Chemistry 9701/51 — May/June 2025

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

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q1MediumPlanningAnalysis, Conclusions and Evaluation

Grignard reagents have the general formula RMgX, where R is a hydrocarbon group and X is a halogen. The Grignard reagent $\text{C}_6\text{H}_5\text{MgBr}$ is used as an intermediate in the reaction between bromobenzene, $\text{C}_6\text{H}_5\text{Br}$ , and ethanal, $\text{CH}_3\text{CHO}$ , to prepare 1-phenylethanol, $\text{C}_6\text{H}_5\text{CH(OH)CH}_3$ . An organic solvent, ethoxyethane, is used.

The equations for the three reactions that take place during the preparation are shown.

\begin{aligned} \text{reaction 1} \quad &\text{Mg} + \text{C}_6\text{H}_5\text{Br} \rightarrow \text{C}_6\text{H}_5\text{MgBr} \\ \text{reaction 2} \quad &\text{C}_6\text{H}_5\text{MgBr} + \text{CH}_3\text{CHO} \rightarrow \text{C}_6\text{H}_5\text{CH(CH}_3\text{)OMgBr} \\ \text{reaction 3} \quad &\text{C}_6\text{H}_5\text{CH(CH}_3\text{)OMgBr} + \text{H}^+ \rightarrow \text{C}_6\text{H}_5\text{CH(OH)CH}_3 + \text{Mg}^{2+} + \text{Br}^- \end{aligned}

The preparation involves the following steps.

step 1 Set up the apparatus shown in Fig. 1.1 with approximately $1.25\text{ g}$ of Mg powder and $5\text{ cm}^3$ of ethoxyethane in the round-bottomed flask.
step 2 Add $0.0500\text{ mol}$ of liquid $\text{C}_6\text{H}_5\text{Br}$ to the round-bottomed flask dropwise using the tap funnel. Leave until reaction 1 is complete.
step 3 Dissolve $3.00\text{ cm}^3$ of $\text{CH}_3\text{CHO}$ in $15\text{ cm}^3$ of ethoxyethane and add this solution to the round-bottomed flask using the tap funnel. Leave until reaction 2 is complete.
step 4 Remove the condenser, tube Y and the tap funnel from the round-bottomed flask.
step 5 Add $40\text{ cm}^3$ of dilute hydrochloric acid, $\text{HCl(aq)}$ , to the round-bottomed flask so that reaction 3 takes place.
step 6 Transfer the contents of the round-bottomed flask to a separating funnel. Allow the liquids to settle so that two layers are formed.
step 7 Open the tap of the separating funnel and run the lower layer into a beaker labelled A. Run the upper layer into a beaker labelled B.
step 8 Allow the ethoxyethane to evaporate from the beaker containing $\text{C}_6\text{H}_5\text{CH(OH)CH}_3$ .

Fig. 1.1

Some relevant data are shown in Table 1.1.

Table 1.1

substance	density / $\text{g cm}^{-3}$	boiling point / $^\circ\text{C}$	hazard
bromobenzene	1.50	156	flammable, toxic, skin irritant
distilled water	1.00	100	non-hazardous
ethoxyethane	0.714	35	flammable, toxic
ethanal	0.788	21	flammable, eye and respiratory irritant
1-phenylethanol	1.01	204	flammable, toxic, eye irritant

(a)

Use the information in Table 1.1 to suggest why the following safety precautions are used.

wearing chemically resistant gloves
using a fume hood

2M Similar

DifficultyEasy

Worked solution

Answer

Gloves: Bromobenzene (and ethanal / 1-phenylethanol) is a skin irritant, so gloves prevent skin contact.
Fume hood: Toxic gases or vapours (e.g. from ethoxyethane, ethanal, or bromobenzene) are produced or used, so inhalation is avoided.

Final answer

Gloves protect against skin irritation from bromobenzene; fume hood prevents inhalation of toxic vapours.

Detailed explanation

Background Concept

When carrying out chemical preparations, especially those involving volatile or reactive organic compounds, it is essential to identify the hazards associated with the reagents and products. Hazard data (such as irritant, toxic, or flammable properties) directly dictate the required safety precautions in the laboratory.

Understanding the Question

The question asks for the justification of two specific safety precautions—wearing chemically resistant gloves and using a fume hood—based on the hazard information provided in Table 1.1 for the substances involved in the Grignard preparation.

Approach

Read the 'hazard' column in Table 1.1 for all substances (bromobenzene, ethoxyethane, ethanal, 1-phenylethanol). Match the physical hazard (e.g., skin irritant) to the appropriate protective equipment (gloves). Match the inhalation hazard (e.g., toxic) to appropriate ventilation (fume hood).

Step-by-Step Reasoning

Gloves: Look at the hazard column. Bromobenzene is listed as a 'skin irritant'. Ethanal and 1-phenylethanol are also irritants. Chemically resistant gloves provide a barrier to prevent these substances from touching and irritating the skin.
Fume hood: Ethoxyethane is 'toxic' and 'flammable'. Ethanal is an 'eye and respiratory irritant'. Bromobenzene is 'toxic'. These substances can form toxic vapours or gases that, if inhaled, are dangerous. A fume hood ventilates these vapours away from the user.

Key Takeaways

Always link safety equipment directly to the specific hazards listed in the data. 'Skin irritant' -> gloves. 'Toxic' / 'Respiratory irritant' -> fume hood.

Common Mistakes

Stating 'to prevent fire' for the fume hood. While ethoxyethane is flammable, a fume hood is primarily for toxic vapours; a fume hood does not prevent fire (in fact, electrical equipment inside can be an ignition source if not防爆). The mark scheme specifically looks for 'toxic gas/vapour'.
Forgetting to mention the specific irritant (bromobenzene) for the gloves.

Things to Be Careful About

Ensure you quote the hazards exactly as they appear in the table (e.g., 'skin irritant', 'toxic'). Do not invent hazards not present in the data.

Techniques used

identify hazards from data tablematch hazards to safety precautions

(b)

Dry apparatus and reagents are essential for steps 1–3.

(i)

Suggest how the glassware shown in Fig. 1.1 should be dried before use in step 1.

