1.

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.)

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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