When reaction is carried out at standard states then, at the equilibrium
When reaction is carried out at standard states then, at the equilibrium
- $\Delta H^{\circ}=0$
- $\Delta S^{\circ}=0$
- equilibrium constant $(k)=0$
- equilibrium constant $(k)=1$
Solution
We know,
$
\Delta G^{\circ}=-R T \ln \frac{[k]}{[q]}
$
where, $q=$ reaction quotient
$
k=\text { equilibrium constant. }
$
$\because$ At equilibrium,
$
\begin{gathered}
k=q \\
\left.\Delta G^{\circ}=0 \quad \text { \{for standard states }\right\}
\end{gathered}
$
So, equilibrium constant
$
k=1 .
$
Asked in: AP EAMCET 2021 (23 Aug Shift 1)
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