Chemistry 9701/33 — October/November 2020

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Qualitative Analysis · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsQualitative AnalysisAnalysis, Conclusions and EvaluationFree sample

In acidic solutions iron(III) ions are reduced by iodide ions to form iron(II) ions. The iodide ions are oxidised to iodine.

2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})

The rate of this reaction can be investigated by using starch indicator, which turns blue-black in the presence of iodine. Sodium thiosulfate is added to the reaction mixture to react with iodine as it is formed. The blue-black colour is seen when all the thiosulfate has reacted.

\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})

You will investigate how the rate of reaction is affected by changing the concentration of the iodide ions.

FA 1 is $0.0500\text{ mol dm}^{-3}$ potassium iodide, KI.
FA 2 is $0.0500\text{ mol dm}^{-3}$ acidified iron(III) chloride, $\text{FeCl}_3$ .
FA 3 is $0.00500\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
FA 4 is starch indicator.

(a)

Method

Prepare a table on page 4 for your results. You will need to include the volume of FA 1, volume of water, reaction time and rate of reaction for each of five experiments.

Experiment 1

Fill the burette labelled FA 1 with FA 1.
Run $20.00\text{ cm}^3$ of FA 1 into the $100\text{ cm}^3$ beaker.
Use the $50\text{ cm}^3$ $50 cm^{3}$ measuring cylinder to add the following to the same $100\text{ cm}^3$ $100 cm^{3}$ beaker:
- $20.0\text{ cm}^3$ of FA 3
- $10.0\text{ cm}^3$ of FA 4.
Use the $25\text{ cm}^3$ measuring cylinder to measure $10.0\text{ cm}^3$ of FA 2.
Add this FA 2 into the same $100\text{ cm}^3$ beaker and start timing immediately.
Stir once and place the beaker on the white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second.
Wash out the beaker and dry it with a paper towel.

Experiment 2

Fill the second burette with distilled water.
Run $10.00\text{ cm}^3$ of FA 1 into the $100\text{ cm}^3$ beaker.
Run $10.00\text{ cm}^3$ of distilled water into the beaker containing FA 1.
Use the $50\text{ cm}^3$ $50 cm^{3}$ measuring cylinder to add the following to the same $100\text{ cm}^3$ $100 cm^{3}$ beaker:
- $20.0\text{ cm}^3$ of FA 3
- $10.0\text{ cm}^3$ of FA 4.
Use the $25\text{ cm}^3$ measuring cylinder to measure $10.0\text{ cm}^3$ of FA 2.
Add the FA 2 to the same $100\text{ cm}^3$ beaker and start timing immediately.
Stir once and place the beaker on the white tile.
Stop timing as soon as the solution turns blue-black.
Record this reaction time to the nearest second.
Wash out the beaker and dry it with a paper towel.

Experiments 3–5

Carry out three further experiments to investigate how the reaction time changes with different volumes of potassium iodide, FA 1.
The combined volume of FA 1 and distilled water must always be $20.00\text{ cm}^3$ .
Do not use a volume of FA 1 that is less than $6.00\text{ cm}^3$ .

Results

The rate of reaction can be calculated as shown:

\text{rate} = \frac{1000}{\text{reaction time}}

10M

DifficultyMedium

Worked solution

Answer

Volume of FA 1 / cm³	Volume of water / cm³	Time / s	Rate / s⁻¹
20.00	0.00	45	22.22
15.00	5.00	80	12.50
12.00	8.00	125	8.00
10.00	10.00	180	5.56
8.00	12.00	281	3.56

Ratio of time (FA 1 = 10) to time (FA 1 = 20): $180 / 45 = 4.00$

Final answer

See working. Representative data provided.

Detailed explanation

Background Concept

This question investigates the rate of a redox reaction using a clock reaction method. The main reaction produces iodine slowly:
$2\text{Fe}^{3+}(\text{aq}) + 2\text{I}^-(\text{aq}) \rightarrow 2\text{Fe}^{2+}(\text{aq}) + \text{I}_2(\text{aq})$
A secondary, very fast reaction immediately consumes the iodine as it forms, using sodium thiosulfate:
$\text{I}_2(\text{aq}) + 2\text{S}_2\text{O}_3^{2-}(\text{aq}) \rightarrow 2\text{I}^-(\text{aq}) + \text{S}_4\text{O}_6^{2-}(\text{aq})$
Starch indicator is added to the mixture. It does not react with iodine until all the thiosulfate has been used up. At that moment, any newly formed iodine reacts with the starch to produce a sudden blue-black colour. The time taken for this colour change is a measure of the initial rate of the main reaction.

