Doc Brown's Chemistry Advanced A Level Notes

Theoretical–Physical Advanced Level Chemistry – Equilibria – Chemical Equilibrium Revision Notes PART 1

1. Examples of dynamic molecular chemical equilibrium and Le Chatelier's principle

What is a dynamic equilibrium? How does Le Chatelier's Principle help us predict the shift in equilibrium position on change of temperature, change in concentration or change in pressure? What effect does a catalyst have on the position of an equilibrium?

Chemical Equilibrium Notes Index

Part 1 sub–index: 1.1 Reversible reactions * 1.2 The equilibrium concept

1.3 Le Chatelier's Principle – equilibrium rules * 1.4 Applying the equilibrium rules

Part 1. Introduction - examples of reversible reactions described, dynamic equilibrium and Le Chatelier's Principle

Part 2a. How to write equilibrium expressions and the effect of temperature on the equilibrium constant K

Part 2b. How to write K_c equilibrium expressions, units and exemplar K_c calculations including esterification

Part 2c. How to write K_p expressions, units, partial pressures and exemplar K_p equilibrium calculations

Part 3. Applying Le Chatelier's Principle and equilibrium expressions to Industrial Processes

In contrast with kinetics, which is a study of how quickly reactions occur, a study of equilibria indicates how far reactions will go and Le Chatelier’s principle is used to predict the effects of changes in temperature, pressure and concentration on the yield of a reversible reaction. Not surprisingly, this has important consequences for many industrial processes.

• A reversible reaction is a chemical change in which the products can be converted back to the original reactants under suitable conditions. A reversible reaction is shown by the sign , a half–arrow to the right (forward reaction, L to R), and, a half–arrow to the left (backward reaction, R to L).
• Reactions which are not reversible (irreversible) have the usual complete arrow only pointing to the right.
• -
• Example 1.1.1 - Thermal decomposition of ammonium salts
  • On heating strongly, the white solid, ammonium chloride, thermally decomposes into a mixture of two colourless gases, ammonia and hydrogen chloride. On cooling the reaction is reversed and solid ammonium chloride reforms. This is also an example of sublimation and involves both a physical of state as well as a chemical change.
  When a substance sublimes it changes directly from a solid into a gas without melting and on cooling reforms the solid without condensing to form a liquid.

  • Ammonium chloride + heat ammonia + hydrogen chloride

  • NH₄Cl_(s) NH_3(g) + HCl_(g)

  • Note: Reversing the reaction conditions reverses the direction of chemical change, typical of a reversible reaction.

  • Thermal decomposition means using 'heat' to 'break down' a molecule into smaller ones. The decomposition of NH₄Cl is endothermic, ΔH +ve (heat absorbed, taken in from the surroundings) and the formation of NH₄Cl is exothermic, ΔH –ve (heat released, given out to the surroundings).

  • This means if the direction of chemical change is reversed, the energy change sign must also be reversed but its numerical value stays the same.
  • -
• Example 1.1.2 Thermal decomposition of hydrated copper(II) sulfate crystals
  • On heating the blue solid, hydrated copper(II) sulphate, steam is given off and the white solid of anhydrous copper(II) sulphate is formed. When the white solid is cooled and water added, blue hydrated copper(II) sulphate is reformed.

  • blue hydrated copper(II) sulphate + heat white anhydrous copper(II) sulphate + water

  • CuSO₄.5H₂O_(s) CuSO_4(s) + 5H₂O_(g)

  • Note: The crystal structure is broken down on heating and the water of crystallisation is given off. Thermal decomposition is endothermic (ΔH +ve) as heat is absorbed to drive off the water. The reverse reaction is exothermic (ΔH +ve) i.e. on adding water to cold white anhydrous copper(II) sulphate the mixture heats up as the blue crystals reform.
• These are typical examples you encounter at an earlier study level, but it begs the question, "is it possible to have a situation, under suitable conditions, in which the reaction does not completely go in one direction or the other and both reactants and products co–exist?", and the answer is yes! and the situation is called a dynamic chemical equilibrium. The word dynamic is used because the 'forward' (L to R) and 'backward' (R to L) reactions do not cease but match each other in rate so the concentrations of reactants and products are constant when the equilibrium is established.
• -

