Using the graph, estimate the initial temperature of the solution.

Determine the maximum temperature reached in the experiment by analysing the graph.

Calculate the concentration of ethanoic acid, CH₃COOH, in mol dm^–3.

Determine the heat change, q, in kJ, for the neutralization reaction between ethanoic acid and sodium hydroxide.

Assume the specific heat capacities of the solutions and their densities are those of water.

Calculate the enthalpy change, ΔH, in kJ mol^–1, for the reaction between ethanoic acid and sodium hydroxide.

Explain the shape of curve X in terms of the collision theory.

Suggest one possible reason for the differences between curves X and Y.

21.4 °C

Accept values in the range of 21.2 to 21.6 °C.

29.0 «°C»

Accept range 28.8 to 29.2 °C.

ALTERNATIVE 1

«volume CH₃COOH =» 26.0 «cm³»

«[CH₃COOH] = 0.995 mol dm^–3 \( \times \frac{{50.0\,{\text{cm³}}}}{{26.0\,{\text{cm³}}}} = \)» 1.91 «mol dm⁻³»

ALTERNATIVE 2

«n(NaOH) =0.995 mol dm^–3 x 0.0500 dm³ =» 0.04975 «mol»

«[CH₃COOH] = \(\frac{{0.04975}}{{0.0260}}\) dm³ =» 1.91 «mol dm^–3»

Accept values of volume in range 25.5 to 26.5 cm³.

Award [2] for correct final answer.

«total volume = 50.0 + 26.0 =» 76.0 cm³ AND «temperature change 29.0 – 21.4 =» 7.6 «°C»

«q = 0.0760 kg x 4.18 kJ kg^–1 K^–1 x 7.6 K =» 2.4 «kJ»

Award [2] for correct final answer.

«n(NaOH) = 0.995 mol dm^–3 x 0.0500 dm³ =» 0.04975 «mol»

OR

«n(CH₃COOH) = 1.91 mol dm^–3 x 0.0260 dm³ =» 0.04966 «mol»

«ΔH = \( - \frac{{2.4\,{\text{kJ}}}}{{0.04975\,{\text{mol}}}} = \)» –48 / –49 «kJ mol^–1»

Award [2] for correct final answer.

Negative sign is required for M2.

«initially steep because» greatest concentration/number of particles at start

OR

«slope decreases because» concentration/number of particles decreases

volume produced per unit of time depends on frequency of collisions

OR

rate depends on frequency of collisions

mass/amount/concentration of metal carbonate more in X

OR

concentration/amount of CH₃COOH more in X