Chemistry 9701/51 — May/June 2024

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Analysis, Conclusions and Evaluation · Planning

Q1MediumAnalysis, Conclusions and EvaluationPlanning

Titration can be used to determine the concentration of dissolved oxygen in samples of river water.

The procedure for the experiment is given.

step 1 Use five $50\text{ cm}^3$ graduated syringes, A, B, C, D and E, to collect five separate $30.0\text{ cm}^3$ samples of river water.

step 2 In the laboratory, carefully add $5.0\text{ cm}^3$ of $0.220\text{ mol dm}^{-3}$ manganese(II) sulfate, $\text{MnSO}_4\text{(aq)}$ , into syringe A and mix well.

step 3 Add $5.0\text{ cm}^3$ of alkaline aqueous potassium iodide into syringe A and mix well.

step 4 Add $10.0\text{ cm}^3$ of dilute sulfuric acid into syringe A and mix well.

step 5 Transfer the contents of syringe A into a $150\text{ cm}^3$ conical flask. Rinse syringe A using $10\text{ cm}^3$ of distilled water and add washings to the conical flask.

step 6 Carry out one accurate titration of all the contents in the conical flask with $0.00200\text{ mol dm}^{-3}$ aqueous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ , using starch indicator.

Repeat steps 2–6 for the samples in syringes B–E.

(a)

Aqueous sodium thiosulfate can be prepared from $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O(s)}$ .

(i)

Determine the mass, in g, of $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O(s)}$ required to prepare $500.0\text{ cm}^3$ of $0.00200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

n(\text{Na}_2\text{S}_2\text{O}_3) = 0.00200 \times \frac{500.0}{1000} = 1.00 \times 10^{-3}\ \text{mol}

M_r(\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O}) = 2(23.0) + 2(32.1) + 3(16.0) + 5(18.0) = 248.2

\text{mass} = 1.00 \times 10^{-3} \times 248.2 = 0.2482\ \text{g}

Answer

0.2482 g (0.248 g)

Final answer

0.2482 g

Detailed explanation

Background Concept

A standard solution is prepared by dissolving a known mass of solute in a known total volume. The amount of solute is found from $n = cV$ , and the mass is then $m = nM_r$ . When the solute is a hydrated salt, the water of crystallisation is part of the formula unit and must be included in the molar mass.

Understanding the Question

The question asks for the mass of hydrated sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O}$ , needed to make $500.0\ \text{cm}^3$ of $0.00200\ \text{mol dm}^{-3}$ solution. The volume must be converted to $\text{dm}^3$ before using $n = cV$ .

Approach

Convert $500.0\ \text{cm}^3$ to $0.5000\ \text{dm}^3$ .
Calculate moles of $\text{Na}_2\text{S}_2\text{O}_3$ : $n = cV$ .
Calculate the molar mass of the hydrated salt.
Multiply moles by molar mass to get the required mass.

Step-by-Step Reasoning

$n(\text{Na}_2\text{S}_2\text{O}_3) = 0.00200 \times 0.5000 = 1.00 \times 10^{-3}\ \text{mol}$ .
Molar mass of $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O}$ $Na_{2} S_{2} O_{3} \cdot 5 H_{2} O$ :
- $2 \times 23.0 = 46.0$
- $2 \times 32.1 = 64.2$
- $3 \times 16.0 = 48.0$
- $5 \times 18.0 = 90.0$
- Total $= 248.2\ \text{g mol}^{-1}$ .
Mass $= 1.00 \times 10^{-3} \times 248.2 = 0.2482\ \text{g}$ .

The mark scheme accepts $0.248\ \text{g}$ or $0.2482\ \text{g}$ .

Key Takeaways

Always include water of crystallisation in the molar mass of a hydrated salt.
Convert volume to $\text{dm}^3$ before using $n = cV$ .
The mass needed is small because the concentration is very dilute.

Common Mistakes

Using the anhydrous molar mass of $\text{Na}_2\text{S}_2\text{O}_3$ ( $158.1$ ) instead of $248.2$ .
Forgetting to divide the volume by 1000.
Quoting the answer without units.

Things to Be Careful About

Use $0.5000\ \text{dm}^3$ , not $500.0$ .
Keep at least three significant figures in the final answer.
The final answer should be in grams.

Techniques used

calculate amount in moles from concentration and volumeuse molar mass to convert moles to mass

(ii)

Identify the piece of apparatus that should be used to prepare $500.0\text{ cm}^3$ of $0.00200\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ after the required mass of $\text{Na}_2\text{S}_2\text{O}_3\cdot5\text{H}_2\text{O(s)}$ has been weighed out.

1M Similar

(b)

The graduations on each syringe are every $1.0\text{ cm}^3$ .

(i)

Calculate the percentage error in the measurement of $5.0\text{ cm}^3$ of alkaline aqueous potassium iodide by the syringe.

Show your working.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{absolute uncertainty} = 2 \times 0.5 = 1.0\ \text{cm}^3

\text{percentage error} = \frac{1.0}{5.0} \times 100 = 20.0\%

Answer

20.0%

Final answer

20.0%

Detailed explanation

Background Concept

Percentage error compares the absolute uncertainty of a measurement with the size of the measured quantity:

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

For a scale with divisions every $1.0\ \text{cm}^3$ , each reading has an uncertainty of half a division, $\pm 0.5\ \text{cm}^3$ .

Understanding the Question

The syringe graduations are every $1.0\ \text{cm}^3$ . The student measures $5.0\ \text{cm}^3$ of potassium iodide. The question asks for the percentage error in this measurement.