Understanding the Question

You are asked to set up a results table and record data for five experiments. The independent variable is the volume of potassium iodide (FA 1). To ensure a fair test, the total volume of the reaction mixture must remain constant. This is achieved by replacing the volume of FA 1 with an equal volume of distilled water. You must calculate the rate of reaction using the formula provided: $\text{rate} = \frac{1000}{\text{reaction time}}$ .

Approach

Design the table: Include columns for volume of FA 1, volume of water, time, and rate. Ensure units are clearly stated in the headings.
Select volumes: Start with 20.00 cm³ of FA 1 and 0.00 cm³ of water. Decrease FA 1 in steps (e.g., 15.00, 12.00, 10.00, 8.00 cm³), increasing the water volume to keep the sum at exactly 20.00 cm³.
Record times: In a real exam, you would record your own times. Here, we use representative times that reflect the expected kinetics (approximately second order in iodide, so halving the concentration roughly quadruples the time).
Calculate rates: Apply the formula $\text{rate} = \frac{1000}{t}$ and round to 2–4 significant figures.

Step-by-Step Reasoning

Experiment 1: 20.00 cm³ FA 1, 0.00 cm³ water. Total volume = 60.0 cm³. Suppose time = 45 s. Rate = $1000 / 45 = 22.22 \text{ s}^{-1}$ .
Experiment 2: 10.00 cm³ FA 1, 10.00 cm³ water. Concentration of iodide is halved. Suppose time = 180 s. Rate = $1000 / 180 = 5.56 \text{ s}^{-1}$ .
Experiments 3–5: Fill in intermediate and lower volumes (15.00, 12.00, 8.00 cm³), ensuring water volumes sum to 20.00 cm³. Calculate corresponding rates.
Ratio calculation: The mark scheme asks for the ratio of time at 10 cm³ to time at 20 cm³. $180 / 45 = 4.00$ . This confirms the rate is not simply inversely proportional to concentration (which would give a ratio of 2), but rather the rate is proportional to the square of the iodide concentration.

Key Takeaways

In clock reactions, the total volume must be kept constant by using water as a solvent to replace the variable reactant. This ensures that only the concentration of the varied reactant changes, not the concentrations of the other reagents.
Always include units in table headings (e.g., / cm³, / s, / s⁻¹).
Rates should be calculated to an appropriate number of significant figures (usually 2–4 sf).

Common Mistakes

Forgetting the water column: If you don't add water to keep the total volume at 20.00 cm³, you are changing the concentration of all reagents, not just the iodide ions.
Incorrect rate calculation: Using $\text{rate} = \frac{\text{time}}{1000}$ instead of $\frac{1000}{\text{time}}$ .
Missing units: Writing just numbers in the table without specifying cm³ or s in the headings.

Things to Be Careful About

Precision: Volumes from a burette should be recorded to 2 decimal places (e.g., 20.00 cm³). Volumes from measuring cylinders are typically recorded to 1 decimal place (e.g., 10.0 cm³), but for the FA 1/water mixture, use the burette precision (2 decimal places) to show careful measurement.
Time recording: Times must be recorded to the nearest second as whole numbers (e.g., 45 s, not 45.0 s).
Ratio calculation: Ensure you divide the larger time by the smaller time to get a value > 1. The mark scheme expects a ratio between 3.20 and 4.80 for this specific reaction system.

Techniques used

set up a results table for a clock reactioncalculate rate from reaction timemaintain constant total volume

(b)

On the grid opposite, plot a graph of rate of reaction ( $y$ -axis) against volume of FA 1 ( $x$ -axis). Include the origin, (0,0), in your scales. Circle any points you consider anomalous and draw a line of best fit.

Graph Grid

DifficultyMedium

Worked solution

Answer

Graph Description:

x-axis: Volume of FA 1 / cm³, scale 0 to 20 (e.g., 2 cm per major division).
y-axis: Rate of reaction / s⁻¹, scale 0 to 25 (e.g., 1 s⁻¹ per major division).
Points plotted: (20.00, 22.22), (15.00, 12.50), (12.00, 8.00), (10.00, 5.56), (8.00, 3.56).
Line of best fit: A smooth curve passing through or near all points, starting from the origin (0,0). The curve should have an increasing gradient (concave up), reflecting that rate increases more than proportionally with concentration.

Final answer

See graph description and diagram.

Detailed explanation

Background Concept

Plotting a graph of rate against concentration (or a proxy like volume, when total volume is constant) reveals the kinetic order of the reaction with respect to that reactant.

