DP Chemistry (first assessment 2025)

Question 19M.2.HL.TZ1.4

Date	May 2019	Marks available	[Maximum mark: 14]	Reference code	19M.2.HL.TZ1.4
Level	HL	Paper	2	Time zone	TZ1
Command term	Calculate, Comment, Determine, Draw, Explain, Sketch, State, Suggest	Question number	4	Adapted from	N/A

[Maximum mark: 14]

19M.2.HL.TZ1.4

This question is about the decomposition of hydrogen peroxide.

Hydrogen peroxide decomposes to water and oxygen when a catalyst such as potassium iodide, KI, is added.

2H₂O₂ (aq) $\xrightarrow{{{\text{KI (aq)}}}}$ O₂ (g) + 2H₂O (l)

(a)

Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.

[1]

Markscheme

decomposes in light [✔]

Note: Accept “sensitive to light”.

Examiners report

There were a couple of comments claiming that this NOS question on “why to store hydrogen peroxide in brown bottles” is not the syllabus. Most candidates were quite capable of reasoning this out.

(b(i))

In a laboratory experiment solutions of potassium iodide and hydrogen peroxide were mixed and the volume of oxygen generated was recorded. The volume was adjusted to 0 at t = 0.

The data for the first trial is given below.

Plot a graph on the axes below and from it determine the average rate of
formation of oxygen gas in cm³ O₂ (g) s⁻¹.

Average rate of reaction:

[3]

Markscheme

points correctly plotted [✔]

best fit line AND extended through (to) the origin [✔]

Average rate of reaction:
«slope (gradient) of line =» 0.022 «cm³ O₂ (g) s⁻¹» [✔]

Note: Accept range 0.020–0.024cm³ O₂ (g) s⁻¹.

Examiners report

Most candidates could plot a best fit line and find the slope to calculate an average rate of reaction.

(b(ii))

Two more trials (2 and 3) were carried out. The results are given below.

Determine the rate equation for the reaction and its overall order, using your answer from (b)(i).

Rate equation:

Overall order:

[2]

Markscheme

Rate equation:
Rate = k[H₂O₂] × [KI] [✔]

Overall order:
2 [✔]

Note: Rate constant must be included.

Examiners report

Good performance but with answers that either typically included only [H₂O₂] with first or second order equation or even suggesting zero order rate equation.

(b(iii))

Additional experiments were carried out at an elevated temperature. On the axes below, sketch Maxwell–Boltzmann energy distribution curves at two temperatures T₁ and T₂, where T₂ > T₁.

[2]

Markscheme

peak of T₂ to right of AND lower than T₁ [✔]

lines begin at origin AND T₂ must finish above T₁ [✔]

Examiners report

Fair performance; errors including not starting the two curves at the origin, drawing peak for T2 above T1, T2 finishing below T1 or curves crossing the x-axis.

(b(iv))

Apart from a greater frequency of collisions, explain, by annotating your graphs in (b)(iii), why an increased temperature causes the rate of reaction to increase.

[2]

Markscheme

E_a marked on graph [✔]

explanation in terms of more “particles” with E ≥ E_a

greater area under curve to the right of E_a in T₂ [✔]

Examiners report

The majority of candidates earned at least one mark, many both marks. Errors included not annotating the graph with E_a and referring to increase of kinetic energy as reason for higher rate at T2.

(b(v))

MnO₂ is another possible catalyst for the reaction. State the IUPAC name for MnO₂.

[1]

Markscheme

manganese(IV) oxide

manganese dioxide [✔]

Note: Accept “manganese(IV) dioxide”.

Examiners report

A well answered question. Very few candidates had problem with nomenclature.

(c)

Comment on why peracetic acid, CH3COOOH, is always sold in solution with ethanoic acid and hydrogen peroxide.

H₂O₂ (aq) + CH₃COOH (aq) ⇌ CH₃COOOH (aq) + H₂O (l)

[1]

Markscheme

moves «position of» equilibrium to right/products [✔]

Note: Accept “reactants are always present as the reaction is in equilibrium”.

Examiners report

One teacher suggested that “stored” would have been better than “sold” for this question. There were a lot of irrelevant answers with many believing the back reaction was an acid dissociation.

(d)

Sodium percarbonate, 2Na₂CO₃•3H₂O₂, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.

M_r (2Na₂CO₃•3H₂O₂) = 314.04

Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.

[2]

Markscheme

M( H₂O₂) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g» [✔]

«% H₂O₂ = 3 × $\frac{{34.02}}{{314.04}}$ × 100 =» 32.50 «%» [✔]

Note: Award [2] for correct final answer.

Examiners report

It is recommended that candidates use the relative atomic masses given in the periodic table.