Approach

Identify the uncertainty in one reading, then account for the fact that a volume is obtained from two readings (the difference between initial and final positions), so the uncertainties add. Then apply the percentage error formula.

Step-by-Step Reasoning

One scale division is $1.0\ \text{cm}^3$ , so one reading has uncertainty $\pm 0.5\ \text{cm}^3$ .
The measured volume is a difference of two readings, so total uncertainty $= 2 \times 0.5 = 1.0\ \text{cm}^3$ .
Percentage error $= \frac{1.0}{5.0} \times 100 = 20.0\%$ .

The mark scheme requires the working to be shown.

Key Takeaways

Absolute uncertainty of a scale is half the smallest division.
When a volume is found by difference, uncertainties of the two readings add.
Percentage error decreases as the measured volume increases.

Common Mistakes

Using only one reading uncertainty ( $0.5\ \text{cm}^3$ ) and getting 10%.
Forgetting to multiply by 100.
Quoting the answer without the % sign.

Things to Be Careful About

The mark scheme specifically uses $2 \times 0.5$ ; show this step.
Keep the units consistent: $\text{cm}^3$ cancels in the ratio.

Techniques used

determine absolute uncertainty from the scale divisioncalculate percentage error

(ii)

Place one tick (✓) in each row in Table 1.1 to show the effect, if any, of using a larger volume of alkaline aqueous potassium iodide.

Table 1.1

	greater effect	no effect	smaller effect
uncertainty of the measurement
percentage error of the measurement

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

	greater effect	no effect	smaller effect
uncertainty of the measurement		✓
percentage error of the measurement			✓

Final answer

Uncertainty: no effect; percentage error: smaller

Detailed explanation

Background Concept

The absolute uncertainty of a measurement depends on the apparatus and its scale division, not on the amount being measured. The percentage error, however, is the absolute uncertainty divided by the measured value, so it depends on the size of the measurement.

Understanding the Question

If a larger volume of alkaline potassium iodide is used, how does this affect (1) the uncertainty of the measurement and (2) the percentage error? One tick must be placed in each row.

Approach

Absolute uncertainty: unchanged, because the syringe scale is the same.
Percentage error: since the denominator increases while the numerator stays the same, the percentage error becomes smaller.

Step-by-Step Reasoning

The syringe graduations remain every $1.0\ \text{cm}^3$ , so the absolute uncertainty stays the same.
A larger measured volume gives a smaller value of $\frac{\text{uncertainty}}{\text{volume}} \times 100$ .
Therefore: uncertainty has no effect; percentage error is smaller.

Key Takeaways

Absolute uncertainty is a property of the apparatus.
Percentage error is relative and decreases for larger measurements.

Common Mistakes

Thinking a larger volume makes the uncertainty larger.
Confusing "greater effect" with "smaller effect".
Placing more than one tick in a row.

Things to Be Careful About

Exactly one tick per row.
The answer must distinguish the two rows clearly.

Techniques used

compare effect of volume on uncertainty and percentage error

(c)

The sample in the conical flask and the prepared solution of sodium thiosulfate are provided.
Describe the following procedures for the experiment using syringe A.

(i)

Preparing the clean burette before taking any readings.

2M Similar

(ii)

Carrying out the one accurate titration in step 6.

2M Similar

(d)

Suggest why the reaction mixture is mixed well in steps 2–4.

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(e)

Draw a table for recording the titration results for the five samples in syringes A–E.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

	A	B	C	D	E
final burette reading / cm³
initial burette reading / cm³
titre / cm³

Final answer

Table with columns A–E and rows for final/initial readings and titre, units cm³.

Detailed explanation

Background Concept

A titration results table must record the initial and final burette readings and the titre for each titration. Headings must include units so that the numbers in the cells are pure values. Here there are five samples, so five columns are needed.

Understanding the Question

The question asks for a table to record the titration results for the five samples in syringes A–E.

Approach

Create a table with five columns labelled A–E and rows for final reading, initial reading and titre. Include units in the headings.

Step-by-Step Reasoning

Each column represents one syringe/sample.
The rows are: final burette reading, initial burette reading, titre.
Units are written in the heading as "/ cm³" or "in cm³", not repeated in every cell.
The titre is the difference between final and initial readings.

Key Takeaways

Units belong in headings, not in each cell.
A titration table needs initial, final and titre rows.
Five samples require five columns.

Common Mistakes

Missing units in headings.
Only one column for all samples.
Omitting the titre row.
Writing units in every cell.

Things to Be Careful About

The mark scheme gives one mark for correct headings and units and one mark for five columns A–E.
Use "titre" or "volume of sodium thiosulfate used".

Techniques used

construct a results table with headings and unitsinclude five columns for replicate samples

(f)

The overall reaction taking place in the experiment is shown.

\text{O}_2\text{(aq)} + 4\text{S}_2\text{O}_3^{2-}\text{(aq)} + 4\text{H}^+\text{(aq)} \rightarrow 2\text{S}_4\text{O}_6^{2-}\text{(aq)} + 2\text{H}_2\text{O(l)}

A student carries out the experiment and determines the mean titre to be $12.65\text{ cm}^3$ .

Calculate the concentration, in $\text{mol dm}^{-3}$ , of dissolved oxygen in the river water.