There are other examples on my GCSE chemistry notes page on reversible reactions

• Although most reactions you have encountered at an earlier academic level did go to 100% completion, it is a fact that many reactions do NOT go to completion i.e. 100% yield from the forward reaction.
• If ammonium chloride were heated in a closed system, over a certain temperature range, some of the NH₄Cl will be sublimed into the gases NH₃ and HCl and some of the solid salt remains.
  • A closed system means nothing can enter or leave the system.
  • When a reversible reaction occurs in a closed system, depending on conditions, a chemical equilibrium is formed, in which the original reactants and products formed coexist. In other words the reaction (i.e. from left to right as the equation is written) never goes to completion.
  • Eventually the 'system settles down' and the net concentrations of the reactants and products remain constant i.e. a state of concentration balance exists.
  • BUT the reactions don't stop! Reactants are continually forming products, and the products are continually re–forming the original reactants, hence the term dynamic equilibrium.
  • In terms of kinetics ('rates of reaction'), it means that the
    • rate of formation of product = rate of re–formation of reactants,
    • or the rate of the forward reaction = rate of the backward reaction
    • -
• Example 1.2.1 The formation/decomposition of hydrogen iodide.
  • hydrogen + iodine hydrogen iodide (2 mol gas ==> 2 mol gas)
  • H_2(g) + I_2(g) 2HI_(g) (all gases above 200^oC)
  • L to R forward reaction: If you start with pure hydrogen and pure iodine, so much of them combines to form hydrogen iodide.
  • R to L backward reaction: If you start with pure hydrogen iodide, some, but not all of it,  will decompose into hydrogen and iodine.
  • Starting with the same total number of moles of either H₂ + I₂ or  HI, the final equilibrium concentrations will be the same at the same temperature, volume and pressure. This is illustrated in the diagram below showing the fate of 2 mol of reacting gases.
  • 
  • Graph lines (1) and (2) show what happens if you start with 2.0 mol of pure hydrogen iodide which decomposes 50%, for the sake of argument and mathematical simplicity, into hydrogen and iodine.
    • Graph line (1) shows the gradual 50% reduction of HI from 2 mol to 1 mol.
    • Graph line (2) shows the gradual formation, from 0 mol of each, of 0.5 mol H₂ and 0.5 mol I₂.
  • Graph lines (3) and (4) show what happens if you start with 1.0 mol of hydrogen plus 1.0 mol of iodine and no hydrogen iodide.
    • Graph line (3) shows the 50% reduction of 1.0 mol of H₂ or I₂ to 0.5 mol of each.
    • Graph line (4) shows the formation of 1.0 mol of HI from the net reaction of 0.5 mol H₂ and 0.5 mol I₂.
  • Note:
    • The final equilibrium composition is the same in each case no matter which direction you started from for the same total moles of gas.
    • Where the graph lines first become horizontal, meaning no further net change in concentration, the equilibrium point was first reached i.e. here, after about 32 minutes.
    • See also K_c expression and application of Le Chatelier's Principle
    • -
• Example 1.2.2 The formation and hydrolysis of the ester ethyl ethanoate.
  • ethanoic acid + ethanol ethyl ethanoate + water (hydrolysis <==> esterification)
  • CH₃COOH_(l) + CH₃CH₂OH_(l) CH₃COOCH₂CH_3(l) + H₂O_(l)
  • L to R forward reaction: If equimolar amounts of pure ethanoic acid and pure ethanol are refluxed with a few drops of conc. H₂SO_4(l) catalyst, about 2/3rds of the initial reactants are converted into the ester.
  • The forward or backward reactions are slow at room temperature without heating and employing a catalyst, but equilibrium would eventually be reached at room temperature even in the absence of a catalyst.
  • R to L backward reaction: If equimolar amounts of pure ethyl ethanoate and pure water are mixed, eventually about 1/3rd of the ester reverts back to ethanoic acid and ethanol.
  • Note: If the ester is refluxed with lots of acidified water, the ester is 100% hydrolysed back to the original acid and alcohol.
  • See also Equilibria Part 2 K_c expressions
  • -
• Example 1.2.3 The formation/decomposition of ammonia.
  • nitrogen + ammonia ammonia (Haber synthesis)
  • N_2(g) + 3H_2(g) 2NH_3(g)
  • L to R forward reaction: This is the direction of interest to the chemical industry as ammonia is an important chemical for fertiliser and nitric acid manufacture.
  • For reasons explained later (The Haber Synthesis in Equilibria Part 3) a mere 8% yield (L to R) is commercially acceptable, even if it doesn't sound much.
  • See also Equilibria Part 2 K_c expression, or K_p expression and Le Chatelier's Principle (this page)
  • -

