Chem HL Data Lab

Question 1

Magnesium and hydrochloric acid

Reactivity 2.2 · Tool 1

A student added 0.055 g of magnesium ribbon to 50.0 cm³ of 1.00 mol dm⁻³ HCl(aq) (excess) and collected the hydrogen in a gas syringe (±0.5 cm³). The room was at 298 K and 100 kPa.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

Table 1 · Volume of hydrogen collected

Time / s ±1	Volume H₂ / cm³ ±0.5
0	0.0
10	14.0
20	25.5
30	34.0
40	40.5
60	48.5
80	52.5
100	54.0
120	54.5

Graph 1 · Volume against time

(a)Determine the initial rate of reaction, in cm³ s⁻¹, by drawing a tangent at t = 0.[2]
Mark scheme
- Tangent drawn at origin; gradient = Δvolume ÷ Δtime from the tangent (not from two data points).
- Initial rate ≈ 1.5 cm³ s⁻¹ (accept 1.3–1.7) with units.
Using only 14.0 ÷ 10 = 1.4 gives an average rate over the first 10 s, which underestimates the initial rate. Examiners look for the tangent.
(b)Explain, in terms of collision theory, why the gradient of the curve decreases with time.[2]
Mark scheme
- Concentration of H⁺/HCl decreases and the surface area of Mg decreases as reactants are used up.
- So the frequency of (successful) collisions decreases. Do not accept "fewer collisions" without reference to frequency or rate.
(c)Calculate the theoretical volume of hydrogen, in cm³, under the room conditions. Use sections 1 and 2 of the data booklet.[2]
Mark scheme
- n(H₂) = n(Mg) = 0.055 ÷ 24.31 = 2.26 × 10⁻³ mol.
- V = nRT/p = (2.26 × 10⁻³ × 8.31 × 298) ÷ (1.00 × 10⁵) = 5.60 × 10⁻⁵ m³ = 56.0 cm³.
Common slip: forgetting that pV = nRT gives m³ when p is in Pa. Multiply by 10⁶ for cm³.
(d)Calculate the percentage uncertainty in the final volume reading, and compare the experimental final volume with your answer to (c). Suggest one reason for the difference.[3]
Mark scheme
- Percentage uncertainty = 0.5 ÷ 54.5 × 100 = 0.9 %.
- Percentage error = (56.0 − 54.5) ÷ 56.0 × 100 = 2.7 %, which is larger than the uncertainty, so a systematic error is present.
- Any one: oxide layer on Mg reduces mass of Mg reacting; some H₂ escapes before the bung is fitted; syringe plunger friction; some H₂ dissolves.

Question 2

Activation energy from an Arrhenius plot

Reactivity 2.2 AHL · Tool 3

The rate constant, k, of a second-order reaction was measured at six temperatures. The Arrhenius equation in its linear form is ln k = −E_a/R · (1/T) + ln A.

Table 2 · Rate constant against temperature

T / K	k / dm³ mol⁻¹ s⁻¹	(1/T) / 10⁻³ K⁻¹	ln k
293	0.0155	3.413	−4.17
303	0.0329	3.300	−3.41
313	0.0651	3.195	−2.73
323	0.108	3.096	?
333	0.236	3.003	−1.44
343	0.414	2.915	−0.88

Graph 2 · ln k against 1/T (line of best fit excludes one point)

