The enthalpy of neutralisation of nitric acid is the enthalpy change when one mole of nitric acid is neutralised as shown in the equation:
HNO3 + NaOH --> NaNO3 + H2O
To find this we must first calculate the number of moles of nitric acid used:
moles = concentration * volume in dm^3 (dm^3 = litres) = 1 mol/l * 0.15 l
= 0.15 mol
We must now calculate the energy released by this reaction. Since the solution temperature increased by 6.81˚C, assuming the solution has the same density as water (1 g/mL) and the same specific heat capacity as water (4.18 J/gK) we can use the equation:
q = mcΔT = 300 g * 4.18 J/gK * 6.81 K = 8540 J = 8.54 kJ
Since this is only for the neutralisation of 0.15 mol of nitric acid, we can calculate its molar enthalpy of neutralisation:
ΔH = 8.54 kJ / 0.15 mol = 56.9 kJ/mol
Hope this helps!
