Question

\[{\Delta _{\text{f}}}{\text{G}}^\circ \] at 500 K for substance 'S' in liquid state and gaseous state are \[ + 110.7{\text{ kcal mo}}{{\text{l}}^{ - 1}}\] and \[ + 103{\text{ kcal mo}}{{\text{l}}^{ - 1}}\], respectively.
Vapour pressure of liquid 'S' at 500 K is approximately equal to: \[{\text{(R}} = 2{\text{ cal }}{{\text{K}}^ - }{\text{ mo}}{{\text{l}}^{ - 1}}{\text{)}}\]
A) 100 atm
B) 1 atm
C) 10 atm
D) $0.1$ atm

Answer 1

Hint:
The change in standard Gibbs free for the two phases that is liquid and gas can be calculated, which is to be used in the given formula. The value of the equilibrium constant will be the same as that of pressure as liquid has no pressure.

Formula used:
\[\Delta {\text{G}}^\circ = - {\text{RT}}\ln {{\text{K}}_{\text{p}}}\]
Here \[\Delta {\text{G}}^\circ \] is change in Gibbs free energy, R is universal gas constant, T is temperature in Kelvin and \[{{\text{K}}_{\text{p}}}\] is equilibrium constant in terms of pressure.

Complete step by step solution:
Basically a physical change is occurring that a substance in liquid state is changing into its gaseous state, a general reaction can be written as:
\[{\text{ Substance (l)}} \to {\text{ Substance (g)}}\]
The equilibrium constant is the ratio of pressure of product to the pressure of reactants in terms of pressure. Since the liquids does not possess any pressure so \[{{\text{K}}_{\text{p}}}\] will be equal to pressure of substance on the product side that is gas. Let the pressure is P. so \[{{\text{K}}_{\text{p}}} = {\text{P}}\]
Now, we will calculate the change in standard Gibbs free energy by subtracting the vapor phase Gibbs energy with liquid phase:
  \Delta {\text{G}}^\circ = + 110.7{\text{ kcal mo}}{{\text{l}}^{ - 1}} - + 103{\text{ kcal mo}}{{\text{l}}^{ - 1}} = 2.3{\text{kcal mo}}{{\text{l}}^{ - 1}} \\
Substituting the known values in formula we will get,
\[2.3 = - \dfrac{2}{{1000}}{\text{ }} \times 500 \times \ln {\text{P}}\]
We have added this 1000 to convert cal to kilocalorie
Rearranging and solving we will get,
\[\ln {\text{P}} = - 2.3\]
Using antilog both side we will get, \[{\text{P}} = 0.1{\text{ atm}}\]

Hence, the correct option is D.

Note:
The formula for \[\Delta {\text{G}}\] is \[\Delta {\text{G}} = \Delta {\text{H}}^\circ - {\text{T}}\Delta {\text{S}}\] . In the above conversion from a liquid to a gas heat is supplied and hence \[\Delta {\text{G}} = \Delta {\text{H}}^\circ - {\text{T}}\Delta {\text{S}}\] is always positive and entropy also increases form a liquid to a gas. So entropy change is also positive. So if value of \[\Delta {\text{H}}\] is more than \[{\text{T}}\Delta {\text{S}}\] then the \[\Delta {\text{G}}\] value is positive.