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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