A-LevelChemistryChemical energetics9701/33 · May/June 2025 · Q210 Marks
You will now determine the concentration of a solution of hydrogen peroxide by a different method. Hydrogen peroxide decomposes slowly into water and oxygen at room temperature. This reaction is exothermic. When a catalyst is added, the decomposition is fast and there is a measurable temperature rise. H2O2(aq) → H2O(l) + ½O2(g) ΔΗ = −98.2 kJ mol-1 FA 1 is aqueous hydrogen peroxide, H2O2. FA 5 is manganese(IV) oxide, MnO2. (a) Method Experiment 1 • Support one of the cups in the 250 cm³ beaker. • Use the 50 cm³ measuring cylinder to add 25.0cm³ of FA 1 to the cup. • Place the thermometer in the FA 1 and tilt the cup, if necessary, so that the bulb of the thermometer is fully covered. Record the temperature in the space for results. • Add a heaped spatula measure of FA 5 to the solution in the cup. • Stir constantly until the maximum temperature is reached. Record this temperature. • Calculate and record the temperature rise. • Rinse and dry the thermometer. Experiment 2 • Support the second cup in the 250 cm³ beaker. • Use the 50 cm³ measuring cylinder to add 40.0 cm³ of FA 1 to the second cup. • Measure and record the initial temperature of the solution. • Add a heaped spatula measure of FA 5 to the solution in the second cup. • Stir constantly until the maximum temperature is reached. Record this temperature. • Calculate and record the temperature rise. Keep FA 5 for use in Question 3.
- (a) ResultsResults I II III IV V[5]
- (b)(i)Calculate the energy change, in J, in Experiment 2. energy change = ______________ J [1][1]
- (b)(ii)Use the information given and your answer to (b)(i) to calculate the concentration, in mol dm⁻³, of hydrogen peroxide in FA 1. concentration of H2O2 in FA 1 = ______________ moldm⁻³ [2][2]
- (c)(i)Heat loss is a large source of error when carrying out the method for Question 2. Describe and explain the effect of heat loss on the value of the concentration of hydrogen peroxide calculated.[1]
- (c)(ii)A student suggests that calculating the concentration of hydrogen peroxide using the method in Question 2 would be less accurate when the concentration is lower. Suggest whether the student is correct. Explain your answer.[1]
Mark scheme
Marking points from the official mark scheme (May/June 2025). The mark or mark type is shown in brackets.
- [1] I Recording of results in the space for results; Unambiguous headings AND units for all recorded results: initial / FA 1 temperature / temp / T; final / maximum temperature / temp / T; temp(erature) change / rise / ΔΤ; Units: (°C) // °C with each heading OR °C with each piece of data.
- [1] II initial and final temperatures recorded for both experiments. AND clear indication of Experiments 1 and 2.
- [1] III All thermometer readings recorded to nearest .5 °C AND temperature changes correctly calculated.
- [1] IV Difference in temperature changes for Exps 1 and 2 are within the following ranges of the candidate's mean ΔΤ: Cand mean ΔΤ < 5.0, |ΔΤ1 – ΔΤ2| <0.5; Cand mean ΔΤ 5.25-10.0, |ΔΤ1 – ΔΤ2| <1.0; Cand mean ΔΤ 10.25-15.0, |ΔΤ1 – ΔΤ2| <1.5; Cand mean ΔΤ > 15.25, |ΔΤ1 – ΔΤ2| <2.0
- [1] V Difference between mean temperature change for candidate and supervisor rounded to the nearest 0.5 °C, δ: Sup mean ΔΤ < 5.0, δ <1.0; Sup mean ΔΤ 5.25-10.0, δ <1.5; Sup mean ΔΤ 10.25-15.0, δ <2.5; Sup mean ΔΤ 15.25-20.0, δ <3.5; Sup mean ΔΤ 20.25-25.0, δ <4.5
- [1] Correctly calculates Energy change = 40 × 4.18 × ΔΤ for Exp 2. AND answer given to 2–4 sf.
- [1] M1 Correctly uses [H2O2] = (b)(i) / (98.2 × 40)
- [1] M2 Correct answer with some working shown OR some correct working which includes: (b)(i) / 98.2 OR (b)(i) / 98200 OR (b)(i) / 3928 AND answer given to 2–4 sf
- [1] Concentration will be lower AND two of: ΔΤ / maximum temp will be lower; (calculated / apparent) energy released / change will be lower; amount / moles (of FA 1 / H2O2) will be lower.
- [1] The student is correct AND ΔT will be lower AND (%) error will be greater OR The student is correct AND the experiment will be slower AND more time for heat to escape OR The student is not correct AND ΔT will be lower AND heat loss will be less OR Student is not correct AND ΔT will be lower AND this is cancelled out by a slower reaction.
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