Question

The value of $\Delta G^{\circ }$ for a reaction in aqueous phase having $K_c=1$, would be:

• A. $-RT$
• B. $-1$
• C. $0$
• D. $+RT$

Answer 1

The correct option is C $0$

Solution:- (C) $0$

As we know that,

$\Delta G^0=-RT\ln K_c$

Given that $K_c=1$

$\therefore \Delta G^0=-RT\left(\ln 1\right)=0[\because \ln 1=0]$

Hence, the value of $\Delta G^0$ will be $0$.