A straight line through the origin indicates first order (rate $\propto$ concentration).
A curve with an increasing gradient (parabolic shape) indicates second order (rate $\propto$ concentration²).

Understanding the Question

You must plot the rate of reaction (calculated in part a) on the y-axis against the volume of FA 1 on the x-axis. You must include the origin (0,0), identify any anomalous points, and draw a line of best fit.

Approach

Set up axes: Choose linear scales that use at least half the grid in both directions. Label axes clearly with quantity and unit.
Plot points: Plot the five (volume, rate) pairs from your table.
Identify anomalies: If any point is far from the trend, circle it. (In representative data, all points may fit well, but in a real exam, one might be slightly off).
Draw the line of best fit: Draw a smooth curve that minimizes the distance to all uncircled points. Since the ratio of times was ~4 when concentration halved, the graph will be a curve, not a straight line.

Step-by-Step Reasoning

Axis scales: x-axis from 0 to 20 cm³. y-axis from 0 to 25 s⁻¹. Both are linear.
Plotting:
- (20.00, 22.22)
- (15.00, 12.50)
- (12.00, 8.00)
- (10.00, 5.56)
- (8.00, 3.56)
Line of best fit: Connect the points with a smooth curve. The curve should pass close to (0,0) because at zero iodide concentration, the rate is zero. The shape will be concave upwards (like $y = kx^2$ ).

Key Takeaways

Always include the origin (0,0) if it is chemically meaningful (rate is zero when concentration is zero).
A curve indicates a non-linear relationship (higher order kinetics). A straight line indicates first-order kinetics.
Points must be plotted within half a small square of the correct position.

Common Mistakes

Non-linear scales: Using a logarithmic or square-root scale on a standard grid without adjusting the labels.
Drawing a straight line: Forcing a straight line through points that clearly form a curve, or drawing a straight line that doesn't pass near the origin.
Poor scaling: Making the graph too small (using only a quarter of the grid) or making the axes too large (points clustered in a tiny corner).

Things to Be Careful About

Line of best fit: Do not join the points with straight line segments (a 'zig-zag' line). Use a smooth curve or a straight line, depending on the trend.
Anomalous points: If you circle a point, your line of best fit must ignore it. Don't let one outlier pull the entire curve towards it.

Techniques used

plot a graph of rate against concentrationidentify anomalous resultsdraw a line of best fit

(c)

Use your graph to calculate the time that the reaction would have taken if $5.00\text{ cm}^3$ of FA 1 had been used. Show on the graph how you obtained your answer.

DifficultyMedium-Easy

Worked solution

Working

Draw a vertical line up from $x = 5.00$ cm³ on the x-axis to the line of best fit.
From the intersection, draw a horizontal line to the y-axis to read the rate.
From the graph, rate at 5.00 cm³ $\approx 1.40 \text{ s}^{-1}$ .
Calculate time: $\text{time} = \frac{1000}{\text{rate}} = \frac{1000}{1.40} = 714 \text{ s}$ .

Answer

Time $\approx 714 \text{ s}$ (Acceptable range: 700–730 s depending on graph reading).

Final answer

714 s

Detailed explanation

Background Concept

Graphs allow you to interpolate (find values within the range of your data) or extrapolate (find values outside the range). Here, 5.00 cm³ is outside the range of your experimental data (minimum was 8.00 cm³), so you are extrapolating. Extrapolation is less reliable than interpolation, but for a clear kinetic trend, it is acceptable.

Understanding the Question

You must find the reaction time if 5.00 cm³ of FA 1 was used. The graph gives the rate, so you must use the formula $\text{time} = \frac{1000}{\text{rate}}$ to convert it back to time.

Approach

Locate 5.00 on the x-axis.
Move vertically to the line of best fit.
Move horizontally to read the corresponding rate on the y-axis.
Calculate time using the given formula.

Step-by-Step Reasoning

Reading the graph: At $x = 5.00$ cm³, the curve is at approximately $y = 1.40 \text{ s}^{-1}$ .
Calculation: $\text{time} = \frac{1000}{1.40} = 714.28... \text{ s}$ .
Rounding: Round to 3 significant figures or the nearest whole number: 714 s.
Consistency check: At 8.00 cm³, time was 281 s. At 5.00 cm³, concentration is $5/8 = 0.625$ of the 8.00 cm³ case. If rate $\propto$ concentration², rate should be $(0.625)^2 = 0.39$ of the 8.00 cm³ rate. $0.39 \times 3.56 = 1.39 \text{ s}^{-1}$ . Time = $1000 / 1.39 = 719 \text{ s}$ . This is very close to our graph reading, confirming the extrapolation is valid.

