A neutralisation reaction occurs when NaOH(aq) is added to H2SO4(aq). equation 1 2NaOH(aq) + H2SO4(aq) → Na2SO4(aq) + 2H2O(l)
- (a)Define enthalpy change of neutralisation, ΔHneut:[2]
- (b)In an experiment, 50.0cm³ of 2.00moldm⁻³ NaOH(aq) is added to 60.0cm³ of 1.00moldm⁻³ H2SO4(aq) in a polystyrene cup and stirred. Both solutions have a temperature of 21.4°C before mixing. The maximum temperature of the mixture is measured.
- (b)(i)Use equation 1 to calculate the amount, in mol, of H2SO4(aq) that is neutralised in the experiment. amount of H2SO4 neutralised = ...................................... mol[1]
- (b)(ii)Calculate the theoretical maximum temperature of the mixture in this experiment. Assume that: • enthalpy change of neutralisation, ΔHneut, of NaOH(aq) and H2SO4(aq) is −57.1 kJ mol⁻¹ • full dissociation of H2SO4(aq) occurs • the specific heat capacity of the final solution is 4.18 J g⁻¹ K⁻¹ • 1.00 cm³ of the final solution has a mass of 1.00 g • there is no heat loss to the surroundings • the experiment takes place at constant pressure. Show your working. theoretical maximum temperature = ...................................... °C[3]
- (c)The enthalpy change of neutralisation of CH3COOH(aq) and NaOH(aq) is –55.2 kJ mol⁻¹.
- (c)(i)Complete the equation for the reaction. ... .CH3COOH + .......NaOH → ..........................................................................................[1]
- (c)(ii)Values for the enthalpy change of neutralisation, ΔHneut, are shown in Table 3.1. Table 3.1 reagents ΔHneut / kJ mol⁻¹ NaOH + HCl -57.1 NaOH + CH3COOH -55.2 Suggest why the value for ΔHneut of the weak acid, CH3COOH, reacting with NaOH is different to the value obtained using the strong acid, HCl. Assume that the values are determined under the same conditions.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.
- [2] M1 (energy change) when 1 mole of water is formed M2 from an (aqueous) acid and an alkali / aqueous base
- [1] 0.05 mol (H2SO4(aq))
- [3] M1 correct value for Q (using 0.1 mol H2O / OR 2× (b)(i)) (and Q in Joules) M2 calculate ΔT (using their Q AND 110) Q/mc=ΔT = (Q) ÷ (4.18 × 110) = 12.42 (°C) M3 use ΔT and initial temperature to find temperature of final solution 21.4 + M2 = 33.8 (°C) [M3 ΔT + 21.4 = Tmax]
- [1] CH3COOH + NaOH → CH3COONa + H2O
- [1] (additional) energy is involved / required to fully ionise / dissociate (weak / ethanoic acid) OR (additional) energy is involved / required to donate/ release H+ ions
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