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The heats of neutralization of HCl with NH₄ O H and that of NaOH with CH₃COOH are respectively -51.4 and -50.6 kJ eq^-1. The heat of neutralization of acetic acid with NH₄OH will be

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$- 57.4 kJ {eq}^{- 1}$

$- 51.4 kJ {eq}^{- 1}$

$- 44.6 kJ {eq}^{- 1}$

$- 50.6 kJ {eq}^{- 1}$

answer is A.

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Detailed Solution

$\begin{aligned} Δ H = (- 51.4) + (- 50.6) - (- 57.4) k J \\ = - 51.4 - 50.6 + 57.4 = - 44.6 k J \end{aligned}$

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