Question

At ${490}^{o}C$ the value of equilibrium constant, $K_p$ is $45.9$ the reaction, $H_2\left(g\right)+I_2\left(g\right)\rightleftharpoons 2HI\left(g\right)$.
Calculate the value of $\Delta G^o$ for the reaction at that temperature?

• A. $-3.5kcal$
• B. $3.5kcal$
• C. $5.8kcal$
• D. $-5.8kcal$

Answer 1

The correct option is D $-5.8kcal$

Solution:- (D) $-5.79kcal$

As we know that the relation between $\Delta G^0$ and the value of the equilibrium constant $K_p$ is given as-

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

Given:-

$T=490℃=(273+490)K=763K$

$K_p=45.9$

$R=2\times {10}^{-3}kcal/mol.K$

$\Delta G^0=-2.303\times 2\times {10}^{-3}\times 763\times \log 45.9$

$\Rightarrow \Delta G^0=-3.5\times \log 45.9=-5.8kcal$

Hence the value of $\Delta G^0$ for the reaction is $-5.8kcal$.