Question

At 320 K, a gas $A_2$ is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J $mo{l}^{-1}$ is approximately: (R=8.318 J$K^{-}mo{l}^{-1}$; In 2=0.693; In 3=1.098)

• A. 1844
• B. 2068
• C. 4281
• D. 4763

Answer 1

The correct option is C 4281

Solution:- (C) $4281$

$\underset{\text{At equilibrium :}}{\underset{\text{Initially :}}{}}\underset{1-\alpha }{\underset{1}{A_2}}\rightleftharpoons \underset{2\alpha }{\underset{0}{2A}}$

Given that $20\%A_2$ is dissociated, i.e.,

$\alpha =\frac{20}{100}=0.2$

Therefore,

$\left[A_2\right]=1-0.2=0.8M$

$\left[A\right]=2\times 0.2=0.4$

Therefore, for the give reaction,

$K_c=\frac{{\left[A\right]}^2}{\left[A_2\right]}$

$\Rightarrow K_c=\frac{{\left(0.4\right)}^2}{0.8}=\frac{0.16}{0.8}=2\times {10}^{-1}$

Now,

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

Given:- $T=320K$

$\therefore \Delta G^0=-2.303\times 8.314\times 320\times \log (2\times {10}^{-1})$

$\Rightarrow \Delta G^0=-6127.08(\log 2+\log {10}^{-1})$

$\Rightarrow \Delta G^0=4282.16\approx 4281J$

The standard free energy change at $320K$ and $1atm$ in $J/mol$ is approximately $4281$.