# 6.20  The equilibrium constant for a reaction is 10. What will be the value of                $\Delta G ^ \ominus ? R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$

M manish

Given values are,
$R = 8.314 JK ^{-1} mol ^{-1} , T = 300 K$
and equilibrium constant  = 10

It is known that,
$\Delta G^o = -2.303\ RT \log K_{eq}$
$\\=-2.303(8.314)(300)\log 10\\ =-5744.14 Jmol^{-1}\\ =-5.744 kJmol^{-1}$

Hence the value of $\Delta G^o$ is -5.744 kJ/mol

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