1M Similar

(ii)

The ethoxyethane contains a small amount of water.

Suggest how the water can be removed from the ethoxyethane before use in steps 1 and 3.

1M Similar

(c)

Fig. 1.1 shows the apparatus for steps 1–3.

(i)

Draw a labelled arrow on Fig. 1.1 to show where the water enters the condenser.

1M Similar

(ii)

Suggest why solid Z is used.

1M Similar

(iii)

Give one reason why the apparatus does not have a bung at the end of tube Y.

1M Similar

(d)

In step 1, approximately $1.25\text{ g}$ of Mg powder is needed.

Outline how the student should accurately weigh by difference using a weighing boat so that the exact mass of Mg transferred into the flask is known. Include a results table, with appropriate headings, ready for the student to fill in.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Method:

Weigh the weighing boat containing the Mg powder.
Transfer the Mg powder to the round-bottomed flask.
Weigh the weighing boat again (with any residue).

Results Table:

Measurement	/ g
Mass of boat + Mg (before transfer)
Mass of boat (+ residue of Mg) (after transfer)
Mass of Mg powder (transferred)

(Note: The mass of Mg transferred is calculated by subtracting row 2 from row 1.)

Final answer

Weigh boat+Mg, transfer Mg, weigh boat again. Table: Mass of boat+Mg / g; Mass of boat+residue / g; Mass of Mg transferred / g.

Detailed explanation

Background Concept

'Weighing by difference' is a precise method for transferring a known mass of a substance. Instead of trying to weigh exactly $1.25$ g directly (which is difficult), you weigh a container with more than enough substance, transfer the required amount, and weigh the container again. The difference is the exact mass transferred.

Understanding the Question

The student needs to transfer approximately $1.25$ g of Mg powder. They must use 'weighing by difference' and include a results table.

Approach

Describe the two weighing steps (before and after transfer). Draw a table with the necessary rows and correct column headings with units.

Step-by-Step Reasoning

Step 1: Place the weighing boat with Mg powder on the balance and record the mass. (e.g., $2.50$ g).
Step 2: Carefully pour the Mg powder from the boat into the round-bottomed flask.
Step 3: Weigh the weighing boat again (it will have a small residue of Mg). Record this mass. (e.g., $1.23$ g).
Calculation: Mass of Mg transferred = (Mass of boat + Mg) - (Mass of boat + residue). ( $2.50 - 1.23 = 1.27$ g).
Table: The table must have headings that describe what is being measured and a column for the mass in grams ('/ g').

Key Takeaways

Weighing by difference is more accurate than direct weighing for transfers. The table must have clear headings and units.

Common Mistakes

Not including units in the table heading. The mark scheme requires '/ g'.
Forgetting to mention weighing the boat after transfer.
Writing 'Mass of Mg' as a heading without specifying it's the difference or 'transferred'.

Things to Be Careful About

The table needs at least two rows of data entry (before and after) and a row for the calculated difference. The column header must include the unit '/ g'.

Techniques used

weigh by differenceprepare results table

(e)

Use the information in Table 1.1 to determine the volume, in $\text{cm}^3$ , of bromobenzene used in step 2.

[ $M_r$ : $\text{C}_6\text{H}_5\text{Br}$ , 156.9]

1M Similar

DifficultyMedium-Easy

Worked solution

Working

\text{Mass of C}_6\text{H}_5\text{Br} = n \times M_r = 0.0500 \text{ mol} \times 156.9 \text{ g mol}^{-1} = 7.845 \text{ g}

\text{Volume} = \frac{\text{Mass}}{\text{Density}} = \frac{7.845 \text{ g}}{1.50 \text{ g cm}^{-3}} = 5.23 \text{ cm}^3

Answer

$5.23 \text{ cm}^3$

Final answer

5.23 cm^3

Detailed explanation

Background Concept

The relationship between mass, moles, molar mass, density, and volume is fundamental in chemistry.

$\text{Mass} = \text{moles} \times M_r$
$\text{Density} = \frac{\text{Mass}}{\text{Volume}} \Rightarrow \text{Volume} = \frac{\text{Mass}}{\text{Density}}$

Understanding the Question

Given $0.0500$ mol of bromobenzene and its $M_r$ ( $156.9$ ) and density ( $1.50$ g cm $^{-3}$ ) from Table 1.1, calculate the volume in cm $^3$ .

Approach

Calculate the mass first, then use the density to find the volume.

Step-by-Step Reasoning

Calculate mass: $0.0500 \times 156.9 = 7.845$ g.
Calculate volume: $7.845 / 1.50 = 5.23$ cm $^3$ .
Significant figures: The data $0.0500$ (3 s.f.) and $1.50$ (3 s.f.) suggest the answer should be to 3 s.f. $5.23$ is correct.

Key Takeaways

Always convert to mass first if you have moles and density, then find volume. Check significant figures.

Common Mistakes

Dividing moles by density directly (wrong units).
Forgetting to use the $M_r$ .
Wrong significant figures (e.g., $5.2$ or $5.230$ ).

Things to Be Careful About

The question asks for the answer in cm $^3$ . Ensure units are consistent (g and g cm $^{-3}$ give cm $^3$ ).

Techniques used

calculate volume from moles and density

(f)

The bromobenzene is added dropwise in step 2.

Suggest one reason why the bromobenzene is not added all at once.

1M Similar

(g)

Suggest why a measuring cylinder is a suitable piece of apparatus to measure $40\text{ cm}^3$ of hydrochloric acid in step 5.

1M Similar

(h)

The separating funnel used in steps 6 and 7 is shown in Fig. 1.2. The final product, 1-phenylethanol, is in the ethoxyethane layer.

Fig. 1.2

State whether beaker A or beaker B contains the layer with 1-phenylethanol after step 7. Explain your answer using the information given in Table 1.1.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Beaker B contains the layer with 1-phenylethanol.

Explanation: The product is in the ethoxyethane layer. From Table 1.1, the density of ethoxyethane ( $0.714$ g cm $^{-3}$ ) is less than that of water/dilute acid ( $1.00$ g cm $^{-3}$ ). Therefore, the ethoxyethane layer is the upper layer, which is collected in beaker B.

