Question

The equilibrium constant $K_c$ for a reaction is 10. Calculate the value of change in standard Gibbs free energy $\Delta G^o\text{in}kJmo{l}^{-1}$ at 300 $K$.

• A. 5.744
• B. -5.744
• C. -2.744
• D. 2.744

Answer 1

The correct option is B -5.744

We know that, the relation of Gibbs free energy and equilibrium constant at equilibrium conditions is given by $\Delta G^o=-2.303RTlog{K}_c$
Given,
$T=300K$
$K_c=10$
$R=8.314Jmo{l}^{-1}{K}^{-1}$
Substituting the values we get,
$\Delta G^o=-2.303\times 8.314\times 300\times log10$
$\Delta G^o=-5744.14\times log10$
$\Delta G^o=-5744.14J/mol(\because log10=1)$
$\Delta G^o=-5.744kJ/mol$


Theory:

Consider for the given reaction:
$aA\left(aq\right)+bB\left(aq\right)\rightleftharpoons cC\left(aq\right)+dD\left(aq\right)$
Gibbs free energy change is given by: $\Delta G=\Delta G^{\circ }+RTlnQ$
Where, $Q$= ReactionQuotient

$\Delta G=\Delta G^{\circ }+\mathrm{2.303.}RTlo{g}_{10}Q$
Where, $\Delta G^{\circ }$ is standard Gibbs free energy change

At equilibrium, $\Delta G=0$ and $Q=K_c$
$\Delta G^{\circ }=-2.303RTlo{g}_{10}{K}_c$
$\Delta G^{\circ }=-RTln{K}_c$