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Heat of neutralisation of HCl with $NH_4OH$ and NaOH with $CH_3COOH$ are respectively -51.4 and -50.6 kJ $eq^{-1}$. What will be the heat of neutralisation of acetic acid with $NH_4OH$?


(a) -44.6 kJ $eq^{-1}$
(b) -50.6 kJ $eq^{-1}$
(c) -51.4 kJ $eq^{-1}$
(d) -70.2 kJ $eq^{-1}$

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Answer: -44.6 kJ $eq^{-1}$
(i) $HCl + NH_4OH \longrightarrow NH_4Cl + H_2O$ ; $\Delta H = -51.4 kJ$
(ii) $CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$ ; $\Delta H = -50.6 kJ$
(ii) $CH_3COOH + NH_4OH \longrightarrow CH_3COONH_4 + H_2O$ ; $\Delta H =?$
We know,
(iv) $HCl + NaOH \longrightarrow NaCl + H_2O$ ; $\Delta H = -57.4 kJ$
To get (iii) add (i) and (ii) and subtract (iv), keeping in view than salts formed in case (i), (ii) and (iii) are completely ionised.
$\therefore \Delta H = (-51.4) + (-50.6) - (-57.4) kJ = -51.4 - 50.6 + 57.4 = -44.6 kJ$
answered Feb 26, 2014 by mosymeow_1
 

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