The enthalpy of formation of H2O (l) is -290 kJ/mol and the enthalpy of neutralization of a strong acid with a strong alkali is - 56 kJ/mol. What is the enthalpy of formation of OH- (aq)? Given ΔHf(H⁺)(aq) = 0.
Correct answer: B. - 234 kJ/mol
- A. - 342 kJ/mol
- B. - 234 kJ/mol
- C. - 346 kJ/mol
- D. - 178 kJ/mol
Explanation
The neutralization reaction is H⁺(aq) + OH⁻(aq) → H₂O(l). The enthalpy change is ΔHneut = ΔHf (H₂O) - [ΔHf (H⁺) + ΔHf (OH⁻)]. Plugging in the values gives -56 = -286 - [0 + ΔHf (OH⁻)]. Solving for ΔHf (OH⁻) gives -230 kJ/mol.
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About Thermochemistry and Energetics of Chemical Reactions
Thermochemistry measures energy changes in chemical reactions and distinguishes exothermic from endothermic processes. Work covers systems, surroundings and state functions, internal energy, the first law of thermodynamics, enthalpy and Hess's law, including the sign conventions used when heat enters or leaves a system.
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