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Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 3

 3. Application of chemical equilibrium concepts to industrial processes

Examples of how to apply the principles of chemical equilibrium to industrial chemical processes. The equation and equilibrium express is given and the predictions from Le Chatelier's Principle discussed for the production of lime, Haber synthesis of ammonia, Contact process in the manufacture of sulfuric acid and the synthesis of methanol.

(c) doc b KS4 Science GCSE/IGCSE reversible reactions, chemical equilibrium and ammonia notes

Part 3 sub–index : 3.1 Lime production * 3.2 Haber Synthesis of ammonia * 3.3 The Contact Process in H2SO4 production * 3.4 Methanol production *

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations *  Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces

3. Introduction to industrial processes

  • For industrial processes, it is important to maximise the concentration of the desired products and minimise the 'leftover' reactants. Le Chatelier's Principle and the principles of reaction kinetics can both be used to design the best reaction conditions to give the highest possible yield of product in an economic way.

  • Ideas on 'economic production' are described in most detail for the Haber Synthesis of ammonia, but equilibrium and kinetic factors are continually mentioned in the other examples too.

  • Equilibria Parts 1.1–1.4 and Part 2.1b should be studied before working through this page.

3.1 Lime production from limestone

  • calcium carbonate (limestone) (c) doc b calcium oxide (lime) + carbon dioxide

  • CaCO3(s) (c) doc b CaO(s) + CO2(g) (ΔH = +178 kJ mol–1)

  • The concentration of the solids is constant, so the only effective equilibrium expression is:
    • Kp = pCO2 atm or Pa
  • The forward reaction is endothermic, 178kJ of heat energy is absorbed (taken in) for every mole of calcium oxide formed and so CaO formation is favoured by high temperature. In fact the reaction is not viable/feasible until at least 900oC, and at 1000oC in a limekiln the equilibrium position is very much on the RHS.
  • One mole of gas is formed in the process, so there is a net increase in the moles of gas in lime formation, since there are no gaseous reactants. Therefore decreasing the pressure will favour the formation of more gas molecules if possible, so more carbon dioxide formed, and hence more lime.
  • However, the reaction is carried out at normal atmospheric pressure in a lime kiln that that is well ventilated. This reduces the partial pressure of carbon dioxide and so reduces the un–desired backward reaction. It means that a true equilibrium is never established because the reaction is not in a closed system and so as CO2 is vented, the forward reaction is promoted.
  • Old kilns essentially worked on a batch process i.e. coal fuelled furnace heating adjacent limestone, but modern tilted rotary kilns run continually with limestone fed in at one end and lime extracted at the other end.



