Transition metals as catalystsEdexcel International A Level Chemistry: Subtopic test

1. 1
   In the Contact process, sulfur dioxide and oxygen are passed over a solid vanadium(V) oxide catalyst at about 450 °C to make sulfur trioxide: SO₂(g) + ½O₂(g) ⇌ SO₃(g).

   (a)

   Which statement correctly describes this catalyst?

   [1 mark]

   • AIt is homogeneous, because all the species are in the same container
   • BIt is homogeneous, because vanadium(V) oxide is not used up
   • CIt is heterogeneous, because the catalyst is a solid and the reactants are gases
   • DIt is heterogeneous, because the catalyst is in the same phase as the products

   (b)

   The mechanism involves SO₂ reducing V₂O₅ to V₂O₄. What is the change in the oxidation number of vanadium in this step?

   [1 mark]

   • A+5 to +4
   • B+4 to +5
   • C+5 to +3
   • D+10 to +8

   (c)

   Write two equations to show how vanadium(V) oxide acts as a catalyst, and state the oxidation number change of vanadium in each.

   [2 marks]

   Total for question 1: 4 marks

2. 2
   A car exhaust system contains a catalytic converter, a ceramic honeycomb coated with a thin layer of platinum and rhodium. It converts carbon monoxide and nitrogen monoxide in the exhaust gases into carbon dioxide and nitrogen.

   (a)

   What is the first step in the action of the catalyst on carbon monoxide and nitrogen monoxide?

   [1 mark]

   • AThe gases dissolve in the metal
   • BThe bonds in CO and NO break in the gas phase before reaching the surface
   • CCarbon dioxide and nitrogen are desorbed from the surface
   • DThe gases are adsorbed on the metal surface, which weakens the bonds within them

   (b)

   Why is the metal spread as a thin layer over a honeycomb structure?

   [1 mark]

   • ATo reduce the activation energy to zero
   • BTo give a large surface area from a small mass of expensive metal
   • CTo make the catalyst dissolve in the exhaust gases
   • DTo make the reaction homogeneous

   (c)

   Write an equation for the reaction of carbon monoxide with nitrogen monoxide in the converter, and explain why it is important that the products leave the surface.

   [2 marks]

   Total for question 2: 4 marks

3. 3
   The reaction between iodide ions and peroxodisulfate ions, S₂O₈²⁻ + 2I⁻ → 2SO₄²⁻ + I₂, is very slow at room temperature. When a few drops of aqueous iron(II) sulfate are added, the reaction becomes much faster.

   (a)

   Write two equations to show how Fe²⁺ ions catalyse this reaction and state why iron is suited to this role.

   [3 marks]

   (b)

   Explain why the uncatalysed reaction is very slow and why iron(II) ions make it faster.

   [4 marks]

   Total for question 3: 7 marks

4. 4
   A student titrates 25.0 cm³ of 0.0500 mol dm⁻³ sodium ethanedioate (Na₂C₂O₄), acidified with dilute sulfuric acid and warmed to about 60 °C, against 0.0200 mol dm⁻³ potassium manganate(VII). The overall equation is 2MnO₄⁻ + 16H⁺ + 5C₂O₄²⁻ → 2Mn²⁺ + 8H₂O + 10CO₂. The first few drops of manganate(VII) lose their colour slowly, but later drops lose their colour almost instantly.

   (a)

   Explain these observations. Your answer should identify the catalyst, name the type of catalysis, and give equations for the catalytic steps.

   [6 marks]

   (b)

   The student repeats the titration but adds a small volume of aqueous manganese(II) sulfate to the ethanedioate solution before starting. Calculate the volume of potassium manganate(VII) solution needed to reach the end point, and explain what difference the manganese(II) sulfate makes to the experiment.

   [6 marks]

   Total for question 4: 12 marks

End of questions