Key Takeaways

Always show how you obtained the answer on the graph (draw the construction lines).
Remember to convert rate back to time using the inverse relationship.

Common Mistakes

Reading the wrong axis: Reading the x-value from the y-axis or vice versa.
Forgetting to invert: Writing the rate (1.40 s⁻¹) as the final answer instead of calculating the time.
Poor construction lines: Drawing lines that don't clearly show the reading process.

Things to Be Careful About

Extrapolation limits: Don't extrapolate too far beyond your data range. 5.00 cm³ is a reasonable extrapolation from 8.00 cm³, but 0.00 cm³ would be unreliable.
Significant figures: The answer should be to 2–4 sf or a whole number of seconds.

Techniques used

read a value from a graphcalculate time from rate

(d)

(i)

Using data from Experiments 1 and 2, show by calculation that the volume of aqueous potassium iodide, FA 1, used was directly proportional to the concentration of iodide ions.

DifficultyMedium

Worked solution

Answer

Experiment 1:
Total volume = $20.00 + 0.00 + 20.0 + 10.0 + 10.0 = 60.0 \text{ cm}^3$
Concentration of I⁻ = $\frac{20.00 \times 10^{-3} \times 0.0500}{60.0 \times 10^{-3}} = 1.67 \times 10^{-2} \text{ mol dm}^{-3}$

Experiment 2:
Total volume = $10.00 + 10.00 + 20.0 + 10.0 + 10.0 = 60.0 \text{ cm}^3$
Concentration of I⁻ = $\frac{10.00 \times 10^{-3} \times 0.0500}{60.0 \times 10^{-3}} = 0.83 \times 10^{-2} \text{ mol dm}^{-3}$

Conclusion:
$\frac{20.00}{10.00} = 2$ and $\frac{1.67 \times 10^{-2}}{0.83 \times 10^{-2}} \approx 2$
Since the ratio of volumes equals the ratio of concentrations, the volume of FA 1 is directly proportional to the concentration of iodide ions.

Final answer

See working.

Detailed explanation

Background Concept

When mixing solutions, the concentration of a solute in the final mixture depends on the volume of the stock solution used and the total final volume. The formula is:
$C_1 V_1 = C_2 V_2 \implies C_2 = \frac{C_1 V_1}{V_2}$
where $C_1$ and $V_1$ are the concentration and volume of the stock solution, and $C_2$ and $V_2$ are the concentration and total volume of the mixture.

Understanding the Question

You must prove that the volume of potassium iodide (FA 1) you add is directly proportional to the concentration of iodide ions in the final reaction mixture. You do this by calculating the concentration for Experiment 1 (20.00 cm³ FA 1) and Experiment 2 (10.00 cm³ FA 1).

Approach

Calculate the total volume of the reaction mixture for both experiments (it should be constant at 60.0 cm³).
Calculate the concentration of I⁻ in each experiment using $C = \frac{n}{V_{\text{total}}}$ .
Compare the ratio of volumes to the ratio of concentrations.

Step-by-Step Reasoning

Total volume: In both experiments, the sum of all reagents is $20.00 \text{ (FA 1 + water)} + 20.0 \text{ (FA 3)} + 10.0 \text{ (FA 4)} + 10.0 \text{ (FA 2)} = 60.0 \text{ cm}^3$ .
Expt 1 concentration: Moles of I⁻ = $20.00 \times 10^{-3} \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 1.00 \times 10^{-3} \text{ mol}$ . Concentration = $\frac{1.00 \times 10^{-3}}{60.0 \times 10^{-3}} = 0.0167 \text{ mol dm}^{-3}$ .
Expt 2 concentration: Moles of I⁻ = $10.00 \times 10^{-3} \text{ dm}^3 \times 0.0500 \text{ mol dm}^{-3} = 5.00 \times 10^{-4} \text{ mol}$ . Concentration = $\frac{5.00 \times 10^{-4}}{60.0 \times 10^{-3}} = 0.00833 \text{ mol dm}^{-3}$ .
Proportionality: Volume ratio = $20.00 / 10.00 = 2$ . Concentration ratio = $0.0167 / 0.00833 = 2.00$ . Since the ratios are equal, they are directly proportional.

Key Takeaways

Always calculate the total volume of the mixture, not just the volume of the reactant.
Direct proportionality means $\frac{y_1}{y_2} = \frac{x_1}{x_2}$ .

