Chemistry 9701/36 — October/November 2015

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

questions

marks

120

minutes

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

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

You will investigate the rate of reaction between iron(III) ions, $\text{Fe}^{3+}$ , and iodide ions, $\text{I}^-$ .

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

The iodine, $\text{I}_2$ , produced can be reacted immediately with thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ .

\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})

When all the thiosulfate has been used, the iodine produced will turn starch indicator blue-black. The rate of the reaction can therefore be measured by finding the time for the blue-black colour to appear.

FB 1 is aqueous iron(III) chloride, $\text{FeCl}_3$ .
FB 2 is aqueous potassium iodide, $\text{KI}$ .
FB 3 is $0.0060\text{ mol dm}^{-3}$ sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3$ .
starch indicator

You are advised to read the instructions before starting any practical work and draw a table for your results in the space on page 3.

(a)

Method

Experiment 1

Fill a burette with FB 1.
Run $20.00\text{ cm}^3$ of FB 1 into a $100\text{ cm}^3$ beaker.
Use the measuring cylinder to place the following in a second $100\text{ cm}^3$ $100 cm^{3}$ beaker.
- $10\text{ cm}^3$ of FB 2
- $20\text{ cm}^3$ of FB 3
- $10\text{ cm}^3$ of starch indicator
Add the contents of the second beaker to the first beaker and start timing.
Stir the mixture once and place the beaker on the white tile.
The mixture turns brown and then yellow before turning a blue-black colour. Stop timing when this blue-black colour appears.
Record in your table the volume of FB 1 used, the volume of distilled water used and the time to the nearest second for the blue-black colour to appear.
Wash both beakers.

For each of Experiments 2-6 you should complete your results table to show the volume of FB 1 used, the volume of distilled water used and the time taken to the nearest second for the blue-black colour to appear.

Experiment 2

Fill the other burette with distilled water.
Run $10.00\text{ cm}^3$ of FB 1 into a $100\text{ cm}^3$ beaker.
Run $10.00\text{ cm}^3$ of distilled water into the same beaker.
Use the measuring cylinder to place the following in a second $100\text{ cm}^3$ $100 cm^{3}$ beaker.
- $10\text{ cm}^3$ of FB 2
- $20\text{ cm}^3$ of FB 3
- $10\text{ cm}^3$ of starch indicator
Add the contents of the second beaker to the first beaker and start timing.
Stir the mixture once and place the beaker on the white tile.
Stop timing when a blue-black colour appears.
Wash both beakers.

Experiments 3-6

Carry out four further experiments to investigate the effect of changing the concentration of $\text{Fe}^{3+}(\text{aq})$ by altering the volume of aqueous $\text{FeCl}_3$ , FB 1, used.

You should not use a volume of FB 1 that is less than $6.00\text{ cm}^3$ and the total volume of the reaction mixture must always be $60\text{ cm}^3$ .

DifficultyMedium

Worked solution

Answer

Construct a single table with the following headings and units:

Experiment	Volume of FB 1 / cm³	Volume of water / cm³	Time / s
1	20.00	0.00	t₁
2	10.00	10.00	t₂
3	12.00	8.00	t₃
4	14.00	6.00	t₄
5	8.00	12.00	t₅
6	6.00	14.00	t₆

All volumes must be recorded to 0.05 cm³ (e.g., 20.00, 10.00, 12.00).
All times must be recorded to the nearest second.
The volume of FB 1 ranges from 6.00 cm³ to 20.00 cm³, with intervals of at least 2 cm³.
The total volume of FB 1 and water is constant at 20.00 cm³ in each experiment.
Times must increase as the volume (and concentration) of FB 1 decreases (t₆ > t₅ > t₄ > t₂ > t₁).

Final answer

See working

Detailed explanation

Background Concept

In kinetics experiments, the rate of reaction is often investigated by varying the concentration of one reactant while keeping the concentrations of all other reactants constant. To do this accurately, the total volume of the reaction mixture must be kept constant by adding a solvent (usually distilled water). This ensures that any change in rate is due solely to the change in concentration of the variable reactant, not a change in the total volume.

When recording data, precision is critical. Volumes measured from a burette are read to 0.05 cm³ (e.g., 20.00 cm³), while times from a stopwatch are typically recorded to the nearest second. A well-designed results table must clearly state the quantities being measured, their units, and the required precision.

Understanding the Question

The question asks you to design and record results for a clock reaction investigating the effect of Fe³⁺ concentration on the rate of reaction. You are given the method for Experiment 1 (20.00 cm³ of FB 1) and Experiment 2 (10.00 cm³ of FB 1 + 10.00 cm³ water). You must plan four further experiments (3-6) and draw a results table.

Constraints given:

Volume of FB 1 must not be less than 6.00 cm³.
Total volume of the reaction mixture must always be 60 cm³. (Note: FB 2 is 10 cm³, FB 3 is 20 cm³, starch is 10 cm³. Total fixed volume = 40 cm³. Thus, FB 1 + water must always equal 20 cm³).
You need four further experiments with intervals not less than 2 cm³, covering values both below and above 10 cm³.