3.2 The Synthesis of ammonia – The Haber Process

  • Ammonia gas is synthesised in the chemical industry by reacting nitrogen gas with hydrogen gas.
  • The three raw principal materials are (i) air, (ii) coal–coke/methane or higher hydrocarbons and (iii) water.
  • The nitrogen is obtained from air (80% N2) by two principal methods:
    1. Air is cooled and compressed under high pressure to form liquid air (liquefaction). The liquid air is fractionally distilled at low temperature to separate oxygen (used in welding, hospitals etc.), nitrogen (for making ammonia), Noble Gases e.g. argon for light bulbs, helium for balloons).
    2. By burning natural gas in air and separating the nitrogen from the water and carbon dioxide combustion products.
  • Synthesis gas is the primary source of hydrogen, though 'syn–gas' or 'syngas' is a mixture of hydrogen and carbon monoxide and is made by three main methods. (Hydrogen can also be obtained as a by–product from oil–cracking and the electrolysis of brine.)
    1. Historically synthesis gas was made by passing steam over white hot coke at 800–1200oC. Coke is made by heating coal at high temperature and is mainly carbon. This process is still used in many parts of the world and if coal is available in a country without oil, then it will be cheaper than using hydrocarbons from oil.
      • C(s) + H2O(g) (c) doc b CO(g) + H2(g) (ΔH = +131 kJ mol–1)
      • The very high temperature favours the endothermic production of hydrogen and of course considerably increases the rate of reaction. Water and carbon show little reaction at lower temperatures.
      • The reaction can be carried out at normal pressure since higher pressures favour the reverse reaction (1 gas mol <== 2 gas mol).
    2. Methane can be partially oxidised by reaction with the oxygen in air.
      • CH4(g) + 1/2O2(g) ==> CO(g) + 2H2(g) (ΔH = –36 kJ mol–1)
      • Low pressure (1.5 gas mol ==> 3.0 gas mol) and low temperatures (exothermic) favour this reaction.
      • This reaction can be carried out with nearly 100% conversion at atmospheric pressure (so low pressure) at 800–1000oC using cobalt/nickel based catalysts e.g. Ni–ThO2 or Ni–SiO2. Very little water or carbon dioxide is formed if the reaction conditions are carefully controlled.
      • Although the 'unfavourable' high temperature is needed (plus catalyst) to break the strong C–H or O=O bonds in the reactant molecules and get an effective rate of reaction, there is quite an entropy increase via the doubling of moles of gas, and quite clearly the RHS is favoured (its probably irreversible?).
      • The reaction also produces the required 1:2 ratio of CO:H2 for methanol production.
    3. These days hydrogen is primarily made by reacting methane (natural gas) and water (steam), and the process is called steam–methane reforming.
      • CH4(g) + H2O(g) (c) doc b 3H2(g) + CO(g) (ΔH = +206 kJ mol–1)
      • Kp =

         pH23 pCO
        pCH4 pH2O
      • The reaction is carried out at 700–1000oC, over a nickel catalyst, at a pressure of 1–2 x 106 Pa (1–2MPa, 10–20 atm).
    • This reaction is favoured by high temperature (endothermic, heat absorbing) and low pressure (2 gas moles ==> 4 gas moles).

    • In practice a high temperature is indeed used, and combined with a catalyst,  both factors will increase the rate of reaction to ensure the process is economic.

    • However, although a low pressure is advantageous, in practice a moderately high pressure is used, which effectively increases the concentration of the reactants and provides a greater bulk of material passing through the reactor in a given time.

    • The yield of hydrogen can be further increased by using the 'shift' reaction.

      • CO(g) + H2O(g) (c) doc b CO2(g) + H2(g) (ΔH = –41 kJ mol–1)

      • Since 2 gas mol ==> 2 gas mol, changing pressure offers no advantage.

      • However, the lower the temperature, the greater the yield of this exothermic reaction.

      • But, too low a temperature leads to an uneconomic low rate of reaction, so research continues into low temperature catalysts, usually based on transition metals and their oxides e.g. Cr–Fe oxides (e.g. Fe3O4) are effective at 400–500oC, Cu–ZnO–Al2O3 at 200–400oC and CuO–ZnO at 175–325oC.

      • The carbon dioxide can be absorbed under pressure into water or potassium carbonate solution.

  • TOP and LINKSThe Bosch–Haber synthesis equation for ammonia is
  • N2(g) + 3H2(g) (c) doc b 2NH3(g) (ΔH = –92 kJ mol–1

  • (see also Le Chatelier example 1.4.2 and calculation 2.2b.2)

