The heats of neutralization of HCl with NH4OH and that of NaOH with CH3 COOH are respectively -51.4 and 1 50.6kJ eq-1 then the heat of neutralisation of acetic acid with NH4OH will be

HCl+NH4OH ;ΔH=−51.4⇒△Hionisation =57.4−51.4=6kJNaOH+CH3COOH ΔH=−50.6⇒△Hionization =57.4−50.6=6.8kJCH3COOH+NH4OH ΔH=−57.4+6+6.8=−44.6kJ

The heats of neutralization of HCl with NH4OH and that of NaOH with CH3 COOH are respectively -51.4 and 1 50.6kJ eq-1 then the heat of neutralisation of acetic acid with NH4OH will be

HCl+NH4OH ;ΔH=−51.4⇒△Hionisation =57.4−51.4=6kJNaOH+CH3COOH ΔH=−50.6⇒△Hionization =57.4−50.6=6.8kJCH3COOH+NH4OH ΔH=−57.4+6+6.8=−44.6kJ

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The heats of neutralization of $HCl with {NH}_{4} OH$ and that of NaOH with CH₃ COOH are respectively -51.4 and 1 50.6kJ eq^-1 then the heat of neutralisation of acetic acid with NH₄OH will be

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$- 44 .6 {kJeq}^{- 1}$

$- 50 .6 kJ {eq}^{- 1}$

$- 57 .4 kJ {eq}^{- 1}$

$- 51 .4 {kJeq}^{- 1}$

answer is A.

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Detailed Solution

$\begin{array}{l} HCl + {NH}_{4} OH; ΔH = - 51 .4 \\ \Rightarrow △ H_{ionisation} = 57 .4 - 51 .4 = 6 kJ \\ NaOH + {CH}_{3} COOH ΔH = - 50 .6 \\ \Rightarrow △ H_{ionization} = 57 .4 - 50 .6 = 6 .8 kJ \\ {CH}_{3} COOH + {NH}_{4} OH ΔH = - 57 .4 + 6 + 6 .8 = - 44 .6 kJ \end{array}$