Question

At $320K$, a gas $A_2$ is $20$% dissociated to $A\left(g\right)$. The standard free energy change at $320K$ and $1$ atm in $J$ ${mol}^{-1}$ is approximately:

$[R=8.314J{K}^{-1}{mol}^{-1};\ln 2=0.693;\ln 3=1.098]$

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

Answer 1

The correct option is C $4281$

Given that,

$A_2\rightleftharpoons 2A$

Initially, suppose $\left[A_2\right]=1M$ and $\left[A\right]=0M$

After 20 % dissociation, 80% of $A_2$ remains.

$\left[A_2\right]=1\times \frac{80}{100}=0.8M$

20% of 1 M is $1\times \frac{20}{100}=0.2$. $\left[A\right]=2\times 0.2=0.4M$

The equilibrium constant

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

$K=\frac{[0.4{]}^2}{\left[0.8\right]}=0.2$

$\Delta G^0=-RTlnK=-8.314J{K}^{-1}{mol}^{-1}\times 320K\times ln0.2=4281J/mol$