2M Similar

DifficultyMedium

Worked solution

Working

n(\text{Na}_2\text{S}_2\text{O}_3) = \frac{12.65}{1000} \times 0.00200 = 2.53 \times 10^{-5}\ \text{mol}

n(\text{O}_2) = \frac{2.53 \times 10^{-5}}{4} = 6.325 \times 10^{-6}\ \text{mol}

[\text{O}_2] = \frac{6.325 \times 10^{-6}}{30.0/1000} = 2.11 \times 10^{-4}\ \text{mol dm}^{-3}

Answer

2.11 × 10⁻⁴ mol dm⁻³

Final answer

2.11 × 10^-4 mol dm^-3

Detailed explanation

Background Concept

In a titration, the amount of titrant used is found from $n = cV$ . The balanced equation gives the stoichiometric ratio between the titrant and the analyte. Here the equation shows that 1 mol of $\text{O}_2$ reacts with 4 mol of $\text{S}_2\text{O}_3^{2-}$ .

Understanding the Question

The mean titre of sodium thiosulfate is $12.65\ \text{cm}^3$ . Each sample was $30.0\ \text{cm}^3$ of river water. We need the concentration of dissolved oxygen in $\text{mol dm}^{-3}$ .

Approach

Calculate moles of thiosulfate from the titre.
Use the 4:1 ratio to find moles of oxygen.
Divide by the sample volume in $\text{dm}^3$ to get concentration.

Step-by-Step Reasoning

$n(\text{Na}_2\text{S}_2\text{O}_3) = 0.00200 \times \frac{12.65}{1000} = 2.53 \times 10^{-5}\ \text{mol}$ .
From the equation, $n(\text{O}_2) = \frac{2.53 \times 10^{-5}}{4} = 6.325 \times 10^{-6}\ \text{mol}$ .
Sample volume $= 30.0\ \text{cm}^3 = 0.0300\ \text{dm}^3$ .
$[\text{O}_2] = \frac{6.325 \times 10^{-6}}{0.0300} = 2.108 \times 10^{-4}\ \text{mol dm}^{-3} \approx 2.11 \times 10^{-4}\ \text{mol dm}^{-3}$ .

The mark scheme gives the method mark for the moles of thiosulfate and the second mark for dividing by 4 and by the sample volume.

Key Takeaways

Always use the stoichiometric ratio from the balanced equation.
Convert all volumes to $\text{dm}^3$ before calculating concentration.
The sample volume is $30.0\ \text{cm}^3$ , not the titre volume.

Common Mistakes

Using a 1:1 ratio instead of 4:1.
Dividing by $12.65\ \text{cm}^3$ instead of $30.0\ \text{cm}^3$ .
Forgetting to convert $\text{cm}^3$ to $\text{dm}^3$ .
Quoting the answer without units.

Things to Be Careful About

Give the final answer to three significant figures.
The units are $\text{mol dm}^{-3}$ .
Show the division by 4 explicitly to earn the stoichiometry mark.

Techniques used

calculate moles from titre and concentrationapply stoichiometric ratio from the balanced equationcalculate concentration from moles and volume

(g)

Freshly distilled water does not contain any dissolved oxygen.

A student decides to run the procedure on a sample of freshly distilled water and at the end obtains a value of $2.26 \times 10^{-5}\text{ mol dm}^{-3}$ dissolved oxygen.

(i)

Suggest why the student did not get a value of $0\text{ mol dm}^{-3}$ . Assume the procedure was carried out correctly.

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(ii)

Suggest how the value of $2.26 \times 10^{-5}\text{ mol dm}^{-3}$ could be used to improve the answer in (f).

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(h)

Suggest why this method is unsuitable for samples of tap water that have been purified by chlorination and so contain $\text{Cl}_2\text{(aq)}$ .

1M Similar

Q2MediumPlanningAnalysis, Conclusions and Evaluation

The activation energy, $E_A$ , for the reaction between dilute hydrochloric acid, $\text{HCl(aq)}$ , and aqueous sodium thiosulfate, $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ , can be determined by an initial rates method.

2\text{HCl(aq)} + \text{Na}_2\text{S}_2\text{O}_3\text{(aq)} \rightarrow 2\text{NaCl(aq)} + \text{H}_2\text{O(l)} + \text{S(s)} + \text{SO}_2\text{(g)}

The solid sulfur formed is seen as a white suspension in the reaction mixture. The reactants are mixed and the time, $t$ , for a fixed quantity of sulfur to be formed is recorded.

A measure of the initial rate of the reaction is $\frac{1}{t}$ .

Standard solutions of $0.100\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ and $0.500\text{ mol dm}^{-3}$ $\text{HCl(aq)}$ are supplied.

Measurements are taken for a series of temperatures using the following procedure.

step 1 A thermostatically controlled water bath is set up.

step 2 A $100\text{ cm}^3$ conical flask is labelled A and a second $100\text{ cm}^3$ conical flask is labelled B.

step 3 $10.00\text{ cm}^3$ of $0.100\text{ mol dm}^{-3}$ $\text{Na}_2\text{S}_2\text{O}_3\text{(aq)}$ is added to flask A. Flask A is placed in the water bath.

step 4 $10\text{ cm}^3$ of $0.500\text{ mol dm}^{-3}$ $\text{HCl(aq)}$ is added to flask B. Flask B is placed in the same water bath.

step 5 Wait for 10 minutes.

step 6 Flask A is removed from the water bath and placed on a tile marked with a black cross.

step 7 The contents of flask B are added to flask A and a timer started.

step 8 The timer is stopped when the black cross is no longer visible. The time is recorded.

(a)

Suggest a reason why it is necessary to wait for 10 minutes in step 5.

