(a) -44.6 kJ $eq^{-1}$

(b) -50.6 kJ $eq^{-1}$

(c) -51.4 kJ $eq^{-1}$

(d) -70.2 kJ $eq^{-1}$

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.

$\therefore \Delta H = (-51.4) + (-50.6) - (-57.4) kJ = -51.4 - 50.6 + 57.4 = -44.6 kJ$

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