Common Mistakes

Using the wrong total volume: Forgetting to add the volumes of FA 2, FA 3, and FA 4.
Calculating moles only: The mark scheme allows marks for calculating moles if concentration isn't fully shown, but concentration is the direct proof.

Things to Be Careful About

Units: Convert cm³ to dm³ when calculating concentration, or keep both in cm³ as long as you are consistent (the $10^{-3}$ cancels out).
Significant figures: Use at least 3 sf for intermediate concentration values.

Techniques used

calculate concentration from volume and total volumeverify direct proportionality

(ii)

Explain, by referring to your graph or your table of results, how the rate of reaction is affected by an increase in the concentration of aqueous potassium iodide, FA 1.

DifficultyMedium

Worked solution

Answer

Observation from graph/table:

The graph is a curve with an increasing gradient (not a straight line through the origin).
Alternatively, comparing Experiment 1 and 2: the concentration of iodide ions was halved (ratio 2:1), but the rate decreased by a factor of 4 (ratio 1:4, or time increased by factor of 4).

Explanation:
The rate of reaction is not directly proportional to the concentration of iodide ions. As the concentration of iodide ions increases, the rate increases more than proportionally (the rate is proportional to the square of the iodide concentration). This is shown by the curve having a steeper gradient at higher concentrations.

Final answer

See explanation.

Detailed explanation

Background Concept

The rate law for a reaction is $\text{rate} = k[A]^m[B]^n$ . The exponent $m$ is the order with respect to A.

If $m=1$ , rate $\propto$ [A]. Graph of rate vs [A] is a straight line.
If $m=2$ , rate $\propto$ [A]². Graph of rate vs [A] is a parabola (curve with increasing gradient).

Understanding the Question

You must explain how the rate changes when the concentration of potassium iodide increases, using evidence from your graph or table.

Approach

Describe the shape of the graph or compare the ratios from the table.
State the conclusion about the relationship (directly proportional vs. not directly proportional / second order).

Step-by-Step Reasoning

Graph evidence: The line of best fit is a smooth curve curving upwards. A straight line would indicate first order. The curve indicates that as x (concentration) increases, y (rate) increases at an accelerating rate.
Table evidence: In Expt 1 and 2, volume of FA 1 (and thus [I⁻]) was halved (20.00 to 10.00). The time increased from 45 s to 180 s (factor of 4). Since rate $\propto 1/\text{time}$ , the rate decreased by a factor of 4. Halving concentration reduces rate by a factor of 4 ( $2^2 = 4$ ), proving second-order kinetics.
Conclusion: The rate is not directly proportional to the concentration of iodide ions; it is proportional to the square of the concentration.

Key Takeaways

A curve on a rate vs concentration graph means the reaction is not first order in that reactant.
Always back up your explanation with specific data (ratios from the table or shape of the graph).

Common Mistakes

Saying 'directly proportional': If the graph is a curve, it is not directly proportional.
Vague explanations: Just saying 'the rate increases' is not enough; you must say how it increases (more than proportionally, or give the factor).

Things to Be Careful About

Line through origin: If you drew a straight line, it must pass within 5 small squares of (0,0) to be considered directly proportional. If it doesn't, or if it's a curve, it's not directly proportional.

Techniques used

interpret a rate vs concentration graphdetermine kinetic order from graph shape

(e)

Thiosulfate ions can reduce iron(III) ions and also react with acid to form sulfur, sulfur dioxide and water.

(i)

Write an ionic equation for the reaction between thiosulfate ions and hydrogen ions in aqueous solution. Include state symbols.

DifficultyMedium-Easy

Worked solution

Answer

$\text{S}_2\text{O}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{S}(\text{s}) + \text{SO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l})$

Final answer

S2O3^2-(aq) + 2H^+(aq) -> S(s) + SO2(aq) + H2O(l)

Detailed explanation

Background Concept

Sodium thiosulfate ( $\text{Na}_2\text{S}_2\text{O}_3$ ) is unstable in acidic conditions. It decomposes to form a pale yellow precipitate of sulfur ( $\text{S}$ ) and sulfur dioxide gas ( $\text{SO}_2$ ), which can dissolve in water to form sulfurous acid. This reaction is often used as a clock reaction (the 'cross and thiosulfate' experiment) because the sulfur precipitate makes the solution cloudy, obscuring a marked cross on paper beneath the flask.

Understanding the Question

You are asked to write the ionic equation for the reaction between thiosulfate ions and hydrogen ions (from the acidified FA 2). You must include state symbols.

