Question

The enthalpy of neutralisation of $N{H}_{4}OH$ with $HCl$ is $-51.46kJmo{l}^{-1}$ and the enthalpy of neutralisation of $NaOH$ with $HCl$ is $-55.90kJmo{l}^{-1}$. The enthalpy of ionisation of $N{H}_{4}OH$ is

• A. $-10.36kJmo{l}^{-1}$
• B. $-4.44kJmo{l}^{-1}$
• C. $+107.36kJmo{l}^{-1}$
• D. $+4.44kJmo{l}^{-1}$

Answer 1

Answer: (D)

$HCl\to H^{+}+C{l}^{-}$
Strong acid (Complete ionisation) ...(i)
$N{H}_{4}OH\rightleftharpoons N{H}_4^{+}+O{H}^{-}$
Weak base $\Delta H=xkJmo{l}^{-1}$ ...(ii)
$H+O{H}^{-}\to +H_{2}O$
$\Delta H=-55.90kJmo{l}^{-1}$ ...(iii)
(from neutralisation of strong acid and strong base)
From equation (i), (ii) and (iii)
$N{H}_{4}OH+HCl\to N{H}^{+}+C{l}^{-}+H_{2}O$
$\Delta H=-51.46kJmo{l}^{-1}$
$\therefore x+(-55.90)=-51.46$
$x=-51.46+55.90$
$=4.44kJmo{l}^{-1}$
$\therefore$ Enthalpy of ionisation of $N{H}_{4}OH=4.44kJmo{l}^{-1}$