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Calculate the free energy change when 1 mole of NaCl is dissolved in water at 298 K. (Given: Lattice energy of NaCl = - 777.8 kJ mol-1), Hydration energy = -774.1 kJ mol-1 and ΔS = 0.043 kJK-1 mol-1 at 298 K.) |
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Answer» ΔH = Hydration energy – Lattice energy ΔH = -774.1 kJ mol-1 (-777.8 kJ mol-1) = 3.7 kJ mol-1 ΔG = ΔH - TΔS = +3.7 kJ- 298 × 0.043 kJ = +3.7 kJ- 12.81 kJ ΔG = -9.11 kJ moH |
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