Question

the enthalpy of vaporisation of benzene is +35.3 kJ/mol at its boiling point of 80 degree celsius. The entropy change in the transition of vapour to liquid at its boiling point is ( in J/kJ mol) :-

a -100
b 100
c -342
d 342

Answer 1

Answer:

- 100 J K⁻¹ mol⁻¹

a. option is correct.

Explanation:

Given :

Enthalpy of vaporisation of benzene = + 35.3 kJ / mol

Temperature = 80 C

T in term of Kelvin

T = 80 + 273 K

T = 353 K

For vapour to liquid Δ H = - 35.3 k J / mol = - 35.3 × 10³ J / mol

We know :

Δ S = Δ H / T

Δ S =  - 35.3 × 10³ J / mol / 353 K

Δ S = - 100 J K⁻¹ mol⁻¹

Therefore , entropy change  in the transition of vapour to liquid is  - 100 J K⁻¹ mol⁻¹