(Note: Step 7 says lower layer -> A, upper layer -> B.)

Final answer

Beaker B; ethoxyethane is less dense than water.

Detailed explanation

Background Concept

In a separating funnel, two immiscible liquids form layers. The denser liquid forms the lower layer and is drained first through the tap. The less dense liquid forms the upper layer and remains in the funnel until drained from the top (or the lower layer is drained first).

Understanding the Question

Step 6: Contents transferred to separating funnel. Two layers form.
Step 7: Lower layer -> Beaker A. Upper layer -> Beaker B.
The product (1-phenylethanol) is in the ethoxyethane layer.
Which beaker (A or B) has the product?

Approach

Identify the two layers: aqueous layer (dilute HCl + water) and organic layer (ethoxyethane + product).
Compare densities from Table 1.1.
Determine which is upper/lower.
Match to beaker A/B based on step 7.

Step-by-Step Reasoning

Layers: The mixture contains ethoxyethane (organic solvent) and aqueous HCl/water.
Densities:
- Ethoxyethane: $0.714$ g cm $^{-3}$ (Table 1.1)
- Distilled water/dilute acid: $\approx 1.00$ g cm $^{-3}$ (Table 1.1 says distilled water is $1.00$ ; dilute HCl is slightly denser but $\approx 1.0$ )
- 1-phenylethanol: $1.01$ g cm $^{-3}$ (but it's dissolved in ethoxyethane, so the organic layer's density is dominated by ethoxyethane, which is much less dense than water).
Position: Ethoxyethane layer (organic, less dense) is the upper layer. Aqueous layer (water, more dense) is the lower layer.
Collection:
- Lower layer (aqueous) -> Beaker A.
- Upper layer (organic, contains product) -> Beaker B.
Conclusion: Beaker B contains 1-phenylethanol.

Key Takeaways

Less dense organic solvents (like ethoxyethane, density $< 1$ ) float on water. Denser organic solvents (like dichloromethane, density $> 1$ ) sink below water. Always check densities!

Common Mistakes

Assuming organic layers are always on top. (True for most, but not all, e.g., halogenated solvents are denser than water).
Forgetting to read step 7 carefully (lower -> A, upper -> B).

Things to Be Careful About

The explanation must mention 'ethoxyethane is less dense than water' (or similar) to get the mark.

Techniques used

determine layer position based on density

(i)

The overall reaction can be represented as shown in Fig. 1.3.

Fig. 1.3

(i)

At the end of step 8, $2.17\text{ g}$ of 1-phenylethanol is obtained.

Determine whether bromobenzene or ethanal is the limiting reagent and hence calculate the percentage yield of 1-phenylethanol. Show your working.

[ $M_r$ : $\text{C}_6\text{H}_5\text{CH(OH)CH}_3$ , 122.0]

2M Similar

DifficultyMedium

Worked solution

Working

Moles of reactants:

$n(\text{bromobenzene}) = 0.0500 \text{ mol}$ (given)
Mass of ethanal = volume $\times$ density = $3.00 \text{ cm}^3 \times 0.788 \text{ g cm}^{-3} = 2.364 \text{ g}$
$M_r(\text{CH}_3\text{CHO}) = 12.0 + 3(1.0) + 12.0 + 1.0 + 16.0 = 44.0$
$n(\text{ethanal}) = \frac{2.364}{44.0} = 0.0537 \text{ mol}$

Limiting reagent:
The stoichiometry is 1:1 (from the overall equation or reaction steps).
$0.0500 < 0.0537$ , so bromobenzene is the limiting reagent.

Theoretical yield:

Moles of 1-phenylethanol expected = moles of limiting reagent = $0.0500 \text{ mol}$
$M_r(\text{C}_6\text{H}_5\text{CH(OH)CH}_3) = 122.0$ (given)
Theoretical mass = $0.0500 \text{ mol} \times 122.0 \text{ g mol}^{-1} = 6.10 \text{ g}$

Percentage yield:

\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\% = \frac{2.17 \text{ g}}{6.10 \text{ g}} \times 100\% = 35.6\%

Answer

Limiting reagent: bromobenzene. Percentage yield: $35.6\%$ .

Final answer

Bromobenzene is limiting; percentage yield is 35.6%.

Detailed explanation

Background Concept

Percentage yield is a measure of the efficiency of a reaction.

$\text{Percentage yield} = \frac{\text{Actual yield}}{\text{Theoretical yield}} \times 100\%$
Theoretical yield is calculated from the limiting reagent.
Limiting reagent is the reactant that produces the least amount of product (or is completely consumed first).

Understanding the Question

Actual yield = $2.17$ g of 1-phenylethanol.
Given: $0.0500$ mol bromobenzene, $3.00$ cm $^3$ ethanal.
Need to: Find limiting reagent, calculate theoretical yield, calculate percentage yield.

Approach

Calculate moles of ethanal (need mass first using density).
Compare moles of bromobenzene and ethanal (1:1 ratio).
Identify limiting reagent.
Calculate theoretical mass of product.
Calculate percentage yield.

Step-by-Step Reasoning

Moles of bromobenzene: Given as $0.0500$ mol.
Moles of ethanal:
- Density = $0.788$ g cm $^{-3}$ , Volume = $3.00$ cm $^3$ .
- Mass = $3.00 \times 0.788 = 2.364$ g.
- $M_r(\text{CH}_3\text{CHO}) = 2(12.0) + 4(1.0) + 16.0 = 44.0$ .
- $n = 2.364 / 44.0 = 0.053727... \approx 0.0537$ mol.
Limiting reagent:
- Reaction ratio is 1:1 (1 mol bromobenzene reacts with 1 mol ethanal).
- $0.0500$ mol bromobenzene requires $0.0500$ mol ethanal.
- We have $0.0537$ mol ethanal, which is more than enough.
- Therefore, bromobenzene is limiting.
Theoretical yield:
- 1 mol bromobenzene -> 1 mol 1-phenylethanol.
- $0.0500$ mol bromobenzene -> $0.0500$ mol 1-phenylethanol.
- $M_r(\text{product}) = 122.0$ .
- Theoretical mass = $0.0500 \times 122.0 = 6.10$ g.
Percentage yield:
- $(2.17 / 6.10) \times 100 = 35.573... \approx 35.6\%$ .

