Question

$CsOH+HCl⟶CsCl+H_{2}O;\Delta H=-13.4Kcalmo{l}^{-1}....\left(i\right)$
$CsOH+HF⟶CsF+H_{2}O;\Delta H=-16.4Kcalmo{l}^{-1}....\left(ii\right)$
Calculate $\Delta H$ for the ionisation of $HF$ in $H_{2}O$

• A. $\Delta H=-3.0Kcal$
• B. $\Delta H=+3.0Kcal$
• C. $\Delta H=-\mathrm{29.7.0}Kcal$
• D. None of these

Answer 1

Answer: (A)

$\Delta H=-3.0Kcal$

Sol. The ionisation of HF is as shown.
$HF⟶H^{\oplus }+F^{\ominus }....\Delta H=?$

The neutralisation of CsOH with HCl is as shown
$CsOH+HCl⟶CsCl+H_{2}O$ or

$\stackrel{\ominus }{O}H\left(\text{Strong base}\right)+H^{\oplus }\left(\text{acid}\right)⟶H_{2}O.....\Delta H_2=-13.4Kcal$ ......(i)

The neutralisation of CsOH with HF is as shown.
$\stackrel{\ominus }{O}H\left(\text{Strong base}\right)+HF\left(\text{Weak acid}\right)⟶H_{2}O....\Delta H_2=-16.4Kcal$ ......(ii)

The enthalpy change for the ionisation of HF is
$\Delta H=\Delta H_2-\Delta H_1$ $=-16.4-(-13.4)$

$=-3.0Kcal$