Consider the following reaction at 298 K. $$\frac{3}{2}O_{2(g)} \rightleftharpoons O_{3(g)}$$, $$K_p = 2.47 \times 10^{-29}$$. $$\Delta_r G^0$$ for the reaction is _________ kJ. (Given $$R = 8.314 \text{ JK}^{-1}\text{mol}^{-1}$$) Round off your answer to the nearest integer.

$$\Delta_r G° = -RT\ln K_p = -2.303RT\log K_p$$.

$$= -2.303 \times 8.314 \times 298 \times \log(2.47 \times 10^{-29})$$.

$$\log(2.47 \times 10^{-29}) = \log 2.47 - 29 = 0.393 - 29 = -28.607$$.

$$\Delta_r G° = -2.303 \times 8.314 \times 298 \times (-28.607) = 2.303 \times 8.314 \times 298 \times 28.607$$.

$$= 5705.8 \times 28.607 ≈ 163250$$ J ≈ 163 kJ.

The answer is $$\boxed{163}$$.