Approach

Identify reactants: $\text{S}_2\text{O}_3^{2-}$ and $\text{H}^+$ .
Identify products: $\text{S}$ , $\text{SO}_2$ , $\text{H}_2\text{O}$ .
Balance the equation for mass and charge.

Step-by-Step Reasoning

Reactants: $\text{S}_2\text{O}_3^{2-}(\text{aq})$ and $\text{H}^+(\text{aq})$ .
Products: Sulfur is a solid precipitate ( $\text{s}$ ). Sulfur dioxide is a gas but can be aqueous in solution ( $\text{aq}$ or $\text{g}$ ). Water is liquid ( $\text{l}$ ).
Balancing:
- Sulfur: 2 on left, 1 in S + 1 in SO₂ = 2 on right. Balanced.
- Oxygen: 3 on left, 2 in SO₂ + 1 in H₂O = 3 on right. Balanced.
- Hydrogen: Need 2 H⁺ to form 1 H₂O.
- Charge: Left = -2 + 2(+1) = 0. Right = 0. Balanced.
Final equation: $\text{S}_2\text{O}_3^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) \rightarrow \text{S}(\text{s}) + \text{SO}_2(\text{aq}) + \text{H}_2\text{O}(\text{l})$

Key Takeaways

Thiosulfate decomposes in acid to give sulfur (yellow ppt) and sulfur dioxide.
Always include state symbols: (aq) for ions, (s) for sulfur precipitate, (l) for water.

Common Mistakes

Wrong products: Writing $\text{SO}_4^{2-}$ instead of $\text{SO}_2$ .
Missing state symbols: Forgetting (s) for sulfur or (aq) for ions.
Unbalanced equation: Not balancing the hydrogen or charge.

Things to Be Careful About

SO2 state: $\text{SO}_2$ can be written as (g) or (aq). In this aqueous mixture, (aq) is often preferred, but (g) is also accepted by CIE.
Charge balance: Ensure the total charge on the left equals the total charge on the right (both should be 0 here).

Techniques used

write an ionic equation for acid decompositioninclude state symbols

(ii)

A student carries out the same investigation as in (a) but the solutions are mixed in a different order. The student places FA 1 and an appropriate volume of distilled water in one beaker and all the other reactants in a second beaker. The student then transfers the mixture from the second beaker to the first and starts timing.

Tick the box for the statement you consider correct. Explain your answer.

The student's method is better than that in (a).
The two methods are equally good.
The student's method is not as good as that in (a).

reason

DifficultyMedium

Worked solution

Answer

Correct statement: The student's method is not as good as that in (a).

Reason:
In the student's method, FA 2 (acidified iron(III)) is mixed with FA 3 (thiosulfate) and FA 4 (starch) before the reaction is started. Iron(III) ions can oxidise thiosulfate ions, and thiosulfate also reacts with acid. This side reaction will consume some of the thiosulfate before the main reaction with iodide begins. As a result, less thiosulfate is available to react with the iodine produced, meaning the blue-black colour will appear earlier (time will be shorter) than it should be, leading to an inaccurate rate.

(Alternative acceptable answer: The student's method is better because it allows all reagents to be transferred to the first beaker at once, reducing the time taken to mix and ensuring the timer starts at the exact moment of mixing. However, the 'not as good' answer is the primary expected response due to the side reaction issue.)

Final answer

Not as good; thiosulfate reacts with Fe3+ or acid before the main reaction starts, consuming it and reducing the time.

Detailed explanation

Background Concept

In a clock reaction, the thiosulfate acts as a limiting reagent that is consumed at a known, constant rate by the iodine produced. The time measured is the time taken to consume all the thiosulfate. If any thiosulfate is consumed by other reactions (side reactions) before the main clock starts, the measured time will be artificially short, and the calculated rate will be artificially high.

Understanding the Question

The student mixes FA 1 (iodide) and water in one beaker, and FA 2 (iron(III)), FA 3 (thiosulfate), and FA 4 (starch) in a second beaker. Then they pour the second mixture into the first and start timing.
In the original method (a), FA 2 is added last, immediately before timing starts. This ensures the clock only starts when the main reactants are combined.

Approach

Identify what happens when FA 2 and FA 3 are mixed before the main reaction.
Consider the side reactions mentioned in part (e)(i) and the redox chemistry.
Conclude whether this affects the accuracy of the time measurement.

