Delta {G^\theta } = - RT\ln {K_{eq}}
Where, ΔG\Delta G represents the change in the free energy during the reaction
Keq{K_{eq}} represents the equilibrium constant of the reaction.
And, TT represents the temperature of the reaction mixture.

Complete step by step solution
We are given the reaction, H2O(l)⇌H2O(g){H_2}O(l) \rightleftharpoons {H_2}O(g)
We can write the equilibrium constant for the reaction as K=aH2O(g)aH2O(l)K = \dfrac{{{a_{{H_2}O(g)}}}}{{{a_{{H_2}O(l)}}}}
Since water as liquid is pure, thus, its concentration or activity will be equal to unity. Thus, the equilibrium constant will be , K=aH2O(g)K = {a_{{H_2}O(g)}} .
The activity of water in vapour form will be equal to the ratio of its partial pressure to the total pressure, which in this case would be equal to the atmospheric pressure. So we can write the equilibrium constant in terms of partial pressure as Kp=K=pH2OP{K_p} = K = \dfrac{{{p_{{H_2}O}}}}{P}
We know the relation between equilibrium constant and partial pressure of gas as ΔGθ=−RTln⁡Keq\Delta {G^\theta } = - RT\ln {K_{eq}}
Or we can write, K=e−ΔGRT{\text{K}} = {{\text{e}}^{\dfrac{{ - {{\Delta G}}}}{{{\text{RT}}}}}} .
⇒pH2O=P.e−ΔGRT\Rightarrow {p_{{H_2}O}} = P.{e^{\dfrac{{ - {{\Delta G}}}}{{{\text{RT}}}}}}
Substituting:
⇒101325Pa.e−8600J8.314×298K\Rightarrow 101325Pa.{{\text{e}}^{\dfrac{{ - 8600{\text{J}}}}{{8.314 \times 298{\text{K}}}}}}
Solving this, we get:
P=3155 Pa=23.17 torrP = 3155{\text{ Pa}} = 23.17{\text{ torr}}
Thus, the correct answer is C.

Note
The numerical calculation for equilibrium constant is done by allowing the chemical reaction to reach the equilibrium and then measuring the concentrations of each constituent at that point. Since the concentrations of these constituents are measured at equilibrium point, thus equilibrium constant will always have the same value for a given reaction no matter what is the initial concentration of the reactants.