Question

The enthalpy change for the transition of liquid water to steam at temperature 373K is 40.8 UJ mol–1. Calculate the entropy change for the process.

Answer 1

ΔG⁰= ΔH⁰ - TΔS⁰

Given:
ΔH vap= 40.8
ΔG⁰= 0 at equilibrium

so we get

ΔS vap = ΔH vap /T
= 40.8/373 
= 40.8 x 1000/ 373
= 109.38 J mol-1 K-1

Answer 2

Answer: The entropy change for the process is 0.109 kJ/molK.

Explanation:

$H_2O(l)\rightleftharpoons H_2O(g)$

Using Gibbs Helmholtz equation:

$\Delta G=\Delta H-T\Delta S$

For a pahse change, the reaction remains in equilibrium, thus $\Delta G=0$

$\Delta H=T\Delta S$

Given: Temperature = 373 K

$\Delta H$= 40.8 kJ/mol

Putting the values  the equation:

$40.8kJ/mol=373\times \Delta S$

$\Delta S=0.109kJ/molK$