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Q.

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

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a

−44.6kJeq−1

b

−50.6kJeq−1

c

−57.4kJeq−1

d

−51.4kJeq−1

answer is A.

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Detailed Solution

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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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