Question

From the values of ΔΗ and ΔS, predict which of the following reactions would be spontaneous at 25^oC. If the reaction is nonspontaneous at 25^oC, at what temperature would it become spontaneous?

A) ΔH = 10.5 kJ/mol, ΔS = 30 J/K * mol

B) ΔΗ = 1.8 kJ/mol, ΔS= -113 J/K * mol

Answer 1

for the spontaneous process,DG0 = -ve, or at least DG = 0

A) DG0 = DH0-TDS0

at T = 25 c = 298 k

DG0 = (10.5*10^3)-(298*30)

      = +1.56 kj

it is non-spontaneous.

if DG0 = 0 ,

0 = (10.5*10^3)-(T*30)

T = 350 k

for the process to be spontaneous,at least T = 350 k (or) 76.85 C

B) DG0 = DH0-TDS0

at T = 25 c = 298 k

DG0 = (1.8*10^3)-(298*-113)

      = +35.474 kj

it is non-spontaneous.

if DG0 = 0 ,

0 = (1.8*10^3)-(T*-113)

T = -15.93 (it is not possible practically)

so that, when DH = +ve,DS= - ve, the process is non-spontaneous at all temperatures.