  • Kp =

    pN2 pH23
  • Since a dynamic equilibrium will form, there is no chance of 100% yield even if you use, as you actually do, the theoretical stoichiometric reactant volume/mole ratio of N2 :H2 of 1 : 3 !
  • In forming ammonia, heat energy is given out to the surroundings i.e. exothermic, 46kJ of heat released per mole of ammonia formed.
  • Four moles of 'reactant' gas form two moles of 'product' gas, so there is net decrease in gas molecules on forming ammonia.
  • So applying the equilibrium rules, the formation of ammonia is favoured by  ...
    • (a) Using high pressure because you are going from 4 to 2 gas molecules (the high pressure also speeds up the reaction because it effectively increases the concentration of the gas molecules) and allows a greater bulk flow rate material through the reactor, but very high pressure means more dangerous and more costly engineering to address the health and safety issues involved.
    • (b) Carrying out the reaction at a low temperature, because it is an exothermic reaction favoured by low temperature, but this may produce too slow a rate of reaction,
    • So, the idea is to use a set of optimum conditions to get the most efficient yield of ammonia and this may well, as in this case, be a low % yield (e.g. 8% conversion) but fast. Described below are the conditions to give the most economic production of ammonia.
    • these arguments make the point that the yield* of an equilibrium reaction depends on the conditions used.
      • * The word 'yield' means how much product you get compared to the theoretical maximum possible if the reaction goes to 100% completion.
      • For more basic ideas on chemical economics see GCSE notes on Extra Industrial Chemistry.
  • The above chart completely illustrates and vindicates the predictions from Le Chatelier's Principle.
  • In industry moderate–high pressures of 2.5–25 x 106 Pa (2.5–5MPa, 25–250 atm) in line with the theory but to high to raise H&S engineering costs.
  • TOP and LINKSTheoretically a low temperature would give a high yield of ammonia BUT ...
    • Nitrogen is very stable molecule and not very reactive i.e. chemically inert and the rate of reaction at high % yield low temperatures is much to slow to be economic.
    • To speed up the reaction an iron oxide catalyst (Fe3O4?) is used as well as a higher temperature (e.g. 400 – 450oC).
    • The higher temperature is an economic compromise, i.e. it is more economic to get a low yield very fast, than a high yield very slowly!
    • Note: a catalyst does NOT affect the yield of a reaction, i.e. the equilibrium position BUT you do get there faster!
  • The above chart illustrates the dilemma caused by the % ammonia formed at equilibrium at different temperatures compared to the relative rate of reaction at four different activation energies. It also illustrates the very positive effect of using a catalyst to lower the activation energy, Ea, to give a faster rate of production at a lower temperature, and the added benefit of a higher % ammonia formed at equilibrium at this lower temperature.
  • At the end of the process, when the gases emerge from the iron catalyst reaction chamber, the gas mixture is cooled under high pressure, when only the ammonia liquefies and is so can be removed and stored in cylinders.
  • Any unreacted nitrogen or hydrogen (NOT liquified), is recycled back through the reactor chamber, nothing is wasted!
    • Nitrogen (bpt –196oC) and hydrogen (bpt –252oC) have much lower boiling points than ammonia (bpt –33oC). Boiling points increase with increase in pressure but these normal atmospheric pressure bpt values offer a fair comparison.
  • To sum up: A low % yield of ammonia is produced quickly at moderate high temperatures and pressures, and is more economic than getting a higher % equilibrium yield of ammonia at a more costly high pressure or slower lower temperature reaction.
  • Mathematical computer modelling of the Haber Synthesis of Ammonia (used to produce the two charts)
  • Detailed AS–A2 notes on "Kinetics – rates of reaction" for further reading.
  • AND there are some more general notes on the GCSE Chemical Economics section on the Extra Industrial Chemistry page.



3.3 The Contact Process in the manufacture of sulphuric acid

  • The basics of sulphuric acid uses and its manufacture via the Contact Process is covered on the GCSE Extra Industrial Chemistry Notes.

  • The mechanism of the Contact Process is discussed on the AS–A2 Transition Metals page but its the application of equilibrium and kinetic concepts which is important here.

  • The Contact Process of sulphur trioxide production must be economically efficient for the manufacture of the important industrial chemical sulphuric acid.

