Example Question - 2

50.0 cm³ of 1.00 mol dm^-3 hydrochloric acid was mixed with 50.0 cm³ of 1.00 mol dm^-3 sodium hydroxide in a polystyrene cup. The temperature rose by 6.8 K. Calculate the enthalpy change of neutralisation. Assume the solution has a density of 1.00 g cm^-3 and a specific heat capacity of 4.18 J g^-1 K^-1.Step 1: the total volume is 100 cm³, so the mass of solution heated is 100 g. q = mcΔT = 100 × 4.18 × 6.8 = 2842 J = 2.842 kJ.Step 2: HCl + NaOH → NaCl + H₂O. Moles of HCl = 1.00 × 0.0500 = 0.0500 mol, which reacts with 0.0500 mol of NaOH to form 0.0500 mol of water.Step 3: Δ_neutH = -q ÷ n = -2.842 ÷ 0.0500 = -56.8 kJ mol^-1.The enthalpy change of neutralisation is defined per mole of water formed. A polystyrene cup with a lid is used because it is a good insulator and reduces heat loss. The temperature change is most accurately found by plotting a temperature-time graph and extrapolating back to the time of mixing.