(a)Calculate the missing value of ln k at 323 K, and identify which data point is anomalous.[2]
Mark scheme
- ln 0.108 = −2.23 (2 d.p. to match the column).
- The 323 K point is anomalous: it lies well below the line through the other five points.
(b)Determine the activation energy, E_a, in kJ mol⁻¹, from the gradient of the line of best fit.[3]
Mark scheme
- Gradient from two well-separated points on the line, e.g. (−0.88 − (−4.17)) ÷ ((2.915 − 3.413) × 10⁻³) ≈ −6.6 × 10³ K.
- E_a = −gradient × R = 6.6 × 10³ × 8.31.
- E_a ≈ 55 kJ mol⁻¹ (accept 52–58), positive, with units in kJ.
Watch the 10⁻³ on the 1/T axis. Forgetting it gives an answer 1000 times too small.
(c)Suggest one experimental reason why the rate constant measured at 323 K is lower than expected.[1]
Mark scheme
- Any one: reaction mixture not at 323 K when timing began (reagents not pre-equilibrated in the water bath); water bath temperature fell during the run; error in concentration of one reactant.
(d)State how the pre-exponential factor, A, could be obtained from the graph, and explain why this is less reliable than finding E_a.[2]
Mark scheme
- y-intercept at 1/T = 0 equals ln A, so A = e^intercept (or substitute a point and E_a into the equation).
- Requires a long extrapolation far outside the data range, so a small change in gradient gives a large change in intercept (and A is exponential in it).

Time / min	Temperature / °C ±0.1
0	21.0
1	21.0
2	21.1
3	21.0
4	mixed
5	27.2
6	27.0
7	26.8
8	26.6
9	26.4
10	26.2

Question 4

Calibration curve for copper(II) ions

Structure 1.3 · Tool 2, 3

The absorbance of standard solutions of CuSO₄(aq) was measured with a colorimeter at the wavelength of maximum absorbance, in a 1.00 cm cuvette. A solution of unknown concentration gave an absorbance of 0.58.

Table 4 · Absorbance of standard solutions

[Cu²⁺] / mol dm⁻³	Absorbance
0.000	0.02
0.020	0.23
0.040	0.45
0.060	0.66
0.080	0.89
0.100	1.10

Graph 4 · Absorbance against concentration

(a)Determine the concentration of the unknown solution from the graph.[1]
Mark scheme
- 0.052 mol dm⁻³ (accept 0.050–0.054), read from the best-fit line at A = 0.58.
(b)The line does not pass through the origin. State the type of error this indicates and suggest its cause.[2]
Mark scheme
- Systematic error (all readings shifted by about +0.02).
- Colorimeter not zeroed with a blank of distilled water / scratched or dirty cuvette / fingerprints on the optical faces.
(c)Explain why the measurements are made at the wavelength of maximum absorbance.[2]
Mark scheme
- Gives the largest change in absorbance per unit change in concentration (steepest calibration line), so the method is most sensitive.
- Absorbance changes least with small wavelength variations at the peak, so readings are more precise / percentage uncertainty is smaller.
(d)A second unknown gave an absorbance of 1.65. Explain why its concentration should not be found by extending the line, and state what the student should do.[2]
Mark scheme
- 1.65 is outside the calibrated range; linearity (Beer–Lambert behaviour) is not established there, as absorbance often deviates from linearity at high values.
- Dilute the sample by a known factor (e.g. 1:2 with a volumetric pipette and flask), measure again, and multiply by the dilution factor.

Question 5

Thermal decomposition of calcium carbonate

Reactivity 1.4 AHL · Tool 3

The standard Gibbs energy change for CaCO₃(s) → CaO(s) + CO₂(g) was calculated at several temperatures from experimental equilibrium data. Assume ΔH^⦵ and ΔS^⦵ do not vary with temperature.

Table 5 · ΔG^⦵ against temperature

T / K	ΔG^⦵ / kJ mol⁻¹
800	+49.0
900	+33.4
1000	+16.8
1100	+1.2
1200	−15.5
1300	−31.1

