Chemistry 9701/53 — May/June 2020

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

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Topics Planning · Analysis, Conclusions and Evaluation

Q1PlanningAnalysis, Conclusions and EvaluationFree sample

Trichloromethane and propanone are both organic liquids. The molecules within each liquid are attracted to each other by relatively weak permanent dipole-dipole interactions.

When trichloromethane is mixed with propanone a strong electrostatic attraction forms between the two different molecules.

Fig. 1.1

A student plans to perform an experiment to investigate the strength of this electrostatic attraction by finding the temperature change when equal volumes of trichloromethane and propanone are mixed together.

(a)

(i)

State and explain your prediction for the temperature change for this experiment.

DifficultyMedium-Easy

Worked solution

Answer

Temperature increases.
The electrostatic attraction between trichloromethane and propanone molecules is stronger than the permanent dipole-dipole interactions within each pure liquid, so energy is released when the new attractions form.

Final answer

Temperature increases; the electrostatic attraction between the different molecules is stronger than the attractions within each pure liquid.

Detailed explanation

Background Concept

When two liquids are mixed, the temperature change depends on the balance of energy required to break the intermolecular forces within the pure liquids and the energy released when new intermolecular forces form between the different molecules. If the new attractions are stronger, more energy is released than is absorbed, resulting in an exothermic process and a temperature increase. If they are weaker, the process is endothermic and the temperature decreases.

Understanding the Question

The question asks for a prediction of the temperature change when trichloromethane and propanone are mixed, along with an explanation. The stem tells us that pure liquids have 'relatively weak permanent dipole-dipole interactions', but mixing them forms a 'strong electrostatic attraction' between the different molecules.

Approach

Compare the strength of the intermolecular forces in the pure liquids with those in the mixture. Stronger forces in the mixture mean energy is released as heat.

Step-by-Step Reasoning

Prediction: The mixture forms a strong electrostatic attraction (shown in Fig. 1.1 between the H of trichloromethane and the O of propanone). Because this new attraction is stronger than the original dipole-dipole interactions in the pure liquids, the formation of these new bonds releases net energy to the surroundings.
Explanation: Energy is released when strong intermolecular attractions form. Since the attraction between unlike molecules is stronger than the attractions within each pure liquid, the overall process is exothermic, causing the temperature of the mixture to increase.

Key Takeaways

Mixing liquids that form stronger intermolecular attractions than those present in the pure liquids results in an exothermic process and a rise in temperature.

Common Mistakes

Predicting a temperature decrease because 'mixing is usually endothermic'. This is not a universal rule; it depends entirely on the relative strengths of the intermolecular forces.
Forgetting to explain why the temperature changes by simply stating 'heat is released' without referencing the relative strengths of the intermolecular attractions.

Things to Be Careful About

Ensure the explanation explicitly compares the strength of the new intermolecular forces to the old ones. Mark schemes require both the prediction (temperature increases) and the comparative explanation to score full marks.

Techniques used

predict temperature change based on intermolecular forcesevaluate experimental outcome

(ii)

The student is given only the following equipment and chemicals for the experiment.

1 × 25 cm³ beaker
2 × thermometers
2 × 25 cm³ measuring cylinders
50 cm³ trichloromethane
50 cm³ propanone

Outline the method the student should use in this one experiment to find the temperature change when trichloromethane is mixed with propanone. Give details of the volumes of liquids used and any readings taken.

volumes used

readings taken

method used

DifficultyMedium

Worked solution

Answer

volumes used:
12.5 cm³ of trichloromethane and 12.5 cm³ of propanone (any two equal volumes totalling less than 25 cm³, e.g., 10 cm³ each).

readings taken:
Initial temperature of each liquid before mixing, and the maximum temperature reached after mixing.

method used:
Measure the equal volumes using the measuring cylinders and leave them to equilibrate to room temperature. Pour both liquids into the 25 cm³ beaker. Insert a thermometer, stir the mixture, and record the temperature every minute until it reaches a maximum and begins to fall.

