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Calculate the free energy change on dissolving one mole (58.5 g) of sodium chloride at 25°C. Lattice energy = 777.8 kJ mol-1, hydration energy = -774.1 kJ mol-1 of NaCl, ΔS at 25°C = 0.043 kJ mol-1 K-1. |
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Answer» ΔHsolution = ΔHlatticet ΔHhydration = 777.8 kJ mol-1 + (-774.1 kJ mol-1) = 777.8 kJ mol-1 - 774.1 kJ mol-1 = 3.7 kJ mol-1 ΔG = ΔH - TΔS = 3.7 kJ mol-1 = (273 + 25) K x 0.043 kJ mol-1 K-1 = (3.7 kJ mol-1) - (298 K) x (0.043 kJ K-1 mol-1) = 3.7 kJ mol-1 - 12.814 kJ mol-1 = -9.114 kJ mol-1 |
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