Chemistry 9701/33 — February/March 2016

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

questions

marks

120

minutes

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

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

You will determine the enthalpy change, $\Delta H$ , of the reaction between magnesium and hydrochloric acid. To do this you will measure the change in temperature when a piece of magnesium ribbon reacts with an excess of hydrochloric acid.

\text{Mg(s)} + 2\text{HCl(aq)} \rightarrow \text{MgCl}_2\text{(aq)} + \text{H}_2\text{(g)}

FA 1 is hydrochloric acid, $\text{HCl}$ .
FA 2 is magnesium ribbon, $\text{Mg}$ .

(a)

Method

Weigh the FA 2 and record the mass in the space below.
Support the plastic cup in the $250\text{ cm}^3$ beaker.
Coil the FA 2 so that it will fit into the bottom of the plastic cup then remove it.
Use the measuring cylinder to transfer $25\text{ cm}^3$ of FA 1 into the plastic cup.
Place the thermometer in the acid and tilt the cup if necessary so that the bulb of the thermometer is fully covered. Record the temperature at time = 0 in the table of results.
Start timing and do not stop the clock until the whole experiment has been completed at time = 8 minutes.
Record the temperature of the acid every half minute for 2 minutes.
At time = $2\frac{1}{2}$ minutes carefully drop the coil of FA 2 into the acid and stir the mixture.
Record the temperature of the mixture at time = 3 minutes and complete the table by recording the temperature every half minute. Stir the mixture between thermometer readings.

Results

mass of FA 2 = ............................. g

time/minutes	0	$\frac{1}{2}$	1	$1\frac{1}{2}$	2	$2\frac{1}{2}$	3	$3\frac{1}{2}$	4
temperature/°C						[shaded]

time/minutes	$4\frac{1}{2}$	5	$5\frac{1}{2}$	6	$6\frac{1}{2}$	7	$7\frac{1}{2}$	8
temperature/°C

DifficultyEasy

Worked solution

Answer

As this is a practical experiment, the values below are representative examples. A candidate would record their own readings.

mass of FA 2 = 0.300 g

time/minutes	0	0.5	1	1.5	2	2.5	3	3.5	4
temperature/°C	20.0	19.5	19.5	19.5	19.5	25.0	32.0	31.5	31.0

time/minutes	4.5	5	5.5	6	6.5	7	7.5	8
temperature/°C	30.5	30.0	29.5	29.0	28.5	28.0	27.5	27.0

All thermometer readings and the mass of FA 2 are recorded. Mass of FA 2 is < 0.50 g. All temperatures are recorded to the nearest 0.5 °C.

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In enthalpy change experiments, the candidate must accurately record the initial conditions and track the temperature changes over time. The reaction between magnesium and hydrochloric acid is exothermic, so the temperature of the mixture will rise after the magnesium is added, reach a maximum, and then gradually fall as heat is lost to the surroundings.

Understanding the Question

Part (a) asks the candidate to follow the method and record the mass of the magnesium ribbon (FA 2) and the temperature of the hydrochloric acid (FA 1) at half-minute intervals from time = 0 to 8 minutes. The marking scheme requires the mass to be recorded and to be less than 0.50 g, and all temperatures to be recorded to 0.5 °C.

Approach

Since the actual data depends on the candidate's experiment, the solution provides a representative data set that follows the marking scheme's conventions. The mass must be < 0.50 g. The initial temperature should be stable (perhaps slightly decreasing due to room temperature cooling). After adding Mg at 2.5 minutes, the temperature should rise sharply and then cool gradually.