Graph 5 · ΔG^⦵ against T

(a)Using ΔG^⦵ = ΔH^⦵ − TΔS^⦵, determine ΔS^⦵ in J K⁻¹ mol⁻¹ from the graph.[2]
Mark scheme
- Gradient = (−31.1 − 49.0) ÷ (1300 − 800) = −0.160 kJ K⁻¹ mol⁻¹; gradient = −ΔS.
- ΔS^⦵ = +160 J K⁻¹ mol⁻¹ (accept 155–165), converted to J.
(b)Determine ΔH^⦵ in kJ mol⁻¹.[1]
Mark scheme
- Extrapolated intercept at T = 0, or substitute a point: 49.0 + 800 × 0.160 = +177 kJ mol⁻¹ (accept 172–184).
(c)Determine the minimum temperature at which the decomposition becomes spontaneous.[1]
Mark scheme
- Where ΔG^⦵ = 0: about 1110 K (accept 1100–1120 K, from the graph or T = ΔH ÷ ΔS).
(d)Explain the sign of ΔS^⦵ and why the reaction is only spontaneous at high temperature.[2]
Mark scheme
- ΔS is positive because a gas is produced from a solid (more ways of distributing energy/particles).
- ΔH is positive, so ΔG is negative only when TΔS > ΔH, i.e. at high T.

Question 6

pH curve for ethanoic acid

Reactivity 3.1 AHL · Tool 1

25.00 cm³ of ethanoic acid solution of unknown concentration was pipetted (±0.03 cm³) into a conical flask and titrated with 0.100 mol dm⁻³ NaOH(aq) from a burette (±0.05 cm³ per reading). The pH was measured with a calibrated pH probe.

Table 6 · pH during the titration

Volume NaOH / cm³ ±0.10	pH ±0.01
0.00	2.90
5.00	4.22
10.00	4.66
15.00	5.06
20.00	5.66
22.00	6.40
22.50	8.72
23.00	11.02
25.00	11.70
30.00	12.13
35.00	12.32

Graph 6 · pH against volume of NaOH added

(a)Determine the volume of NaOH at the equivalence point and calculate the concentration of the ethanoic acid.[2]
Mark scheme
- Equivalence at the centre of the steep section: 22.50 cm³.
- n(NaOH) = 0.100 × 0.02250 = 2.25 × 10⁻³ mol = n(CH₃COOH); c = 2.25 × 10⁻³ ÷ 0.02500 = 0.0900 mol dm⁻³.
(b)Use the graph to determine pK_a and K_a of ethanoic acid. Explain your method.[2]
Mark scheme
- At half-equivalence (11.25 cm³) [CH₃COOH] = [CH₃COO⁻], so pH = pK_a.
- pK_a ≈ 4.76 (accept 4.7–4.8); K_a = 10^−4.76 = 1.7 × 10⁻⁵.
(c)Explain why the pH at the equivalence point is greater than 7, and identify a suitable indicator using section 18 of the data booklet.[2]
Mark scheme
- At equivalence the solution contains CH₃COO⁻, a weak base: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻.
- Phenolphthalein, because its colour-change range (about pH 8–10) lies within the steep section. Methyl orange is unsuitable as it changes at pH 3–4.
(d)Calculate the absolute uncertainty in the concentration of ethanoic acid, ignoring any uncertainty in the NaOH concentration.[2]
Mark scheme
- Titre is a difference of two readings: ±0.10 cm³ → 0.44 %; pipette 0.03 ÷ 25.00 → 0.12 %; total 0.56 %.
- 0.56 % of 0.0900 = ±0.0005 mol dm⁻³, so c = 0.0900 ± 0.0005 mol dm⁻³.

Question 7

Dinitrogen tetroxide and nitrogen dioxide

Reactivity 2.3 · 1.4 AHL

0.100 mol of N₂O₄(g) was placed in a sealed 1.00 dm³ flask at 373 K. At t = 60 s the volume of the flask was suddenly halved, at constant temperature.