Key Takeaways

Always calculate moles of all reactants to find the limiting reagent. Use the $M_r$ of the product to find theoretical mass. Check significant figures (3 s.f. here).

Common Mistakes

Using the wrong $M_r$ for ethanal ( $44.0$ , not $46.0$ which is ethanol).
Forgetting to convert volume of ethanal to mass using density before finding moles.
Calculating percentage yield as (theoretical/actual) instead of (actual/theoretical).
Rounding intermediate values too early (keep $0.053727$ for calculation, round at end).

Things to Be Careful About

The mark scheme shows the working: $n(\text{ethanal}) = 3.00 \times 0.788 / 44.0 = 0.0537$ . Ensure you show this substitution. The final answer $35.6\%$ is to 3 s.f.

Techniques used

determine limiting reagentcalculate percentage yield

(ii)

Suggest why the infrared spectrum of the product detected the presence of a $\text{C}=\text{O}$ peak.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The product contains unreacted ethanal (which has a C=O group).

Final answer

The product contains unreacted ethanal.

Detailed explanation

Background Concept

Infrared (IR) spectroscopy identifies functional groups by their absorption of specific frequencies. A C=O (carbonyl) peak appears around $1700-1750$ cm $^{-1}$ .

1-phenylethanol: Contains an -OH group (broad peak $3200-3600$ cm $^{-1}$ ) and C-O, but no C=O.
Ethanal: Contains a C=O group (sharp peak $\sim 1720$ cm $^{-1}$ ).

Understanding the Question

The IR spectrum of the final product (after evaporation of ethoxyethane) shows a C=O peak. The pure product (1-phenylethanol) should not have a C=O peak. Why is it there?

Approach

Think about what impurities could be present. The reaction uses ethanal. If the reaction is not 100% complete, unreacted ethanal will remain.

Step-by-Step Reasoning

Pure 1-phenylethanol has an -OH group and a benzene ring, but no carbonyl (C=O) group.
The reactant ethanal ( $\text{CH}_3\text{CHO}$ ) has a C=O group.
If the IR spectrum shows a C=O peak, it means a carbonyl-containing compound is present.
The most likely source is unreacted ethanal that was not completely converted or was not removed during purification (evaporation of ethoxyethane might not remove all volatile ethanal, or it could be co-evaporating, though ethanal bp is $21 ^\circ\text{C}$ and ethoxyethane is $35 ^\circ\text{C}$ , so ethanal should evaporate first... wait. If ethanal is volatile (bp $21$ ), it should evaporate. However, it might be trapped or the question implies incomplete reaction. The mark scheme says 'the product contains unreacted ethanal'. This is the standard answer: impurity from starting material).
Actually, ethanal is very volatile (bp $21 ^\circ\text{C}$ ). If it's unreacted, it should evaporate in step 8. But perhaps it's trapped in the solid/liquid or the question simply asks for the chemical reason a C=O peak could appear, which is the presence of unreacted ethanal. The mark scheme is simple: 'the product contains unreacted ethanal'.

Key Takeaways

IR spectroscopy is used to check purity. A peak for a functional group not in the product indicates an impurity containing that group (usually unreacted starting material).

Common Mistakes

Saying 'the product has a C=O group'. No, 1-phenylethanol is an alcohol.
Saying 'ethoxyethane has a C=O'. No, it's an ether (C-O-C).
Not naming the specific impurity (ethanal).

Things to Be Careful About

The mark scheme is brief: 'the product contains unreacted ethanal'. State this clearly.

Techniques used

interpret IR spectrum for impurities

Q2MediumPlanningAnalysis, Conclusions and Evaluation

An experiment is carried out to determine the rate constant, $k$ , for the hydrolysis of ethyl ethanoate, $\text{CH}_3\text{COOC}_2\text{H}_5$ , using a hydrochloric acid, $\text{HCl(aq)}$ , catalyst.

The equation for the reaction is shown.

\text{CH}_3\text{COOC}_2\text{H}_5 + \text{H}_2\text{O} \xrightarrow{\text{H}^+} \text{CH}_3\text{COOH} + \text{C}_2\text{H}_5\text{OH}

The rate equation for this reaction is shown.

\text{rate of reaction} = k [\text{CH}_3\text{COOC}_2\text{H}_5]

The progress of the reaction is followed by determining how the concentration of acid changes with time.

A portion of the reaction mixture is removed every 5 minutes and titrated with sodium hydroxide, $\text{NaOH(aq)}$ . A final titration is carried out after 180 minutes.

A student carries out the following steps.

step 1 Add $70\text{ cm}^3$ of iced water to seven separate small conical flasks. Add a few drops of phenolphthalein indicator to each flask. Phenolphthalein is pink in alkaline conditions and colourless in acidic conditions.
step 2 Use a measuring cylinder to transfer $100\text{ cm}^3$ of $0.200\text{ mol dm}^{-3}\ \text{HCl(aq)}$ into a large conical flask.
step 3 Add $5.00\text{ cm}^3$ of $\text{CH}_3\text{COOC}_2\text{H}_5$ to the large conical flask and swirl the flask to mix the contents. Start a stopwatch.
step 4 Transfer $10.00\text{ cm}^3$ of reaction mixture to one of the small conical flasks containing the iced water and indicator. Record the time. Shake the small flask.
step 5 Carry out a single titration of the mixture in the small conical flask using $0.15\text{ mol dm}^{-3}\ \text{NaOH(aq)}$ .
step 6 Repeat steps 4 and 5 at the times shown in Table 2.1 using a different small conical flask for each titration.

(a)

Give two reasons that explain why the use of iced water in step 4 decreases the rate of reaction.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The iced water dilutes the reaction mixture, decreasing the concentration of reactants and thus the rate of reaction.
The iced water reduces the temperature of the reaction mixture, decreasing the kinetic energy of the particles and thus the rate of reaction.

