Calculate the proton affinity of NH 3 (g) from the following data: ΔH dissociation H 2 = 436 kJ mole -1 ΔH dissociation Cl 2 = 244 kJ mole -1 ΔH f 0 of NH 3 (g) = -46 kJ mole -1 ΔH f 0 of NH 4 Cl(s) =

Question

Calculate the proton affinity of NH 3 (g) from the following data: ΔH dissociation H 2 = 436 kJ mole -1 ΔH dissociation Cl 2 = 244 kJ mole -1 ΔH f 0 of NH 3 (g) = -46 kJ mole -1 ΔH f 0 of NH 4 Cl(s) = -314 kJ mole -1 Ionization energy of H = 1312 kJ mole -1 , E.A. of Cl(g) = -349 kJ mole -1 Lattice energy of NH 4 Cl(s) = -683 kJ mole -1

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NH 3 (g) + 1/2H 2 + 1/2Cl 2 → NH 4 Cl(s) has ΔH = -314 - (-46) = -268 kJ. Steps: atomize: 436/2 = 218 and 244/2 = 122; ionize H: +1312; Cl electron gain: -349; NH 3 + H + → NH 4 + : PA; lattice: -683. So -268 = 218 + 122 + 1312 - 349 + PA - 683 = 620 + PA, giving PA = -888 kJ mole -1 . This is option (b).

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