0620/51 May/June 2025 Worked Answers (IGCSE Chemistry 0620 Extended)

21 questions · 40 marks · 75 minutes

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1. Question 1(a)

   4 marksEffect of concentration on reaction time
   Step 1: A higher concentration of acid has more acid particles in the same volume, so there are more frequent collisions and the reaction is faster. Step 2: As the concentration falls from 2.0 to 0.5 mol/dm³ the reaction gets slower, so the magnesium takes longer to disappear. Step 3: So the time increases as the concentration decreases, and each time is recorded in whole seconds.
   Method:

   Link concentration to collision frequency, then to rate, then to the time recorded.
   Examiner tips

   • Concentration up = rate up = time down; concentration down = rate down = time up.

2. Question 1(b)

   4 marksPlotting a graph of reaction time against concentration
   Step 1: A good graph uses a linear scale (equal steps) chosen so the plotted points fill more than half of the available grid. Step 2: Every result is plotted accurately, then a single smooth line of best fit is drawn through the trend. Step 3: So the correct approach is a linear y-axis scale filling over half the grid, all points plotted accurately and one line of best fit.
   Method:

   Recall the rules for a good line graph: linear scale, points spread out, single best-fit line.
   Examiner tips

   • Choose a scale so the points use more than half the grid; draw one best-fit line.

3. Question 1(c)

   2 marksReading a value by interpolation from a best-fit line
   Step 1: To read an intermediate value, find 1.3 mol/dm³ on the concentration axis. Step 2: Draw a vertical construction line up to the line of best fit, then a horizontal line across to the time axis. Step 3: The value where the horizontal line meets the time axis is the answer, so you draw a vertical line up from 1.3 then read horizontally across to the time axis.
   Method:

   Use construction lines to interpolate the time at the stated concentration.
   Examiner tips

   • Interpolation: up to the line, then across to the axis.

4. Question 1(d)(i)

   2 marksCalculating mean rate of reaction with units
   Step 1: Mean rate = length ÷ time = 5 cm ÷ 20 s. Step 2: 5 ÷ 20 = 0.25, and the unit is length over time, so cm/s. Step 3: So the mean rate of reaction is 0.25 cm/s.
   Method:

   Substitute the 5 cm length and the time into rate = length ÷ time and attach the unit cm/s.
   Examiner tips

   • rate = length ÷ time; keep the unit as cm/s.

5. Question 1(d)(ii)

   1 marksIdentifying the slowest reaction
   Step 1: A lower concentration of acid gives fewer collisions per second and so a slower reaction. Step 2: Experiment 5 uses the lowest concentration, 0.5 mol/dm³. Step 3: So the mean rate of reaction is slowest in Experiment 5.
   Method:

   Match the slowest rate to the experiment with the lowest acid concentration.
   Examiner tips

   • Lowest concentration = slowest mean rate.

6. Question 1(e)

   1 marksReason for repeating an experiment
   Step 1: Repeating an experiment lets you compare the results from each run. Step 2: Close (concordant) results show the data is repeatable, and any value that differs a lot can be spotted as an anomaly and left out. Step 3: So repeating each experiment lets you check the results are repeatable and identify anomalous values.
   Method:

   Recall why repeat readings improve the reliability of experimental data.
   Examiner tips

   • Repeats improve reliability and reveal anomalies.

7. Question 1(f)(i)

   2 marksCalculating a temperature change
   Step 1: The temperature change is the final temperature minus the initial temperature. Step 2: That is 35.0 °C − 21.0 °C = 14.0 °C, written to the same one decimal place as the readings. Step 3: So the temperature change is 14.0 °C.
   Method:

   Subtract the initial temperature from the final temperature, keeping consistent decimal places.
   Examiner tips

   • Always do final minus initial, keeping the same decimal places.

8. Question 1(f)(ii)

   1 marksWhy temperature must be controlled in a rate experiment
   Step 1: Temperature is a factor that affects the rate of reaction: a higher temperature speeds the reaction up. Step 2: If the temperature varied between experiments, the rate would change for that reason as well as the concentration, so the comparison would be unfair. Step 3: So keeping the temperature constant is an improvement because a change in temperature would change the rate of reaction.
   Method:

   Recall that temperature is a rate factor that must be controlled to keep the test fair.
   Examiner tips

   • Control temperature so only the concentration is varied.

9. Question 1(f)(iii)

   1 marksWhy an insulated cup does not keep temperature constant
   Step 1: The reaction of magnesium with acid is exothermic, so it releases heat. Step 2: Polystyrene is a good insulator, so it prevents that heat escaping to the surroundings. Step 3: With the heat trapped, the temperature still rises, so a polystyrene cup does not keep the temperature constant because it prevents heat escaping.
   Method:

   Link the insulating property of polystyrene to trapped heat and a rising temperature.
   Examiner tips

   • Insulation traps heat, so the temperature is not controlled.

10. Question 1(f)(iv)

    1 marksUsing a water bath to control temperature
    Step 1: To keep a temperature constant you surround the reaction vessel with something held at that temperature. Step 2: A water bath kept at a fixed temperature absorbs or supplies heat to hold the acid at that temperature. Step 3: So you stand the conical flask in a water bath kept at a constant temperature.
    Method:

    Recall that a water bath at a fixed temperature is used to control temperature.
    Examiner tips

    • Use a thermostatically controlled water bath to fix the temperature.

