Calculate delta_f H° for aqueous chloride ion from the following data: 1/2 (g) + 1/2 (g) - (g), delta_f H° = -92.4 kJ; (g) + nH2O(l) - H+(aq) + Cl-(aq), δ H° = -74.8 kJ; delta_f H°(H+, aq) = 0.0 kJ

Question

Calculate delta_f H° for aqueous chloride ion from the following data: 1/2 $H_2$(g) + 1/2 $Cl_2$(g) -> $HCl$(g), delta_f H° = -92.4 kJ; $HCl$(g) + nH2O(l) -> H+(aq) + Cl-(aq), δ H° = -74.8 kJ; delta_f H°(H+, aq) = 0.0 kJ

Prashant Kotian · Accepted Answer

Correct answer: D. Adding the two reactions: 1/2 (g) + 1/2 (g) -> H+(aq) + Cl-(aq); δ H° = -92.4 + (-74.8) = -167.2 kJ. Since delta_f H°(H+, aq) = 0.0 kJ, the entire enthalpy of formation of the hydrochloric acid solution is attributed to the chloride ion: delta_f H°(Cl-, aq) = -167.2 kJ.

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