Step-by-Step Reasoning

Side reaction 1: Iron(III) is an oxidising agent. Thiosulfate is a reducing agent. $2\text{Fe}^{3+} + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{Fe}^{2+} + \text{S}_4\text{O}_6^{2-}$ . This consumes thiosulfate.
Side reaction 2: Thiosulfate reacts with acid (from acidified FA 2) to form sulfur and SO₂ (as shown in e(i)). This also consumes thiosulfate.
Effect: By the time the student pours the mixture and starts the timer, some thiosulfate has already been destroyed. The amount of thiosulfate left to react with iodine is less than the intended amount. Therefore, the iodine will turn the solution blue-black sooner. The recorded time will be too short, and the rate too high.
Conclusion: The student's method is not as good because it introduces a systematic error (thiosulfate loss) before the reaction is properly timed.

Key Takeaways

In clock reactions, timing must start the moment the main reactants are combined. Pre-mixing reagents that can react with the 'clock' reagent (thiosulfate) will invalidate the results.
Always consider side reactions when evaluating methods.

Common Mistakes

Saying 'better' without reason: While transferring all reagents at once can be faster, the chemical error (thiosulfate consumption) outweighs the timing benefit.
Not mentioning the specific side reaction: Vague answers like 'they react' don't score. You must specify that thiosulfate reacts with iron(III) or acid.

Things to Be Careful About

Mark scheme allowances: The mark scheme does accept 'same' or 'better' if the reason is valid (e.g., 'concentration of thiosulfate is very small so side reaction is negligible' or 'quicker to mix all at once'). However, the most chemically sound answer is 'not as good' due to the side reactions.

Techniques used

evaluate experimental methodidentify sources of error in mixing

(f)

Another student investigates the effect of iron(III) concentration on the rate of this reaction. The student carries out another experiment, Experiment 6, and the rate is compared to that of Experiment 2. In Experiment 2, the volumes used were:

reagent	volume / $\text{cm}^3$
FA 1	10.00
FA 2	10.0
FA 3	20.0
FA 4	10.0
distilled water	10.00

(i)

Suggest the volumes the student could use for Experiment 6.

reagent	volume / $\text{cm}^3$
FA 1
FA 2
FA 3
FA 4
distilled water

DifficultyMedium

Worked solution

Answer

reagent	volume / cm³
FA 1	10.00
FA 2	15.0
FA 3	20.0
FA 4	10.0
distilled water	5.00

(Any volumes where FA 1, FA 3, FA 4 are unchanged from Experiment 2, and FA 2 + water = 20.00 cm³, with FA 2 ≠ 10.00 cm³)

Final answer

See table. FA 1=10.00, FA 3=20.0, FA 4=10.0, FA 2=15.0, water=5.00.

Detailed explanation

Background Concept

To investigate the effect of iron(III) concentration on the rate, you must change the volume of FA 2 (iron(III)) while keeping everything else constant. This is a control of variables technique. The total volume of the mixture must remain constant (60.0 cm³) so that the concentrations of the other reagents (iodide, thiosulfate) don't change.

Understanding the Question

Experiment 2 used: 10.00 cm³ FA 1, 10.0 cm³ FA 2, 20.0 cm³ FA 3, 10.0 cm³ FA 4, 10.00 cm³ water.
You need to propose volumes for Experiment 6 to change the iron(III) concentration. This means changing FA 2 and adjusting the water volume to keep the total constant.

Approach

Keep FA 1, FA 3, FA 4 exactly the same as Experiment 2.
Change the volume of FA 2 (e.g., increase to 15.0 cm³ or decrease to 5.0 cm³).
Calculate the new water volume: Water = 20.00 - FA 2 (since FA 1 + water must be 20.00 cm³ to keep the total volume 60.0 cm³).

Step-by-Step Reasoning

Constant volumes: FA 1 = 10.00 cm³, FA 3 = 20.0 cm³, FA 4 = 10.0 cm³.
Variable volume: Let FA 2 = 15.0 cm³ (an increase from 10.0 cm³).
Water volume: The sum of FA 1 + water must be 20.00 cm³ (to match the total volume of the other reagents + FA 2 + water = 60.0 cm³). So water = 20.00 - 15.0 = 5.00 cm³.
Check total: 10.00 + 15.0 + 20.0 + 10.0 + 5.00 = 60.0 cm³. Correct.

Key Takeaways

When investigating the effect of one reactant, vary only that reactant's volume and use water to compensate.
The total volume must be constant to ensure fair testing.

Common Mistakes

Changing FA 1: If you change FA 1, you are investigating iodide concentration, not iron(III).
Forgetting to adjust water: If you increase FA 2 to 15.0 cm³ but keep water at 10.00 cm³, the total volume becomes 65.0 cm³, diluting all other reagents and ruining the experiment.

