Date | November 2018 | Marks available | 1 | Reference code | 18N.2.hl.TZ0.1 |
Level | HL | Paper | 2 | Time zone | TZ0 |
Command term | State | 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.
The mass of copper obtained experimentally was 0.872 g. Calculate the percentage yield of copper.
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.
State another assumption you made in (b)(i).
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.
Sketch a graph of the concentration of iron(II) sulfate, FeSO4, against time as the reaction proceeds.
Outline how the initial rate of reaction can be determined from the graph in part (c)(i).
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.
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.
Markscheme
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.
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.
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.
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”.
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.
initial concentration is zero AND concentration increases with time ✔
decreasing gradient as reaction proceeds ✔
«draw a» tangent to the curve at time = 0 ✔
«rate equals» gradient/slope «of the tangent» ✔
Accept suitable diagram.
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”.
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.