Question

For a particular reaction at 247.1 °C, AG = -699.04 kJ, and AS = 244.36 J/K. Calculate AG for this reaction at - 78.9 °C. AG =

Answer 1

Ans :-

Given, gibb's free energy change = ΔG = -699.04 KJ

Temperature = T = 247.1 0C = 247.1 + 273.15 = 520.25 K

and

Entropy change = ΔS = 244.36 J/K = 244.36 x 10-3 KJ/K , because 1 J = 10-3 KJ

From Gibb;s-Helmholtz equation :

ΔG = ΔH - TΔS

Therefore,

Change in enthalpy = ΔH = ΔG + TΔS

ΔH = -699.04 KJ + (520.25 K) x (244.36 x 10-3 KJ/K)

ΔH = -699.04 KJ + 127.12829 KJ

ΔH = -571.91 KJ

Again using Gibb;s-Helmholtz equation when temperature -78.90C = 194.25 K and ΔH = -571.91 KJ :

ΔG = ΔH - TΔS

ΔG = -571.91 KJ - (194.25 K)((244.36 x 10-3 KJ/K)

ΔG = -571.91 KJ - 47.467 KJ

ΔG = -619.38 KJ

Hence,  ΔG = - 619.38 KJ