Things to Be Careful About

FA 2 + water = 20.00 cm³: This is the key constraint. FA 1 is fixed at 10.00 cm³, so water must adjust to keep the sum 20.00 cm³.
Significant figures: Record FA 2 and water to 1 or 2 decimal places as appropriate (burette or measuring cylinder).

Techniques used

design an experiment to test effect of one variablecontrol variables in kinetics

(ii)

This student records a time of $178\text{ s}$ for Experiment 2.

The rate of reaction is directly proportional to the concentration of iron(III) ions.

Suggest how long it would take the reaction mixture proposed for Experiment 6 in (f)(i) to turn blue-black. Assume that Experiment 6 is carried out at the same temperature as Experiment 2.

Do not carry out Experiment 6.

DifficultyMedium

Worked solution

Working

Rate is directly proportional to concentration of iron(III) ions.
Since total volume is constant, rate $\propto$ volume of FA 2.

$\frac{\text{rate}_6}{\text{rate}_2} = \frac{V_{\text{FA 2, 6}}}{V_{\text{FA 2, 2}}}$

Since $\text{rate} = \frac{1000}{t}$ :

$\frac{t_2}{t_6} = \frac{V_{\text{FA 2, 6}}}{V_{\text{FA 2, 2}}}$

$t_6 = t_2 \times \frac{V_{\text{FA 2, 2}}}{V_{\text{FA 2, 6}}}$

Using the proposed volumes from (f)(i): $V_{\text{FA 2, 6}} = 15.0 \text{ cm}^3$ , $V_{\text{FA 2, 2}} = 10.0 \text{ cm}^3$ .

$t_6 = 178 \times \frac{10.0}{15.0} = 118.67 \text{ s} \approx 119 \text{ s}$

Answer

Time for Experiment 6 $\approx 119 \text{ s}$

Final answer

119 s

Detailed explanation

Background Concept

If the rate of reaction is directly proportional to the concentration of a reactant (first order), then:
$\text{rate} \propto [\text{reactant}]$
Since concentration $= \frac{\text{moles}}{\text{total volume}}$ and total volume is constant, concentration is proportional to the volume of the stock solution added. Therefore:
$\text{rate} \propto V_{\text{stock}}$
Also, for a clock reaction where the amount of 'clock' reagent (thiosulfate) is constant:
$\text{rate} \propto \frac{1}{\text{time}}$
Combining these:
$\frac{1}{t} \propto V_{\text{stock}} \implies t \propto \frac{1}{V_{\text{stock}}}$

Understanding the Question

You are told the rate is directly proportional to [Fe³⁺]. Experiment 2 took 178 s with 10.0 cm³ of FA 2. You must calculate the time for Experiment 6 (which has 15.0 cm³ of FA 2, based on your answer in f(i)).

Approach

Establish the relationship: $t_1 V_1 = t_2 V_2$ (inverse proportionality between time and volume/concentration).
Substitute the known values.
Calculate the new time.

Step-by-Step Reasoning

Proportionality: Rate $\propto$ [Fe³⁺] $\propto$ $V_{\text{FA 2}}$ . Time $\propto 1/\text{rate} \propto 1/V_{\text{FA 2}}$ .
Equation: $t_2 \times V_{\text{FA 2, 2}} = t_6 \times V_{\text{FA 2, 6}}$
Substitution: $178 \text{ s} \times 10.0 \text{ cm}^3 = t_6 \times 15.0 \text{ cm}^3$
Calculation: $t_6 = \frac{178 \times 10.0}{15.0} = 118.67 \text{ s}$
Rounding: Round to 3 significant figures: 119 s.

Key Takeaways

For first-order reactions in a clock experiment, time is inversely proportional to the volume of the varied reactant (when total volume is constant).
Always use the formula $t_1 V_1 = t_2 V_2$ for quick calculations.

Common Mistakes

Direct proportionality error: Thinking time is directly proportional to volume (i.e., $t_6 = 178 \times 1.5 = 267 \text{ s}$ ). This is wrong; higher concentration means faster rate, so shorter time.
Using wrong volumes: Using the volume of FA 1 or water instead of FA 2.

Things to Be Careful About

Consistency with (f)(i): Your calculation must use the volume of FA 2 you proposed in part (f)(i). If you proposed 15.0 cm³, use 15.0 in the calculation.
Units: Time is in seconds, volume in cm³. The units cancel correctly in the ratio.

Techniques used

use rate proportionality to calculate timeapply stoichiometric ratios to kinetics

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Q2Qualitative Analysis · Manipulation, Measurement and Observation16M

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