Questions on catalysis: heterogeneous, homogeneous and autocatalytic

Question 1

Explain, with equations, why the reaction between peroxodisulfate ions and iodide ions is slow, how Fe²⁺ ions catalyse it, and how the reaction profile of the catalysed reaction differs from that of the uncatalysed one.

Accepted Answer

Indicative content

both reactants are negative ions, which repel each other, so the activation energy of the direct reaction is high and it is slow
step 1: S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺, in which the iron(II) is oxidised to iron(III)
step 2: 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂, in which the iron(III) is reduced back to iron(II) and the catalyst is regenerated
each step is between oppositely charged ions, which attract, so each has a lower activation energy than the direct reaction
the uncatalysed profile has one high hump; the catalysed profile has two lower humps with a trough between them for the intermediate, Fe³⁺
the reactants and the products are at the same energy levels on both profiles, because the catalyst does not change ΔH

Levels

Level 3 (5–6 marks) the repulsion between the two negative ions is given as the reason for the slow direct reaction, both steps of the catalysis are written with correct equations and the oxidation state changes, the lower activation energy of each step is explained through the attraction of opposite charges, and the two profiles are compared with the intermediate and the unchanged ΔH
Level 2 (3–4 marks) the repulsion of the ions and both equations are given; the regeneration of Fe²⁺, why each step is faster, or the comparison of the two profiles is thin
Level 1 (1–2 marks) the ions are said to repel or Fe²⁺ to be a catalyst; the catalysed profile is often drawn with a different ΔH, and one equation at most is given

Question 2

Explain how vanadium(V) oxide acts as a catalyst in the Contact process. Include the two equations for the catalysed route.

Accepted Answer

B1 SO₂ + V₂O₅ → SO₃ + V₂O₄; allow multiples B1 V₂O₄ + ½O₂ → V₂O₅; allow 2V₂O₄ + O₂ → 2V₂O₅ B1 vanadium(V) oxide oxidises sulfur dioxide and is reduced to vanadium(IV) oxide, +5 to +4 B1 oxygen then oxidises vanadium(IV) oxide back to vanadium(V) oxide, +4 to +5, so the catalyst is regenerated B1 the route depends on vanadium's variable oxidation state; 'it lowers the activation energy' alone is not enough

Question 3

Calculate E°cell for each of the two steps in the Fe²⁺-catalysed reaction between S₂O₈²⁻ and I⁻. E°(S₂O₈²⁻/SO₄²⁻) = +2.01 V, E°(Fe³⁺/Fe²⁺) = +0.77 V and E°(I₂/I⁻) = +0.54 V.

Accepted Answer

M1 step 1, S₂O₈²⁻ oxidising Fe²⁺: E°cell = +2.01 − (+0.77) A1 E°cell for step 1 = +1.24 V M1 step 2, Fe³⁺ oxidising I⁻: E°cell = +0.77 − (+0.54) A1 E°cell for step 2 = +0.23 V, to two significant figures; both positive, so both steps are feasible

Question 4

Calculate the mean rate of the autocatalysed reaction of manganate(VII) with ethanedioate over the first 60 s and over the interval from 180 s to 240 s, using the readings marked on the graph.

Accepted Answer

M1 mean rate from 0 to 60 s = (0.0200 − 0.0195) ÷ 60 A1 8.3 × 10⁻⁶ mol dm⁻³ s⁻¹, to two significant figures M1 mean rate from 180 to 240 s = (0.0140 − 0.0070) ÷ 60 A1 1.2 × 10⁻⁴ mol dm⁻³ s⁻¹, to two significant figures, about fourteen times the first

Question 5

Compare the way vanadium(V) oxide catalyses the Contact process with the way Fe²⁺ ions catalyse the reaction between S₂O₈²⁻ and I⁻.

Accepted Answer

B1 vanadium(V) oxide is a heterogeneous catalyst, a solid with gaseous reactants; Fe²⁺ is homogeneous, in the same solution as the reactants B1 the vanadium reactions happen on the solid surface, whereas the iron reactions happen between ions in solution B1 both work by changing oxidation state and back: vanadium between +5 and +4, iron between +2 and +3 B1 both provide an alternative route with a lower activation energy and are regenerated at the end

Question 6

Describe the three steps by which a solid catalyst speeds up a reaction between gases.

Accepted Answer

B1 adsorption: reactant molecules form weak bonds to active sites on the catalyst surface, which weakens bonds within them B1 reaction: the adsorbed molecules, held close together in a favourable orientation, react on the surface B1 desorption: the product molecules break away from the surface, freeing the active sites

Question 7

Explain why a good heterogeneous catalyst must adsorb reactants neither too strongly nor too weakly.

