When 1M ${\mathrm{H}}_2{\mathrm{SO}}_4$  is completely neutralized by sodium hydroxide, the heat liberated is 114.64 KJ. What is the enthalpy of neutralization?

The reaction of neutralization of Sulfuric acid with Sodium hydroxide is given below:

${\mathrm{H}}_2{\mathrm{SO}}_4+2\mathrm{NaOH}\to {\mathrm{Na}}_2{\mathrm{SO}}_4+2{\mathrm{H}}_2\mathrm{O}$ 

The enthalpy for this reaction will be -114.64 KJ because the heat is liberated in the reaction.

When 1M of Sulfuric acid is neutralized the 2 moles of water is released. So, for 1 mole of water the enthalpy will be:

$\mathrm{Enthalpy}=\frac{-114.64}{2}=-57.32\text{ KJ}$ 

So, the enthalpy will be -57.32 KJ.

Therefore, the correct option is C.