1M Similar

DifficultyEasy

Worked solution

Answer

To ensure the solutions in flasks A and B reach thermal equilibrium with the water bath (and are at the same temperature) before mixing.

Final answer

To ensure the solutions in flasks A and B reach thermal equilibrium with the water bath (and are at the same temperature) before mixing.

Detailed explanation

Background Concept

In kinetics experiments where the initial rate is measured at a specific temperature, the temperature of the reactants must be precisely controlled and known at the moment of mixing. If reactants are at different temperatures, the initial temperature of the mixture is ambiguous, and the reaction will proceed at a rate corresponding to a changing temperature rather than a fixed initial temperature.

Understanding the Question

The question asks why the two flasks containing the reactants are left in the thermostatically controlled water bath for 10 minutes before being mixed. This is a standard step in initial rates experiments.

Approach

Think about what happens when a liquid is placed in a temperature-controlled environment. Heat transfer will occur until thermal equilibrium is reached. The 10-minute wait ensures this equilibrium is achieved for both reactants independently.

Step-by-Step Reasoning

Step 5 involves leaving flasks A and B in the water bath.
The water bath is thermostatically controlled, meaning it maintains a constant temperature.
Leaving the flasks for 10 minutes allows sufficient time for heat to transfer between the water bath and the solutions in both flasks.
This ensures both solutions reach the exact temperature of the water bath and are at thermal equilibrium with each other.
When they are mixed in step 7, the initial temperature of the reaction mixture is known and uniform, allowing the time $t$ to be meaningfully related to that specific temperature.

Key Takeaways

Always allow reactants to equilibrate to the reaction temperature before mixing in initial rates experiments. This ensures the measured rate corresponds to a well-defined, constant initial temperature.

Common Mistakes

Saying "to heat the reactants": this is incomplete and doesn't explain why they need to be heated (to reach equilibrium).
Saying "to ensure they are the same temperature": while true, it misses the crucial link to the water bath temperature (thermal equilibrium).

Things to Be Careful About

Use precise terminology: "thermal equilibrium" or "same temperature as the water bath".
Do not say "to prevent the reaction from starting early"; the reaction only starts when they are mixed.

Techniques used

ensure thermal equilibrium before mixing

(b)

The procedure does not mention how a value for the temperature of the mixture during the reaction is obtained.

(i)

State the temperature measurements that should be taken and at which stage in the procedure they should be taken.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Measure the temperature of the mixture in flask A:

At the start of the reaction (immediately before step 7 / after step 6).
At the end of the reaction (when the black cross is no longer visible / immediately after step 8).

Final answer

Measure the temperature of the mixture in flask A at the start of the reaction (immediately before step 7) and at the end of the reaction (when the cross is no longer visible).

Detailed explanation

Background Concept

In exothermic or endothermic reactions, the temperature of the mixture changes as the reaction proceeds. For an initial rates experiment, we need a single representative temperature for the reaction. Since the reaction is relatively fast and the mixture is not perfectly insulated, the temperature will drift during the time $t$ .

Understanding the Question

The procedure does not specify how to determine the temperature during the reaction. We need to identify when to take temperature readings to capture the temperature at the start and end of the measured time interval.

Approach

Identify the time interval being measured (from step 7 to step 8). To get an accurate temperature for this interval, we should measure the temperature at the beginning and end of this interval.

Step-by-Step Reasoning

The reaction starts at step 7 when flask B is added to flask A.
The reaction ends at step 8 when the cross is no longer visible.
To capture the temperature range during this interval, measure the temperature of the mixture in flask A:
1. At the start: Immediately before step 7 (or after step 6, once mixed but before the timer starts). This gives the initial temperature.
2. At the end: Immediately after step 8 (when the cross is no longer visible). This gives the final temperature.
These two measurements bracket the reaction time $t$ .

Key Takeaways

When a reaction causes a temperature change, measure the temperature at the start and end of the reaction interval to establish the temperature range during which the reaction occurred.

Common Mistakes

Suggesting to measure the temperature of the water bath: the water bath temperature is already known; we need the mixture temperature.
Only measuring at one point: a single measurement doesn't account for any temperature change during the reaction.

Things to Be Careful About

Specify which flask or mixture: the temperature of the reaction mixture in flask A after mixing, not the separate reactants.
Be precise about when: "before step 7" and "after step 8" are clearer than "at the beginning and end".

Techniques used

measure initial and final temperatures of the reaction mixture

(ii)

State how to use the temperature measurements to determine an accurate value for the temperature of the mixture during the reaction.

1M Similar

DifficultyEasy

Worked solution

Answer

Calculate the mean (average) of the two temperature measurements.

Final answer

Calculate the mean (average) of the two temperature measurements.

Detailed explanation

Background Concept

When a quantity changes linearly (or approximately linearly) over a time interval, the average value over that interval is well-approximated by the arithmetic mean of the initial and final values.

Understanding the Question

Having taken two temperature measurements (start and end of reaction), how do we combine them to get a single accurate value for the temperature during the reaction?

Approach

Use basic statistics: the mean of two values gives the midpoint, which is a good estimate for the average temperature during the interval.

Step-by-Step Reasoning

Let $T_{\text{start}}$ be the temperature at the beginning of the reaction.
Let $T_{\text{end}}$ be the temperature at the end of the reaction.
The temperature during the reaction is assumed to change approximately linearly between these two values.
The mean temperature is $T_{\text{mean}} = \frac{T_{\text{start}} + T_{\text{end}}}{2}$ .
This mean value is used as the representative temperature for calculating $\frac{1}{T}$ and plotting the graph.

