Identify the formula and state symbol of X.

Suggest why the experiment should be carried out in a fume hood or in a well-ventilated laboratory.

The precipitate of sulfur makes the mixture cloudy, so a mark underneath the reaction mixture becomes invisible with time.

10.0 cm³ of 2.00 mol dm^-3 hydrochloric acid was added to a 50.0 cm³ solution of sodium thiosulfate at temperature, T1. Students measured the time taken for the mark to be no longer visible to the naked eye. The experiment was repeated at different concentrations of sodium thiosulfate.

Show that the hydrochloric acid added to the flask in experiment 1 is in excess.

Draw the best fit line of \(\frac{1}{{\rm{t}}}\) against concentration of sodium thiosulfate on the axes provided.

A student decided to carry out another experiment using 0.075 mol dm^-3 solution of sodium thiosulfate under the same conditions. Determine the time taken for the mark to be no longer visible.

An additional experiment was carried out at a higher temperature, T₂.

(i) On the same axes, sketch Maxwell–Boltzmann energy distribution curves at the two temperatures T₁ and T₂, where T₂ > T₁.

(ii) Explain why a higher temperature causes the rate of reaction to increase.

Suggest one reason why the values of rates of reactions obtained at higher temperatures may be less accurate.

H₂O AND (l)
Do not accept H₂O (aq).

SO₂ (g) is an irritant/causes breathing problems
OR
SO₂ (g) is poisonous/toxic

Accept SO₂ (g) is acidic, but do not accept “causes acid rain”.
Accept SO₂ (g) is harmful.
Accept SO₂ (g) has a foul/pungent smell.

n(HCl) = «\(\frac{{10.0}}{{1000}}\)dm³ × 2.00 mol dm^-3 =» 0.0200 / 2.00 × 10^-2«mol»
AND
n(Na₂S₂O₃) = «\(\frac{{50}}{{1000}}\)dm³ × 0.150 mol × dm^-3 =» 0.00750 / 7.50 × 10^-3 «mol»

0.0200 «mol» > 0.0150 «mol»
OR
2.00 × 10^-2«mol» > 2 × 7.50 × 10^-3 «mol»
OR
\(\frac{1}{2}\) × 2.00 × 10^-2 «mol» > 7.50 × 10^-3 «mol»

Accept answers based on volume of solutions required for complete reaction.
Award [2] for second marking point.
Do not award M2 unless factor of 2 (or half) is used.

five points plotted correctly
best fit line drawn with ruler, going through the origin

22.5 × 10^-3 «s^-1»

«Time = \(\frac{1}{{22.5 \times {{10}^{ - 3}}}}\) =» 44.4 «s»

Award [2] for correct final answer.
Accept value based on candidate’s graph.
Award M2 as ECF from M1.
Award [1 max] for methods involving taking mean of appropriate pairs of \(\frac{1}{{\rm{t}}}\) values.
Award [0] for taking mean of pairs of time values.
Award [2] for answers between 42.4 and 46.4 «s».

(i)

correctly labelled axes
peak of T₂ curve lower AND to the right of T₁ curve

Accept “probability «density» / number of particles / N / fraction” on y-axis.

Accept “kinetic E/KE/E_K” but not just “Energy/E” on x-axis.

(ii)

greater proportion of molecules have E ≥ E_a or E > E_a
OR
greater area under curve to the right of the E_a

greater frequency of collisions «between molecules»
OR
more collisions per unit time/second

Accept more molecules have energy greater than E_a.
Do not accept just “particles have greater kinetic energy”.
Accept “rate/chance/probability/likelihood/” instead of “frequency”.
Accept suitably shaded/annotated diagram.
Do not accept just “more collisions”.

shorter reaction time so larger «%» error in timing/seeing when mark disappears

Accept cooling of reaction mixture during course of reaction.