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Date	November 2018	Marks available	2	Reference code	18N.2.hl.TZ0.1
Level	HL	Paper	2	Time zone	TZ0
Command term	Outline	Question number	1	Adapted from	N/A

Question

3.26 g of iron powder are added to 80.0 cm³ of 0.200 mol dm⁻³ copper(II) sulfate solution. The following reaction occurs:

Fe (s) + CuSO₄ (aq) → FeSO₄ (aq) + Cu (s)

Determine the limiting reactant showing your working.

[2]

a.i.

The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.

[2]

a.ii.

The reaction was carried out in a calorimeter. The maximum temperature rise of the solution was 7.5 °C.

Calculate the enthalpy change, ΔH, of the reaction, in kJ, assuming that all the heat released was absorbed by the solution. Use sections 1 and 2 of the data booklet.

[2]

b.i.

State another assumption you made in (b)(i).

[1]

b.ii.

The only significant uncertainty is in the temperature measurement.

Determine the absolute uncertainty in the calculated value of ΔH if the uncertainty in the temperature rise was ±0.2 °C.

[2]

b.iii.

Sketch a graph of the concentration of iron(II) sulfate, FeSO₄, against time as the reaction proceeds.

[2]

c.i.

Outline how the initial rate of reaction can be determined from the graph in part (c)(i).

[2]

c.ii.

Explain, using the collision theory, why replacing the iron powder with a piece of iron of the same mass slows down the rate of the reaction.

[2]

c.iii.

A student electrolyzed aqueous iron(II) sulfate, FeSO₄ (aq), using platinum electrodes. State half-equations for the reactions at the electrodes, using section 24 of the data booklet.

[2]

Markscheme

n_CuSO4 «= 0.0800 dm³ × 0.200 mol dm^–3» = 0.0160 mol AND

n_Fe « $\frac{{3.26\,{\text{g}}}}{{55.85\,{\text{g}}\,{\text{mo}}{{\text{l}}^{ - 1}}}}$ » = 0.0584 mol ✔

CuSO₄ is the limiting reactant ✔

Do not award M2 if mole calculation is not shown.

a.i.

ALTERNATIVE 1:
«0.0160 mol × 63.55 g mol^–1 =» 1.02 «g» ✔

« $\frac{{0.872\,{\text{g}}}}{{1.02\,{\text{g}}}} \times 100 =$ » 85.5 «%» ✔

ALTERNATIVE 2:
« $\frac{{0.872\,{\text{g}}}}{{63.55\,{\text{g}}\,{\text{mo}}{{\text{l}}^{--1}}}} =$ » 0.0137 «mol» ✔

« $\frac{{0.0137\,{\text{mol}}}}{{0.0160\,{\text{mol}}}} \times 100 =$ » 85.6 «%» ✔

Accept answers in the range 85–86 %.

Award [2] for correct final answer.

a.ii.

ALTERNATIVE 1:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«per mol of CuSO₄ = $\frac{{ - 2.5\,{\text{kJ}}}}{{0.0160\,{\text{mol}}}} = - 1.6 \times {10^2}$ kJ mol^–1»

«for the reaction» ΔH = –1.6 × 10² «kJ» ✔

ALTERNATIVE 2:

q = «80.0 g × 4.18 J g^–1 K^–1 × 7.5 K =» 2.5 × 10³ «J»/2.5 «kJ» ✔

«n_Cu = $\frac{{0.872}}{{63.55}}$ = 0.0137 mol»

«per mol of CuSO₄ = $\frac{{ - 2.5\,{\text{kJ}}}}{{0.0137\,{\text{mol}}}} = - 1.8 \times {10^2}$ kJ mol^–1»

«for the reaction» ΔH = –1.8 × 10² «kJ» ✔

Award [2] for correct final answer.

b.i.

density «of solution» is 1.00 g cm⁻³

specific heat capacity «of solution» is 4.18 J g⁻¹ K⁻¹/that of «pure» water

reaction goes to completion

iron/CuSO₄ does not react with other substances ✔

The mark for “reaction goes to completion” can only be awarded if 0.0160 mol was used in part (b)(i).

Do not accept “heat loss”.

b.ii.

ALTERNATIVE 1:

« $0.2\,^\circ {\text{C}} \times \frac{{100}}{{7.5\,^\circ {\text{C}}}} =$ » 3 %/0.03 ✔

«0.03 × 160 kJ» = «±» 5 «kJ» ✔

ALTERNATIVE 2:

« $0.2\,^\circ {\text{C}} \times \frac{{100}}{{7.5\,^\circ {\text{C}}}} =$ » 3 %/0.03 ✔

«0.03 × 180 kJ» = «±» 5 «kJ» ✔

Accept values in the range 4.1–5.5 «kJ».

Award [2] for correct final answer.

b.iii.

initial concentration is zero AND concentration increases with time ✔

decreasing gradient as reaction proceeds ✔

c.i.

«draw a» tangent to the curve at time = 0 ✔

«rate equals» gradient/slope «of the tangent» ✔

Accept suitable diagram.

c.ii.

piece has smaller surface area ✔

lower frequency of collisions

fewer collisions per second/unit time ✔

Accept “chance/probability” instead of “frequency”.

Do not accept just “fewer collisions”.

c.iii.

Anode (positive electrode):

2H₂O (l) → O₂ (g) + 4H⁺ (aq) + 4e^– ✔

Cathode (negative electrode):

2H₂O (l) + 2e^– → H₂ (g) + 2OH^– (aq)
OR
2H⁺ (aq) + 2e^– → H₂ (g) ✔

Accept “4OH^– (aq) → O₂ (g) + 2H₂O (l) + 4e^–” OR “Fe²⁺ (aq) → Fe³⁺ (aq) + e^–” for M1.

Accept “Fe²⁺ (aq) + 2e^– → Fe (s)” OR “SO₄^2- (aq) 4H⁺ (aq) + 2e^– → 2H₂SO₃(aq) + H₂O (l)”
for M2.

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

b.iii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

Syllabus sections

Core » Topic 6: Chemical kinetics » 6.1 Collision theory and rates of reaction

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