Question
Calculate proton affinity of NH3(g) from the following data.
ΔHdissociation H2 = 218 kJ mole–1
ΔHdissociation Cl2 = 124 kJ mole–1
ΔHf0 of NH3(g) = −46 kJ mole–1
ΔHf0 of NH4Cl(s) = −314 kJ mole–1
Ionization energy of H = 1310 kJ mole–1
E.A. of Cl(g) = −348 kJ mole–1
Lattice energy of NH4Cl(s) = −683 kJ mole–1
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Solution
Given thermochemical equations are
$(i)H_{2}(g) \rightarrow 2H(g);\Delta H = 218\text{ kJ} $$${(ii)Cl_{2}(g) \rightarrow 2Cl(g);\Delta H = 124\text{ kJ} }{(iii)\frac{1}{2}N_{2}(g) + \frac{3}{2}H_{2}(g) \rightarrow NH_{3}(g);\Delta H = - 46\text{ kJ} }{(iv)\frac{1}{2}N_{2}(g) + 2H_{2}(g) + \frac{1}{2}Cl_{2}(g) \rightarrow NH_{4}Cl(s);\Delta H = - 314\text{ kJ} }{(v)H(g) \rightarrow H^{+}(g) + e^{-};\Delta H = 1310\text{ kJ} }{(vi)Cl(g) + e^{-} \rightarrow Cl^{-}(g);\Delta H = - 348\text{ kJ} }{(vii)NH_{4}Cl(s) \rightarrow NH_{4}^{+}(g) + Cl^{-}(g);\Delta H = 683\text{ kJ}}$$
Required thermochemical equations are
$NH_{3}(g) + H^{+}(g) \rightarrow NH_{4}^{+}(g);\Delta H = ? $$${\text{From }(vii) + (iv) - (iii) - \frac{1}{2}(ii) - \frac{1}{2}(i) - (v) - (vi) }{\Delta H_{\text{required}} = (683) + ( - 314) - ( - 46) - \frac{1}{2} \times (124) - \frac{1}{2} \times (218) - (1310) - ( - 348) }{ = - 718\text{ kJ/mol}}$$
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