N₂O₄(g) ⇌ 2NO₂(g) ΔH = +57 kJ mol⁻¹

Table 7 · Concentrations against time

Time / s	[N₂O₄] / mol dm⁻³	[NO₂] / mol dm⁻³
0	0.100	0.000
10	0.072	0.056
20	0.055	0.090
30	0.046	0.108
40	0.042	0.116
50	0.040	0.120
60 (before)	0.040	0.120
60 (after)	0.080	0.240
70	0.094	0.212
80	0.101	0.198
90	0.103	0.194
100	0.104	0.193

Graph 7 · Concentration against time

(a)Write the equilibrium constant expression and calculate K_c at 373 K using the data at 50 s.[2]
Mark scheme
- K_c = [NO₂]² ÷ [N₂O₄].
- K_c = 0.120² ÷ 0.040 = 0.36 (no units required).
(b)Explain the changes in both concentrations at and after t = 60 s, referring to the reaction quotient, Q.[3]
Mark scheme
- Halving the volume doubles both concentrations instantly.
- Q = 0.240² ÷ 0.080 = 0.72, which is greater than K_c.
- So the position shifts left (towards fewer gas moles) until Q = K_c: [NO₂] falls by twice as much as [N₂O₄] rises.
(c)Show that the data at 100 s are consistent with K_c being unchanged.[1]
Mark scheme
- 0.193² ÷ 0.104 = 0.358 ≈ 0.36; K_c depends only on temperature, which was constant.
(d)Calculate ΔG^⦵ for the reaction at 373 K, and comment on its sign.[2]
Mark scheme
- ΔG^⦵ = −RT ln K = −8.31 × 373 × ln 0.36 = +3.17 × 10³ J mol⁻¹ (+3.2 kJ mol⁻¹).
- Positive because K < 1: at equilibrium reactants are favoured (the position lies slightly to the left).

Question 8

Making ethyl ethanoate

Structure 3.2 · Reactivity 1.1, 2.1

Ethanoic acid and ethanol were heated under reflux with a few drops of concentrated sulfuric acid, then the ester was separated by distillation. Other groups in the class refluxed for different times using the same quantities.

CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O

Table 8a · One group's results (60 min reflux)

Substance	Mass / g ±0.01	M / g mol⁻¹
Ethanoic acid	6.00	60.06
Ethanol	6.90	46.08
Ethyl ethanoate (product)	5.12	88.12

Table 8b · Class results

Reflux time / min	Yield / %
0	0
15	31
30	47
45	55
60	58
90	60
120	61

Graph 8 · Percentage yield against reflux time

(a)Identify the limiting reactant and calculate the theoretical mass of ethyl ethanoate.[2]
Mark scheme
- n(acid) = 6.00 ÷ 60.06 = 0.0999 mol; n(ethanol) = 6.90 ÷ 46.08 = 0.150 mol; ethanoic acid is limiting (1:1 ratio).
- Theoretical mass = 0.0999 × 88.12 = 8.80 g.
(b)Calculate the percentage yield for this group.[1]
Mark scheme
- 5.12 ÷ 8.80 × 100 = 58.2 %.
(c)Explain the shape of the graph, including why the yield levels off at about 61 %.[2]
Mark scheme
- Rate of forward reaction is highest at the start, when reactant concentrations are greatest, then decreases.
- The reaction is reversible: after about 90 min equilibrium is reached, so further heating does not increase the yield. Not because a reactant runs out.
(d)Calculate the atom economy of this reaction.[1]
Mark scheme
- 88.12 ÷ (60.06 + 46.08) × 100 = 83.0 %.
Atom economy uses the balanced equation only; it does not depend on the masses used or the yield.
(e)Suggest two changes to the method that would increase the percentage yield, and explain each.[2]
Mark scheme
- Use a larger excess of ethanol, or remove water/ester as it forms (e.g. distil off the ester): shifts the equilibrium to the right.
- Reduce losses during separation: e.g. cool the receiver to reduce evaporation of the volatile ester, rinse apparatus, careful transfer. Adding more catalyst does not score, as it does not change the equilibrium position.

Question 9

Identifying the product by spectroscopy

Structure 3.2 AHL · Tool 2

The distillate from Question 8 has molecular formula C₄H₈O₂. Its mass spectrum, infrared spectrum and ¹H NMR spectrum are shown. Use section 20 and 21 of the data booklet.