Step-by-Step Reasoning

Mass of FA 2: Weigh the magnesium ribbon. The mark scheme rejects masses > 0.50 g because a large mass would require more acid to react completely or cause excessive heat loss during the slow reaction. Record it to 3 decimal places (e.g., 0.300 g).
Initial temperatures (0 to 2 minutes): Record the temperature of the acid every 0.5 minutes. The acid may cool slightly due to the room being cooler than the acid, or remain stable. Record to 0.5 °C (e.g., 20.0, 19.5, 19.5, 19.5, 19.5).
Reaction and cooling (2.5 to 8 minutes): At 2.5 minutes, add the Mg. The temperature rises rapidly. Record at 3.0, 3.5, 4.0 minutes. The peak temperature will be around 3.0–3.5 minutes. After the peak, the temperature falls as heat is lost to the surroundings. Record every 0.5 minutes up to 8.0 minutes (e.g., 32.0, 31.5, 31.0, 30.5, 30.0, 29.5, 29.0, 28.5, 28.0, 27.5, 27.0).

Key Takeaways

Always record mass and temperature to the precision required by the apparatus (balance to 0.001 g or 0.01 g, thermometer to 0.5 °C).
Ensure the mass of the limiting reagent is small enough to ensure the acid is in excess and the reaction completes reasonably quickly.

Common Mistakes

Recording the mass of Mg as > 0.00 g. Recording the mass of Mg as > 0.50 g (mark scheme explicitly rejects this).
Recording temperatures to 1 °C instead of 0.5 °C.
Not recording the mass of FA 2 at all.

Things to Be Careful About

The mark scheme awards marks for all readings being recorded and the mass being < 0.50 g. Do not omit any time points.
Ensure temperature readings are to the nearest 0.5 °C. If using a thermometer with 1 °C graduations, estimate to 0.5 °C.

Techniques used

record mass to correct precisionmeasure initial temperaturerecord temperature changes at regular intervals

(b)

Plot a graph of temperature on the $y$ -axis against time on the $x$ -axis on the grid below.
The scale for the temperature axis should extend $10\text{ }^\circ\text{C}$ greater than the maximum temperature you recorded.
You will use the graph to determine the theoretical maximum temperature rise at $2\frac{1}{2}$ minutes.

Fig. 1.1

Draw two lines of best fit through the points on your graph, the first for the temperature before adding FA 2 and the second for the cooling of the mixture once the reaction is complete.
Extrapolate the two lines to $2\frac{1}{2}$ minutes, draw a vertical line between the two and determine the theoretical rise in temperature at this time.

theoretical rise in temperature at $2\frac{1}{2}$ minutes = ........................ °C

DifficultyMedium-Easy

Worked solution

Answer

Graph Setup:

$y$ -axis: temperature / °C, uniform scale covering the range of recorded temperatures plus 10 °C above the maximum.
$x$ -axis: time / minutes, uniform scale from 0 to 8.

Plotting and Lines:

Plot all recorded data points (minimum 9 points: 0, 0.5, 1, 1.5, 2, 2.5, 3, 3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8 min).
Draw a line of best fit (straight line or smooth curve) through the pre-reaction points (0 to 2 min). This line should account for any slight cooling of the acid.
Draw a second line of best fit through the post-reaction cooling points (after the peak temperature, e.g., 3.5 to 8 min).
Extrapolate both lines to time = 2.5 minutes.
Draw a vertical line at 2.5 minutes between the two extrapolated lines.
The vertical distance between the two lines at 2.5 minutes is the theoretical rise in temperature ( $\Delta T$ ).

theoretical rise in temperature at 2.5 minutes = 30.0 °C

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

In exothermic reactions, heat is lost to the surroundings continuously. The maximum temperature recorded in the experiment is lower than the theoretical maximum because the reaction takes time to complete and heat is lost during that time. To find the theoretical maximum temperature rise at the moment of mixing (2.5 minutes), we extrapolate the cooling curve backwards and the pre-reaction baseline forwards to that time.

Understanding the Question

Part (b) requires the candidate to plot a temperature-time graph, draw two lines of best fit (one for the pre-reaction baseline, one for the post-reaction cooling), extrapolate them to 2.5 minutes, and read off the theoretical temperature rise ( $\Delta T$ ).

