Question

Heat of neutralisation of NaOH and HCl is $-57.3\, kJ\, mol^{-1}$. The heat of ionization of water will be

• A. $-57.3\, kJ \, mol^{-1}$
• B. $-114.6\, kJ \, mol^{-1}$
• C. $+57.3\, kJ\, mol^{-1}$
• D. $+ 114.6\, kJ \,mol^{-1}$

Answer 1

Answer: (C)

$NaOH + HC1 \rightarrow NaCl + H_2O$
or $H^+ + OH^- \rightarrow H_2O ; \Delta H = -57.3$ kJ
$\therefore \, H_2O \rightarrow H^+ + OH^- ; \Delta H = +57.3$ kJ
This is according to Laplace law