Some important outcomes from experimentally studying dynamic equilibrium reactions

1. It doesn't matter whether you start with the 'reactants' or the 'products', either way, if the conditions are suitable, both are present when a state of equilibrium exists.
2. Eventually the net concentrations of the reactants and products remain the same BUT the forward and backward reactions don't stop.
3. When a dynamic equilibrium is achieved there is a state of balance between the constant concentrations of the reactants and products because the rate at which the reactants change into products is exactly equal to the rate at which the products change back to the original reactants.
4. However, the actual relative amounts of the original reactants left, and products formed, at equilibrium, depend on the particular reaction and reaction conditions e.g. the initial concentrations, temperature and pressure (if gaseous reactants or products are involved) and the value of the equilibrium constant (see on this page Le Chatelier's Principle below and Equilibria Part 2).
5. A catalyst does not affect the position of the equilibrium, i.e. the final constant concentrations are the same with or without a catalyst, you simply get to the equilibrium point faster with a catalyst!
6. In some cases you can adjust reaction conditions sufficiently to make the reaction go virtually 100% in one direction e.g. example 1.2.2, the hydrolysis of an ester.
7. At a given constant temperature, all the final equilibrium concentrations are mathematically governed by the equilibrium expression and the equilibrium constant and these are dealt with in detail in Equilibria Part 2.

1.3 Le Chatelier's Principle

• This is initially described without reference to equilibrium constants and equilibrium expressions which are fully explained and described in Equilibria Part 2.

• Le Chatelier's Principle states that if an 'instantaneous' change is imposed on an equilibrium, the position of the equilibrium will further change to minimise the 'enforced' change.
• In other words, if a change is 'instantaneously' imposed, the equilibrium attempts to restore the original situation, but it cannot do this completely BUT the change 'trend' can be predicted. Also, when considering the equilibrium rules outlined below, any change affects BOTH the rates of the forward and backward reactions.
  • Rule 1 – Temperature and energy changes (ΔH, enthalpy change)
    • 1a. Raising the temperature favours the endothermic direction (ΔH +ve). The system absorbs the heat energy from the surroundings to try to minimise the temperature increase.
    • 1b. Decreasing the temperature favours the exothermic direction (ΔH –ve). The system releases heat energy to the surroundings to try to minimise the temperature decrease.
    • -
  • Rule 2 – Gas pressure changes at constant temperature (ΔV, change in volume)
    • 2a. Increasing the pressure favours the side of the equilibrium with the least number of gaseous molecules as indicated by the balanced symbol equation. The system attempts to reduce the number of gas molecules present to reduce the pressure increase.
    • 2b. Decreasing the pressure favours the side of the equilibrium with the most number of gaseous molecules as indicated by the balanced symbol equation. The system attempts to increase the number of gas molecules to minimise the pressure decrease.
    • NOTE:
      • States symbols _(g/l/s/aq) are particularly important when considering equilibrium equations, if no (g) the pressure rule doesn't apply since solids and liquids are virtually incompressible.
      • Rule 2 ONLY applies to a reaction with one or more gaseous reactants or products because pressure has no real effect on the 'concentration' on the virtually incompressible liquids or solids.
      • If there is NO net change in the number of gas molecules, changing total gas pressure has NO effect on the position of the equilibrium, though pressure increase effectively increases gas concentration so both the forward and backward reactions will be speeded up.
      • Over and above rule 2, all the individual partial pressures of the gases, must comply with the mathematics of the K_p equilibrium expression described and explained in Equilibria Part 2.1b (also explains what partial pressures are).
      • The rest of the rules 1, 3 and 4 apply to ANY reaction, whatever the physical states of the reactants and products.
      • -
  • Rule 3 – Concentration changes at constant temperature
    • 3a. If the concentration of a reactant (on the left) is increased, then some of it must change to the products (on the right) to maintain a balanced equilibrium position.
    • 3b. If the concentration of a reactant (on the left) is decreased, then some of  the products (on the right) must change back to reactants to maintain a balanced equilibrium position.
    • This means if you change ANY concentration, all the other concentrations must change too (see Example 1.4.2 below).
    • Also, any net concentration changes must comply with the K_c equilibrium expression (fully explained in Equilibria Part 2.1a)
    • -
  • Rule 4 – Using a catalyst
    • A catalyst does NOT affect the position of an equilibrium.
    • BUT it does is enable you get to the point of equilibrium faster!
    • A catalyst speeds up both the forward and reverse reactions by providing a mechanistic pathway with a lower activation energy, but there is no way it can influence the final 'balanced' concentration ratios.
    • The importance of a catalyst lies with economics of chemical production e.g. bringing about reactions with high activation energies at lower temperatures and so reducing energy requirements and time, and both reductions save money!

