Question

If Kp>1 for a reaction, comment on the sign of standard free energy change of the reaction.

Answer 1

The equilibrium constant (K_p) is related to the standard Gibbs free energy change of reaction ($\Delta$ G):

$\Delta$ G = - R * T * ln(K_p);

If K_p = >1, then ln(K_p) = >0 (or +, possitive).

If ln(K_p) = >0 (or +) then the standard free energy change of reaction: $\Delta$ G = - R * T * ln(K_p).

Therefore, the sign of the standard free energy change of reaction is negative.

Answer: negative (-).