Question

Heats of neutralization $\left(△H\right)$ of $N{H}_{4}OH$ and $HF$ are $-51.5$ and $68.6kJ$ respectively. Calculate their heats of dissociation.

Answer 1

$HCl+NaOH\left(aq\right)\to NaCl\left(aq\right)+H_{2}O$ $△H=-57.3KJ$

$HCl\left(aq\right)+N{H}_{4}OH\left(aq\right)\to N{H}_{4}Cl\left(aq\right)+H_{2}O$ $△H=-51.5KJ$

Weak base

$\therefore$ Heat of dissociation of $N{H}_{4}OH$

$△H=-51.5-(-57.3)=5.8KJ$

Similarly, we have

$HF\left(aq\right)+NaOH\left(aq\right)\to NaF\left(aq\right)+H_{2}O$ $△H=+68.6KJ$

$\therefore$ Heat of dissociation of $HF$,

$△H=+68.6-(-57.3)=125.9KJ$