Question

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. 4763
• B. 2068
• C. 1844
• D. 4281

Answer 1

Answer: (D)

$<br/>A_2\rightleftharpoons 2A<br/>$
Initially, suppose $\left[A_2\right]=1M$ and $[A]=0M$
After $20\%$ dissociation, $80\%$ of $A_2$ remains.
$<br/>\left[A_2\right]=1\times \frac{80}{100}=0.8M<br/>$
$20\%$ of $1M$ is $1\times \frac{20}{100}=0.2$. [A] $=2\times 0.2=0.4M$
The equilibrium constant
$<br/>K=\frac{[A{]}^2}{\left[A_2\right]}<br/>$
$<br/>K=\frac{[0.4{]}^2}{[0.8]}=0.2<br/>$
$<br/>\Delta G^0=-RT\ln K=-8.314J{K}^{-1}mo{l}^{-1}\times 320K\times \ln 0.2=4281J/mol<br/>$