Key Takeaways

The arithmetic mean of initial and final temperatures provides a reasonable estimate for the average temperature during a reaction interval, assuming no significant heat loss or gain.

Common Mistakes

Taking the temperature of the water bath instead: this ignores any temperature change in the reaction mixture.
Using a weighted average without justification: the simple mean is the standard and expected method.

Things to Be Careful About

Ensure the temperatures are in the same units (usually $^\circ\text{C}$ ) before calculating the mean.
The mean temperature is then converted to Kelvin for use in the Arrhenius equation.

Techniques used

calculate the mean of initial and final temperatures

(c)

A student carries out the procedure at three different temperatures and records the measurements in Table 2.1.

Complete Table 2.1. Record values for temperature to the nearest whole number and the values for $\frac{1}{t}$ to four decimal places.

Table 2.1

temperature, $T / ^\circ\text{C}$	time, $t / \text{s}$	temperature, $T / \text{K}$	$\frac{1}{t} / \text{s}^{-1}$
15	176
24	92
32	62

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Conversions:

$T(\text{K}) = T(\text{\textdegree C}) + 273$
$\frac{1}{t} = \frac{1}{\text{time in seconds}}$

Calculations:

For $15\text{\textdegree C}$ : $T = 15 + 273 = 288\text{ K}$ ; $\frac{1}{t} = \frac{1}{176} = 0.005681... \approx 0.0057\text{ s}^{-1}$
For $24\text{\textdegree C}$ : $T = 24 + 273 = 297\text{ K}$ ; $\frac{1}{t} = \frac{1}{92} = 0.010869... \approx 0.0109\text{ s}^{-1}$
For $32\text{\textdegree C}$ : $T = 32 + 273 = 305\text{ K}$ ; $\frac{1}{t} = \frac{1}{62} = 0.016129... \approx 0.0161\text{ s}^{-1}$

Answer

temperature, $T / \text{\textdegree C}$	time, $t / \text{s}$	temperature, $T / \text{K}$	$\frac{1}{t} / \text{s}^{-1}$
15	176	288	0.0057
24	92	297	0.0109
32	62	305	0.0161

Final answer

See working

Detailed explanation

Background Concept

The Arrhenius equation relates the rate constant $k$ to temperature $T$ :
$k = A e^{-E_A / RT}$
Taking the logarithm (base 10) of both sides:
$\log k = \log A - \frac{E_A}{2.303 RT}$
Since the initial rate is proportional to $\frac{1}{t}$ (where $t$ is the time for a fixed amount of product to form), we can use $\log\left(\frac{1}{t}\right)$ as a proxy for $\log k$ . The equation becomes:
$\log\left(\frac{1}{t}\right) = -\frac{0.434 E_A}{R} \cdot \frac{1}{T} + \text{constant}$
This is a linear equation of the form $y = mx + c$ , where $y = \log\left(\frac{1}{t}\right)$ , $x = \frac{1}{T}$ , and the gradient $m = -\frac{0.434 E_A}{R}$ .

Understanding the Question

We are given temperatures in $\text{\textdegree C}$ and times $t$ in seconds. We need to convert the temperatures to Kelvin (required for the Arrhenius equation) and calculate $\frac{1}{t}$ to four decimal places (as specified).

Approach

Convert $T(\text{\textdegree C})$ to $T(\text{K})$ using $T(\text{K}) = T(\text{\textdegree C}) + 273$ .
Calculate $\frac{1}{t}$ for each time value and round to four decimal places.

Step-by-Step Reasoning

Row 1: $T = 15\text{\textdegree C}$ , $t = 176\text{ s}$

$T(\text{K}) = 15 + 273 = 288\text{ K}$
$\frac{1}{t} = \frac{1}{176} = 0.0056818... \rightarrow 0.0057\text{ s}^{-1}$ (to 4 d.p.)

Row 2: $T = 24\text{\textdegree C}$ , $t = 92\text{ s}$

$T(\text{K}) = 24 + 273 = 297\text{ K}$
$\frac{1}{t} = \frac{1}{92} = 0.0108695... \rightarrow 0.0109\text{ s}^{-1}$ (to 4 d.p.)

Row 3: $T = 32\text{\textdegree C}$ , $t = 62\text{ s}$

$T(\text{K}) = 32 + 273 = 305\text{ K}$
$\frac{1}{t} = \frac{1}{62} = 0.0161290... \rightarrow 0.0161\text{ s}^{-1}$ (to 4 d.p.)

Key Takeaways

Always convert temperatures to Kelvin when using the Arrhenius equation. Pay close attention to significant figures as specified in the question.

Common Mistakes

Forgetting to convert to Kelvin: using $\text{\textdegree C}$ in $\frac{1}{T}$ will give completely wrong results.
Incorrect rounding: $\frac{1}{176} = 0.005681...$ rounds to $0.0057$ , not $0.0056$ .
Using natural log ( $\ln$ ) instead of base-10 log ( $\log$ ): the equation given uses $\log$ (base 10), and the constant $0.434$ comes from $\frac{1}{\ln 10} \approx 0.434$ .

Things to Be Careful About

The question asks for $\frac{1}{t}$ to four decimal places, not four significant figures. $0.0057$ has two significant figures but four decimal places.
Ensure the calculator is set to the correct log base when plotting later (though here we just need $\log_{10}$ ).

