The standard enthalpy and standard entropy of decomposition of $$N_{2}O_{4}$$ to $$NO_{2}$$ are 55.0 kJ $$mol^{-1}$$ and 175.0 J/K/mol respectively. The standard free energy change for this reaction at $$25^{\circ}C$$ in J $$mol^{-1}$$ is ______ (Nearest integer)

We use the Gibbs free energy equation:

$$ \Delta G^0 = \Delta H^0 - T\Delta S^0 $$

The standard enthalpy change is given by $$\Delta H^0 = 55.0$$ kJ mol$$^{-1} = 55000$$ J mol$$^{-1}$$, the standard entropy change is $$\Delta S^0 = 175.0$$ J K$$^{-1}$$ mol$$^{-1}$$, and the temperature is $$T = 25°C = 298$$ K.

Substituting these values into the Gibbs free energy equation yields:

$$ \Delta G^0 = 55000 - 298 \times 175.0 $$

$$ \Delta G^0 = 55000 - 52150 $$

$$ \Delta G^0 = 2850 \text{ J mol}^{-1} $$

The answer is 2850 J mol$$^{-1}$$.