11. Question 2(a)(i)

    3 marksObservations when acid reacts with a carbonate
    Step 1: A carbonate reacts with dilute acid to give a salt, water and carbon dioxide, so you see effervescence as the gas bubbles off and a colourless solution as the solid dissolves. Step 2: Carbon dioxide turns limewater milky. Step 3: So the observations are effervescence, a colourless solution as the solid dissolves, and the gas turning limewater milky.
    Method:

    Apply the acid + carbonate reaction and the limewater test for the gas.
    Examiner tips

    • Carbonate + acid → effervescence + colourless solution + CO₂ (limewater milky).

12. Question 2(a)(ii)

    1 marksIdentifying a gas that turns limewater milky
    Step 1: Limewater turning milky is the standard test for carbon dioxide. Step 2: The other gases give different results, for example hydrogen pops and oxygen relights a splint. Step 3: So the gas that turns limewater milky is carbon dioxide.
    Method:

    Match the limewater-milky test result to its gas.
    Examiner tips

    • Carbon dioxide is the only common gas that turns limewater milky.

13. Question 2(b)(i)

    1 marksTesting for sulfate ions with barium nitrate
    Step 1: Acidified barium nitrate gives a white precipitate of barium sulfate if sulfate ions are present. Step 2: Here no precipitate forms, so there are no sulfate ions in solution H. Step 3: So the observation is no change, showing that sulfate ions are absent.
    Method:

    Apply the barium nitrate test for sulfate and interpret the absence of a precipitate.
    Examiner tips

    • Barium nitrate + sulfate = white ppt; no ppt = no sulfate.

14. Question 2(b)(ii)

    2 marksEffect of sodium hydroxide on a calcium salt solution
    Step 1: Adding sodium hydroxide to a calcium salt gives a white precipitate of calcium hydroxide. Step 2: Calcium hydroxide is insoluble in excess sodium hydroxide, so the precipitate stays when more is added. Step 3: So a white precipitate forms dropwise and it remains insoluble in excess.
    Method:

    Match the white-precipitate-insoluble-in-excess result to a calcium ion.
    Examiner tips

    • White ppt, insoluble in excess NaOH = calcium (or magnesium).

15. Question 2(b)(iii)

    1 marksReaction of a calcium salt with sodium carbonate
    Step 1: Calcium ions react with carbonate ions to form calcium carbonate, which is insoluble. Step 2: The insoluble calcium carbonate appears as a white precipitate. Step 3: So adding sodium carbonate to solution H gives a white precipitate.
    Method:

    Apply the rule that calcium carbonate is an insoluble white solid.
    Examiner tips

    • Most carbonates are insoluble, so calcium + carbonate gives a white precipitate.

16. Question 2(c)

    2 marksIdentifying a solid from its test results
    Step 1: Fizzing with acid to give carbon dioxide (limewater milky) shows a carbonate ion. Step 2: A white precipitate with sodium hydroxide that is insoluble in excess shows a calcium ion. Step 3: Combining a calcium ion with a carbonate ion gives calcium carbonate, $\text{CaCO}_3$.
    Method:

    Deduce the anion and cation from the tests, then name the compound.
    Examiner tips

    • Identify the cation and the anion separately, then combine them.

17. Question 2(d)

    1 marksFlame test colour for potassium
    Step 1: Each metal ion gives a characteristic flame colour. Step 2: Potassium ions give a lilac flame. Step 3: So the flame colour for solid G is lilac.
    Method:

    Recall the flame colour produced by potassium ions.
    Examiner tips

    • Flame colours: Na yellow, K lilac, Ca orange-red, Cu blue-green.

18. Question 2(e)(i)

    1 marksSilver nitrate test giving a cream precipitate
    Step 1: Acidified silver nitrate gives a precipitate whose colour identifies the halide: white for chloride, cream for bromide, yellow for iodide. Step 2: Solution G gives a cream precipitate, which corresponds to a bromide ion. Step 3: So the observation is a cream precipitate.
    Method:

    Match the silver nitrate precipitate colour to the halide ion present.
    Examiner tips

    • Silver nitrate: chloride white, bromide cream, iodide yellow.

19. Question 2(e)(ii)

    1 marksDisplacement of bromine by chlorine
    Step 1: Chlorine is more reactive than bromine, so it displaces bromide ions from solution. Step 2: The displaced bromine colours the solution yellow to orange-brown. Step 3: So adding aqueous chlorine to the bromide solution turns it yellow to orange-brown.
    Method:

    Apply the halogen reactivity order to predict displacement and the colour change.
    Examiner tips

    • A more reactive halogen displaces a less reactive one from its salt.

20. Question 2(f)

    2 marksIdentifying a solid from flame and halide tests
    Step 1: A lilac flame shows a potassium ion. Step 2: A cream precipitate with silver nitrate shows a bromide ion. Step 3: Combining a potassium ion with a bromide ion gives potassium bromide, KBr.
    Method:

    Deduce the cation from the flame test and the anion from the silver nitrate test, then name the salt.
    Examiner tips

    • Identify the metal from the flame test and the halide from the silver nitrate colour.

21. Question 3

    6 marksPlanning a comparison of two endothermic reactions
    Step 1: To compare fairly, use the same known mass of sodium hydrogencarbonate and the same measured volume of each acid, measured with suitable apparatus in a suitable container. Step 2: Measure the temperature of the acid before mixing, add and stir, then measure the temperature after the reaction. Step 3: Endothermic reactions take in heat, so the temperature falls; the reaction with the larger temperature decrease is the more endothermic.
    Method:

    Design a fair test that measures the temperature change and links the larger decrease to the more endothermic reaction.
    Examiner tips

    • Fair test: same mass, same volume; the bigger temperature decrease is more endothermic.