C3.5 The Production of Ammonia

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  • Created by: Fiona S
  • Created on: 13-05-15 02:01

Haber Process

N2(g) + 3H2(g) <--> 2NH3(g)

Ammonia is mostly used as a fertiliser but it is also used in explosive and textiles.

(http://www.chemguide.co.uk/physical/equilibria/haberflow.gif)

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Making Ammonia

Temperature 450 degrees

  • Lower Temperature --> more NH3
  • Reaction Exothermic
  • Higher Temperature --> faster
  • Compromise between yield and rate

Pressure 200 atm 

  • Higher Pressure --> more NH3
  • 4 Molecules(more pressure) --> 2 Molecules(less pressure)
    • N2 + 3H2 --> 2NH3
  • Higher Pressure --> very expensive
  • Comprimise between yield and cost

Catalyst Iron

  • Speeds up reaction
  • No effect on yield of NH3
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Making Ammonia

  • NH3 is removed by cooling (forms a liquid)
  • Less than 30% yield, so the unreacted H2 and N2 are recycled

It condenses as it has a higher boiling point than the nitrogen and hydrogen.

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Equilibrium

When a reversable reaction takes place in a closed system there will be a point at which the concentrations do not change anymore. The forward and backward reactions are still taking place, just they are taking place at the same rate so there is no overall change.

(http://everythingmaths.co.za/science/grade-12/08-chemical-equilibrium/pspictures/cc73d2ae9ad9505db854bd5e484e2290.png)

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Equilibrium

If a reversible reaction takes place in a closed system then at the same point the forward reaction will have the same rate as the backward reaction. At this point the concentration of the reactants and products will no longer change. This is known as (dynamic) equilibrium.

  • System must be closed
  • The reaction has NOT stopped
  • The concentration of reactant and products is not the same, it simply remains constant

We can change the conditions to 'shift equilibrium' to the left or the right to promote either the forwards or backwards reaction.

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Equilibrium

Le Chatelier

Le Chatelier stated that a system will shift equilibrium to oppose any change placed upon the system.

  • 1) Temperature

For every reversible reaction one direction will always be exothermic (-ve ΔH) and the other way will always be endothermic (+ve ΔH).

If we increase the temperature equilibrium will shift to oppose this change. The endothermic reaction will be favoured as the extra energy can then be absorbed.

If we decrease the temperature equilibrium will shift to oppose this change. The exothermic reaction will be favoured as extra energy can be released and the temperature will increase.

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Equilibrium

  • 2) Pressure

If we increase the pressure then equilibrium will shift to oppose this change. The reaction which results in the fewest number of moles will be favoured as this lead to a reduction in pressure.

Remember, pressure is caused by particles hitting the sides of the container. Fewer particles = lower pressure.

If we decrease the pressure the equilibrium will shift to oppose this change. the reaction which results in the greatest number of moles will be favoured as this will lead to an increase in pressure.

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Compromise

Temperature

Low Temperature

  • Using low temperature will shift the equilibrium to favour the exothermic reaction. The forward reaction is exothermic  (-ve ΔH) so more products will be produced, this will oppose the change.

However, too low a temperature means the reaction rate will be very slow

Pressure

High Pressure

  • This will shift equilibrium to favour forwards reaction which leads to fewer number of moles.

However, too high a pressure is expensive

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