HI has a normal boiling point of -35.4°C, and its ΔHvap is 21.16 kJ/mol. Calculate the molar entropy of vaporization (ΔSvap).

Answer 1

Given:

T = -35.4C + 273.15 = 237.75K

∆Hvap = 21.16 kJ/mol

∆G = 0 since vaporization is an equilibrium state and ∆G = 0 at equilibrium

Asked to find ∆S

Equation: ∆G = ∆H - T∆S

0 = 21.16 kJ/mol - (237.75K)(∆S)

∆S = 21.16 kJ/mol / 237.75 K = 0.08900 kJ/molK and converting this to Joules/K mol we have...

∆S = 89.00 J/K-mol