Question

Calculate ∆Gº at 25ºC for the reaction where ∆Hº=24.6 kJ kJ and ∆Sº=132 J/K:
   -3.93x104 kJ
   -14.7 kJ
   -3280 kJ
   21.3 kJ

Keq = 32.5 for a reaction. Which of the following must be true?
   ∆Gº<0
   ∆Gº>1
   ∆Gº=1
   ∆Gº=0

Answer 1

1. Given ∆H° = 24.6 KJ ,

• T = 25°C = 298 K
• ∆S° = 132 J/K = 0.132 KJ/ K

∆G° = ∆H° - T∆S°

Putting the values in above equation we get-

∆G° = 24.6 - (298 × 0.132)

= 24.6 - 39.34

= - 14.74 KJ

~ 14.7 KJ is answer.

2.

∆G° = - RT ln K​​​​​eq

R = 8.314 , Keq = 32.5 , T = 298K

All the values are positive so ∆G° is negative.