Approach

Table Structure: Create a table with columns for Experiment number, Volume of FB 1, Volume of water, and Time. Include correct units (cm³ and s) in the headings.
Experimental Design: Choose volumes for FB 1 that span the required range (6.00 to 20.00 cm³) with at least 2 cm³ between each. Ensure at least one is < 10 cm³ and one is > 10 cm³.
Control Variable: Calculate the volume of water needed for each experiment so that Volume of FB 1 + Volume of water = 20.00 cm³.
Data Recording: Use placeholder times (t₁ to t₆) that logically follow the trend: as concentration decreases, time increases.

Step-by-Step Reasoning

Headings and Units: The mark scheme requires correct headings and units. Use 'Volume of FB 1 / cm³', 'Volume of water / cm³', and 'Time / s'. Volumes from a burette are read to two decimal places (0.05 cm³), so write 20.00, 10.00, etc., not 20 or 10.
Range and Intervals: Experiment 1 uses 20.00 cm³. Experiment 2 uses 10.00 cm³. We need four more. Let's use 14.00, 12.00, 8.00, and 6.00 cm³. The intervals are 2 cm³, which satisfies 'not less than 2 cm³'. The range is 6.00 to 20.00 cm³, satisfying 'no volume less than 6 cm³'. We have values less than 10 (8.00, 6.00) and greater than 10 (14.00, 12.00).
Total Volume: The fixed reagents (FB 2, FB 3, starch) total 10 + 20 + 10 = 40 cm³. The total mixture is 60 cm³, so FB 1 + water = 20 cm³. For 14.00 cm³ FB 1, add 6.00 cm³ water. For 6.00 cm³ FB 1, add 14.00 cm³ water.
Trend: The reaction rate is proportional to [Fe³⁺]. Lower [Fe³⁺] means a slower rate, so it takes longer for the blue-black colour to appear. Thus, time for 6.00 cm³ > time for 8.00 cm³ > time for 10.00 cm³ > time for 20.00 cm³.

Key Takeaways

Always control the total volume when investigating concentration effects by adding solvent.
Burette readings must be recorded to the correct precision (0.05 cm³ or two decimal places).
Experimental design must cover a sensible range with appropriate intervals to establish a clear trend.

Common Mistakes

Wrong precision: Writing volumes as '20' or '10' instead of '20.00' or '10.00'. Burettes read to 0.05 cm³.
Ignoring total volume: Not adding water to keep the FB 1 + water volume constant at 20 cm³. This changes the total volume and thus the concentrations of all other reactants (I⁻, S₂O₃²⁻), invalidating the experiment.
Poor range selection: Choosing all volumes above 10 cm³, or using intervals smaller than 2 cm³.

Things to Be Careful About

Ensure the table has only one set of headings at the top, not repeated for each row.
The mark scheme awards marks for the ratio of times (e.g., time for exp 2 / time for exp 1) being close to the supervisor's ratio. This means your recorded times must be physically realistic and consistent with the chemical kinetics.

Techniques used

construct a results table with correct headings and unitsplan an experiment with a suitable range and intervals of the independent variablecontrol variables by maintaining a constant total volume

(b)

Calculations

The rate of reaction can be found by calculating the change in concentration of $\text{Fe}^{3+}(\text{aq})$ that occurred when enough iodine was produced to change the colour of the indicator to blue-black.

Use your data and the equations on page 2 to carry out the following calculations.

Show your working and appropriate significant figures in the final answer to each step of your calculations.

(i)

Calculate the number of moles of thiosulfate ions, $\text{S}_2\text{O}_3^{2-}$ used in each experiment in (a).

\text{moles } \text{S}_2\text{O}_3^{2-} = \text{...................... mol}

DifficultyEasy

Worked solution

Working

\text{moles of } \text{S}_2\text{O}_3^{2-} = \text{concentration} \times \text{volume}

\text{moles} = 0.0060 \text{ mol dm}^{-3} \times \frac{20.00 \text{ cm}^3}{1000 \text{ cm}^3 \text{ dm}^{-3}}

\text{moles} = 1.2 \times 10^{-4} \text{ mol}

Answer

1.2 \times 10^{-4} \text{ mol}

Final answer

1.2 x 10^-4 mol

Detailed explanation

Background Concept

The amount of substance (moles, $n$ ) can be calculated from its concentration ( $c$ ) and volume ( $V$ ) using the equation:

n = c \times V

where $c$ is in mol dm⁻³ and $V$ is in dm³. If volume is given in cm³, it must be divided by 1000 to convert to dm³.

Understanding the Question

Part (b)(i) asks for the number of moles of thiosulfate ions ( $\text{S}_2\text{O}_3^{2-}$ ) used in each experiment. The volume of FB 3 (sodium thiosulfate) is constant at 20 cm³, and its concentration is given as 0.0060 mol dm⁻³.

