$$40\%$$ of HI undergoes decomposition to $$H_2$$ and $$I_2$$ at $$300$$ K. $$\Delta G^\circ$$ for this decomposition reaction at one atmosphere pressure is ______ J mol$$^{-1}$$ [nearest integer]
(Use $$R = 8.31$$ J K$$^{-1}$$ mol$$^{-1}$$; $$\log 2 = 0.3010$$, $$\ln 10 = 2.3$$, $$\log 3 = 0.477$$)

We need to find $$\Delta G^\circ$$ for the decomposition of HI at 300 K, given that 40% of HI decomposes.

The reaction under consideration is $$2HI(g) \rightleftharpoons H_2(g) + I_2(g)$$.

Let the initial amount of HI be 1 mol at 1 atm total pressure. If 40% of HI decomposes, then 0.4 mol of HI undergoes decomposition.

At equilibrium, 0.6 mol of HI remain, and 0.2 mol each of H_2 and I_2 are formed.

The total number of moles at equilibrium is $$0.6 + 0.2 + 0.2 = 1.0$$.

With a total pressure of 1 atm and total moles of 1.0, the partial pressures are $$p_{HI} = \frac{0.6}{1.0}\times1 = 0.6$$ atm, $$p_{H_2} = \frac{0.2}{1.0}\times1 = 0.2$$ atm, and $$p_{I_2} = \frac{0.2}{1.0}\times1 = 0.2$$ atm.

The equilibrium constant in terms of partial pressures is $$K_p = \frac{p_{H_2}\times p_{I_2}}{p_{HI}^2} = \frac{0.2\times0.2}{0.6^2} = \frac{0.04}{0.36} = \frac{1}{9}$$.

The standard Gibbs energy change is given by $$\Delta G^\circ = -RT\ln K_p$$, which becomes $$\Delta G^\circ = -RT\ln\left(\frac{1}{9}\right) = RT\ln9$$.

Expressing the natural logarithm in terms of the common logarithm yields $$\Delta G^\circ = RT\times2.3\times\log9$$. Since $$\log9 = \log3^2 = 2\times0.477 = 0.954$$, it follows that $$\Delta G^\circ = 8.31\times300\times2.3\times0.954$$.

Evaluating this product gives $$\Delta G^\circ = 2493\times2.3\times0.954 = 2493\times2.1942 = 5470.1$$ J mol$$^{-1}$$ for the reaction $$2HI \rightleftharpoons H_2 + I_2$$.

For the decomposition of one mole of HI, represented by $$HI(g) \rightleftharpoons \frac{1}{2}H_2(g) + \frac{1}{2}I_2(g)$$, the standard Gibbs energy change is half of this value, namely $$\frac{5470.1}{2}\approx2735$$ J mol$$^{-1}$$.

Therefore, the answer is 2735 J mol$$^{-1}$$.