    • So remember a catalyst cannot affect the position of the equilibrium constant or the value of the equilibrium constant K (K_c or K_p).

    • -

1.4 Applying Le Chatelier's Principle and the equilibrium rules

• These are initially described without reference to equilibrium constants and equilibrium expressions which are fully explained and described in Equilibria Part 2 (where some examples are repeated with their K_c or K_p equilibrium expressions)

• Example 1.4.1 - limestone decomposition

  • The thermal decomposition of calcium carbonate (limestone) to make calcium oxide (quicklime):

  • CaCO_3(s) CaO_(s) + CO_2(g)  (ΔH = +178 kJ mol^–1)

    • Note: By convention, the ΔH value quoted corresponds to the forwards reaction (L to R), reversing the sign gives the ΔH for the backward reaction (R to L).

    • For more details of the industrial process see Equilibria Part 3.1

  • Rule 1 – temperature and energy change (ΔH)

    • Increasing temperature favours the endothermic direction (RHS) so more quicklime is formed.

  • Rule 2 – gas pressure (ΔV)

    • Decreasing the partial pressure of carbon dioxide increases the yield of quicklime (0 ==> 1 mole gas).

  • Rule 3 – concentration

    • Not applicable to the reactants because you can't decrease or increase the concentration of a solid but you can reduce the concentration of carbon dioxide by venting the gases to increase the yield of quicklime.

  • Rule 4 – catalyst: Not applicable.

  • -

• Example 1.4.2 - Synthesis of ammonia

  • The synthesis of ammonia: nitrogen + hydrogen ammonia

  • N_2(g) + 3H_2(g) 2NH_3(g)  (ΔH = –92 kJ mol^–1)

  • Rule 1 – temperature and energy change (ΔH)
    • The forward, and desirable reaction, to form ammonia, is exothermic, so lowering the temperature favours its formation.
  • Rule 2 – gas pressure (ΔV)
    • Increase in pressure favours ammonia formation since 4 mol of gaseous reactants ==> 2 mol gaseous products.
  • Rule 3 – concentration: In terms of enforced change => system response
    • If the nitrogen or hydrogen concentration was increased, some of this 'extra' gases would change to ammonia.
    • If the nitrogen or hydrogen concentration was decreased, some of ammonia would change to nitrogen and hydrogen.
      • A 1:3 ratio N₂:H₂ mixture is used in industry, but for academic reasoning practice, specifically ...
      • Increasing nitrogen ==> decreases hydrogen and increases ammonia.
      • Increasing hydrogen ==> decreases nitrogen and increases ammonia.
      • Decreasing ammonia ==> decreases nitrogen and hydrogen.
      • Decreasing nitrogen ==> increases hydrogen and decreases ammonia.
      • Decreasing hydrogen ==> increases nitrogen and decreases ammonia.

  • Rule 4 – catalyst: An iron oxide catalyst is used, time = money for industrial chemical production!

  • More details of the Haber process are given in Equilibria Part 3.

  • -

• Example 1.4.3 - A historic classic equilibrium study by Max Bodenstein, first published in 1899 !!

  • He made some of the first accurate calculations of equilibrium constants.

  • The formation of hydrogen iodide from hydrogen and iodine:

  • H_2(g) + I_2(g) 2HI_(g) (ΔH = –10 kJ mol^–1, iodine gaseous above 200^oC)

  • Rule 1 – temperature and energy change (ΔH)

    •  Increasing temperature favours the LHS, i.e. increases the endothermic decomposition of hydrogen iodide.

  • Rule 2 – gas pressure (ΔV):

    • No effect on position of equilibrium, 2 mol gas ==> 2 mol gas, no net change in gas moles.

  • Rule 3 – concentration

    •  e.g. if more iodine was added to a constant volume container, the hydrogen concentration or partial pressure would decrease as some reacts with added iodine to give more hydrogen iodide as the system tries to minimise the iodine increase. Please note that there would still be an overall increase in iodine at the new equilibrium point.

  • Rule 4 – catalyst: Not applicable.

