中文（繁體）
中文（简体）
English
CIE A-Level Chemistry 9701 Paper 5 Practice Paper (Planning, Analysis and Evaluation) – 30 marks, 1 h 15 min, with answers and mark scheme
TutorZone original practice paper – CIE A-Level Chemistry 9701 Paper 5 (Planning, Analysis and Evaluation)
2 questions, 11 parts – 30 marks – 1 hour 15 minutes – calculator allowed
Paper comparison
| Paper | Time | Marks | Role |
|---|---|---|---|
| Paper 5 (planning, analysis and evaluation) | 1 h 15 min | 30 | this paper |
| Paper 4 (A Level structured) | 2 hours | 100 | other A Level paper |
| Paper 3 (AS practical skills) | 2 hours | 40 | AS paper |
| Paper 2 (AS structured) | 1 h 15 min | 60 | AS paper |
Data and formulae
| Quantity | Value |
|---|---|
| Gas constant R | 8.31 J K-1 mol-1 |
| Avogadro constant L | 6.02 x 1023 mol-1 |
| Planck constant h | 6.63 x 10-34 J s |
| Speed of light c | 3.00 x 108 m s-1 |
| Elementary charge e | 1.60 x 10-19 C |
| Relative atomic masses H, C, O, Na, S | 1.0, 12.0, 16.0, 23.0, 32.1 |
Questions
Total: 30 marks · 1 h 15 min · calculator allowed
Q 1 · Planning: effect of temperature on the rate of the sodium thiosulfate / acid reaction [15]
(a) A student investigates how the rate of the reaction between sodium thiosulfate solution and dilute hydrochloric acid changes with temperature. State the independent variable, the dependent variable and three variables that must be kept constant. [2]
(b) Describe how the reaction mixture is held steadily between 20.0 C and 60.0 C, in steps of 10 C, and how the temperature is read accurately. [3]
(c) Give the full procedure for timing the disappearance of the cross, including the volumes used, how the cross is observed, and how many repeats are taken. [4]
(d) Draw a table for the results, with every column heading and unit. [2]
(e) Explain how the activation energy Ea is found from these results: state what is plotted against what, and how the gradient gives Ea. [3]
(f) State two safety precautions for this experiment. [1]
Q 2 · Analysis and evaluation: activation energy from rate data [15]
(a) A student timed the disappearance of a black cross in the same reaction at five temperatures; the results are in the table above. Complete the 103/T and ln (1/t) columns (three significant figures). [4]
(b) Plot a suitable graph of the data and explain how you judge whether the points lie on a straight line. [4]
(c) Find the gradient of the line, with its unit. [2]
(d) Hence calculate the activation energy Ea in kJ mol-1 (R = 8.31 J K-1 mol-1). [2]
(e) Give two sources of error in the experiment, each with an improvement. [3]
Answers and mark scheme
| Q | Part | Answer | Mark points | marks |
|---|---|---|---|---|
| 1 | (a) | Independent variable: the temperature (20 C to 60 C). Dependent variable: the rate of reaction, measured by the time t for the black cross to disappear (rate is proportional to 1/t). Control variables: the concentration and volume of sodium thiosulfate solution, the concentration and volume of dilute hydrochloric acid, the depth of the cross and the observer eye level, the same conical flask. | 1 mark: temperature independent, rate (time t or 1/t) dependent 1 mark: three controls (concentration, volumes, depth of the cross) |
2 |
| 1 | (b) | Heat a thermostatted water bath to the target temperature. Stand the tube of sodium thiosulfate solution and the tube of dilute acid in the bath for 5 minutes so both reach the bath temperature. Read the temperature with a thermometer graduated to 0.5 C dipped in the solution, and mix only when the reading is within 0.5 C of the target. Use five temperatures: 20, 30, 40, 50 and 60 C. | 1 mark: thermostatted water bath, 5 min equilibration 1 mark: thermometer in the solution, read to within 0.5 C 1 mark: at least five temperatures between 20 and 60 C |
3 |
| 1 | (c) | Pipette 25.0 cm3 of sodium thiosulfate solution into a conical flask and add 25.0 cm3 of distilled water; stand the flask on white paper carrying a black cross. Measure 5.0 cm3 of dilute acid in a measuring cylinder and equilibrate it with the flask in the water bath. Pour the acid into the flask, start the stopwatch at the same moment, cover with a watch glass and swirl; look vertically down through the flask and record the time t when the cross just disappears. Repeat three times at each temperature and take the mean. | 1 mark: fixed volumes measured accurately 1 mark: acid added and stopwatch started together 1 mark: viewed vertically from above 1 mark: three repeats averaged at each temperature |
