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Q. The equilibrium constant for a reaction is $10$. What will be the value of $\Delta G^°$ at $300\, K$ ?

Thermodynamics

Solution:

$\Delta G = -2.303\,RT\,log\,K$
$= -2.303 \times 8.314 \times 300 \,log \,10$
$= -5744.1\,J$
$\Rightarrow -5.74\,kJ$