The enthalpy of formation of NaCl(s) from the following data is (i) NaOH(aq) + HCl(aq) → NaCl(aq) + H 2 O(l); ΔH = -13.7 kcal (ii) H 2 (g) + 1/2O 2 (g) → H 2 O(l); ΔH = -68.4 kcal (iii) 1/2H 2 (g) + 1

Question

The enthalpy of formation of NaCl(s) from the following data is

(i) NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l); ΔH = -13.7 kcal

(ii) H₂(g) + 1/2O₂(g) → H₂O(l); ΔH = -68.4 kcal

(iii) 1/2H₂(g) + 1/2Cl₂(g) + aq → HCl(aq); ΔH = -39.3 kcal

(iv) Na(s) + 1/2O₂(g) + 1/2H₂(g) + aq → NaOH(aq); ΔH = -112.0 kcal

(v) NaCl(s) + aq → NaCl(aq); ΔH = +1.0 kcal

TestHub · Accepted Answer

The target reaction is: Na(s) + 1/2Cl 2 (g) → NaCl(s) To achieve this, we add reactions (iv), (iii), and (i), then reverse reactions (ii) and (v). All other species cancel out. The enthalpy change (ΔH) is calculated as follows: ΔH = -112.0 - 39.3 - 13.7 - (-68.4) - (+1.0) = -97.6 kcal/mol This corresponds to option (a).

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