| Date | November 2017 | Marks available | 2 | Reference code | 17N.2.sl.TZ0.1 |
| Level | SL | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate | Question number | 1 | Adapted from | N/A |
Question
A student titrated an ethanoic acid solution, CH3COOH (aq), against 50.0 cm3 of 0.995 mol dm–3 sodium hydroxide, NaOH (aq), to determine its concentration.
The temperature of the reaction mixture was measured after each acid addition and plotted against the volume of acid.
Curves X and Y were obtained when a metal carbonate reacted with the same volume of ethanoic acid under two different conditions.
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.
Markscheme
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