Final answer

The iced water dilutes the mixture (decreasing concentration) and reduces the temperature.

Detailed explanation

Background Concept

The rate of a chemical reaction depends on the frequency of effective collisions between reactant particles. Two key factors that influence this frequency are concentration and temperature. Increasing concentration increases the number of particles per unit volume, leading to more frequent collisions. Increasing temperature increases the average kinetic energy of the particles, meaning a greater proportion of them have energy exceeding the activation energy, and they also move faster, increasing collision frequency.

Understanding the Question

The student adds iced water to conical flasks before transferring samples of the reaction mixture into them. This step is intended to 'quench' or stop the reaction so that the titration can be performed. The question asks why this specific choice (iced water) decreases the rate of reaction.

Approach

We need to identify the two physical properties of iced water that affect reaction kinetics: its effect on concentration (dilution) and its effect on temperature (cooling).

Step-by-Step Reasoning

Dilution: The reaction mixture (ethyl ethanoate, water, HCl catalyst) is transferred into flasks already containing 70 cm³ of water. This significantly increases the total volume, thereby diluting the reactants and the catalyst. A lower concentration means fewer particles per unit volume, reducing the collision frequency and the rate.
Temperature reduction: 'Iced' water is at approximately 0 °C. The reaction is likely being carried out at room temperature or higher. Transferring the mixture into iced water rapidly cools it down. Lower temperature means particles have less kinetic energy, resulting in fewer particles exceeding the activation energy and a slower rate.

Key Takeaways

Quenching a reaction often involves dilution and cooling. Both actions serve to slow or stop the reaction by reducing concentration and temperature, respectively.

Common Mistakes

Stating 'the water stops the reaction' without explaining why (must mention concentration or temperature).
Forgetting that the catalyst (H⁺) is also diluted, which reduces its concentration and thus the rate.

Things to Be Careful About

Ensure both reasons are distinct: one must relate to concentration/dilution and the other to temperature. Do not just say 'it cools it down' without linking temperature to rate.

Techniques used

explain effect of dilution on reaction rateexplain effect of temperature on reaction rate

(b)

Describe the observation that is used to determine the end-point of the titrations.

1M Similar

DifficultyEasy

Worked solution

Answer

Appearance of a permanent pink colour.

Final answer

appearance of a permanent pink colour

Detailed explanation

Background Concept

Phenolphthalein is a common acid-base indicator. It is colourless in acidic solutions (pH < 8.2) and pink in alkaline solutions (pH > 8.2). In this titration, the conical flask contains the quenched acidic reaction mixture (excess HCl from the catalyst) and indicator. The burette contains NaOH(aq), an alkali.

Understanding the Question

The student titrates the acidic sample with NaOH. The question asks for the observation that signals the end-point (the point at which the acid has been exactly neutralised and a slight excess of alkali is present).

Approach

Consider the initial and final conditions of the solution in the conical flask. Initially, it is acidic (colourless). As NaOH is added, it neutralises the acid. At the end-point, the acid is consumed, and the next drop of NaOH makes the solution slightly alkaline, changing the indicator's colour.

Step-by-Step Reasoning

Initially, the solution in the flask is acidic due to the HCl catalyst, so phenolphthalein is colourless.
As NaOH is added from the burette, it reacts with the HCl: $\text{H}^+ + \text{OH}^- \rightarrow \text{H}_2\text{O}$ .
At the end-point, all the $\text{H}^+$ has been neutralised. The next drop of NaOH makes the solution slightly alkaline.
Phenolphthalein turns pink in alkaline conditions.
The observation must be 'permanent' pink to ensure the end-point has been truly reached and not just a localised colour change that disappears on shaking.

Key Takeaways

Always specify 'permanent' or 'first permanent' when describing titration end-points with indicators to show you understand the need for a stable colour change.

Common Mistakes

Saying 'the solution turns pink' without specifying 'permanent'.
Confusing the indicator colour change direction (e.g., saying colourless to yellow, which is for methyl orange).

Things to Be Careful About

The question states phenolphthalein is pink in alkaline and colourless in acidic. Use this information to deduce the correct colour change direction (colourless to pink).

Techniques used

identify titration end-point observation

(c)

The results of the experiment are shown in Table 2.1.

$V_t$ is the titre at a given time, $t$ .
$V_\infty$ is the final titre at $t = 180\text{ min}$ when the reaction is assumed to be complete.

Table 2.1

1	2	3	4
time, $t$ / min	titre, $V_t$ / $\text{cm}^3$	$(V_\infty - V_t)$ / $\text{cm}^3$	$\log[(V_\infty - V_t) / \text{cm}^3]$
0	12.00
5	16.40
10	22.15
15	23.45
20	26.30
25	28.70
180	45.70

(i)

Complete Table 2.1.

Give your answers in column 3 to two decimal places and your answers in column 4 to four significant figures.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

$V_\infty = 45.70 \text{ cm}^3$ (from the final titration at $t = 180$ min)

Column 3: $(V_\infty - V_t) = 45.70 - V_t$
Column 4: $\log[(V_\infty - V_t) / \text{cm}^3]$

Time, $t$ / min	Titre, $V_t$ / cm³	$(V_\infty - V_t)$ / cm³	$\log[(V_\infty - V_t) / \text{cm}^3]$
0	12.00	33.70	1.528
5	16.40	29.30	1.467
10	22.15	23.55	1.372
15	23.45	22.25	1.347
20	26.30	19.40	1.288
25	28.70	17.00	1.230
180	45.70	—	—

Answer

See completed table above.

Final answer

33.70, 1.528; 29.30, 1.467; 23.55, 1.372; 22.25, 1.347; 19.40, 1.288; 17.00, 1.230

Detailed explanation

Background Concept

In kinetics experiments, we often measure a quantity that is proportional to the concentration of a reactant or product over time. Here, the titre $V_t$ is proportional to the amount of acid present. Since the reaction produces acid (ethanoic acid) from ethyl ethanoate, the total acid increases over time. The quantity $(V_\infty - V_t)$ is proportional to the amount of acid remaining to be produced, which is directly proportional to the concentration of the unreacted ethyl ethanoate. Taking the logarithm of this quantity linearises the first-order rate equation.

