Question

∆G for the conversion of 2 mol of C6H6(l) at 80°C
(normal boiling point) to vapour at the same
temperature and a pressure of 0.2 atm is​

Answer 1

Thus the value of Gibbs free energy is ΔG = = - 9.44689 Kcal/mol

Explanation:

Gibbs free energy = The energy associated with a chemical reaction that can be used to do work.

Mathematically:

ΔG = ΔH - TΔS

ΔH = Change in enthalpy

T = temperature

ΔS = Change in entropy

ΔG  = 0 at boiling point

ΔH = TbΔS

ΔG = -nRT ln K

ΔG = - 2 x 8.314 x 353 x ln (0.2)

ΔG = - 9446.89 cal/mol

ΔG = - 9.44689 Kcal/mol

Thus the value of Gibbs free energy is ΔG = = - 9.44689 Kcal/mol

Answer 2

ΔG for the conversion is -1.1364 Kcal/mol

Explanation:

C₆H₆ (l) → C₆H₆ (s)

From question, the pressure = P = 0.2 atm

n = 2 moles

T = 80° C = 80 + 273 = 353 K

ΔG = RT ln Kp

Where,

R = 2 cal/mol.K

Kp = 0.2 atm

Now, the ΔG is:

ΔG = 2.303 RT log Kp

ΔG = 2.303 × 2 × 353 log (0.2)

ΔG = -1136.4 cal/mol

∴ ΔG = -1.1364 Kcal/mol