Approach

Set up the axes with appropriate labels and units. The $y$ -axis must extend 10 °C above the highest recorded temperature.
Plot all data points accurately.
Draw the pre-reaction line through the points from 0 to 2 minutes. This line may have a slight negative gradient due to cooling.
Draw the post-reaction line through the cooling points after the peak (ignore the rising part and the peak itself, as these are affected by the ongoing reaction and heat loss).
Extrapolate both lines to $t = 2.5$ min and read the difference in temperature.

Step-by-Step Reasoning

Axes: Label $y$ -axis "temperature / °C" and $x$ -axis "time / minutes". Choose a scale that uses more than half of each axis. For example, if temperatures range from 19.5 to 32.0 °C, the $y$ -axis could go from 15 to 45 °C (which is > 10 °C above 32.0).
Plotting: Plot all 17 data points from the table in part (a).
Pre-reaction line: Draw a straight line through the points at 0, 0.5, 1, 1.5, and 2 minutes. Balance the points on either side. This line represents the temperature trend before the reaction.
Post-reaction line: Draw a straight line (or smooth curve) through the cooling points, e.g., from 3.5 to 8 minutes. Points on the rising part (2.5 to 3.5 min) should be ignored or ringed as anomalous for this line.
Extrapolation: Extend both lines to $x = 2.5$ min. Read the $y$ -values. The difference is $\Delta T$ .

Key Takeaways

Extrapolation is essential in enthalpy experiments to correct for heat loss during the reaction.
The post-reaction line should only use points after the peak temperature, as the reaction is complete and only cooling is occurring.

Common Mistakes

Forgetting to label axes with units.
Not extending the $y$ -axis 10 °C above the maximum recorded temperature.
Drawing a single line of best fit through all points instead of two separate lines.
Extrapolating the wrong sections (e.g., using the rising part for the post-reaction line).

Things to Be Careful About

The lines must be straight or smooth curves; jagged lines connecting points are not accepted.
Points not on the line must be balanced on either side. If a point is clearly anomalous, ring it and ignore it.
The theoretical $\Delta T$ must be read accurately from the graph, within 0.5 °C of the correct value.

Techniques used

plot temperature-time graphdraw lines of best fitextrapolate lines to find theoretical maximum temperature rise

(c)

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

(i)

Use your answer to (b) to calculate the heat energy, in joules, given out when FA 2 is added to FA 1.
(Assume $4.2\text{ J}$ of heat energy raises the temperature of $1.0\text{ cm}^3$ of the mixture by $1.0\text{ }^\circ\text{C}$ .)

heat energy evolved = ............................. J

DifficultyMedium-Easy

Worked solution

Working

The heat energy evolved is calculated using the formula:

Q = m \times c \times \Delta T

where:

$m$ = volume of mixture = 25 cm $^3$ (assume density = 1 g/cm $^3$ , so mass = 25 g)
$c$ = specific heat capacity = 4.2 J g $^{-1}$ °C $^{-1}$
$\Delta T$ = theoretical rise in temperature from (b) = 30.0 °C

Q = 25 \times 4.2 \times 30.0

Q = 3150 \text{ J}

Answer

heat energy evolved = 3150 J

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

The heat energy ( $Q$ ) absorbed or released by a solution can be calculated using the equation $Q = mc\Delta T$ , where $m$ is the mass of the solution, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. For aqueous solutions, it is standard to assume the density is 1 g/cm $^3$ (so mass in grams equals volume in cm $^3$ ) and the specific heat capacity is 4.2 J g $^{-1}$ °C $^{-1}$ (same as water).

Understanding the Question

Part (c)(i) asks the candidate to calculate the heat energy in joules given out when FA 2 is added to FA 1, using the theoretical $\Delta T$ from part (b) and the given assumption that 4.2 J raises the temperature of 1.0 cm $^3$ of the mixture by 1.0 °C.

