The value of DeltaG^(@) for a reaction is 7.97 KJ//"mole". Its equilibrium constant K_(c) at 298K is 4 xx 10^(-x) What is x?

The value of DeltaG^(@) for a reaction is 7.97 KJ//"mole". Its equilib

1.	The value of DeltaG^(@) for a reaction is 7.97 KJ//"mole". Its equilibrium constant K_(c) at 298K is 4 xx 10^(-x) What is x?
Answer» Solution :`Delta G^(@)= -RTlnK` `LOG K = -1.4= -2+0.6=log(4 XX 10^(-2)) IMPLIES K=2`