Question

The equilibrium constant for a reaction is 10 at ${27}^{o}C$. Calculate the value of $\Delta G^o$ at ${27}^{o}C$.

• A. -5982.63 J
• B. -5422.28 J
• C. -5743.10 J
• D. none of these

Answer 1

The correct option is C -5743.10 J

Solution:- (C) $-5743.1J$

As we know that, the relation between $\Delta G^0$ and equillibrium constant $K_c$ is given as-

$\Delta G^0=-2.303RT\log K_c$

Given:-

$K_c=10$

$T=27℃=(27+273)=300K$

$R=8.314J$

$\therefore \Delta G^0=-2.303\times 8.314\times 300\times \log 10$

$\Rightarrow \Delta G^0=-5744.14J\approx -5743.10J$

Hence the value of $\Delta G^0$ is $-5743.10J$.