Question

At ${60}^o$ $N_2{O}_4$ is $50$% dissociated at this temperature and one atmosphere pressure the standard free energy change is:

• A. $-863.8KJ$ ${mol}^{-1}$
• B. $-963.8KJ$ ${mol}^{-1}$
• C. $-973.8KJ$ ${mol}^{-1}$
• D. none of these

Answer 1

The correct option is D none of these

$N_2{O}_{4\left(g\right)}\rightleftharpoons 2N{O}_{2\left(g\right)}$

$10⟶Initial$

$1-0.52\times 0.5⟶Eqn$ ($50\%$ Dissociation.)

$0.5mol1mol$

$Total=0.5+1=1.5mol$

$P_{N_2{O}_4}=\frac{0.5}{1.5}\times 1=\frac{1}{3}atm$

$P_{N{O}_2}=\frac{1}{1.5}\times 1=\frac{2}{3}atm$

$\therefore K_P=\frac{{\left(P_{N{O}_2}\right)}^2}{P_{N_2{O}_4}}=\frac{{\left(\frac{2}{3}\right)}^2}{\frac{1}{3}}=\frac{2}{3}\times \frac{2}{3}\times 3=\frac{4}{3}atm$

We know, $△G^0=-2.303RT\log K_P=-2.303\times 8.314\times 333\times \log \left(\frac{4}{3}\right)=-796.99J/mol$