Accepted Answer

B1 if adsorption is too weak, as with silver, the reactant molecules leave before they react B1 if adsorption is too strong, as with tungsten, the products do not desorb and the active sites stay blocked B1 intermediate adsorption, as with nickel or platinum, weakens the reactants' bonds but still lets the products leave

Question 8

Explain why the platinum and rhodium in a catalytic converter are spread as a thin layer on a ceramic honeycomb.

Accepted Answer

B1 only the surface atoms of the metal act as active sites B1 the honeycomb gives a very large surface area, so a thin layer on it exposes most of the metal atoms B1 less of the expensive metal is needed for the same catalytic activity, which reduces the cost

Question 9

Explain why a car fitted with a catalytic converter must not run on leaded petrol.

Accepted Answer

B1 lead is adsorbed very strongly onto the platinum and rhodium surface B1 it does not desorb, so it blocks the active sites (poisons the catalyst) B1 fewer active sites are available, so the converter no longer removes CO and NO effectively

Question 10

Suggest whether Mn³⁺/Mn²⁺ (E° = +1.51 V) and V³⁺/V²⁺ (E° = −0.26 V) could each catalyse the reaction between S₂O₈²⁻ and I⁻. E°(S₂O₈²⁻/SO₄²⁻) = +2.01 V and E°(I₂/I⁻) = +0.54 V.

Accepted Answer

B1 a catalyst couple must have E° between +0.54 V and +2.01 V, so that both of its steps are feasible B1 Mn³⁺/Mn²⁺ at +1.51 V lies in the range: S₂O₈²⁻ oxidises Mn²⁺ and Mn³⁺ oxidises I⁻, so it could catalyse the reaction B1 V³⁺/V²⁺ at −0.26 V lies outside it: V³⁺ cannot oxidise I⁻ (E°cell −0.80 V), so it cannot complete the cycle

Question 11

Describe the shape of a graph of [MnO₄⁻] against time for the reaction of acidified manganate(VII) with ethanedioate ions.

Accepted Answer

B1 at first the concentration falls only slowly, so the curve is nearly flat B1 the curve then becomes steep, the rate reaching a maximum part way through B1 finally the curve levels off towards zero as the manganate(VII) is used up, giving an inverted S shape

Question 12

Explain why the reaction between manganate(VII) and ethanedioate ions speeds up after it has started.

Accepted Answer

B1 Mn²⁺, a product of the reaction, catalyses it, so the reaction is autocatalytic B1 at the start there is little Mn²⁺, so the reaction is slow because both reactants are negative ions B1 as Mn²⁺ builds up, more of the reaction goes by the catalysed route through Mn³⁺, so the rate increases

Question 13

Suggest how the graph of [MnO₄⁻] against time changes when a little manganese(II) sulfate is added to an acidified ethanedioate solution before the manganate(VII), and why.

Accepted Answer

B1 the slow, nearly flat start disappears and the reaction is fast from the beginning B1 the catalyst, Mn²⁺, is present from the start instead of having to build up B1 the curve becomes steepest at the start and less steep throughout, like an ordinary reaction

Question 14

Calculate the number of platinum atoms in a catalytic converter that contains 1.50 g of platinum. Ar of Pt = 195.1, L = 6.022 × 10²³ mol⁻¹.

Accepted Answer

M1 moles of Pt = 1.50 ÷ 195.1 = 7.69 × 10⁻³ mol M1 number of atoms = 7.69 × 10⁻³ × 6.022 × 10²³ A1 4.63 × 10²¹ atoms, to three significant figures

Question 15

Explain how nitrogen dioxide acts as a homogeneous catalyst for the oxidation of sulfur dioxide in the atmosphere. Include equations.

Accepted Answer

B1 NO₂ + SO₂ → NO + SO₃, in which nitrogen dioxide oxidises sulfur dioxide and is reduced B1 NO + ½O₂ → NO₂, in which oxygen oxidises nitrogen monoxide back to nitrogen dioxide B1 the nitrogen dioxide is regenerated and is in the same (gas) phase as the reactants, so it is a homogeneous catalyst

Question 16

State the meaning of the terms heterogeneous catalyst and homogeneous catalyst.

Accepted Answer

B1 a heterogeneous catalyst is in a different phase from the reactants B1 a homogeneous catalyst is in the same phase as the reactants

Question 17

Suggest why the hydrogen fed to an ammonia plant is first treated to remove sulfur compounds.

Accepted Answer

B1 sulfur compounds are strongly adsorbed onto the iron catalyst and poison it by blocking the active sites B1 a poisoned catalyst gives a lower rate and must be replaced more often, which raises costs

Question 18

Suggest why nitrogen dioxide can catalyse the oxidation of sulfur dioxide even though nitrogen is not a transition metal.

Accepted Answer

B1 nitrogen has more than one oxidation state, +4 in NO₂ and +2 in NO B1 so it can be reduced in one step and oxidised back in the next, and is regenerated like a transition metal catalyst