Understanding the Question

The student must complete a data table by calculating the difference between the final titre and each intermediate titre, and then calculating the base-10 logarithm of that difference. Column 3 must be to two decimal places, and column 4 to four significant figures.

Approach

Identify $V_\infty$ from the final row (t = 180 min): $V_\infty = 45.70 \text{ cm}^3$ .
For each row, calculate $45.70 - V_t$ and round to two decimal places.
Calculate $\log_{10}$ of the result from step 2 and round to four significant figures.

Step-by-Step Reasoning

t = 0: $45.70 - 12.00 = 33.70$ . $\log(33.70) = 1.52760... \rightarrow 1.528$ (4 s.f.)
t = 5: $45.70 - 16.40 = 29.30$ . $\log(29.30) = 1.46687... \rightarrow 1.467$ (4 s.f.)
t = 10: $45.70 - 22.15 = 23.55$ . $\log(23.55) = 1.37195... \rightarrow 1.372$ (4 s.f.)
t = 15: $45.70 - 23.45 = 22.25$ . $\log(22.25) = 1.34735... \rightarrow 1.347$ (4 s.f.)
t = 20: $45.70 - 26.30 = 19.40$ . $\log(19.40) = 1.28778... \rightarrow 1.288$ (4 s.f.)
t = 25: $45.70 - 28.70 = 17.00$ . $\log(17.00) = 1.23044... \rightarrow 1.230$ (4 s.f.)

Key Takeaways

Always check the required significant figures or decimal places before calculating. For logarithms, the number of significant figures in the result should match the number of significant figures in the original number (here, 4 s.f. is explicitly requested).

Common Mistakes

Forgetting to use $V_\infty = 45.70$ and using a wrong value.
Rounding the logarithm to the wrong number of significant figures (e.g., 3 decimal places instead of 4 s.f.). Note that 1.230 has 4 s.f., not 3.

Things to Be Careful About

The final row (t = 180) does not require calculation for columns 3 and 4 because $V_\infty - V_\infty = 0$ , and $\log(0)$ is undefined. Leave these blank or put a dash.

Techniques used

calculate difference from final titrecalculate logarithm to correct significant figures

(ii)

Plot a graph on the grid in Fig. 2.1 to show the relationship between $\log[(V_\infty - V_t)/\text{cm}^3]$ and time, $t$ . Use a cross ( $\times$ ) to plot each data point.

Draw a straight line of best fit.

Fig. 2.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot the following six points on the grid in Fig. 2.1:

(0, 1.528)
(5, 1.467)
(10, 1.372)
(15, 1.347)
(20, 1.288)
(25, 1.230)

Draw a straight line of best fit that passes as close as possible to these points, with roughly equal numbers of points above and below the line.

Final answer

See graph with six points plotted and a straight line of best fit drawn.

Detailed explanation

Background Concept

For a first-order reaction, the integrated rate law is $[A] = [A]_0 e^{-kt}$ . Taking the logarithm gives $\log[A] = \log[A]_0 - \frac{kt}{2.303}$ . This is in the form $y = mx + c$ , where $y = \log[A]$ , $x = t$ , and the gradient $m = -\frac{k}{2.303}$ . Plotting $\log[(V_\infty - V_t)]$ against time should yield a straight line with a negative gradient.

Understanding the Question

The student must plot the data from Table 2.1 (columns 1 and 4) on the provided grid and draw a straight line of best fit.

Approach

Use the time values (column 1) as the x-axis coordinates.
Use the log values (column 4) as the y-axis coordinates.
Plot each point accurately using a sharp pencil and a cross (×) as specified.
Draw a straight line of best fit. Do not join the points with a straight line; the line should represent the overall trend.

Step-by-Step Reasoning

Plotting: Ensure the axes are read correctly. The x-axis ranges from 0 to 30 min. The y-axis ranges from 1.10 to 1.55. Plot each point carefully. For example, at t = 10, y = 1.372. This point is slightly above the line that would be drawn through the other points (it will be identified as anomalous in part (iii)).
Line of best fit: The line should minimize the distance to all points. Since the point at t = 10 is anomalous, the line should pass close to the other five points and may not pass through the t = 10 point. The line should extend across the range of the data.

Key Takeaways

A line of best fit is not a 'join-the-dots' line. It must reflect the overall trend, balancing points above and below the line. Anomalous points are excluded from determining the line's position.

Common Mistakes

Joining the points with a straight line or a curved line through all points.
Failing to plot the anomalous point (t = 10) or plotting it incorrectly.
Using a thick line that makes it hard to judge which side of the line the points are on.

Things to Be Careful About

The question explicitly asks to use a cross (×) to plot each data point. Do not use dots or circles. The line must be straight.

Techniques used

plot data points on graphdraw line of best fit

(iii)

Circle the one point on the graph that you consider to be most anomalous.

Suggest one reason to explain the anomalous point you have circled.

Assume no error was made in the experimental value of the titre.

1M Similar

DifficultyMedium

Worked solution

Answer

The anomalous point is at t = 10 min.
Reason: The actual time the sample was removed from the reaction mixture was greater than 10 minutes (the reaction continued for longer before being quenched by the iced water).

Final answer

t = 10 min; actual time was > 10 mins

Detailed explanation

Background Concept

In a kinetics experiment, samples are removed at specific time intervals and immediately quenched to stop the reaction. If a sample is not quenched promptly, the reaction continues in the sample tube, altering the concentration of reactants and products before titration.

Understanding the Question

The student must identify the point on the graph that deviates most from the line of best fit and suggest a procedural reason for this deviation, assuming the titre reading itself is correct.

Approach

Examine the graph or the data table to find the point that is furthest from the expected linear trend.
Analyze whether the titre $V_t$ is too high or too low for that time.
Relate the titre value to the amount of acid present and deduce what procedural error could cause this.

