If K<1, ∆G
<0
=1
>0
=0
Calculate ∆Gº at 25ºC for the reaction where ∆Hº=23.7 kJ kJ and
∆Sº=52.4 J/K:
-1290 kJ
22.4 kJ
-1.56x104kJ
8.08 kJ
1)
∆G = -RT*ln K
K<1
So,
ln K < 0
So,
RT*ln K < 0
This means,
-RT*ln K > 0 and hence ∆G > 0
Answer: > 0
2)
ΔHo = 23.7 KJ
ΔSo = 52.4 J/K
= 0.0524 KJ/K
T= 25.0 oC
= (25.0+273) K
= 298 K
use:
ΔGo = ΔHo - T*ΔSo
ΔGo = 23.7 - 298.0 * 0.0524
ΔGo = 8.08 KJ
Answer: 8.08 KJ
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