Keq = 127 for a reaction. Which of the following must be true?
∆Gº>0 | |
∆Gº=1 | |
∆Gº=0 | |
∆Gº<0 |
Answer:
Standard Gibbs energy change (Go) and Keq is related as,
Go = - 2.303RT log(Keq) ............ (1)
R = 8.314 J.K-1mol-1, T = 298.15 K, Keq = 127
Placing these values in eq.(1),
Go = - 2.303 * 8.314 J.K-1mol-1 * 298.15 K * ln(127)
Go = - 12010 J.mol-1.
Go = -12.01 kJ.mol-1.
I.e. Go < 0
Answer Option (4) Go < 0
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