The entropy of vaporization of a liquid is 100 J mol⁻¹K⁻¹ and it boils at 300K. What would be the value of ∆Hvap?

Question

Vimal Swarnkar · Accepted Answer

(b) 30.0 kJ mol⁻¹. At boiling point, ΔSvap = ΔHvap/Tb → ΔHvap = 100 × 300 = 30,000 J mol⁻¹ = 30.0 kJ mol⁻¹ → (b).

Answer

(a) 3.0 kJ mol⁻¹

Answer

(c) 300 kJ mol⁻¹

Answer

(d) 3000 kJ mol⁻¹

Geo-Scientist Prelims 2020 Chemistry Q55: The entropy of vaporization of a liquid is 100 J mol⁻¹K⁻¹ and it…