Transition metals: Heterogeneous catalysis

Edexcel A level Chemistry (2017)

Topic 15: Principles of transition metal chemistry

Learning Objectives related to heterogeneous catalysis

15/29. To know that transition metals and their compounds can act as heterogeneous and homogeneous catalysts

15/30. To know that a heterogeneous catalyst is in a different phase from the reactants and that the reaction occurs at the surface of the catalyst

15/31. To understand, in terms of oxidation number, how V2O5 acts as a catalyst in the contact process

15/32. To understand how a catalytic converter decreases carbon monoxide and nitrogen monoxide emissions from internal combustion engines by:

i adsorption of CO and NO molecules onto the surface of the catalyst 


ii weakening of bonds and chemical reaction 


iii desorption of CO2 and N2 product molecules from the surface of the catalyst

Transition metals as catalysts (2)

Catalysts are substances that change the rate of chemical reaction and remain unchanged at the end of that reaction. 

They act so as to reduce the activation energy of a chemical change.

The reduction of the activation energy of a reaction means that more molecules within that reaction possess the activation energy.

A greater percentage of collisions between these molecules can be effective if the activation energy is lower.  It is this increase in the number of effective collisions that leads to an increase in the reaction rate.  

With transition metals catalysts can be either homogenous or heterogeneous.

Heterogeneous transition metal catalysts

Heterogeneous catalysts are not in the same state as the reagents in the chemical change.

More often than not heterogeneous catalysts are solids over which reactants in the gaseous or liquid state pass. 

Since the catalyst is in a different phase to the reactants the reaction occurs on the surface of the catalyst. 

There are several examples of this type of catalysis. 

First off we will examine the Contact process for the manufacture of sulphuric acid.

 

You probably learned about this process for the manufacture of sulphuric acid in a previous chemistry course. If you did, you will remember that in a key part of the process a catalyst of vanadium(V)oxide is used.

The Contact process

Stage 1:  Combustion

In Stage 1 sulphur is burned in oxygen

S_{8 (l)}   +     8O_2(g)       ⟶    8SO_2(g)

Stage 2: Oxidation

In Stage 2 sulphur dioxide is oxidised to sulphur trioxide.  Oxidation takes place in the presence of vanadium(V)oxide (V₂O₅) catalyst.

2SO_2(g)    +    O_2(g)    ⟶    2SO_3(g)

Stage 3: Dissolution

In Stage 3, sulphur trioxide is added to concentrated sulphuric acid to form oleum.  Sulphur trioxide is better dissolved in sulphuric acid than in water since the dissolution of the gas in water is so exothermic a fog is formed which is very difficult to condense.

  2SO_3(g)     +     H₂SO_4(l)     ⟶   H₂S₂O_7(l)    

Stage 4: Dilution

In Stage 4, oleum is diluted with water to form concentrated sulphuric acid doubling the number of moles of the acid we started with. 

   H₂O_(l)      +      H₂S₂O_7(l)    ⟶       2H₂SO_4(l)                    

The role of the catalyst vanadium(V)oxide.

Sulfur dioxide and oxygen react like this:

2SO_2(g) + O_2(g) ⇌ 2SO_3(g) : ΔH = -197 kJ·mol⁻¹

Since this reaction is exothermic, a lower temperature would shift the chemical equilibrium to the right.  That would increase the percentage yield.

But too low a temperature lowers the rate of formation to an uneconomical level. So to increase the rate, high temperatures (450 °C), medium pressures (1-2 atm), and  vanadium(V)oxide (V₂O₅) are used.  These conditions give a 95% conversion.

All the catalyst does is increase the rate of reaction.   

It does not change the position of equilibrium.

The catalyst acts in two steps:

Step 1:  Oxidation of SO₂ into SO₃ by V⁵⁺:

2SO₂ + 4V⁵⁺ + 2O²⁻ → 2SO₃ + 4V⁴⁺

Step 2: Oxidation of V⁴⁺ back into V⁵⁺ by oxygen to regenerate the catalyst

4V⁴⁺ + O₂ → 4V⁵⁺ + 2O²⁻

This summarises the action of the V₂O₅ catalyst.

How a catalytic converter works

Between the car engine and the exhaust box and silencer there is another box: the catalytic converter.

It is usually fitted close to the exhaust manifold as it is heated using the hot exhaust gases.

Some need to be at about 700^oC to operate at maximum efficiency.

 

The diagram shows us that inside the steel box there is a ceramic structure, a honeycomb that has a very high surface area roughly equivalent to a couple of soccer pitches!!

On the ceramic surface are particles of noble metal platinum and rhodium.  It is on the surface of these metals that the conversion of carbon monoxide to carbon dioxide and nitrogen oxides to nitrogen will occur.  

You can see that in the diagram above two changes take place in the converter.

First, a reduction involving nitrogen oxides and carbon monoxide that can be summarised in the following equation:

NO(g)    +     CO(g)      ⟶        CO₂(g)     +    ½N₂(g) 

Second, an oxidation involving residual oxygen, unburnt hydrocarbons and carbon monoxide.

CO   +    CH₄   +   2½O₂     ⟶       2CO₂     +     2H₂O

These two reactions take place on the surface of the catalyst.

We can see below how this process works when oxygen and carbon monoxide are involved.

A: Adsorption of the gas to the platinum surface

B: Formation of weak bonds between the gas and the surface platinum atoms.

C: Coming together of the two gases on the platinum surface close enough to collide with a very lower E_a activation energy.

D: Collision and conversion of carbon monoxide to carbon dioxide

E: Desorption of the carbon dioxide product from the surface as the bonds with the Platinum are now too weak to hold it there.