Question 19M.2.SL.TZ1.4
| Date | May 2019 | Marks available | [Maximum mark: 12] | Reference code | 19M.2.SL.TZ1.4 |
| Level | SL | Paper | 2 | Time zone | TZ1 |
| Command term | Calculate, Comment, Determine, Explain, Sketch, State, Suggest | Question number | 4 | Adapted from | N/A |
This question is about peroxides.
Hydrogen peroxide decomposes to water and oxygen when a catalyst such as potassium iodide, KI, is added.
2H2O2 (aq) O2 (g) + 2H2O (l)
Suggest why many chemicals, including hydrogen peroxide, are kept in brown bottles instead of clear colourless bottles.
[1]
decomposes in light [✔]
Note: Accept “sensitive to light”.
The explanation that the brown bottle prevented light causing a decomposition of the chemical was well answered but some incorrectly suggested it helped to stop mixing up of chemicals e.g. acid/water/peroxide.
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]
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.
The graphing was disappointing with a surprising number of students missing at least one mark for failing to draw a straight line or for failing to draw the line passing through the origin. Also some were unable to calculate the gradient.
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]
peak of T2 to right of AND lower than T1 [✔]
lines begin at origin AND T2 must finish above T1 [✔]
The drawing of the two curves at T1 and T2 was generally poorly done.
Apart from a greater frequency of collisions, explain, by annotating your graphs in (b)(ii), why an increased temperature causes the rate of reaction to increase.
[2]
Ea marked on graph [✔]
explanation in terms of more “particles” with E ≥ Ea
OR
greater area under curve to the right of Ea in T2 [✔]
Explaining why temperature increase caused an increase in reaction rate was generally incorrectly answered with most students failing to mention “activation energy” in their answer or failing to annotate the graph.
MnO2 is another possible catalyst for the reaction. State the IUPAC name for MnO2.
[1]
manganese(IV) oxide
OR
manganese dioxide [✔]
Note: Accept “manganese(IV) dioxide”.
Many could correctly name manganese(IV)oxide, but there were answers of magnesium(IV) oxide or manganese(II) oxide.
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]
move «position of» equilibrium to right/products [✔]
Note: Accept “reactants are always present as the reaction is in equilibrium”.
Suggesting why peractic acid was sold in solution was very poorly answered and only a few students mentioned equilibrium and, if they did, they thought it would move to the left to restore equilibrium.
Sodium percarbonate, 2Na2CO3•3H2O2, is an adduct of sodium carbonate and hydrogen peroxide and is used as a cleaning agent.
Mr (2Na2CO3•3H2O2) = 314.04
Calculate the percentage by mass of hydrogen peroxide in sodium percarbonate, giving your answer to two decimal places.
[2]
M (H2O2) «= 2 × 1.01 + 2 × 16.00» = 34.02 «g» [✔]
«% H2O2 = 3 × × 100 =» 32.50 «%» [✔]
Note: Award [2] for correct final answer.
Calculating the % by mass was generally well answered although some candidates started by using rounded values of atomic masses which made their final answer unprecise.