Final answer

Use 12.5 cm³ of each liquid (total < 25 cm³). Measure initial temperatures, mix in the beaker, and record the maximum temperature reached.

Detailed explanation

Background Concept

A simple calorimetry experiment to measure enthalpy changes or temperature changes from mixing involves combining known volumes of reactants in an insulated container and monitoring the temperature. Key considerations include ensuring the container is not overfilled, allowing liquids to reach thermal equilibrium before mixing, and capturing the peak temperature before heat loss to the surroundings causes cooling.

Understanding the Question

The student has a 25 cm³ beaker, two thermometers, two 25 cm³ measuring cylinders, and 50 cm³ of each liquid. The task is to outline a method to find the temperature change, specifying volumes used, readings taken, and the procedure.

Approach

Volumes: Must be equal (as per the experiment design) and must not exceed the beaker's capacity (total < 25 cm³). So, 12.5 cm³ each is ideal.
Readings: Need the initial temperature (baseline) and the final temperature (maximum, since heat will be lost after the peak).
Method: Logical sequence: measure -> equilibrate -> mix -> monitor -> record peak.

Step-by-Step Reasoning

Volumes used: The experiment requires equal volumes. Using 12.5 cm³ of trichloromethane and 12.5 cm³ of propanone gives a total of 25 cm³, which fits in the beaker. (Any equal volumes totalling < 25 cm³, like 10 cm³ each, are acceptable).
Readings taken: Before mixing, measure the initial temperature of both liquids to ensure they are at the same starting temperature (room temperature). After mixing, the temperature will rise and then fall due to heat loss. The maximum temperature reached is the true final temperature for the calculation.
Method used:
1. Use the measuring cylinders to measure 12.5 cm³ of each liquid.
2. Leave them in the room to equilibrate so both are at the same initial temperature.
3. Pour both liquids into the 25 cm³ beaker.
4. Insert a thermometer and stir continuously.
5. Record the temperature every minute until it reaches a peak and starts to fall. The highest reading is the maximum temperature.

Key Takeaways

When designing a simple mixing experiment, ensure the total volume fits the container, allow for thermal equilibration, and record the maximum temperature to account for inevitable heat loss.

Common Mistakes

Suggesting volumes that total more than 25 cm³ (e.g., 20 cm³ each), which would overflow the beaker.
Forgetting to leave the liquids to equilibrate before mixing, leading to an inaccurate initial temperature reading.
Recording the temperature at a fixed time (e.g., 'after 5 minutes') instead of the maximum temperature, which would underestimate the temperature change due to heat loss.

Things to Be Careful About

The mark scheme explicitly requires 'two equal volumes' and 'total volume < 25 cm³'. Be precise in stating the volumes. Also, 'monitor temperature every minute until it starts to fall' or 'read the maximum temperature' are key phrases that earn the final method mark.

Techniques used

design a simple calorimetry experimentcontrol volumes and measure temperature changeidentify the maximum temperature

(b)

The apparatus used leads to significant heat loss.

State one improvement the student could make to the apparatus to reduce heat loss.

DifficultyEasy

Worked solution

Answer

Use a polystyrene cup instead of a glass beaker (or cover the beaker with a lid / insulate the beaker).

Final answer

Use a polystyrene cup instead of a glass beaker (or cover with a lid).

Detailed explanation

Background Concept

In simple calorimetry experiments, heat loss to the surroundings is the largest source of error. Glass beakers are poor insulators; heat is lost through the glass by conduction and through the open top by convection and evaporation. Polystyrene is an excellent thermal insulator, and lids prevent convective heat loss and evaporation.

Understanding the Question

The apparatus (a glass beaker) leads to significant heat loss. State one improvement to reduce this.

Approach

Identify a better insulating container or a way to trap heat at the top.

Step-by-Step Reasoning

Polystyrene cup: Polystyrene contains trapped air and is a much poorer thermal conductor than glass. Replacing the beaker with a polystyrene cup significantly reduces conductive heat loss.
Lid: Covering the beaker with a lid (or a piece of cardboard with holes for the thermometer) reduces heat loss by convection and evaporation from the surface.
Insulation: Wrapping the glass beaker in insulating material (like cotton wool or foil) also reduces heat loss.