Approach

Apply the formula $n = cV$ , ensuring the volume is converted from cm³ to dm³.

Step-by-Step Reasoning

Concentration of $\text{S}_2\text{O}_3^{2-}$ = 0.0060 mol dm⁻³.
Volume of $\text{S}_2\text{O}_3^{2-}$ = 20.00 cm³ = $20.00 / 1000$ dm³ = 0.02000 dm³.
Moles = $0.0060 \times 0.02000 = 1.2 \times 10^{-4}$ mol.
The concentration 0.0060 has 2 significant figures, so the answer should be given to 2 significant figures: $1.2 \times 10^{-4}$ .

Key Takeaways

Always convert volume from cm³ to dm³ by dividing by 1000 when using concentration in mol dm⁻³.
Match the significant figures of the answer to the least precise given value (here, 0.0060 has 2 s.f.).

Common Mistakes

Forgetting to divide the volume by 1000, giving $1.2 \times 10^{-1}$ .
Using the wrong volume (e.g., using the volume of FB 1 instead of FB 3).

Things to Be Careful About

The question asks for the answer to each step with appropriate significant figures. $1.2 \times 10^{-4}$ is correct (2 s.f.).

Techniques used

calculate moles from concentration and volume

(ii)

Calculate the number of moles of iodine, $\text{I}_2$ , that react with the number of moles of $\text{S}_2\text{O}_3^{2-}$ in (i).

\text{moles } \text{I}_2 = \text{...................... mol}

DifficultyEasy

Worked solution

Working

From the second equation:

\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})

The mole ratio of $\text{I}_2$ to $\text{S}_2\text{O}_3^{2-}$ is 1 : 2.

\text{moles of } \text{I}_2 = \frac{\text{moles of } \text{S}_2\text{O}_3^{2-}}{2}

\text{moles of } \text{I}_2 = \frac{1.2 \times 10^{-4}}{2} = 6.0 \times 10^{-5} \text{ mol}

Answer

6.0 \times 10^{-5} \text{ mol}

(or $0.6 \times 10^{-4}$ mol)

Final answer

6.0 x 10^-5 mol

Detailed explanation

Background Concept

Stoichiometry allows us to relate the amounts of reactants and products in a chemical reaction using the balanced chemical equation. The coefficients in the balanced equation represent the mole ratios.

Understanding the Question

Part (b)(ii) asks for the moles of iodine ( $\text{I}_2$ ) that react with the thiosulfate calculated in (i). The relevant equation is the reaction between iodine and thiosulfate.

Approach

Identify the mole ratio from the balanced equation and apply it to the moles from part (i).

Step-by-Step Reasoning

Equation: $\text{I}_2 + 2\text{S}_2\text{O}_3^{2-} \rightarrow 2\text{I}^- + \text{S}_4\text{O}_6^{2-}$
Ratio $\text{I}_2 : \text{S}_2\text{O}_3^{2-}$ is 1 : 2.
Moles of $\text{I}_2$ = (moles of $\text{S}_2\text{O}_3^{2-}$ ) / 2 = $(1.2 \times 10^{-4}) / 2 = 6.0 \times 10^{-5}$ mol.
Given to 2 s.f. as $6.0 \times 10^{-5}$ .

Key Takeaways

Always use the mole ratio from the balanced equation, not the overall reaction.
The clock reaction uses a known amount of thiosulfate to 'consume' the iodine produced until it is depleted.

Common Mistakes

Using the ratio from the first equation ( $\text{Fe}^{3+} : \text{I}_2$ is 2:1) instead of the second equation ( $\text{I}_2 : \text{S}_2\text{O}_3^{2-}$ is 1:2).
Multiplying by 2 instead of dividing.

Things to Be Careful About

The mark scheme accepts $0.6 \times 10^{-4}$ , but standard scientific notation $6.0 \times 10^{-5}$ is preferred.

Techniques used

use stoichiometric ratios from balanced equations

(iii)

Calculate the number of moles of iron(III) ions, $\text{Fe}^{3+}$ , that were used to produce the number of moles of iodine in (ii).

\text{moles } \text{Fe}^{3+} = \text{...................... mol}

DifficultyEasy

Worked solution

Working

From the first equation:

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

The mole ratio of $\text{Fe}^{3+}$ to $\text{I}_2$ is 2 : 1.

\text{moles of } \text{Fe}^{3+} = \text{moles of } \text{I}_2 \times 2

\text{moles of } \text{Fe}^{3+} = 6.0 \times 10^{-5} \times 2 = 1.2 \times 10^{-4} \text{ mol}

Answer

1.2 \times 10^{-4} \text{ mol}

Final answer

1.2 x 10^-4 mol

Detailed explanation

Background Concept

In a clock reaction, the amount of product from the main reaction that has occurred when the colour change happens is determined by the amount of 'clock' reagent (thiosulfate) added. We work backwards from the thiosulfate to find the moles of iodine, and then from the iodine to find the moles of the main reactant ( $\text{Fe}^{3+}$ ).

