Question

The enthalpy of neutralisation of $NH_4OH$ with $HCl$ is $-51.46\, kJ\, mol^{-1}$ and the enthalpy of neutralisation of $NaOH$ with $HCl$ is $-55.90\, kJ\, mol^{-1}$. The enthalpy of ionisation of $NH_4OH$ is

• A. $-10.36\, kJ\, mol^{-1}$
• B. $-4.44\, kJ\, mol^{-1}$
• C. $+107.36\, kJ\, mol^{-1}$
• D. $+4.44\, kJ\, mol^{-1}$

Answer 1

Answer: (D)

$HCl \rightarrow H^+ +Cl^-$
Strong acid (Complete ionisation) ...(i)
$NH_4OH \rightleftharpoons NH^+_4 +OH^-$
Weak base $\Delta H = x\, kJ\, mol^{-1}$ ...(ii)
$H+OH^- \rightarrow +H_2O$
$\Delta H = -55.90\, kJ\, mol^{-1}$ ...(iii)
(from neutralisation of strong acid and strong base)
From equation (i), (ii) and (iii)
$NH_4OH + HCl \rightarrow NH^+ +Cl^- +H_2O$
$\Delta H = -51.46\, kJ\, mol^{-1}$
$\therefore x + (- 55.90) = -51.46$
$x = - 51.46 + 55.90$
$= 4.44\, kJ\, mol^{-1}$
$\therefore $ Enthalpy of ionisation of $NH_4OH=4.44\, kJ\, mol^{-1}$