MCQ
Heat of neutralisation of $NaOH$ and $HCl$ is $-57.46\, kJ/eq.$ Then heat of ionisation of water in $kJ/mol$ is
- A$-57.46$
- ✓$+57.46$
- C$-114.92$
- D$+114.92$
✓
Answer
Correct option: B.
$+57.46$
b
$\Delta H_{heat} = -\Delta H_{ionisation} = H_2O = +57.46\, KJ/eq.$
$\Delta H_{heat} = -\Delta H_{ionisation} = H_2O = +57.46\, KJ/eq.$
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