Question

If for a reversible reaction the equilibrium constant $K_c$ is $5\times {10}^{12}$ at ${25}^{\circ }C$, the value of ${\Delta }_r{G}^o$ at the same temperature will be:

• A. $-8.314J{\text{mol}}^{-1}{K}^{-1}\times 298K\times \ln (2\times {10}^{12})$
• B. $-2.303\times 8.314J{\text{mol}}^{-1}{K}^{-1}\times 298K\times \log (5\times {10}^{12})$
• C. $-8.314J{\text{mol}}^{-1}{K}^{-1}\times 298K\times \log (5\times {10}^{12})$
• D. $-2.303\times 8.314J{\text{mol}}^{-1}{K}^{-1}\times 298K\times \ln (5\times {10}^{12})$

Answer 1

The correct option is B $-2.303\times 8.314J{\text{mol}}^{-1}{K}^{-1}\times 298K\times \log (5\times {10}^{12})$

Given: $T=273+25=298K$
$K_c=5\times {10}^{12}\text{and}R=8.314J{\text{mol}}^{-1}{K}^{-1}$
${\Delta }_r{G}^{\circ }=-2.303RT{\log }_{10}\left(K_c\right)$
$=-2.303\times 8.314{\text{J mol}}^{-1}{K}^{-1}\times 298K\times {\log }_{10}(5\times {10}^{12})$