Approach

Identify the volume of the mixture: 25 cm $^3$ of HCl + mass of Mg (negligible volume, so total volume ≈ 25 cm $^3$ ).
Use the formula $Q = 25 \times 4.2 \times \Delta T$ .
Substitute the $\Delta T$ from part (b) and calculate $Q$ in joules.

Step-by-Step Reasoning

The marking scheme gives the formula implicitly: $Q = 25 \times 4.2 \times \Delta T$ .
Substitute $\Delta T = 30.0$ °C (from the representative graph in part b).
$Q = 25 \times 4.2 \times 30.0 = 3150$ J.
The answer should be given to appropriate significant figures (2-4 sig figs). 3150 J is 3 or 4 sig figs depending on interpretation, but 3150 is acceptable.

Key Takeaways

Always use the theoretical $\Delta T$ from the extrapolated graph, not the maximum recorded temperature.
The mass of the solution is taken as the volume of the acid in cm $^3$ , assuming the density is 1 g/cm $^3$ and the volume of the solid Mg is negligible.

Common Mistakes

Using the maximum recorded temperature instead of the theoretical $\Delta T$ .
Forgetting to multiply by 4.2 or using the wrong value for $c$ .
Calculating in kJ instead of J (the question asks for joules).

Things to Be Careful About

The question asks for the answer in joules, not kJ. Do not divide by 1000 here.
Use the value from (b), not a rounded intermediate value if possible, to avoid compounding errors.

Techniques used

calculate heat energy using Q = mcΔT

(ii)

Use the Periodic Table on page 12 and your answer to (i) to calculate the enthalpy change, in $\text{kJ mol}^{-1}$ , when 1 mole of magnesium, FA 2, reacts with hydrochloric acid, FA 1.

enthalpy change = [sign] [value] ............................. $\text{kJ mol}^{-1}$

DifficultyMedium

Worked solution

Working

Step 1: Calculate moles of Mg reacted

\text{moles of Mg} = \frac{\text{mass of Mg}}{M_r \text{ of Mg}} = \frac{0.300}{24.3} = 0.01235 \text{ mol}

Step 2: Calculate enthalpy change ( $\Delta H$ )

\Delta H = \frac{-Q}{\text{moles of Mg}} = \frac{-3150}{0.01235} = -255061 \text{ J mol}^{-1}

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

\Delta H = -255 \text{ kJ mol}^{-1}

Alternatively, using the combined expression:

\Delta H = \frac{-Q \times 24.3}{\text{mass of Mg} \times 1000} = \frac{-3150 \times 24.3}{0.300 \times 1000} = -255 \text{ kJ mol}^{-1}

Answer

enthalpy change = -255 kJ mol $^{-1}$

Final answer

See working / candidate-dependent

Detailed explanation

Background Concept

Enthalpy change ( $\Delta H$ ) is the heat energy change per mole of reaction. For an exothermic reaction, $\Delta H$ is negative. The heat energy $Q$ calculated in part (i) is the energy released by the reaction of the specific mass of Mg used. To find $\Delta H$ in kJ mol $^{-1}$ , we divide $Q$ (in joules) by the moles of Mg reacted, and then convert to kJ.

Understanding the Question

Part (c)(ii) asks the candidate to calculate the enthalpy change in kJ mol $^{-1}$ when 1 mole of Mg reacts, using the Periodic Table value for the $M_r$ of Mg (24.3) and the answer from (c)(i).

Approach

Calculate the moles of Mg using the mass recorded in part (a) and $M_r = 24.3$ .
Calculate $\Delta H = -Q / \text{moles}$ . The negative sign is required because the reaction is exothermic.
Convert from J mol $^{-1}$ to kJ mol $^{-1}$ by dividing by 1000.
Ensure the final answer is given to 2-4 significant figures and includes the negative sign.