Understanding the Question

Part (b)(iii) asks for the moles of iron(III) ions that produced the iodine calculated in (ii).

Approach

Use the mole ratio from the first equation: $\text{Fe}^{3+} : \text{I}_2 = 2 : 1$ .

Step-by-Step Reasoning

Moles of $\text{I}_2$ = $6.0 \times 10^{-5}$ mol.
Ratio $\text{Fe}^{3+} : \text{I}_2$ = 2 : 1.
Moles of $\text{Fe}^{3+}$ = $6.0 \times 10^{-5} \times 2 = 1.2 \times 10^{-4}$ mol.
Notice this is the same numerical value as the moles of thiosulfate. This is because the stoichiometry works out: 2 mol $\text{Fe}^{3+}$ produces 1 mol $\text{I}_2$ , which reacts with 2 mol $\text{S}_2\text{O}_3^{2-}$ . Thus, moles $\text{Fe}^{3+}$ = moles $\text{S}_2\text{O}_3^{2-}$ .

Key Takeaways

Chain stoichiometric calculations carefully, using the correct equation at each step.
It is a useful check to see if the overall ratio makes sense.

Common Mistakes

Using the wrong mole ratio (e.g., 1:1 instead of 2:1).
Carrying forward a calculation error from part (ii).

Things to Be Careful About

The mark scheme notes this is marked jointly with (ii). Ensure your answer is consistent with your answer in (ii).

Techniques used

use stoichiometric ratios from balanced equations

(iv)

When the moles of $\text{Fe}^{3+}$ that you calculated in (iii) reacted, a change in the concentration of moles of $\text{Fe}^{3+}$ occurred. Calculate this change in concentration.

\text{change in concentration of } \text{Fe}^{3+}(\text{aq}) = \text{...................... mol dm}^{-3}

DifficultyMedium-Easy

Worked solution

Working

\text{change in concentration} = \frac{\text{moles}}{\text{total volume in dm}^3}

The total volume of the reaction mixture is always 60 cm³ = 0.060 dm³.

\text{change in concentration of } \text{Fe}^{3+} = \frac{1.2 \times 10^{-4} \text{ mol}}{0.060 \text{ dm}^3}

\text{change in concentration} = 2.0 \times 10^{-3} \text{ mol dm}^{-3}

Answer

2.0 \times 10^{-3} \text{ mol dm}^{-3}

Final answer

2.0 x 10^-3 mol dm^-3

Detailed explanation

Background Concept

Concentration is defined as moles per unit volume ( $c = n/V$ ). In this experiment, the 'change in concentration' refers to the amount of $\text{Fe}^{3+}$ that reacted to produce the fixed amount of iodine, divided by the total volume of the mixture.

Understanding the Question

Part (b)(iv) asks for the change in concentration of $\text{Fe}^{3+}$ when the moles calculated in (iii) reacted.

Approach

Divide the moles of $\text{Fe}^{3+}$ by the total volume of the reaction mixture in dm³.

Step-by-Step Reasoning

Moles of $\text{Fe}^{3+}$ reacted = $1.2 \times 10^{-4}$ mol.
Total volume = 60 cm³ = $60 / 1000$ dm³ = 0.060 dm³.
Change in concentration = $(1.2 \times 10^{-4}) / 0.060 = 2.0 \times 10^{-3}$ mol dm⁻³.
This value is constant for all experiments because the same amount of thiosulfate (and thus the same amount of iodine and $\text{Fe}^{3+}$ reacted) is used in each, and the total volume is constant.

Key Takeaways

Always use the total volume of the mixture, not the volume of the specific reactant.
Convert volume to dm³ before calculating concentration.

Common Mistakes

Dividing by the volume of FB 1 (e.g., 0.020 dm³) instead of the total volume (0.060 dm³).
Forgetting to convert cm³ to dm³.

Things to Be Careful About

The mark scheme gives the expression $(1.2 \times 10^{-4}) / 0.060$ . Ensure your working shows this clearly.

Techniques used

calculate concentration change from moles and total volume

(v)

The following formula can be used as a measure of the 'rate of reaction'.

\text{'rate of reaction'} = \frac{\text{change in concentration of } \text{Fe}^{3+}(\text{aq})}{\text{reaction time}} \times 10^6

Complete the table to show the volume of FB 1, the reaction time and the rate in Experiments 1-6. You should include units.

If you were unable to calculate a value for the change in concentration of $\text{Fe}^{3+}(\text{aq})$ in (iv), you should assume it is $2.50 \times 10^{-3}\text{ mol dm}^{-3}$ . (Note: this is not the correct value.)

Experiment
1
2
3
4
5
6

DifficultyMedium-Easy

Worked solution

Working

\text{rate} = \frac{\text{change in concentration of Fe}^{3+}}{\text{time}} \times 10^6

Using the change in concentration from (iv) = $2.0 \times 10^{-3}$ mol dm⁻³:

\text{rate} = \frac{2.0 \times 10^{-3}}{t} \times 10^6 = \frac{2000}{t} \text{ mol dm}^{-3} \text{ s}^{-1}

Complete the table using your recorded times ( $t$ ) for Experiments 1-6. Ensure units are included.

