For the reaction ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\to {\mathrm{H}}_2\mathrm{O}$; $\Delta \mathrm{H}=-13.7\mathrm{ }\mathrm{K}$.cal. Then for ${\mathrm{H}}_3{\mathrm{O}}^{+}$ $+{\mathbf{OH}}^{-}\to 2{\mathrm{H}}_2\mathrm{O},\Delta \mathrm{H}$ will be

 If ${\mathrm{H}}^{+}+{\mathrm{OH}}^{-}\leftrightharpoons {\mathrm{H}}_2\mathrm{O}+13.7\mathrm{Kcal}$, then heat of complete neutralization of one

gram mole of ${\mathrm{H}}_2{\mathrm{SO}}_4$ with strong base will be $27.4\mathrm{Kcal}$.

${\mathrm{H}}_2{\mathrm{SO}}_4\leftrightharpoons 2{\mathrm{H}}^{+}+{\mathrm{SO}}_4^{2-}$

$2{\mathrm{H}}^{+}+2{\mathrm{OH}}^{-}\leftrightharpoons 2{\mathrm{H}}_2\mathrm{O}+27.4\mathrm{Kcal}$

$13.7\times 2=27.4$

Hence the correct answer is (A) +27.4 K.Cals.