Step-by-Step Reasoning

Moles of Mg: $\text{mass} = 0.300$ g, $M_r = 24.3$ . $\text{moles} = 0.300 / 24.3 = 0.012346$ mol.
Enthalpy change: $Q = 3150$ J. $\Delta H = -3150 / 0.012346 = -255158$ J mol $^{-1}$ .
Convert to kJ: $-255158 / 1000 = -255.158$ kJ mol $^{-1}$ .
Significant figures: The mass (0.300) has 3 sig figs, $\Delta T$ (30.0) has 3 sig figs. The answer should be given to 3 sig figs: $-255$ kJ mol $^{-1}$ .

The mark scheme accepts the combined expression: $\Delta H = \frac{-Q \times 24.3}{\text{mass} \times 1000}$ . This avoids intermediate rounding errors.

Key Takeaways

Always include the negative sign for exothermic reactions.
Convert J to kJ at the end, not before, to avoid errors.
Use unrounded intermediate values in calculations to avoid compounding rounding errors.

Common Mistakes

Forgetting the negative sign (enthalpy change for exothermic reactions is negative).
Forgetting to convert J to kJ (giving an answer like -255000 kJ mol $^{-1}$ ).
Rounding intermediate values (e.g., moles of Mg to 0.012) and getting a significantly different final answer.
Using the wrong $M_r$ for Mg (must use 24.3 from the Periodic Table).

Things to Be Careful About

The mark scheme explicitly requires the negative sign and 2-4 significant figures. No rounding to 1 sig fig during calculation.
The expression in the mark scheme is $\frac{-\text{(c)(i)} \times 24.3}{\text{mass in (a)} \times 1000}$ . Ensure you use the $Q$ value in joules, not kJ, in this expression, or adjust the denominator accordingly.

Techniques used

calculate moles of magnesiumcalculate enthalpy change per mole

(d)

A student carried out the same procedure using the same concentration of sulfuric acid, $\text{H}_2\text{SO}_4$ , instead of hydrochloric acid. Before starting the experiment the student predicted that the enthalpy change would be twice that with hydrochloric acid.
Was the student correct? Explain your answer.

DifficultyMedium-Easy

Worked solution

Answer

Incorrect.

The hydrochloric acid was already in excess in the original experiment. Replacing it with sulfuric acid (also in excess) will not change the amount of magnesium that reacts. The net ionic equation for both reactions is the same:

\text{Mg(s)} + 2\text{H}^+(\text{aq}) \rightarrow \text{Mg}^{2+}(\text{aq}) + \text{H}_2(\text{g})

Therefore, the enthalpy change per mole of magnesium will be the same (or very similar), not twice as much.

Final answer

Incorrect, as the acid was in excess already and the net ionic equation is the same.

Detailed explanation

Background Concept

The enthalpy change of a reaction depends on the moles of the limiting reagent that react. In this experiment, the acid is in excess, so magnesium is the limiting reagent. The enthalpy change is determined by the reaction of 1 mole of Mg with 2 moles of H $^+$ ions. Both HCl and H $_2$ SO $_4$ are strong acids and will provide H $^+$ ions in excess.

Understanding the Question

A student predicts that using H $_2$ SO $_4$ instead of HCl will give an enthalpy change twice that of HCl. The candidate must evaluate this prediction and explain why it is correct or incorrect.

Approach

Identify the limiting reagent in both cases (Mg, since acid is in excess).
Write the net ionic equation for both reactions.
Compare the enthalpy changes per mole of Mg.

Step-by-Step Reasoning

Limiting reagent: In both experiments, the acid is in excess, so Mg is the limiting reagent. The amount of heat evolved depends only on the moles of Mg reacted.
Net ionic equation:
- With HCl: $\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2 \implies \text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$
- With H $_2$ SO $_4$ : $\text{Mg} + \text{H}_2\text{SO}_4 \rightarrow \text{MgSO}_4 + \text{H}_2 \implies \text{Mg} + 2\text{H}^+ \rightarrow \text{Mg}^{2+} + \text{H}_2$
Conclusion: Since the net ionic equation is identical, the enthalpy change per mole of Mg will be the same. The student's prediction is incorrect.