  • In the Contact Process reactor the sulphur dioxide is mixed with air (the required stoichiometric volume/mole SO2:O2 ratio is 2:1, in practice 1–2:1 is used) and the mixture passed over a catalyst of vanadium(V) oxide V205 at a relatively high temperature of about 450°C and at a pressure of between 1–2 atm. It is an exothermic oxidation and is known as the .

  • (1) In the reactor the sulphur dioxide is oxidised in the reversible exothermic reaction ...

    •   2SO2(g) + O2(g) (c) doc b 2SO3(g) (ΔH = –196 kJ mol–1)

    • (see also Le Chatelier example 1.4.8)

    • Kp =

      pSO22 pO2
  • The reaction forms sulphur trioxide and the equilibrium is very much to the right hand side because  

    • Despite the reaction being exothermic a relatively high temperature is used which favours the reverse reaction R to L, from the energy change equilibrium rule, i.e. increasing temperature shifts the equilibrium in the endothermic direction. However the value of Kp is high enough to give a 99% yield.

    • The reaction is favoured by high pressure (pressure equilibrium rule, 3 => 2 gas molecules), but only a small increase in pressure is used to give high yields of sulphur trioxide, because the right hand side is energetically very favourable (quite exothermic and high Kp)

    • The use of the V2O5 catalyst ensures a fast reaction without having to use too a higher temperature which would begin to favour the left hand side too much (energy change equilibrium rule), but remember a catalyst does not affect the % yield or equilibrium concentration of SO3, you just get there more economically faster.

    • Multiple reactor beds are used to ensure the maximum % conversion and heat exchange systems are used to control the temperature, and pre–heat incoming reactant gases.

    • Good anti–pollution measures need to be in place since the sulphur oxides are harmful and would cause local acid rain! To help this situation AND help the economics of the process the residual SO2 is kept to the minimum by the reaction conditions describe above.

  • (2) The sulphur trioxide is dissolved in concentrated sulphuric acid to form fuming sulphuric acid (oleum).

    • SO3(g) + H2SO4(l) ==> H2S2O7(l)

  • (3) Water is then carefully added to the oleum to produce concentrated sulphuric acid (98% H2SO4).

    • H2S2O7(l) + H2O(l) ==>  2H2SO4(l) 

    • If the sulphur trioxide is added directly to water an acid mist forms which is difficult to contain because the reaction to form sulphuric acid solution is very exothermic with a big K value!



3.4 Methanol production

  • Methanol is an important alcohol used in fuel mixtures, making methyl esters and oxidation to methanol (formaldehyde) to make urea–formaldehyde resin glues.

  • It is manufactured directly from synthesis gas (CO + H2, see section 3.2 above).

  • CO(g) + 2H2(g) (c) doc b CH3OH(g) (ΔH = –90 kJ mol–1)


    Kp =

    pCO pH22

  • The reaction is carried out at 250oC, over a Cu–ZnO–Al2O3(alumina) catalyst at a pressure of 5–10 x 106 Pa (5–10MPa, 50–100 atm).

  • Theoretically the reaction is favoured by high pressure (3 gas mol ==> 1 gas mol) and low temperature.

  • In practice a high pressure is used to give an acceptable yield in accordance with Le Chatelier's Principle, but a moderately high temperature plus a catalyst, are employed to get an economic production rate.



Part 3 sub–index : 3.1 Lime production * 3.2 Haber Synthesis of ammonia * 3.3 The Contact Process in H2SO4 production * 3.4 Methanol production *

Advanced Equilibrium Chemistry Notes Part 1. Equilibrium, Le Chatelier's Principle–rules * Part 2. Kc and Kp equilibrium expressions and calculations *  Part 4. Partition, solubility product and ion–exchange * Part 5. pH, weak–strong acid–base theory and calculations * Part 6. Salt hydrolysis, Acid–base titrations–indicators, pH curves and buffers * Part 7. Redox equilibria, half–cell electrode potentials, electrolysis and electrochemical series * Part 8. Phase equilibria–vapour pressure, boiling point and intermolecular forces

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