1. Reaction and condition

The dissociation is: $N_2O_4(g) \rightleftharpoons 2NO_2(g)$

Given: degree of dissociation is $50\%$, so $\alpha = 0.5$ at $T = 27^\circ C = 300\,K$ and total pressure $P=1\,\text{atm}$.

We need $\Delta G^\circ = -x\,\text{J mol}^{-1}$.

2. Find equilibrium constant in terms of pressure

Assume initially 1 mole of $N_2O_4$.

At equilibrium:

• moles of $N_2O_4 = 1-\alpha = 0.5$
• moles of $NO_2 = 2\alpha = 1.0$
• total moles $= 1+\alpha = 1.5$

Hence mole fractions: $y_{N_2O_4} = \frac{0.5}{1.5} = \frac{1}{3}$ $y_{NO_2} = \frac{1.0}{1.5} = \frac{2}{3}$

Partial pressures at total pressure $1$ atm: $p_{N_2O_4} = \frac{1}{3}\,\text{atm}, \qquad p_{NO_2} = \frac{2}{3}\,\text{atm}$

So, $K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = \frac{\left(\frac{2}{3}\right)^2}{\frac{1}{3}} = \frac{4}{3} = 1.33$

3. Use relation between $\Delta G^\circ$ and equilibrium constant

$\Delta G^\circ = -RT\ln K_p$

Substitute values: $\Delta G^\circ = -(8.31)(300)\ln(1.33)$

Given: $\log(1.33)=0.1239, \qquad \ln 10 = 2.3$

Therefore, $\ln(1.33) = \log(1.33)\times \ln 10 = 0.1239 \times 2.3$ $\ln(1.33) \approx 0.285$

Now, $\Delta G^\circ = -(8.31)(300)(0.285)$ $\Delta G^\circ \approx -710.5\,\text{J mol}^{-1}$

Nearest integer: $x = 710$

4. Comparison with stored answer

Derived answer: $710$

Stored correct answer: $710$

They agree.