DP Chemistry (last assessment 2024)

Test builder »

Question 18N.2.sl.TZ0.1

Select a Test
Date November 2018 Marks available [Maximum mark: 15] Reference code 18N.2.sl.TZ0.1
Level sl Paper 2 Time zone TZ0
Command term Calculate, Determine, Explain, Outline, Sketch, State Question number 1 Adapted from N/A
1.
[Maximum mark: 15]
18N.2.sl.TZ0.1

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)

(a.i)

Determine the limiting reactant showing your working.

[2]

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.ii)

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

[2]

Markscheme

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.

(b.i)

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]

Markscheme

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.ii)

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

[1]

Markscheme

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.iii)

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]

Markscheme

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.

 

(c.i)

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

[2]

Markscheme

 

 

initial concentration is zero AND concentration increases with time ✔

decreasing gradient as reaction proceeds ✔

(c.ii)

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

[2]

Markscheme

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

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

 

Accept suitable diagram.

(c.iii)

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]

Markscheme

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”.

Syllabus sections