Question

one mole of ice is converted into water at 273 Kelvin the entropy is of H2O solid and H2O liquid are 38.20 and 60.01 respectively the enthalpy change for the conversion is

Answer 1

The enthalpy change for the given conversion is 5.95kJ/mol.

- Given:-

T = 273 K

S(ice) = 38.20

S(water) = 60.01

- To find :- Enthalpy change, i.e., ΔH

- We know that

ΔG = ΔH − TΔS

At equilibrium , ΔG=0

⇒0 = ΔH - TΔS

⇒ ΔH = TΔS

⇒ ΔH = 273 × (60.01−38.20)

⇒ ΔH = 273 × 21.81

⇒ ΔH = 5954.13 J/mol

⇒ ΔH = 5.95 kJ/mol