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NEET CHEMISTRYMedium

The standard enthalpy and standard entropy changes for the oxidation of ammonia at 298 K298\text{ K} are −382.64 kJ mol−1-382.64\text{ kJ mol}^{-1} and −145.6 J K−1 mol−1-145.6\text{ J K}^{-1}\text{ mol}^{-1}, respectively. Standard Gibbs energy change for the same reaction at 298 K298\text{ K} is:

A

−339.3 kJ mol−1-339.3\text{ kJ mol}^{-1}

B

−439.3 kJ mol−1-439.3\text{ kJ mol}^{-1}

C

−523.2 kJ mol−1-523.2\text{ kJ mol}^{-1}

D

−221.1 kJ mol−1-221.1\text{ kJ mol}^{-1}

Step-by-Step Solution

According to the Gibbs equation, the standard Gibbs free energy change (ΔG∘\Delta G^\circ) is given by: ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ Given: Standard enthalpy change, ΔH∘=−382.64 kJ mol−1\Delta H^\circ = -382.64\text{ kJ mol}^{-1} Standard entropy change, ΔS∘=−145.6 J K−1 mol−1=−0.1456 kJ K−1 mol−1\Delta S^\circ = -145.6\text{ J K}^{-1}\text{ mol}^{-1} = -0.1456\text{ kJ K}^{-1}\text{ mol}^{-1} Temperature, T=298 KT = 298\text{ K} Substituting the values into the equation: ΔG∘=−382.64 kJ mol−1−(298 K×−0.1456 kJ K−1 mol−1)\Delta G^\circ = -382.64\text{ kJ mol}^{-1} - (298\text{ K} \times -0.1456\text{ kJ K}^{-1}\text{ mol}^{-1}) ΔG∘=−382.64−(−43.3888)\Delta G^\circ = -382.64 - (-43.3888) ΔG∘=−382.64+43.3888=−339.2512 kJ mol−1≈−339.3 kJ mol−1\Delta G^\circ = -382.64 + 43.3888 = -339.2512\text{ kJ mol}^{-1} \approx -339.3\text{ kJ mol}^{-1}

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