Experiment	Volume of FB 1 / cm³	Time / s	Rate / (mol dm⁻³ s⁻¹)
1	20.00	t₁	$(2000 / t_1)$
2	10.00	t₂	$(2000 / t_2)$
3	12.00	t₃	$(2000 / t_3)$
4	14.00	t₄	$(2000 / t_4)$
5	8.00	t₅	$(2000 / t_5)$
6	6.00	t₆	$(2000 / t_6)$

(Replace $t_1$ to $t_6$ with your actual recorded times and calculate the rates to 2 or 3 significant figures.)

Answer

See working (table completed with calculated rates and units $\text{mol dm}^{-3} \text{ s}^{-1}$ ).

Final answer

See working

Detailed explanation

Background Concept

The rate of reaction can be measured as the change in concentration of a reactant or product per unit time. In a clock reaction, we measure the average rate over the time taken for the colour change to occur.

The formula given is:

\text{rate} = \frac{\Delta[\text{Fe}^{3+}]}{t} \times 10^6

The $\times 10^6$ is a scaling factor to make the numbers easier to work with (avoiding very small decimals like $3.3 \times 10^{-5}$ ).

Understanding the Question

Part (b)(v) asks you to calculate the rate for each experiment using the formula and complete the table with units.

Approach

Use the constant change in concentration ( $2.0 \times 10^{-3}$ mol dm⁻³).
Divide by the recorded time $t$ for each experiment.
Multiply by $10^6$ .
Add the correct units: mol dm⁻³ s⁻¹.

Step-by-Step Reasoning

Change in concentration = $2.0 \times 10^{-3}$ mol dm⁻³.
Rate = $(2.0 \times 10^{-3} / t) \times 10^6 = 2000 / t$ .
For example, if time for Exp 1 is 40 s, rate = $2000 / 40 = 50.0$ mol dm⁻³ s⁻¹.
The table must have three columns: Volume of FB 1, Time, and Rate. The Rate column must have the unit $\text{mol dm}^{-3} \text{ s}^{-1}$ .
Calculate rates to 2 or 3 significant figures (consistent with your time and concentration data).

Key Takeaways

The $\times 10^6$ factor is crucial; without it, your rates will be tiny numbers.
Units for rate are always concentration per time (e.g., mol dm⁻³ s⁻¹).

Common Mistakes

Forgetting the $\times 10^6$ factor.
Using the wrong units (e.g., mol cm⁻³ s⁻¹).
Not including units in the table heading or final column.
Calculating rates to too many or too few significant figures.

Things to Be Careful About

The mark scheme awards marks for 3 correct columns and correct units. Ensure your table is clear and well-formatted.
If you couldn't calculate (iv), you are allowed to assume $2.50 \times 10^{-3}$ mol dm⁻³. In that case, rate = $2500 / t$ . But use your calculated value if possible.

Techniques used

calculate rate of reaction from concentration change and time

(c)

On the grid, plot the rate ( $y$ -axis) against the volume of FB 1 ( $x$ -axis). Draw a line of best fit through the points. You should identify any points you consider anomalous.

Grid 1

DifficultyMedium

Worked solution

Answer

x-axis: Volume of FB 1 / cm³ (range 0 to 20, or at least 6 to 20).
y-axis: Rate / (mol dm⁻³ s⁻¹) (range starting from 0, using calculated rates).
Plot all 6 points accurately within half a small square.
Draw a smooth curve or straight line of best fit through the points.
If any point is clearly off the trend (anomalous), ring or label it and do not include it in the line of best fit.
The graph should show rate increasing as volume of FB 1 increases.

Final answer

See working

Detailed explanation

Background Concept

Graphs are used to show relationships between variables. In kinetics, plotting rate against concentration (or volume, if total volume is constant) often reveals the order of reaction. A straight line through the origin indicates first order; a curve indicates higher order.

When plotting:

Axes: Independent variable (Volume of FB 1) on the x-axis, dependent variable (Rate) on the y-axis.
Scales: Must be uniform (equal spacing represents equal values) and use at least half of each axis. Start from 0,0 if possible.
Plotting: Points must be plotted accurately (within half a small square on graph paper).
Line of best fit: A single straight line (ruler) or smooth curve that passes as close as possible to all points. Points not on the line should be balanced on either side. Anomalous points (clearly outside the trend) are ignored for the line but still plotted and labelled.

Understanding the Question

Part (c) asks you to plot rate against volume of FB 1 on the provided grid and draw a line of best fit.

Approach

Label axes with quantities and units.
Choose scales that use the grid efficiently.
Plot the 6 data points from your table.
Draw the line of best fit, ignoring any anomalous points.