Techniques used

convert Celsius to Kelvincalculate reciprocal of time

(d)

A second student carries out the procedure at six different temperatures and analyses their data to give the results in Table 2.2.

Table 2.2

$\frac{1}{T} / \text{K}^{-1}$	$\log\left(\frac{1}{t}\right)$
0.00353	–2.43
0.00336	–1.99
0.00325	–1.68
0.00314	–1.47
0.00302	–1.21
0.00287	–0.82

(i)

Use the results from Table 2.2 to plot a graph on the grid in Fig. 2.1 to show the relationship between $\log\left(\frac{1}{t}\right)$ and $\frac{1}{T}$ . Use a cross (×) to plot each data point. Draw a line of best fit.

Fig. 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the following points using × marks on the grid in Fig. 2.1:

$(0.00353, -2.43)$
$(0.00336, -1.99)$
$(0.00325, -1.68)$
$(0.00314, -1.47)$
$(0.00302, -1.21)$
$(0.00287, -0.82)$

Draw a straight line of best fit that passes close to all points, balancing the points above and below the line. One point (e.g., $0.00325, -1.68$ ) may be slightly off the line and can be treated as an anomaly.

Final answer

See diagram

Detailed explanation

Background Concept

A graph of $\log\left(\frac{1}{t}\right)$ against $\frac{1}{T}$ should yield a straight line if the reaction follows Arrhenius behavior. The gradient of this line is related to the activation energy $E_A$ by:
$\text{gradient} = -\frac{0.434 E_A}{R}$
Plotting the data correctly is essential for determining an accurate gradient.

Understanding the Question

We are given six pairs of $(x, y)$ values from Table 2.2 and asked to plot them on the provided grid and draw a line of best fit.

Approach

Identify the axes: $x$ -axis is $\frac{1}{T} / \text{K}^{-1}$ , $y$ -axis is $\log\left(\frac{1}{t}\right)$ .
Plot each point carefully, ensuring correct alignment with the grid lines.
Draw a straight line of best fit using a ruler, ensuring it passes as close as possible to all points, with roughly equal numbers of points above and below the line.

Step-by-Step Reasoning

Plotting points:

$x = 0.00353, y = -2.43$ : Near the top right.
$x = 0.00336, y = -1.99$ : Middle right.
$x = 0.00325, y = -1.68$ : Middle.
$x = 0.00314, y = -1.47$ : Middle left.
$x = 0.00302, y = -1.21$ : Lower left.
$x = 0.00287, y = -0.82$ : Bottom left.

Line of best fit:

The points generally follow a linear trend with a negative gradient.
A straight line drawn through the middle of the data cloud will have a gradient of approximately $-2400$ to $-2500$ .
The point $(0.00325, -1.68)$ might lie slightly below the line; this is acceptable as an anomaly.

Key Takeaways

When plotting a graph for a linear relationship, always use a ruler for the line of best fit. Do not connect the dots with a zig-zag line.

Common Mistakes

Plotting points incorrectly: misreading the axis scales (e.g., $0.00353$ is not $0.0035$ ).
Drawing a curve instead of a straight line: the Arrhenius equation gives a linear relationship.
Forcing the line through the origin: the line does not necessarily pass through $(0,0)$ .

Things to Be Careful About

The $x$ -axis scale: increments of $0.0001$ , so $0.00353$ is $3$ small squares to the right of $0.00350$ .
The $y$ -axis scale: increments of $0.2$ , so $-2.43$ is $1.5$ small squares below $-2.4$ .
Use × marks as specified, not dots or circles.

Techniques used

plot data points on a graphdraw a line of best fit

(ii)

Determine the gradient of your line of best fit in Fig. 2.1. State the coordinates of both points you use in your calculation. These must be selected from your line of best fit. Give the gradient to three significant figures.

coordinates 1 .............................................. coordinates 2 ..............................................

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Coordinates 1: $(0.00350, -2.35)$
Coordinates 2: $(0.00290, -0.85)$

Gradient calculation:

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-0.85 - (-2.35)}{0.00290 - 0.00350} = \frac{1.50}{-0.00060} = -2500

Gradient = $-2.50 \times 10^3\text{ K}$ (to 3 significant figures)

Final answer

Gradient = -2.50 x 10^3 K

Detailed explanation

Background Concept

The gradient of the line of best fit is used to calculate the activation energy. It is crucial to use coordinates from the line of best fit, not from the original data points, to minimize the effect of experimental errors.

Understanding the Question

We need to determine the gradient of the line drawn in part (d)(i). We must state two coordinates from the line and show the calculation.

Approach

Choose two points on the line of best fit that are far apart to minimize reading errors.
Ensure the points are at convenient grid intersections or can be read accurately.
Apply the gradient formula: $m = \frac{\Delta y}{\Delta x}$ .

Step-by-Step Reasoning

Selecting points:

Look for points where the line crosses grid intersections clearly.
Point 1: Let's use $(0.00350, -2.35)$ . This is on the line near the top right.
Point 2: Let's use $(0.00290, -0.85)$ . This is on the line near the bottom left.
These points are well-separated, reducing the relative error in the gradient calculation.

Calculating gradient:

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-0.85 - (-2.35)}{0.00290 - 0.00350}

\text{gradient} = \frac{1.50}{-0.00060} = -2500

Significant figures:

The question asks for three significant figures.
$-2500$ can be written as $-2.50 \times 10^3$ to clearly show 3 s.f.
Alternatively, $-2500$ is acceptable if the context implies 3 s.f., but scientific notation is safer.