Key Takeaways

When comparing enthalpy changes for reactions with different acids but the same metal, look at the net ionic equation.
If the acid is in excess, the enthalpy change depends only on the metal reacted.

Common Mistakes

Assuming that because H $_2$ SO $_4$ is diprotic, it will release twice the heat. This is only true if the acid is the limiting reagent, which it is not here.
Forgetting to state that the acid is in excess.

Things to Be Careful About

The mark scheme specifically looks for the point that "the acid was in excess already". Make sure to state this clearly.
The answer is "Incorrect"; do not just say "No".

Techniques used

deduce enthalpy change for different acidrecognize limiting reagent

(e)

The enthalpy change determined in (c)(ii) is only an approximation of the actual value.

Suggest and explain one improvement you would make to the method in (a) to increase the accuracy of the experiment.

DifficultyEasy

Worked solution

Answer

Improvement: Use a lid on the plastic cup (or use extra insulation around the cup).

Explanation: This reduces heat loss to the surroundings by convection (and conduction), resulting in a higher maximum recorded temperature and a more accurate theoretical $\Delta T$ .

(Alternative acceptable answers:)

Use a pipette or burette for FA 1 to reduce the percentage error in measuring the volume of acid.
Use magnesium turnings or powder instead of ribbon so the reaction completes more quickly, reducing heat loss during the reaction.

Final answer

Use a lid on the cup to reduce heat loss by convection.

Detailed explanation

Background Concept

Enthalpy change experiments are prone to heat loss to the surroundings, which causes the measured temperature rise to be lower than the theoretical value. This leads to an underestimation of the enthalpy change. Improvements to the method aim to minimize heat loss or reduce measurement errors.

Understanding the Question

Part (e) asks the candidate to suggest one improvement to the method in part (a) and explain how it increases the accuracy of the experiment.

Approach

Think about the sources of error in the experiment:

Heat loss to surroundings (convection, conduction, radiation).
Inaccurate measurement of volume or mass.
Slow reaction rate causing heat loss during the reaction.
Propose an improvement that addresses one of these and explain the effect.

Step-by-Step Reasoning

Heat loss: The plastic cup is open at the top, allowing heat to escape via convection. Adding a lid (e.g., a piece of cardboard with a hole for the thermometer) reduces this heat loss. Explanation: less heat loss means a higher maximum temperature is recorded, giving a more accurate $\Delta T$ and a more accurate $\Delta H$ .
Volume measurement: A measuring cylinder is used for the acid, which has a higher percentage error than a pipette or burette. Using a pipette or burette would reduce the % error in the volume of acid. Explanation: more accurate volume means more accurate mass of solution, leading to a more accurate $Q$ calculation.
Reaction rate: Magnesium ribbon reacts slowly, and heat is lost while the reaction is still proceeding. Using magnesium turnings or powder increases the surface area, making the reaction faster. Explanation: the reaction completes sooner, reducing the time available for heat loss during the reaction.

Any one of these is acceptable.

Key Takeaways

Always link the improvement to the specific error it reduces.
"Using a lid" is the most common and acceptable improvement for this experiment.
Do not give vague answers like "be more careful" or "repeat the experiment".

Common Mistakes

Suggesting "use a better thermometer" without explaining how it improves accuracy.
Suggesting "insulate the room" or "do it in a vacuum".
Not providing an explanation for the improvement.

Things to Be Careful About

The improvement and explanation must be paired. "Use a lid" alone is not enough; you must explain that it reduces heat loss by convection.
The mark scheme accepts specific improvements: lid/insulation, pipette/burette for FA 1, magnesium turnings/powder, lid/higher walls to reduce acid spray.

Techniques used

suggest and explain an improvement to reduce heat loss or measurement error

The rest of this paper

2 more questions

Q2Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation14M
Q3Qualitative Analysis13M

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