Question

The heats of neutralisation of HCl with $NH_4OH$ and $NaOH$ with $ CH_3COOH $are respectively $-51.4 \, and -50.6 kJ\, eq^{-1}$ The heat of neutralisation of acetic acid with $NH_4OH$ will be

• A. $-44.6\, kJ\, eq^{-1}$
• B. $-50.6\, kJ \,eq^{-1}$
• C. $-51.4\, kJ\, eq^{-1}$
• D. $-57.4\, kJ\, eq^{-1}$

Answer 1

Answer: (A)

(i) $HCI + NH_4OH \rightarrow NH_4Cl + H_2O ; \Delta H = -51.4 \,kJ$
(ii) $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O ; \Delta H = - 50.6 \, kJ$
(iii) $CH_3COOH + NH_4OH \rightarrow CH_3COONH_4 + H_2O ; AH = ?$
We know
(iv) $HCl + NaOH \rightarrow NaCl + H_2O; \Delta H $ = -57-4 kJ
To get (iii) add (i) and (ii) and subtract (iv).
(Please note that salts formed in case (i) and (ii) are almost completely ionised. Also suppose that the salt formed in case (iii) is also ionised).
$\Delta $H = (- 51.4) + (-50.6) - (-57.4) kJ
= _ 51.4 - 50.6 + 57.4 = -44.6 kJ