Key Takeaways

Always use points from the line of best fit, not the data points. Choose points that are far apart to minimize percentage error in the gradient.

Common Mistakes

Using data points instead of line coordinates: this gives the gradient of a line through two specific points, not the best fit.
Incorrect subtraction: $y_2 - y_1$ must be calculated carefully, especially with negative numbers.
Wrong number of significant figures: $-2500$ has 2, 3, or 4 s.f. depending on context; use $-2.50 \times 10^3$ for 3 s.f.

Things to Be Careful About

The coordinates must be explicitly stated as requested.
The gradient is negative; do not lose the sign.
The units of the gradient are $\text{K}$ (since $y$ is dimensionless and $x$ is $\text{K}^{-1}$ ).

Techniques used

calculate gradient from line of best fit

(iii)

An equation relating time and temperature variables is shown.

\log\left(\frac{1}{t}\right) = -\frac{0.434E_A}{RT} + \text{constant}

Determine the activation energy, $E_A$ , of this reaction using this equation and your answer to (d)(ii).

(If you were unable to find the gradient in (d)(ii), then use the value $-3.21 \times 10^3\text{ K}$ . This is not the correct answer.)

Include units in your answer.

Show your working.

3M Similar

DifficultyMedium

Worked solution

Working

Equation relating gradient and $E_A$ :

\log\left(\frac{1}{t}\right) = -\frac{0.434 E_A}{R} \cdot \frac{1}{T} + \text{constant}

Comparing with $y = mx + c$ , the gradient $m = -\frac{0.434 E_A}{R}$ .

Rearranging for $E_A$ :

E_A = -\frac{\text{gradient} \times R}{0.434}

Substitution:

Gradient = $-2.50 \times 10^3\text{ K}$
$R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$

E_A = -\frac{(-2.50 \times 10^3) \times 8.314}{0.434}

E_A = \frac{20785}{0.434} = 47891\text{ J mol}^{-1}

Convert to kJ mol $^{-1}$ :

E_A = 47.9\text{ kJ mol}^{-1}

Answer

$E_A = 47.9\text{ kJ mol}^{-1}$

Final answer

47.9 kJ mol^-1

Detailed explanation

Background Concept

The Arrhenius equation in logarithmic form is:
$\ln k = -\frac{E_A}{RT} + \ln A$
Converting to base-10 logarithm ( $\log_{10}$ ):
$\log k = -\frac{E_A}{2.303 RT} + \log A$
Since $\frac{1}{2.303} \approx 0.434$ , we have:
$\log k = -\frac{0.434 E_A}{R} \cdot \frac{1}{T} + \text{constant}$
For an initial rates experiment, $\text{rate} \propto \frac{1}{t}$ , so $\log\left(\frac{1}{t}\right)$ is proportional to $\log k$ . The gradient of the graph $\log\left(\frac{1}{t}\right)$ against $\frac{1}{T}$ is therefore:
$\text{gradient} = -\frac{0.434 E_A}{R}$

Understanding the Question

We are given the equation and asked to calculate $E_A$ using the gradient from part (d)(ii). We must include units.

Approach

Rearrange the gradient equation to solve for $E_A$ .
Substitute the gradient value and the gas constant $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$ .
Calculate $E_A$ in J mol $^{-1}$ , then convert to kJ mol $^{-1}$ .

Step-by-Step Reasoning

Rearranging the equation:

\text{gradient} = -\frac{0.434 E_A}{R}

E_A = -\frac{\text{gradient} \times R}{0.434}

Substituting values:

Gradient = $-2.50 \times 10^3\text{ K}$ (from part d(ii))
$R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$

E_A = -\frac{(-2.50 \times 10^3) \times 8.314}{0.434}

E_A = \frac{20785}{0.434} = 47891.7\text{ J mol}^{-1}

Converting units:

Divide by 1000 to get kJ mol $^{-1}$ :

E_A = 47.89\text{ kJ mol}^{-1}

Round to 3 significant figures: $47.9\text{ kJ mol}^{-1}$ .

Key Takeaways

The factor $0.434$ comes from $\frac{1}{\ln 10}$ . Always check whether the equation uses $\ln$ or $\log_{10}$ and use the correct constant.

Common Mistakes

Forgetting the negative sign: the gradient is negative, and $E_A$ must be positive.
Using $R = 8.314$ but not converting the final answer to kJ mol $^{-1}$ : the mark scheme requires kJ mol $^{-1}$ .
Using the wrong value for $R$ : $R = 8.314\text{ J K}^{-1}\text{ mol}^{-1}$ is the correct value for energy calculations.
Calculating $\frac{0.434 \times E_A}{R}$ instead of rearranging correctly.

Things to Be Careful About

The gas constant $R$ is $8.314\text{ J K}^{-1}\text{ mol}^{-1}$ , not $0.0821\text{ L atm K}^{-1}\text{ mol}^{-1}$ (which is for ideal gas law).
The gradient has units of K (since $y$ is dimensionless and $x$ is $\text{K}^{-1}$ ). When multiplied by $R$ (J K $^{-1}$ mol $^{-1}$ ), the K cancels, giving J mol $^{-1}$ .
The question provides a fallback gradient of $-3.21 \times 10^3\text{ K}$ if you couldn't calculate it. Using this:

E_A = -\frac{(-3.21 \times 10^3) \times 8.314}{0.434} = 61.5\text{ kJ mol}^{-1}

This would also be accepted if the working is correct.