Step-by-Step Reasoning

Identifying the anomalous point: Looking at the log values, the drop from t=5 to t=10 is 1.467 - 1.372 = 0.095. The drops for other 5-minute intervals are roughly 0.060-0.061. The point at t = 10 min (log = 1.372) is significantly lower than expected. On the graph, it will be the point furthest below the line of best fit.
Analyzing the deviation: A lower log value means $(V_\infty - V_t)$ is smaller, which means $V_t$ is larger than it should be. A larger $V_t$ means more acid was present in the sample than expected at t = 10 min.
Deducing the error: More acid means the reaction produced more ethanoic acid than it should have in 10 minutes. This happens if the reaction was allowed to continue for longer than 10 minutes before being quenched. Therefore, the student took the sample late (e.g., at 12 or 13 minutes instead of 10 minutes), but recorded the time as 10 minutes.

Key Takeaways

When identifying errors from data, work backwards from the result: what would cause the measured value to be higher/lower than expected, and what procedural mistake could lead to that outcome?

Common Mistakes

Suggesting 'human error' or 'misreading the stopwatch' without explaining how it affected the result.
Suggesting the sample was taken too early (this would give a lower titre, not a higher one).
Forgetting to assume the titre reading is correct and instead blaming the titration.

Things to Be Careful About

The question states 'Assume no error was made in the experimental value of the titre'. This means the burette reading is correct, and the error is in the timing or quenching step.

Techniques used

identify anomalous data pointdeduce procedural error causing anomalous result

(iv)

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.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Select two points on the line of best fit that are far apart to minimize error.
Example points: $(0, 1.528)$ and $(25, 1.230)$

\text{gradient} = \frac{\Delta y}{\Delta x} = \frac{1.230 - 1.528}{25 - 0} = \frac{-0.298}{25} = -0.01192

Rounding to three significant figures: $-0.0119 \text{ min}^{-1}$

(Note: Using other points on your line may give a slightly different value, e.g., $-0.0312 \text{ min}^{-1}$ if using the anomalous point or a poorly drawn line. The mark scheme accepts a range based on the drawn line.)

Answer

Coordinates used: (0, 1.528) and (25, 1.230)
Gradient = $-0.0119 \text{ min}^{-1}$

Final answer

Gradient = -0.0119 min^-1 (using points (0, 1.528) and (25, 1.230))

Detailed explanation

Background Concept

The gradient of the line in a first-order kinetics plot of $\log[A]$ against time is equal to $-k / 2.303$ . To find $k$ , we first need the gradient. The gradient is calculated as $\frac{\text{change in y}}{\text{change in x}}$ using two points on the line of best fit, not necessarily the data points themselves.

Understanding the Question

The student must read coordinates from their drawn line of best fit and calculate the gradient to three significant figures.

Approach

Choose two points on the line of best fit that are as far apart as possible (to minimize percentage error in the gradient calculation).
Read the x and y coordinates for these points.
Calculate $\frac{y_2 - y_1}{x_2 - x_1}$ .
Round to three significant figures.

Step-by-Step Reasoning

Selecting points: Use points near the ends of the line. For example, at $x_1 = 0$ , read $y_1 \approx 1.528$ . At $x_2 = 25$ , read $y_2 \approx 1.230$ .
Calculation: $\text{gradient} = \frac{1.230 - 1.528}{25 - 0} = \frac{-0.298}{25} = -0.01192$ .
Significant figures: The question asks for three significant figures, so $-0.0119$ .
Units: The y-axis is dimensionless (log of a volume), and the x-axis is in minutes, so the gradient is in $\text{min}^{-1}$ .

Key Takeaways

Always use points on the line of best fit, not the raw data points (especially anomalous ones). Choose points that are far apart to reduce the impact of reading errors.

Common Mistakes

Using raw data points to calculate the gradient instead of points on the line of best fit.
Using points that are too close together, leading to a large percentage error.
Forgetting the units ( $\text{min}^{-1}$ ).
Rounding to the wrong number of significant figures.

Things to Be Careful About

The coordinates must be read from the line, not the data points. If the line is drawn poorly, the gradient will be wrong, which will affect the calculation of $k$ in part (d). The mark scheme allows for a range of gradients based on different valid lines of best fit.

Techniques used

read coordinates from line of best fitcalculate gradient of straight line

(d)

The equation of the straight line plotted in Fig. 2.1 is shown.

\log[(V_\infty - V_t)/\text{cm}^3] = -\frac{kt}{2.303} + \text{constant}

Use the gradient determined in (c)(iv) to calculate a value for $k$ in $\text{s}^{-1}$ .

[If you were unable to determine an answer to (c)(iv), then use the value $-0.0312\text{ min}^{-1}$ for the gradient. This is not the correct answer.]

2M Similar

DifficultyMedium

Worked solution

Working

The equation of the line is:

\log[(V_\infty - V_t)/\text{cm}^3] = -\frac{kt}{2.303} + \text{constant}

Comparing this to $y = mx + c$ , the gradient $m = -\frac{k}{2.303}$ .

Therefore:

k = -\text{gradient} \times 2.303

From part (c)(iv), gradient $= -0.0119 \text{ min}^{-1}$ .

k = -(-0.0119) \times 2.303 = 0.0274157 \text{ min}^{-1}

Convert to $\text{s}^{-1}$ by dividing by 60:

k = \frac{0.0274157}{60} = 4.57 \times 10^{-4} \text{ s}^{-1}

(If using the given value $-0.0312 \text{ min}^{-1}$ : $k = 0.0312 \times 2.303 / 60 = 1.20 \times 10^{-3} \text{ s}^{-1}$ )

Answer

$k = 4.57 \times 10^{-4} \text{ s}^{-1}$

Final answer

4.57e-4 s^-1

Detailed explanation

Background Concept

For a first-order reaction, the integrated rate law in logarithmic form is $\log[A] = -\frac{kt}{2.303} + \log[A]_0$ . The gradient of the plot of $\log[A]$ against $t$ is $-\frac{k}{2.303}$ . To find the rate constant $k$ , we rearrange this relationship. The units of the gradient will be the inverse of the time unit used on the x-axis (here, $\text{min}^{-1}$ ). The question asks for $k$ in $\text{s}^{-1}$ , so a unit conversion is required.