Key Takeaways

Polystyrene cups and lids are standard improvements for simple calorimetry experiments to minimize heat loss to the surroundings.

Common Mistakes

Suggesting 'use a better thermometer'. This improves accuracy of temperature reading but does not reduce heat loss.
Vague answers like 'insulate the beaker' without specifying how (e.g., 'wrap in cotton wool'). 'Use a polystyrene cup' or 'cover with a lid' are preferred specific answers.

Things to Be Careful About

The question specifically asks to reduce heat loss. Ensure the improvement directly addresses thermal insulation, not measurement precision.

Techniques used

identify heat loss pathways in calorimetrypropose insulation improvements

(c)

Trichloromethane and propanone are both volatile and flammable.

State one relevant precaution that should be taken when carrying out this experiment.

DifficultyEasy

Worked solution

Answer

Ensure there are no naked flames nearby (or use an electric heater instead of a Bunsen burner).

Final answer

Perform the experiment away from naked flames (or use an electric heater).

Detailed explanation

Background Concept

Volatile liquids produce flammable vapours at room temperature. If these vapours mix with air and encounter an ignition source (naked flame, hot surface, spark), they can ignite or explode. Standard safety protocols for flammable liquids require eliminating ignition sources.

Understanding the Question

Both trichloromethane and propanone are volatile and flammable. State one relevant precaution.

Approach

Identify the ignition risk and state the corresponding safety measure.

Step-by-Step Reasoning

Risk: Volatile flammable liquids release vapours that can travel and ignite near a flame.
Precaution: Remove all naked flames (Bunsen burners, matches) from the vicinity of the experiment. If heating is needed (not in this case, but generally), use an electric heater which does not produce a naked flame.

Key Takeaways

Always state 'no naked flames' or 'use an electric heater' when dealing with volatile flammable organic liquids.

Common Mistakes

Suggesting 'wear gloves' or 'wear goggles'. While good general lab practice, these do not address the specific risk of flammability and volatility highlighted in the question.
'Do not breathe the vapours'. While trichloromethane is toxic, the question specifically highlights 'volatile and flammable', so the expected answer relates to fire safety.

Things to Be Careful About

Match the precaution to the specific hazard mentioned in the question stem ('volatile and flammable').

Techniques used

identify fire hazards from volatile flammable liquidsapply safety precautions

(d)

State one change, apart from reducing heat loss, that could be made to improve the accuracy of this experiment.

DifficultyEasy

Worked solution

Answer

Use a burette or graduated pipette instead of a measuring cylinder (or use a more accurate thermometer / use larger volumes of liquids).

Final answer

Use a burette or graduated pipette to measure the volumes (or use a more accurate thermometer).

Detailed explanation

Background Concept

Accuracy in calorimetry depends on the precision of all measurements: volumes, masses, and temperatures. Measuring cylinders have relatively low accuracy (e.g., ±0.5 cm³ for a 25 cm³ cylinder). Burettes and graduated pipettes offer much higher precision (±0.05 cm³ or better). Additionally, using larger volumes increases the total thermal mass, resulting in a larger temperature change for the same enthalpy change, which reduces the percentage uncertainty in the temperature reading.

Understanding the Question

State one change, apart from reducing heat loss, to improve the accuracy of the experiment.

Approach

Think about sources of measurement error: volume measurement and temperature measurement.

Step-by-Step Reasoning

Volume measurement: Measuring cylinders are not very accurate. Using a burette or graduated pipette to measure the 12.5 cm³ volumes will reduce the uncertainty in the amount of reactants.
Temperature measurement: A standard laboratory thermometer might only read to the nearest 0.5 °C or 1 °C. A more accurate thermometer (reading to 0.1 °C) reduces the percentage uncertainty in the temperature change.
Larger volumes: Using larger volumes (e.g., 20 cm³ each, if the container allows) means more molecules are reacting, releasing more total heat. This produces a larger temperature change, making the relative error in the temperature reading smaller.

