Question

The heat of neutralization of $LiOH$ and $HCl$ at ${25}^{\circ }C$ is $-34.868KJmo{l}^{-1}$. The heat ionization of $LiOH$ will be:

• A. $44.674kJ$
• B. $22.232kJ$
• C. $32.684kJ$
• D. $96.464kJ$

Answer 1

The correct option is B $22.232kJ$

Solution:- (B) $22.232kJ/mol$

$NaOH+HCl⟶NaCl+H_{2}O\Delta H=-57.1kJ/mol$

As we know that heat of neutralization of a strong acid by a strong base is heat of formation of water.

Therefore,

$H^{+}+{OH}^{-}⟶H_{2}O\Delta H=-57.1kJ/mol$

$H_{2}O⟶H^{+}+{OH}^{-}\Delta H=57.1kJ/mol.....\left(1\right)$

Given that heat of neutralisation of $LiOH$ and $HCl$ is $-34.868kJ/mol$

Therefore,

$H^{+}+LiOH⟶{Li}^{+}+H_{2}O\Delta H=-34.868kJ/mol.....\left(2\right)$

Adding ${eq}^n\left(1\right)\&\left(2\right)$, we have

$H_{2}O+LiOH+H^{+}⟶H^{+}+{OH}^{-}+{Li}^{+}+H_{2}O;\Delta H=-34.868+57.1$

$LiOH⟶{Li}^{+}+{OH}^{-};\Delta H=22.232kJ/mol$

Hence the heat of ionisation of $LiOH$ is $22.232kJ/mol$.