  • See also section 1.2.1, 2.1a.1 on equilibrium–rates connection and calculation 2.2a.1

  • HISTORIC NOTE:

    • The equilibrium 2 HI (g)  H₂(g) + I₂(g) was the first gas phase equilibrium to be thoroughly studied.  It was investigated by the chemist Max Bodenstein (1871-1942) in the late 1890's in Germany.  He found that the reaction could be studied between about 380 and 500 °C, but that at higher temperatures the reaction was shifted too far to the product side - endothermic decomposition from left to right.

    • To study the reaction, Bodenstein sealed various amounts of HI in glass containers. These were then heated to the desired temperature for various intervals of time in an accurately thermostated oven. The glass containers were then removed and cooled rapidly, broken open and the contents analyzed for all three components.  Bodenstein was able to determine the equilibrium constant for the reaction from the quantities of hydrogen gas, iodine vapor that had formed, and hydrogen iodide that remained.

    • I don't think he actually analyse directly the hydrogen and iodine directly, but you can titrate the hydrogen iodide with sodium hydroxide. From one molar measurement and the starting quantities, you can work out the quantities of the other components in the final equilibrium mixture. See calculation 2.2a.1

  • -

• Example 1.4.4 - Ester formation and hydrolysis

  • Esterification: e.g. ethanoic acid + ethanol ethyl ethanoate + water
  • CH₃COOH_(l) + CH₃CH₂OH_(l) CH₃COOCH₂CH_3(l) + H₂O_(l)  (ΔH = –2 kJ mol^–1)

  • Rule 1 – temperature and energy change (ΔH)

    •  Decrease in temperature favours the ester formation (ethyl ethanoate, RHS) but in practice a small decrease in yield at higher temperatures is acceptable because heating under reflux is needed to efficiently prepare the ester.

  • Rule 2 – gas pressure (ΔV): Not applicable, no gases involved.

  • Rule 3 – concentration

    • Sometimes it is desirable to add a large excess of the alcohol to ensure most of the acid is converted into ester. The alcohol may be cheaper than the acid and the ester and unreacted alcohol separated by distillation and the latter recycled.

  • Rule 4 – catalyst

    • The forward esterification reaction is catalysed by acids e.g. a few drops of conc. sulphuric acid.

    • The reverse reaction i.e. hydrolysis of the ester back to the acid and alcohol is catalysed by dilute acids.

    • -

• Example 1.4.5 - Equilibria involving the oxides of nitrogen (teacher demonstrations in fume cupboard)

  • The equilibria involving oxygen O₂, nitrogen (II) oxide NO, nitrogen(IV) oxide NO₂ and its dimer N₂O₄.

  • Nitrogen dioxide, NO₂ can be made from the irreversible  thermal decomposition of lead(II) nitrate in a pyrex boiling tube connected to a 100 cm³ gas syringe in a fume cupboard.

    • Preparation of nitrogen(IV) oxide

    • You can conveniently prepare a sample of nitrogen(IV) oxide by heating lead(II) nitrate - which crackles!

    • lead(II) nitrate ==> lead(II) oxide + nitrogen(IV) oxide + oxygen

    • 2Pb(NO₃)_2(s) ==> 2PbO_(s) + 4NO_2(g) + O_2(g)

    • Both lead nitrate and nitrogen dioxide are poisonous, for teachers only!

  • (a) The thermal decomposition of nitrogen(IV) oxide

  • 2NO_{2(g, brown)} 2NO_{(g, colourless)} + O_{2(g, colourless)}  (ΔH = +113 kJ mol^–1)

    • Transfer some NO₂ to a pyrex boiling tube and keep the gas in with a wad of mineral wool.

    • The temperature effect can be observed by strongly heating the gases in the pyrex tube above 400^oC.

    • The brown colour should fade at the lowest hottest part of the boiling tube.

    • The equilibrium will shift from left to right, in the endothermic direction - absorbing the heat to counteract the increase in temperature.

    • This experiment can be done after you have done the lower temperature boiling tube experiment below.

  • (b) Playing around with nitrogen(IV) oxide and its dimer ('dinitrogen tetroxide')

  • 2NO_{2(g, brown)} N₂O_{4(g, colourless)}  (ΔH = –58 kJ mol^–1)

    • (b)(i) Again set up a boiling tube of NO₂ plus wad of mineral wool to contain it.

    • Cool the tube in ice and the brown colour is considerably decreased - equilibrium shifts in the exothermic direction and more colourless N₂O₄ formed - heat released to counteract the decrease in temperature.

