Question 18N.2.hl.TZ0.1
| Date | November 2018 | Marks available | [Maximum mark: 17] | Reference code | 18N.2.hl.TZ0.1 |
| Level | hl | Paper | 2 | Time zone | TZ0 |
| Command term | Calculate, Determine, Explain, Outline, Sketch, State | Question number | 1 | Adapted from | N/A |
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]
nCuSO4 «= 0.0800 dm3 × 0.200 mol dm–3» = 0.0160 mol AND
nFe «» = 0.0584 mol ✔
CuSO4 is the limiting reactant ✔
Do not award M2 if mole calculation is not shown.
The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.
[2]
ALTERNATIVE 1:
«0.0160 mol × 63.55 g mol–1 =» 1.02 «g» ✔
«» 85.5 «%» ✔
ALTERNATIVE 2:
«» 0.0137 «mol» ✔
«» 85.6 «%» ✔
Accept answers in the range 85–86 %.
Award [2] for correct final answer.
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]
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 = 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 = = 0.0137 mol»
«per mol of CuSO4 = kJ mol–1»
«for the reaction» ΔH = –1.8 × 102 «kJ» ✔
Award [2] for correct final answer.
State another assumption you made in (b)(i).
[1]
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”.
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]
ALTERNATIVE 1:
«» 3 %/0.03 ✔
«0.03 × 160 kJ» = «±» 5 «kJ» ✔
ALTERNATIVE 2:
«» 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.
Sketch a graph of the concentration of iron(II) sulfate, FeSO4, against time as the reaction proceeds.
[2]
initial concentration is zero AND concentration increases with time ✔
decreasing gradient as reaction proceeds ✔
Outline how the initial rate of reaction can be determined from the graph in part (c)(i).
[2]
«draw a» tangent to the curve at time = 0 ✔
«rate equals» gradient/slope «of the tangent» ✔
Accept suitable diagram.
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]
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”.
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]
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 “SO42- (aq) 4H+ (aq) + 2e– → 2H2SO3(aq) + H2O (l)”
for M2.