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Q.
At 320 K, a gas A2 is 20% dissociated to A(g). The standard free energy change at 320 K and 1 atm in J mol$^{-1}$ is approximately : (R=8.314 JK$^{-1}$ mol$^{-1}$; ln 2=0.693; ln 3=1.098)
$
A _{2} \rightleftharpoons 2 A
$
Initially, suppose $\left[ A _{2}\right]=1 M$ and $[ A ]=0 M$
After $20 \%$ dissociation, $80 \%$ of $A _{2}$ remains.
$
\left[ A _{2}\right]=1 \times \frac{80}{100}=0.8 M
$
$20 \%$ of $1 M$ is $1 \times \frac{20}{100}=0.2$. [A] $=2 \times 0.2=0.4 M$
The equilibrium constant
$
K =\frac{[ A ]^{2}}{\left[ A _{2}\right]}
$
$
K =\frac{[0.4]^{2}}{[0.8]}=0.2
$
$
\Delta G ^{0}=- RT \ln K =-8.314 JK ^{-1} mol ^{-1} \times 320 K \times \ln 0.2=4281 J / mol
$