Step-by-Step Reasoning

Axes: x-axis = 'Volume of FB 1 / cm³', y-axis = 'Rate / (mol dm⁻³ s⁻¹)'.
Scales: x-axis could be 0 to 20 cm³ (e.g., 1 large square = 2 cm³). y-axis depends on your rates (e.g., 0 to 100, 1 large square = 10).
Plotting: Mark each (volume, rate) pair with a clear cross or dot.
Line of best fit: Draw a smooth curve or straight line. Since rate is proportional to [Fe³⁺] (and thus volume of FB 1), a straight line through the origin is expected for a first-order reaction. If it's curved, draw a smooth curve.
Anomalies: If one point is far from the line, ring it and say 'anomalous'. Do not force the line through it.

Key Takeaways

Always label axes with quantity and unit.
Scales must be uniform and use most of the grid.
Line of best fit is not a 'join-the-dots' line; it shows the overall trend.

Common Mistakes

Forgetting units on axes.
Using non-uniform scales (e.g., 0, 1, 2, 5, 10).
Joining the dots with straight line segments instead of a smooth line of best fit.
Including an anomalous point in the line of best fit.

Things to Be Careful About

The mark scheme requires the line to be drawn with a ruler (if straight) or be a smooth curve. No kinks or zig-zags.
Points must be within half a small square of the correct position.

Techniques used

plot a graph with correct axes and scalesdraw a line of best fitidentify anomalous results

(d)

Using your graph, what conclusion can you reach about the effect of changing the concentration of $\text{FeCl}_3$ on the rate of the reaction between $\text{Fe}^{3+}(\text{aq})$ and $\text{I}^-(\text{aq})$ ?

DifficultyMedium-Easy

Worked solution

Answer

As the concentration of $\text{Fe}^{3+}$ (or volume of FB 1) increases, the rate of reaction increases.
The graph is a straight line through the origin (or close to it), indicating that the rate is directly proportional to the concentration of $\text{Fe}^{3+}$ (first order with respect to $\text{Fe}^{3+}$ ).
The results are consistent, as all points lie on or near the line of best fit (or comment on any anomalous point).

Working

From the graph, as volume of FB 1 increases (meaning [Fe³⁺] increases), the rate increases. The shape of the graph (straight line through 0,0) shows proportionality.

Answer

Rate increases with concentration of Fe³⁺; rate is proportional to [Fe³⁺] (first order).

Final answer

Rate increases as concentration of Fe3+ increases; rate is proportional to concentration (straight line through origin).

Detailed explanation

Background Concept

The rate law for a reaction is $\text{rate} = k[\text{A}]^m[\text{B}]^n$ . If we plot rate against [A] and get a straight line through the origin, the reaction is first order with respect to A ( $m=1$ ). If we get a curve, it's higher order.

In this experiment, the volume of FB 1 is proportional to the concentration of Fe³⁺ (since total volume is constant). So plotting rate against volume of FB 1 is equivalent to plotting rate against [Fe³⁺].

Understanding the Question

Part (d) asks for a conclusion about the effect of changing [FeCl₃] on the rate, based on the graph.

Approach

State the trend: as concentration increases, rate increases.
Comment on the shape of the graph to infer the order (straight line = first order / proportional).

Step-by-Step Reasoning

Look at the graph: as x (volume of FB 1) increases, y (rate) increases.
If the line is straight and passes through (0,0), rate $\propto$ [Fe³⁺]. This means the reaction is first order with respect to Fe³⁺.
Also, comment on the reliability: 'all points are on/near the line, so results are consistent' or 'there is one anomalous point'.

Key Takeaways

Always state the trend clearly (A increases as B increases).
Use the graph shape to infer the order of reaction.
Comment on the quality of the data (consistency, anomalies).

Common Mistakes

Saying 'rate increases' but not commenting on the proportionality or order.
Not referring to the graph (e.g., 'the graph shows...').
Saying 'concentration increases rate' without specifying which reactant (though here it's obvious).

Things to Be Careful About

The mark scheme awards 1 mark for the trend and 1 mark for a comment on the graph (e.g., proportional, through 0,0, consistent). Give both.

Techniques used

interpret a graph to draw a conclusion about reaction kinetics

(e)

A student wanted to investigate how changing the concentration of $\text{I}^-$ would affect the rate of reaction. Explain how this investigation could be carried out.

DifficultyMedium-Easy

Worked solution

Answer

Alter the volume of FB 2 (aqueous KI) while keeping the volume of FB 1 (aqueous FeCl₃) constant.
Add distilled water to the FB 2 / KI beaker (or to the mixture) to keep the total volume of the reaction mixture constant at 60 cm³.

Working

To investigate [I⁻], change [I⁻] by changing the volume of FB 2. To keep other concentrations constant, add water to maintain the total volume.

Answer

Change volume of FB 2 (KI), add water to keep total volume constant.

Final answer

Alter volume of FB 2 / KI while keeping volume of FB 1 constant; add water to keep total volume constant.