Techniques used

calculate activation energy from gradient

(iv)

Use your graph to state whether the results from the experiment are reliable. Justify your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Yes, the results are reliable.
Justification: Most points lie on or near the line of best fit, with no significant anomalous points (or only one minor anomaly), indicating a strong linear relationship consistent with the Arrhenius equation.

Final answer

Yes, the results are reliable because most points lie on or near the line of best fit, indicating a strong linear relationship.

Detailed explanation

Background Concept

The reliability of experimental data is assessed by how well the data fits the expected theoretical model. For a linear relationship, we look at:

How close the points are to the line of best fit.
The number of anomalous points (outliers).
The strength of the correlation (how tightly the points cluster around the line).

Understanding the Question

We need to state whether the results are reliable and justify the answer based on the graph.

Approach

Examine the graph from part (d)(i). Count how many points lie on or near the line. Identify any anomalies. Draw a conclusion based on the overall trend.

Step-by-Step Reasoning

Observing the graph:

Five out of six points lie very close to the line of best fit.
One point (e.g., $0.00325, -1.68$ ) may lie slightly off the line, but this is a minor deviation.
The overall trend is strongly linear, which is expected for the Arrhenius equation.

Conclusion:

The results are reliable because the data shows a clear linear relationship with minimal scatter.
The minor deviation of one point could be due to random experimental error (e.g., slight variation in timing or temperature measurement), but it does not invalidate the overall trend.

Alternative answer:

If a student identifies a more significant anomaly or notes that points are widely scattered, they could argue the results are not reliable. However, the mark scheme accepts either answer as long as it is justified by the graph.

Key Takeaways

Reliability is about consistency and fit to the expected model. A few minor anomalies do not necessarily make data unreliable if the overall trend is clear.

Common Mistakes

Saying "yes, because the line is straight": this is circular reasoning; you must explain why a straight line indicates reliability.
Not mentioning the number of anomalous points: the justification must reference the graph.
Saying "yes, because we used a line of best fit": the line is drawn because the data is reliable, not the other way around.

Things to Be Careful About

The justification must be based on the graph, not on general statements about the experiment.
Use terms like "most points lie on or near the line" or "few anomalous points".
If you say "no", you must justify it by pointing to specific anomalies or poor fit.

Techniques used

evaluate reliability of experimental data

(e)

Suggest a change to one controlled variable that the student could make so that the time measured for a given temperature is shorter.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Increase the concentration of one or both of the reactants (e.g., use a more concentrated $\text{HCl}$ or $\text{Na}_2\text{S}_2\text{O}_3$ solution).

Final answer

Increase the concentration of one or both of the reactants.

Detailed explanation

Background Concept

The rate of a reaction depends on the concentration of the reactants (for reactions that are not zero order with respect to a reactant). The rate equation is:
$\text{rate} = k [\text{HCl}]^m [\text{Na}_2\text{S}_2\text{O}_3]^n$
where $m$ and $n$ are the orders of reaction with respect to each reactant.

For this reaction, the rate is typically first order with respect to $\text{Na}_2\text{S}_2\text{O}_3$ and first order with respect to $\text{HCl}$ (or at least not zero order for either). Therefore, increasing the concentration of either reactant will increase the rate.

Since $\text{rate} \propto \frac{1}{t}$ , increasing the rate means decreasing the time $t$ .

Understanding the Question

We want the time $t$ measured for a given temperature to be shorter. This means we want the reaction to proceed faster. We need to suggest a change to a controlled variable that would achieve this.

Approach

Think about what factors affect the rate of reaction:

Temperature (already controlled and fixed for each experiment)
Concentration of reactants (can be changed)
Catalyst (not applicable here)
Surface area (not applicable for solutions)

Since we are looking for a change to a controlled variable (a variable that is kept constant during the experiment but could be changed between experiments), concentration is the most appropriate answer.

Step-by-Step Reasoning

The time $t$ is inversely proportional to the rate: $\text{rate} \propto \frac{1}{t}$ .
To make $t$ shorter, we need to increase the rate.
The rate depends on the concentrations of $\text{HCl}$ and $\text{Na}_2\text{S}_2\text{O}_3$ .
If we increase the concentration of $\text{HCl}$ (e.g., from $0.500\text{ mol dm}^{-3}$ to $1.000\text{ mol dm}^{-3}$ ), the rate will increase, and $t$ will decrease.
Similarly, increasing the concentration of $\text{Na}_2\text{S}_2\text{O}_3$ will increase the rate.
Note: We cannot change the temperature, as that is the independent variable being investigated. We are looking for a change to a variable that is currently controlled (kept constant) but could be adjusted to speed up the reaction.

Key Takeaways

To decrease reaction time (increase rate), increase the concentration of reactants (assuming the reaction is not zero order with respect to them).

Common Mistakes

Suggesting to increase the temperature: temperature is the independent variable; we are asked for a change to a controlled variable for a given temperature.
Suggesting to add a catalyst: there is no catalyst for this reaction, and the question asks for a change to a controlled variable (concentration is more appropriate).
Suggesting to decrease the volume: changing the volume while keeping the same concentration does not change the rate (rate depends on concentration, not volume). However, if the volume is decreased, the total amount of product needed to obscure the cross might also change, which complicates the experiment.

Things to Be Careful About

The question asks for a change to one controlled variable.
The change must result in a shorter time.
Increasing concentration is the most straightforward and correct answer. Be specific: "increase the concentration of HCl" or "increase the concentration of Na2S2O3".

Techniques used

identify controlled variables to modify