The standard free energy change $$\Delta G°$$ for $$50\%$$ dissociation of $$N_2O_4$$ into $$NO_2$$ at $$27°$$C and $$1$$ atm pressure is $$-x$$ J mol$$^{-1}$$. The value of $$x$$ is ______ J. (Nearest Integer)
[Given : $$R = 8.31$$ J K$$^{-1}$$ mol$$^{-1}$$, $$\log 1.33 = 0.1239$$, $$\ln 10 = 2.3$$]

We need to find the standard free energy change for 50% dissociation of $$N_2O_4$$ into $$NO_2$$. Setting up the dissociation equilibrium $$N_2O_4 \rightleftharpoons 2NO_2$$ and starting with 1 mol of $$N_2O_4$$, 50% dissociation corresponds to 0.5 mol dissociating, leaving $$1 - 0.5 = 0.5$$ mol of $$N_2O_4$$ and producing $$2 \times 0.5 = 1$$ mol of $$NO_2$$, which makes a total of $$0.5 + 1 = 1.5$$ mol.

At a total pressure of 1 atm, the partial pressures are given by the mole fractions, namely $$p_{N_2O_4} = \frac{0.5}{1.5} \times 1 = \frac{1}{3}$$ atm and $$p_{NO_2} = \frac{1}{1.5} \times 1 = \frac{2}{3}$$ atm, from which we calculate:

$$K_p = \frac{(p_{NO_2})^2}{p_{N_2O_4}} = \frac{(2/3)^2}{1/3} = \frac{4/9}{1/3} = \frac{4}{3} \approx 1.33$$

To find the standard free energy change, we use the relation $$\Delta G° = -RT \ln K_p = -RT \times 2.303 \times \log K_p$$ at $$T = 27°C = 300$$ K. Thus,

$$\Delta G° = -8.31 \times 300 \times 2.3 \times \log(1.33)$$
$$= -8.31 \times 300 \times 2.3 \times 0.1239$$
$$= -8.31 \times 300 \times 0.28497$$
$$= -8.31 \times 85.491$$
$$= -710.43 \text{ J mol}^{-1}$$

Hence, since $$\Delta G° = -x$$ J mol$$^{-1}$$, we find $$x \approx 710$$ (nearest integer).

The correct answer is $$\mathbf{710}$$.