Detailed explanation

Background Concept

To investigate the effect of one reactant's concentration on the rate, you must vary that reactant's concentration while keeping all others constant. This is done by changing the volume of that reactant's solution and adding solvent (water) to maintain the total volume, so the concentrations of the other reactants don't change due to dilution.

Understanding the Question

Part (e) asks how to modify the method to investigate the effect of [I⁻] instead of [Fe³⁺].

Approach

Identify which solution contains I⁻: FB 2 (KI).
Vary the volume of FB 2.
Keep the volume of FB 1 constant.
Add water to keep the total volume constant.

Step-by-Step Reasoning

In the original method, FB 1 volume was varied and water added to keep FB 1 + water = 20 cm³.
Now, vary FB 2 volume. Keep FB 1 volume constant (e.g., 20 cm³).
Add water to the FB 2 beaker (or to the final mixture) so that the total volume of FB 2 + water = 30 cm³ (since FB 1 is 20, FB 3 is 20, starch is 10, total = 60. So FB 2 + water must be 30 cm³).
This ensures [Fe³⁺], [S₂O₃²⁻], and [starch] remain constant, and only [I⁻] changes.

Key Takeaways

To vary [A], change volume of A's solution and add water to keep total volume constant.
Keep all other reactant volumes constant.

Common Mistakes

Changing the concentration of FB 2 by diluting it (this changes the amount of I⁻ but also changes the concentration of everything else if total volume isn't managed correctly).
Forgetting to add water to keep the total volume constant.
Changing the volume of FB 3 or starch (this would change [S₂O₃²⁻] or the clock reaction timing).

Things to Be Careful About

The mark scheme specifically asks to 'alter volume of FB 2' and 'add water to keep total volume constant'. Use these exact phrases.

Techniques used

design an experiment to investigate the effect of a different reactant's concentration

(f)

It was found, by carrying out experiments similar to those used in (a), that increasing the concentration of $\text{I}^-$ increased the rate of the reaction.

The student suggested modifications to the method as used in (a). In each case, state what the effect would be on the reaction time in Experiment 1 and explain how these changes would affect the possible errors in the measurements.

Suggested modification 1
The reaction was carried out using the same volumes of all reagents but with the concentrations of FB 1 and FB 2 being double their original values.

Suggested modification 2
The reaction was carried out using half the volume of all reagents.

DifficultyMedium

Worked solution

Answer

Modification 1 (double concentrations):

Reaction time: Less (shorter).
Error: Less accurate since there is a larger percentage error in the time (smaller time means larger % error for the same absolute error of ±0.5 s).

Modification 2 (half volumes):

Reaction time: Stays the same (concentrations are unchanged).
Error: Less accurate since there is a greater percentage error in the volume (smaller volumes measured have larger % error).

Working

Mod 1: Doubling [Fe³⁺] and [I⁻] increases the rate, so the time for the colour change is shorter. % error in time = (0.5 / time) × 100%. Smaller time = larger % error.
Mod 2: Halving all volumes keeps concentrations the same (since all are halved, the ratio remains constant). Thus, rate and time are the same. However, measuring smaller volumes (e.g., 5 cm³ instead of 10 cm³) with a measuring cylinder or burette gives a larger percentage error in volume.

Answer

Mod 1: Time less, larger % error in time. Mod 2: Time same, greater % error in volume.

Final answer

Mod 1: Time less, larger % error in time. Mod 2: Time same, greater % error in volume.

Detailed explanation

Background Concept

Percentage error = (absolute error / measured value) × 100%. For a stopwatch, the absolute error is typically ±0.5 s (human reaction time). For a measuring cylinder or burette, the absolute error is ±0.05 cm³ or ±0.5 cm³ depending on the apparatus.

If you measure a smaller value, the percentage error increases, making the measurement less accurate.

Understanding the Question

Part (f) asks to evaluate two modifications:

Double the concentrations of FB 1 and FB 2 (same volumes).
Halve the volumes of all reagents (same concentrations).

For each, state the effect on reaction time and explain the effect on possible errors.

Approach

Mod 1: Higher concentrations → faster reaction → shorter time. Shorter time → larger % error in time.
Mod 2: Same concentrations (all volumes halved, so ratios same) → same rate → same time. Smaller volumes → larger % error in volume.

Step-by-Step Reasoning

Modification 1:

Concentrations are doubled. Rate increases (rate ∝ [Fe³⁺][I⁻] or similar). So the reaction finishes faster. Reaction time is less.
Absolute error in time is still ±0.5 s. If time was 40 s, % error = (0.5/40)×100 = 1.25%. If time is 20 s, % error = (0.5/20)×100 = 2.5%. So larger percentage error in time, making it less accurate.

Modification 2:

All volumes are halved. Concentrations = moles/volume. Moles are halved, volume is halved, so concentration is unchanged. Rate depends on concentration, so rate is the same, and reaction time stays the same.
However, you are now measuring smaller volumes (e.g., 5.00 cm³ instead of 10.00 cm³). The absolute error in volume (e.g., ±0.05 cm³ for a burette, or ±0.5 cm³ for a measuring cylinder) is the same, but the measured value is smaller. So percentage error in volume is greater, making the concentration less accurate.