Key Takeaways

Accuracy can be improved by using more precise measuring instruments (burettes, accurate thermometers) or by increasing the scale of the reaction to produce a larger measurable signal.

Common Mistakes

Suggesting 'repeat the experiment and take an average'. This improves precision (reliability), not accuracy.
Suggesting 'insulate the beaker'. The question explicitly says 'apart from reducing heat loss'.

Things to Be Careful About

Distinguish between accuracy (reducing systematic or random errors in measurement) and reliability (repeating to find anomalies). The question asks for accuracy.

Techniques used

improve volumetric measurement accuracyimprove temperature measurement accuracyincrease signal-to-noise ratio in calorimetry

(e)

In another experiment, a student uses 37.50 g of trichloromethane and 19.75 g of propanone and determines that the energy released is 1.67 kJ.

(i)

Calculate the number of moles of each compound in this mixture.

$M_r$ trichloromethane = 119.5
$M_r$ propanone = 58.0

moles of trichloromethane = .............................. mol

moles of propanone = .............................. mol

DifficultyEasy

Worked solution

Working

\text{moles of trichloromethane} = \frac{37.50}{119.5} = 0.314 \text{ mol}

\text{moles of propanone} = \frac{19.75}{58.0} = 0.341 \text{ mol}

Answer

moles of trichloromethane = 0.314 mol
moles of propanone = 0.341 mol

Final answer

trichloromethane: 0.314 mol; propanone: 0.341 mol

Detailed explanation

Background Concept

The number of moles ( $n$ ) of a substance is calculated by dividing its mass ( $m$ ) in grams by its relative molecular mass ( $M_r$ ):

n = \frac{m}{M_r}

This is a fundamental calculation in stoichiometry and energetics.

Understanding the Question

Given masses of trichloromethane (37.50 g) and propanone (19.75 g), and their $M_r$ values (119.5 and 58.0 respectively), calculate the number of moles of each.

Approach

Apply the formula $n = m / M_r$ for each compound.

Step-by-Step Reasoning

Trichloromethane: $n = 37.50 / 119.5 = 0.313807... \approx 0.314$ mol (to 3 s.f.).
Propanone: $n = 19.75 / 58.0 = 0.340517... \approx 0.341$ mol (to 3 s.f.).

Key Takeaways

Always use the correct $M_r$ and ensure mass is in grams. Round to an appropriate number of significant figures (usually 3 for A-Level data).

Common Mistakes

Using atomic mass instead of molecular mass (not applicable here as $M_r$ is given).
Arithmetic errors in division.
Incorrect significant figures (e.g., 0.3138 instead of 0.314).

Things to Be Careful About

The mark scheme accepts 0.314 and 0.341. Ensure calculations are carried to at least 3 significant figures to avoid rounding errors in subsequent parts.

Techniques used

calculate moles from mass and relative molecular mass

(ii)

Calculate the enthalpy change, $\Delta H$ , of the electrostatic attraction formed between trichloromethane and propanone. You must include a sign in your answer.

\Delta H = \text{.............................. kJ mol}^{-1}

DifficultyMedium

Worked solution

Working

The limiting factor is trichloromethane (0.314 mol), as it is present in fewer moles than propanone (0.341 mol).

Energy released = 1.67 kJ

\Delta H = -\frac{\text{energy released}}{\text{moles of limiting factor}} = -\frac{1.67}{0.314} = -5.32 \text{ kJ mol}^{-1}

Answer

\Delta H = -5.32 \text{ kJ mol}^{-1}

Final answer

-5.32 kJ mol^-1

Detailed explanation

Background Concept

Enthalpy change ( $\Delta H$ ) is the heat energy transferred per mole of reaction. When energy is released (exothermic process), $\Delta H$ is negative. To find $\Delta H$ in kJ mol $^{-1}$ , divide the total energy change by the number of moles of the limiting reagent (the reactant that is completely consumed and determines the extent of the reaction).

