Transition metals as catalystsEdexcel A-Level Chemistry: Subtopic test

1. 1
   In the Contact process, sulfur dioxide gas is oxidised to sulfur trioxide gas by oxygen in the presence of a solid vanadium(V) oxide, V₂O₅, catalyst at about 450 °C: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).

   (a)

   Why is vanadium(V) oxide described as a heterogeneous catalyst in this process?

   [1 mark]

   • AIt is in the same phase as the reactants
   • BIt is not used up in the reaction
   • CIt is a solid in a different phase from the gaseous reactants
   • DIt forms an intermediate species with the reactants in solution

   (b)

   When V₂O₅ reacts with SO₂ to form SO₃ and V₂O₄, what is the change in the oxidation number of vanadium?

   [1 mark]

   • A+5 to +4
   • B+4 to +5
   • C+5 to +3
   • D+3 to +5

   (c)

   Write two equations to show how V₂O₅ acts as a catalyst for the oxidation of SO₂ and is regenerated.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A car's catalytic converter contains platinum and rhodium spread as a thin layer over a ceramic honeycomb. It converts carbon monoxide and nitrogen monoxide in exhaust gases into carbon dioxide and nitrogen.

   (a)

   Which equation shows the main reaction in the catalytic converter?

   [1 mark]

   • ACO + NO → CO₂ + N₂
   • B2CO + 2NO → 2CO₂ + N₂
   • CCO + 2NO → CO₂ + N₂
   • D2CO + 2NO → 2CO₂ + 2N

   (b)

   Which row gives the correct sequence of events for CO and NO at the catalyst surface?

   [1 mark]

   • Adesorption, weakening of bonds, reaction, adsorption
   • Badsorption, desorption, weakening of bonds, reaction
   • Cweakening of bonds, adsorption, desorption, reaction
   • Dadsorption, weakening of bonds, reaction, desorption

   (c)

   Suggest why the platinum and rhodium are used as a thin layer on a ceramic honeycomb rather than as solid blocks of metal.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The reaction between iodide ions and peroxodisulfate ions, S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq), is very slow, but it is catalysed by a few drops of iron(II) sulfate solution. Standard electrode potentials: S₂O₈²⁻ + 2e⁻ ⇌ 2SO₄²⁻, E° = +2.01 V; Fe³⁺ + e⁻ ⇌ Fe²⁺, E° = +0.77 V; I₂ + 2e⁻ ⇌ 2I⁻, E° = +0.54 V.

   (a)

   Explain why the uncatalysed reaction is slow, and why iron(II) ions speed it up. State the type of catalysis involved.

   [3 marks]

   (b)

   Write the two ionic equations for the steps in the catalysed reaction, and use the E° values to show that each step is feasible.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   Transition metal compounds act as catalysts in many reactions. In the Contact process, solid V₂O₅ catalyses the reaction of gaseous sulfur dioxide with oxygen. In a laboratory experiment, acidified potassium manganate(VII) is added to warm sodium ethanedioate solution: the purple colour fades very slowly at first, but then fades rapidly as the reaction proceeds, because Mn²⁺ ions are formed.

   (a)

   Explain the observation in the laboratory experiment. Include the overall equation, the role of Mn²⁺ ions and the equations for the steps in which Mn²⁺ acts as a catalyst.

   [6 marks]

   (b)

   Compare the vanadium(V) oxide catalyst in the Contact process with the Mn²⁺ catalyst in the laboratory experiment, in terms of the type of catalysis, how each catalyst works and the property of transition metals that makes both effective.

   [6 marks]

   Total for question 4: 12 marks

End of questions