Question

The standard enthalpy and standard entropy of decomposition of \( N_2O_4 \) to \( NO_2 \) are 55.0 kJ mol\(^{-1}\) and 175.0 J/mol respectively. The standard free energy change for this reaction at 25°C in J mol\(^{-1}\) is  (Nearest integer)

Hint:

Remember to always convert units when necessary. In this case, converting the enthalpy from kJ to J helped to maintain consistency in the units for entropy (J/mol).

Answer 1

We are given the following data: \[ \Delta H^\circ_{\text{rxn}} = 55 \, \text{kJ/mol}, \quad \Delta S^\circ_{\text{rxn}} = 175 \, \text{J/mol}, \quad T = 298 \, \text{K} \] The formula to calculate the standard free energy change (\( \Delta G^\circ_{\text{rxn}} \)) is: \[ \Delta G^\circ_{\text{rxn}} = \Delta H^\circ_{\text{rxn}} - T \Delta S^\circ_{\text{rxn}} \] Substitute the values into the equation: \[ \Delta G^\circ_{\text{rxn}} = 55,000 \, \text{J/mol} - 298 \times 175 \, \text{J/mol} \] \[ \Delta G^\circ_{\text{rxn}} = 55,000 \, \text{J/mol} - 52,150 \, \text{J/mol} \] \[ \Delta G^\circ_{\text{rxn}} = 2,850 \, \text{J/mol} \] Thus, the standard free energy change for the reaction at 25°C is 2850 J/mol.