The equation for the reaction of sulfuric acid with sodium hydroxide is shown below:
H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l) \text{H}_2\text{SO}_4(\text{aq}) + 2\text{NaOH}(\text{aq}) \rightarrow \text{Na}_2\text{SO}_4(\text{aq}) + 2\text{H}_2\text{O}(\text{l}) H2SO4(aq)+2NaOH(aq)→Na2SO4(aq)+2H2O(l)A volume of 50 cm3 of 0.80 mol dm-3 H2SO4\text{H}_2\text{SO}_4H2SO4 is reacted with excess NaOH\text{NaOH}NaOH. The energy given out is 4.56 kJ. What is the enthalpy change of neutralisation, in kJ mol−1\text{kJ mol}^{-1}kJ mol−1?
−57.0-57.0−57.0
−114.0-114.0−114.0
−28.5-28.5−28.5
−228.0-228.0−228.0