Understanding the Question

Given 37.50 g trichloromethane (0.314 mol) and 19.75 g propanone (0.341 mol), and energy released = 1.67 kJ, calculate $\Delta H$ of the electrostatic attraction. Include a sign.

Approach

Identify the limiting reagent.
Calculate $\Delta H$ using $\Delta H = -q / n$ (negative because energy is released).

Step-by-Step Reasoning

Limiting factor: The reaction forms attractions between trichloromethane and propanone in a 1:1 ratio (one H from trichloromethane interacts with one O from propanone). Since 0.314 mol < 0.341 mol, trichloromethane is the limiting factor. Only 0.314 mol of attractions can form.
Enthalpy change: The energy released is 1.67 kJ. Since energy is released, the process is exothermic, so $\Delta H$ must be negative.

\Delta H = -\frac{1.67 \text{ kJ}}{0.314 \text{ mol}} = -5.318... \approx -5.32 \text{ kJ mol}^{-1}

Key Takeaways

Always identify the limiting reagent before calculating molar enthalpy changes. Remember that 'energy released' means $\Delta H$ is negative.

Common Mistakes

Using the moles of the excess reagent (propanone, 0.341 mol) to calculate $\Delta H$ , giving $-1.67 / 0.341 = -4.90$ kJ mol $^{-1}$ .
Forgetting the negative sign. 'Energy released' explicitly indicates an exothermic process.
Dividing by the total moles (0.314 + 0.341), which is incorrect because the reaction is limited by the smaller amount.

Things to Be Careful About

The mark scheme requires the use of the limiting factor (0.314) for M1, and the correct calculation with sign for M2. Ensure the sign is explicitly included in the final answer.

Techniques used

identify the limiting reagentcalculate enthalpy change per mole of limiting reagentapply correct sign for exothermic process

(f)

Suggest an experiment the student could carry out to test whether the number of moles of trichloromethane affects the temperature change.

DifficultyMedium-Easy

Worked solution

Answer

Keep the volume of trichloromethane the same and vary the volume of propanone, then record the temperature change for each mixture.

Final answer

Keep volume of trichloromethane constant and vary volume of propanone, recording the temperature change.

Detailed explanation

Background Concept

To investigate how the number of moles of one reactant affects a dependent variable (temperature change), the independent variable must be the amount of that reactant. All other variables that could affect the outcome (volume of the other reactant, total volume, initial temperature, apparatus) must be kept constant (controlled variables).

Understanding the Question

Suggest an experiment to test whether the number of moles of trichloromethane affects the temperature change.

Approach

Vary the amount of trichloromethane (independent variable) while keeping the amount of propanone (and other conditions) constant (controlled variables). Measure the temperature change (dependent variable).

Step-by-Step Reasoning

Independent variable: Number of moles of trichloromethane. Since volume is proportional to moles (assuming constant density), vary the volume of trichloromethane.
Dependent variable: Temperature change ( $\Delta T$ ).
Controlled variables: Volume of propanone must be kept constant. Total volume could be kept constant by adjusting volumes, but the simplest correct answer is to keep propanone constant and vary trichloromethane.
Method: Carry out the experiment multiple times. In each trial, use a fixed volume of propanone (e.g., 12.5 cm³) and different volumes of trichloromethane (e.g., 5 cm³, 10 cm³, 15 cm³, 20 cm³). Record the temperature change for each.

Key Takeaways

To test the effect of one variable, vary it while keeping all other relevant variables constant.

Common Mistakes

Varying both volumes simultaneously, which makes it impossible to determine which variable caused the change.
Forgetting to state that the volume of propanone should be kept constant.

Things to Be Careful About

The question asks to test the effect of trichloromethane moles. Therefore, trichloromethane is the independent variable (vary it), and propanone is the controlled variable (keep it constant).

Techniques used

design an experiment to test the effect of one variablecontrol variables in a calorimetry experiment

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