Key Takeaways

Higher concentration → faster reaction → shorter time → larger % error in time.
Same concentrations (scaled volumes) → same time, but larger % error in volume measurements.
Always link the physical change to the mathematical consequence (percentage error formula).

Common Mistakes

Thinking halving volumes changes the concentration (it doesn't, if all are halved proportionally).
Saying 'time is less' for Mod 2.
Not explaining the error correctly (e.g., just saying 'less accurate' without saying why or which measurement has the larger % error).

Things to Be Careful About

The mark scheme requires both the effect on time AND the explanation for the error for each modification. Give two points for each (4 marks total).
Use the terms 'percentage error' or '% error'.

Techniques used

evaluate the effect of method modifications on reaction time and percentage error

(g)

(i)

Which of the experiments you carried out in (a) had the greatest percentage error in the reaction time?

DifficultyEasy

Worked solution

Answer

The experiment with the shortest reaction time (i.e., the experiment with the highest volume of FB 1, Experiment 1) had the greatest percentage error in the reaction time.

Answer

Experiment 1 (or the experiment with the shortest time).

Final answer

Experiment 1 (shortest reaction time)

Detailed explanation

Background Concept

Percentage error = (absolute error / measured value) × 100%. If the absolute error (±0.5 s) is constant, the percentage error is largest when the measured value (reaction time) is smallest.

Understanding the Question

Part (g)(i) asks which experiment had the greatest percentage error in time.

Approach

Identify the experiment with the shortest time.

Step-by-Step Reasoning

As volume of FB 1 decreases, concentration decreases, rate decreases, time increases.
So Experiment 1 (20.00 cm³ FB 1) has the highest concentration, fastest rate, and shortest time.
Shortest time means largest % error (since absolute error is constant at ±0.5 s).
Therefore, Experiment 1 has the greatest percentage error.

Key Takeaways

Greatest % error occurs for the smallest measured value when absolute error is constant.
In kinetics, highest concentration → shortest time → largest % error in time.

Common Mistakes

Choosing the experiment with the longest time (this has the smallest % error).
Not specifying which experiment (e.g., just saying 'the first one' without justification).

Things to Be Careful About

The mark scheme accepts 'experiment with shortest reaction time'. Be specific if possible (e.g., Experiment 1).

Techniques used

identify the experiment with the greatest percentage error in time

(ii)

Calculate this percentage error. Assume that the error in measuring the reaction time is $\pm 0.5\text{ s}$ .

\text{percentage error} = \text{..................... \%}

DifficultyEasy

Worked solution

Working

\text{percentage error} = \frac{\text{absolute error}}{\text{measured value}} \times 100\%

Absolute error in time = $\pm 0.5$ s.
Measured value = smallest reaction time (from your table, e.g., $t_1$ ).

\text{percentage error} = \frac{0.5}{t_1} \times 100\%

(Substitute your actual recorded time for Experiment 1 here. For example, if $t_1 = 40$ s:)

\text{percentage error} = \frac{0.5}{40} \times 100\% = 1.25\%

Answer

\frac{0.5 \times 100\%}{\text{smallest reaction time}}

(e.g., 1.25% if time is 40 s)

Final answer

(0.5 / smallest time) x 100%

Detailed explanation

Background Concept

Percentage error quantifies the uncertainty in a measurement relative to its size. For time measurements with a stopwatch, the absolute error is typically ±0.5 s (due to human reaction time when starting and stopping).

Understanding the Question

Part (g)(ii) asks to calculate the percentage error for the experiment identified in (i), assuming the error in time is ±0.5 s.

Approach

Use the formula: % error = (absolute error / measured value) × 100%.

Step-by-Step Reasoning

Absolute error = 0.5 s.
Measured value = the shortest reaction time from your table (e.g., 40 s for Experiment 1).
% error = (0.5 / 40) × 100 = 1.25%.
The mark scheme gives the expression: $(0.5 \times 100\%) / \text{smallest reaction time}$ .
You must use your actual recorded time to get the numerical answer. If you didn't record data, you can state the formula.

Key Takeaways

Percentage error increases as the measured value decreases (for constant absolute error).
Always show the working: formula, substitution, answer.

Common Mistakes

Using the wrong absolute error (e.g., ±1.0 s instead of ±0.5 s).
Forgetting to multiply by 100 to get a percentage.
Not using the smallest time (the one from part i).

Things to Be Careful About

The mark scheme awards 1 mark for the correct expression. Ensure you write $(0.5 \times 100) / t_{\text{min}}$ or equivalent.
Give your final answer to 2 or 3 significant figures.

Techniques used

calculate percentage error

The rest of this paper

1 more questions

Q2Qualitative Analysis · Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation13M

Loading the full paper…