A-LevelChemistryChemical energetics9701/33 · Feb/Mar 2020 · Q211 Marks

In this experiment you will determine the enthalpy change, AH, for the catalytic decomposition of hydrogen peroxide into water and oxygen. H2O2(aq) → H2O(l) + 2O2(g) FA 3 is aqueous hydrogen peroxide, H2O2. FA 5 is manganese(IV) oxide, MnO2.

(a)Method Experiment 1 • Support one of the plastic cups inside the 250 cm³ beaker. • Use the 50 cm³ measuring cylinder to add 30 cm³ of FA 3 into the plastic cup. • Measure and record the initial temperature of the solution. • Add a heaped spatula measure of FA 5 to the solution in the plastic cup. • Stir constantly until the maximum temperature is reached and record this temperature. • Calculate and record the temperature rise. • Rinse and dry the thermometer. Experiment 2 • Support the second plastic cup inside the 250 cm³ beaker. • Use the 50 cm³ measuring cylinder to add 40 cm³ of FA 3 into the plastic cup. • Measure and record the initial temperature of the solution. • Add a heaped spatula measure of FA 5 to the solution in the plastic cup. • Stir constantly until the maximum temperature is reached and record this temperature. • Calculate and record the temperature rise.[5]
(b)(i)Calculate the energy released in Experiment 1. [Assume that 4.2 J changes the temperature of 1.0 cm³ of solution by 1.0°C.] energy released = .............................................. J[1]
(b)(ii)Use your answer to 1(c)(iv) to calculate the number of moles of hydrogen peroxide used in Experiment 1. (If you were unable to calculate the concentration of H2O2 in FA 3, assume that it is 1.02 mol dm-³. This may not be the correct value.) moles of H2O2 = .............................................. mol[1]
(b)(iii)Calculate the enthalpy change, in kJ mol-1, for the decomposition of 1 mole of hydrogen peroxide into water and oxygen. enthalpy change = .............................................. kJ mol-¹ sign value[1]
(c)(i)A student suggested that the experiment would be more accurate if the same mass of FA 5, manganese(IV) oxide, had been weighed out for each experiment. State and explain whether you agree with the student's suggestion.[2]
(c)(ii)The student also suggested that Experiments 1 and 2 should give the same temperature rise, even though a greater volume of FA 3 was used in Experiment 2. State and explain whether you agree with the student's suggestion.[1]

Mark scheme

Marking points from the official mark scheme (Feb/Mar 2020). The mark or mark type is shown in brackets.

[1] (a) I Table (or two lists) showing unambiguous headings and data for both experiments in the space provided: • two initial thermometer readings + values • two final/highest thermometer readings + values • two temperature rises / changes + values
[1] (a) II Recording of data • correct units 'covering' all thermometer readings • all four readings recorded to .0 or .5 °C • both rises in temperature correctly calculated
[1] (a) III Award this mark based on the table below
[1] (a) IV Award this mark based on the table below
[1] (a) V Award mark if both the candidate's corrected temperature rises are within 1.0 °C of each other
[1] (b)(i) Correctly calculates heat produced = 30 × 4.2 × temp rise for expt 1 AND answer given to 2–4 sf
[1] (b)(ii) Correctly uses number of moles H2O2 = 0.03 × answer to 1(c)(iv) AND answer given to 2–4 sf Answer for default value = 0.0306 / 0.031 mol
[1] (b)(iii) Correct expression for enthalpy change • Enthalpy change = ans (i) / ans (ii) x 1/1000 • Negative sign must be shown • Answer is shown to 2-4 sf
[1] (c)(i) (FA 5 / MnO2 is a) catalyst
[1] (c)(i) (Student is wrong:) the mass of MnO2 used does not alter the heat produced / enthalpy change / temperature change (for decomposition of H2O2)
[1] (c)(ii) Student is correct: energy / heat released is greater because more / greater (moles / molecules of) hydrogen peroxide / FA 3 is used in Experiment 2 AND more moles / molecules / greater amount of water / solution / FA 3 / hydrogen peroxide heated is greater (in the same proportion)