If the enthalpy change for the transition of liquid water to steam is 30 kJ mol⁻¹ at 27⁰C, the entropy change for the process would be
Options
(a) 10 J mol⁻¹ K⁻¹
(b) 1.0 J mol⁻¹ K⁻¹
(c) 0.1 J mol⁻¹ K⁻¹
(d) 100 J mol⁻¹ K⁻¹
Correct Answer:
100 J mol⁻¹ K⁻¹
Explanation:
We know that ΔG = ΔH – TΔS,
0 = ΔH -TΔS [ThereforeΔG=0],
ΔS = ΔH/T = 30 x 10³ / 300 = 100 J mol⁻¹ K⁻¹.