Understanding the Question

The student must use the gradient calculated in part (c)(iv) to find $k$ , and ensure the final answer is in $\text{s}^{-1}$ .

Approach

Rearrange the given equation to solve for $k$ in terms of the gradient.
Substitute the gradient value from part (c)(iv).
Convert the time unit from minutes to seconds.

Step-by-Step Reasoning

Rearrangement: $\text{gradient} = -\frac{k}{2.303} \Rightarrow k = -\text{gradient} \times 2.303$ .
Substitution: Using gradient $= -0.0119 \text{ min}^{-1}$ :
$k = -(-0.0119) \times 2.303 = 0.0274157 \text{ min}^{-1}$ .
Unit conversion: Since $1 \text{ min} = 60 \text{ s}$ , $k \text{ (in s}^{-1}) = \frac{k \text{ (in min}^{-1})}{60}$ .
$k = \frac{0.0274157}{60} = 4.569 \times 10^{-4} \text{ s}^{-1}$ .
Rounding: To three significant figures, $k = 4.57 \times 10^{-4} \text{ s}^{-1}$ .

Key Takeaways

Always check the required units for the final answer. If the gradient is in $\text{min}^{-1}$ and $k$ is required in $\text{s}^{-1}$ , divide by 60. Do not forget to convert.

Common Mistakes

Forgetting to convert from $\text{min}^{-1}$ to $\text{s}^{-1}$ (giving an answer 60 times too large).
Multiplying by 60 instead of dividing.
Using the wrong sign for the gradient (forgetting that $k$ must be positive).

Things to Be Careful About

The question provides a fallback value ( $-0.0312 \text{ min}^{-1}$ ) if you couldn't calculate the gradient. If you use this, your $k$ will be different ( $1.20 \times 10^{-3} \text{ s}^{-1}$ ), but you will still get method marks if your working is correct.

Techniques used

relate gradient to rate constantconvert units from min^-1 to s^-1

(e)

Use your graph in Fig. 2.1 to state whether you consider the results to be reliable. Give a reason for your answer.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The results are not reliable.
Reason: There is an anomalous point (at t = 10 min), which indicates an error in the procedure and reduces the reliability of the results.

Final answer

not reliable; there is an anomalous point

Detailed explanation

Background Concept

Reliability in an experiment refers to the consistency and reproducibility of the results. If there are anomalous points (outliers) that deviate significantly from the expected trend, it suggests that there were errors in the procedure for those specific measurements. This reduces the overall reliability of the data set because the line of best fit may not accurately represent the true relationship.

Understanding the Question

The student must state whether the results are reliable and give a reason based on the graph or data.

Approach

Look at the graph or data table for any anomalous points.
If there is an anomalous point, state that the results are not reliable.
Give the reason: the presence of the anomalous point indicates an error.

Step-by-Step Reasoning

In part (c)(iii), the student identified an anomalous point at t = 10 min.
The presence of this point means that at least one measurement was affected by a procedural error (taking the sample late).
Therefore, the data set is not fully reliable because not all points follow the expected trend.

Key Takeaways

An anomalous point reduces the reliability of the results. Always mention this when evaluating data.

Common Mistakes

Saying the results are reliable because 'most points are on the line'. (The presence of an outlier still reduces reliability).
Giving a vague reason like 'there was an error' without specifying that there is an anomalous point.

Things to Be Careful About

The question asks to 'state whether you consider the results to be reliable'. A simple 'no' is not enough; you must give a reason. The reason must be linked to the anomalous point.

Techniques used

evaluate reliability of experimental data

(f)

$k$ increases with temperature.

A second experiment is carried out at a higher temperature.

Sketch a suggested line of best fit on Fig. 2.1 for the second experiment.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Sketch a straight line on Fig. 2.1 with:

The same y-intercept (starting at the same point at t = 0).
A more negative gradient (steeper line, falling faster).

Final answer

straight line with same intercept and more negative gradient

Detailed explanation

Background Concept

The rate constant $k$ increases with temperature (as described by the Arrhenius equation). From the equation $\log[(V_\infty - V_t)] = -\frac{kt}{2.303} + \text{constant}$ , the gradient is $-\frac{k}{2.303}$ . If $k$ increases, the magnitude of the gradient increases (it becomes more negative), meaning the line is steeper. The y-intercept is $\log[(V_\infty - V_0)]$ , which depends only on the initial concentration of ethyl ethanoate. If the initial concentrations are the same, the y-intercept will be the same.

Understanding the Question

A second experiment is carried out at a higher temperature. The student must sketch the expected line of best fit on the same graph.

Approach

Determine how the y-intercept changes (it shouldn't, as initial conditions are the same).
Determine how the gradient changes (it should become more negative because $k$ increases).
Sketch the new line.

Step-by-Step Reasoning

Intercept: At $t = 0$ , the concentration of ethyl ethanoate is the same in both experiments (assuming the same initial amounts were used). Therefore, $\log[(V_\infty - V_0)]$ is the same. The new line must start at the same y-intercept (approximately 1.528).
Gradient: At a higher temperature, the reaction is faster, so $k$ is larger. The gradient is $-\frac{k}{2.303}$ . A larger $k$ means a more negative gradient (e.g., if the original gradient was $-0.0119$ , the new gradient might be $-0.0200$ ). The line will fall more steeply.
Sketch: Draw a straight line starting at the same y-intercept (0, 1.528) but with a steeper negative slope. It should reach lower log values at the same times compared to the original line.

Key Takeaways

Increasing temperature increases $k$ , which increases the magnitude of the negative gradient on a first-order kinetics plot. The initial concentration (y-intercept) is unaffected by temperature.

Common Mistakes

Changing the y-intercept (this would imply a different initial concentration).
Drawing a line with a less negative gradient (this would imply a slower reaction, which is wrong for higher temperature).
Drawing a curved line (the relationship is still linear for a first-order reaction).

Things to Be Careful About

The sketch should be qualitative. You don't need to calculate the exact new gradient, just show that it is steeper and starts at the same point. Ensure the line is straight.

Techniques used

predict effect of temperature on kinetics graph