Graph 9a · Mass spectrum

Graph 9b · Infrared spectrum

Table 9 · ¹H NMR data

δ / ppm	Integration	Splitting
4.12	2	quartet
2.04	3	singlet
1.26	3	triplet

Graph 9c · ¹H NMR spectrum (multiplet spacing exaggerated)

(a)Calculate the index of hydrogen deficiency (IHD) of C₄H₈O₂ and state what it indicates.[1]
Mark scheme
- IHD = (2 × 4 + 2 − 8) ÷ 2 = 1: one double bond (e.g. C=O) or one ring.
(b)Identify the species responsible for the peaks at m/z = 88, 43 and 29.[2]
Mark scheme
- 88: molecular ion [C₄H₈O₂]⁺.
- 43: CH₃CO⁺; 29: C₂H₅⁺. Fragments must carry a positive charge.
(c)Identify the bonds responsible for the absorptions near 1740 cm⁻¹ and 1240 cm⁻¹. Explain how the IR spectrum shows that the distillate is free of ethanoic acid.[2]
Mark scheme
- 1740 cm⁻¹: C=O; 1240 cm⁻¹: C–O.
- No broad O–H absorption at 2500–3000 cm⁻¹ (carboxylic acid) or 3200–3600 cm⁻¹ (alcohol).
The small sharp dip near 3480 cm⁻¹ is an overtone of the C=O stretch. It is far too weak and narrow to be O–H.
(d)Explain the splitting of the signals at 4.12 ppm and 1.26 ppm, and why the signal at 2.04 ppm is a singlet.[2]
Mark scheme
- Quartet: CH₂ next to a CH₃ (3 + 1 = 4); triplet: CH₃ next to a CH₂ (2 + 1 = 3), so an ethyl group is present.
- Singlet: CH₃ with no H on the adjacent atom (bonded to C=O).
(e)Deduce the structure of the compound, and explain how its NMR spectrum distinguishes it from its isomer methyl propanoate, CH₃CH₂COOCH₃.[2]
Mark scheme
- CH₃COOCH₂CH₃ (ethyl ethanoate): the CH₂ quartet at 4.12 ppm is high because the CH₂ is bonded to the ester O.
- Methyl propanoate would show the singlet (O–CH₃) near 3.7 ppm and the quartet (CH₂–C=O) near 2.3 ppm, the reverse of what is seen.

Reference

Data-handling toolkit

Uncertainty of a reading

Analogue scale: ± half the smallest division. Digital: ± the last digit.
A difference of two readings (ΔT, burette titre) doubles the absolute uncertainty.
Record data to the same decimal place as its uncertainty.

Propagating uncertainty

Adding or subtracting: add absolute uncertainties.
Multiplying or dividing: add percentage uncertainties.
Raising to a power n: multiply percentage uncertainty by n.
% unc = abs unc ÷ value × 100

Error vs uncertainty

% error = |exp − lit| ÷ lit × 100
If % error > total % uncertainty, systematic error must be present.
Random error: scatter around the line; reduce by repeats. Systematic: shifted intercept or consistent bias; repeats do not help.

Graphs

Line of best fit: balance points either side; ignore anomalies and say so.
Gradient from the line, using points far apart, not from data points.
Instantaneous rate = gradient of a tangent. Include units.
Rearrange to y = mx + c: Arrhenius (ln k vs 1/T), ΔG vs T.

Command terms

State: brief answer, no reasons.
Calculate: show working, give units and sensible s.f.
Determine: one valid answer, usually from a graph.
Explain: give reasons or mechanisms.
Evaluate: weigh strengths and limitations, compare values.

Improvements that score

Name the specific weakness, then the fix that addresses it.
Heat loss → lid / insulation. Gas loss → divided flask or add solid via a tube. Timing → data logger.
Avoid generic "be more careful" or "repeat" unless the issue is random error.