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

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

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

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

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

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

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

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

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

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