4 |
| 1 | (d) | Columns: temperature / C; T / K; t1, t2, t3 / s; mean t / s; 1/t or ln (1/t) / s-1; 1/T / K-1. | 1 mark: three timings plus a mean column 1 mark: T/K, 1/T and 1/t (or ln 1/t) with correct units |
2 |
| 1 | (e) | Plot ln (1/t) against 1/T, with 1/T in K-1. From the Arrhenius equation k = A exp(-Ea/RT), ln k = ln A – (Ea/R)(1/T); since the rate is proportional to 1/t, the gradient equals -Ea/R, so Ea = -gradient x R with R = 8.31 J K-1 mol-1, divided by 1000 for kJ mol-1. | 1 mark: ln (1/t) against 1/T 1 mark: gradient = -Ea/R 1 mark: Ea = -gradient x 8.31, converted to kJ mol-1 |
3 |
| 1 | (f) | Dilute hydrochloric acid is corrosive, so wear eye protection; the reaction releases sulfur dioxide, so work in a fume cupboard or a well-ventilated room; take care with the hot water bath. | 1 mark: any two (eye protection / ventilation / hot water) | 1 |
| 2 | (a) | T / K = 293, 303, 313, 323, 333; 103/T = 3.41, 3.30, 3.19, 3.10, 3.00 K-1; t = 95.00, 46.95, 24.27, 13.07, 7.30 s; ln (1/t) = -4.554, -3.849, -3.189, -2.570, -1.988. | 2 marks: all five T/K and 103/T values 2 marks: all five ln (1/t) values (3 s.f.) |
4 |
| 2 | (b) | Plot ln (1/t) on the y axis against 1/T in K-1 on the x axis, x from 3.00 x 10-3 to 3.50 x 10-3 and y from -4.6 to -1.9. The five points should lie close to a straight line; draw the best-fit line with a transparent ruler so the points scatter evenly about it, and re-measure any temperature whose point deviates by more than one small square. | 1 mark: 1/T on x, ln (1/t) on y, axes labelled with units 1 mark: suitable scales, five points plotted 1 mark: best-fit line drawn 1 mark: straightness judged from the scatter |
4 |
| 2 | (c) | Gradient = change in ln (1/t) / change in 1/T = (-1.988 – (-4.554)) / (3.00 x 10-3 – 3.41 x 10-3), which is about -6259 K, i.e. -6.25 x 103 K. | 1 mark: gradient from change in y / change in x 1 mark: unit K, negative sign kept |
2 |
| 2 | (d) | Ea = -gradient x R = 6259 x 8.31 = 52012 J mol-1, which is about 52.0 kJ mol-1. | 1 mark: Ea = -gradient x R substituted 1 mark: about 52 kJ mol-1 with the correct unit |
2 |
| 2 | (e) | The moment of disappearance is judged by eye, so different observers react at different times; use one observer, or a light sensor / time-lapse camera instead of the eye. There is a delay between pouring the acid and starting the stopwatch, and the temperature falls slightly on mixing; equilibrate in the bath, pour quickly and start the watch at once, and record the actual temperature after mixing. Each temperature is only timed once; take three repeats and more temperatures to reduce random error. | 1 mark: one valid source of error 1 mark: a second source 1 mark: a matching improvement |
3 |
Topic breakdown
| Question | Topic | Marks |
|---|---|---|
| 1 | Planning: effect of temperature on the rate of the sodium thiosulfate / acid reaction | 15 |
| 2 | Analysis and evaluation: activation energy from rate data | 15 |
Common lost marks
- Plotting V rather than ln V, which gives a curve and a gradient that is not 1/(RC).
- Omitting the unit s-1 for the gradient, or putting ln V into C = 1/(R x gradient) as if it were a voltage.
- Writing only “measure the frequency” in the planning question without saying how resonance is obtained.
- Counting only the 5% resistor tolerance and forgetting the 3% uncertainty in reading the gradient.
Frequently asked
What does Paper 5 test?
Paper 5 is the planning, analysis and evaluation paper: 1 hour 15 minutes, 30 marks. Question 1 asks you to design an experiment (variables, method, data handling, safety); Question 2 gives data for you to complete a table, plot a graph, find a gradient and an uncertainty, and evaluate the errors.
Can I do this paper without having done the experiment?
Yes. Paper 5 tests experimental design and data handling, not hands-on experience, but you must be fluent with graphs, gradients, logarithmic linearisation and percentage uncertainties.
Why plot ln V instead of V?
Taking logs of V = V0 exp(-t/RC) gives the straight line ln V = ln V0 – t/(RC), whose gradient is exactly -1/(RC) and so gives C. A plot of V against t is a curve from which the constant cannot be read easily.
This paper is written by TutorZone tutors from the public syllabus; all questions and data are original and not endorsed by CIE.
The answer booklet gives mark-scheme points for study use only; please credit the source when reusing.
