Question

For the equilibrium, $\Delta G=-RTlnK$.
If true enter 1 else 0.

• A. 0

Answer 1

The correct option is A 0
The relationship between the Gibbs free energy change $\Delta G$, the standard Gibbs free energy change $\Delta G^o$ and the reaction quotient Q is $\Delta G=\Delta G^o+RTlnQ$
But at equilibrium, $\Delta G=0$ and $Q=K$. Here, K is the equilibrium constant.
Hence, $0=\Delta G^o+RTlnK$
$\Delta G^o=-RTlnK=2.303logK$