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Date November 2018 Marks available 2 Reference code 18N.2.sl.TZ0.1
Level SL 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.

Markscheme

nCuSO4 «= 0.0800 dm3 × 0.200 mol dm–3» = 0.0160 mol AND

nFe « 3.26 g 55.85 g mo 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»  ✔

« 0.872 g 1.02 g × 100 = » 85.5 «%»  ✔

 

ALTERNATIVE 2:
« 0.872 g 63.55 g mo l 1 = » 0.0137 «mol»  ✔

« 0.0137 mol 0.0160 mol × 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 =  2.5 kJ 0.0160 mol = 1.6 × 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» ✔

«nCu = 0.872 63.55 = 0.0137 mol»

«per mol of CuSO4 =  2.5 kJ 0.0137 mol = 1.8 × 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

OR

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

OR

reaction goes to completion

OR

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 C × 100 7.5 C = » 3 %/0.03 ✔

«0.03 × 160 kJ»«±» 5 «kJ» 

 

ALTERNATIVE 2:

« 0.2 C × 100 7.5 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

OR

fewer collisions per second/unit time ✔

 

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

Do not accept just “fewer collisions”.

c.iii.

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.

Syllabus sections

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