    • If you then warm up the boiling tube of NO₂ in hot water, the brown colour intensity increases as some the N₂O₄ dimer is decomposed to NO₂ - the equilibrium shifts in the 'right to left' endothermic direction

  • Rule 1 – temperature and energy change (ΔH)

    • Increases in temperature favours the endothermic decomposition of NO₂ to NO and O₂, so at high temperatures the brown colour fades.

    • Decrease in temperature favours the exothermic formation of the dimer N₂O₄ from NO₂, so the brown colour fades on cooling the gas mixture.

  • Rule 2 – gas pressure (ΔV)

    • (b)(ii) The pressure effect can be observed by sealing the gases at room temperature in the gas syringe and compressing and decompressing it

      • I just doubled up the rubber tubing and clamped it together, BUT you need to hold the gas syringe very firmly when compressing the gaseous contents.

      • You can do this on an OHP, take care, its a bit awkward for all of an A level class to see it at close quarters in the confines of a fume cupboard.

      • 2NO_{2(g, brown)} N₂O_{4(g, colourless)}

    • Increase in pressure favours the RHS, more N₂O₄, because 2 mol gas ==> 1 mol gas, so theoretically the mixture would get lighter in colour - reduction in the number of gas molecules to counteract the increase in pressure.

    • Decrease in pressure favours LHS NO₂ formation, 1 mol gas ==> 2 mol gas, so theoretically the mixture would get darker in colour - increase in the number of gas molecules to counteract the decrease in pressure.

    • (b)(ii) can be demonstrated by compressing/decompressing the gas mixture in the syringe to see the brown colour intensity increase/decrease - but the observations are quite subtle!

    • Compressing the NO₂/N₂O₄ mixture:

      • When you press the gas syringe plunger in, you increase the pressure of the gases.

      • The orange-brown colour becomes more darker because ALL concentrations are increased.

      • BUT, within a few seconds the orange-brown colour fades just a little bit as the equilibrium moves from left to right - a little more N₂O₄ formed - as predicted from the pressure rule.

    • Decompressing the NO₂/N₂O₄ mixture:

      • When you pull the gas syringe plunger out, you decrease the pressure of the gases.

      • The orange-brown colour becomes lighter because ALL concentrations are decreased.

      • BUT, within a few seconds, the colour gets a little bit darker as the equilibrium moves from right to left - a little more NO₂ formed - as predicted from the pressure rule.

    • In fact you can even see the dynamic equilibrium 'kinetics' in directly in action here. There is a time lag of about 1–2 seconds before the new equilibrium position is established as the 'imposed' colour intensity change becomes constant.

      • Both activation energies are relatively low and what you are witnessing is a particle collision frequency of ~10⁸/s kapow!

    • -

• Example 1.4.6 - Ligand displacement reaction involving two cobalt(II) complex ions

  • Pink hexa–aqua cobalt(II) ions form a complex with chloride ions, the blue tetrachlorocobaltate(II) ion.

  • [Co(H₂O)₆]²⁺_{(aq, pink)} + 4Cl^–_(aq) [CoCl₄]^2–_{(aq, blue)} + 6H₂O_(l) (ΔH +ve, don't know value)

  • Rule 1 – temperature and energy change (ΔH)

    •  Increase in temperature favours the endothermic blue complex formation, cooling favours the exothermically form pink ion.

    • If prepared at higher temperature, with just enough chloride added to turn the solution blue, on cooling it becomes pink.

  • Rule 2 – gas pressure (ΔV): Not applicable, no gases involved.

  • Rule 3 – concentration

    • Increase in chloride concentration decreases the pink ion concentration and increases the blue ion concentration i.e. shifts the equilibrium position from left to right.

    • Diluting with water shifts the equilibrium to the left so solution is less blue and more pink.

  • Rule 4 – catalyst: Not applicable here, despite involving transition metal complexes!

  • Similarly if concentrated sodium chloride solution or hydrochloric acid is added to copper(II) sulphate solution the pale yellow–brown tetrachlorocuprate(II) complex ion is formed (seen as green at first due to the mixture of blue from the original Cu²⁺ ion).

    • [Cu(H₂O)₆]²⁺_(aq) + 4Cl^–_(aq) [CuCl₄]^2–_(aq) + 6H₂O_(l)

    • This is quite a good reaction to demonstrate Le Chatelier's equilibrium principles:

      • If you dissolve copper(II) chloride in water you get a greenish–blue solution as both copper(II) complexes are present in equilibrium.

      • By adding water i.e. dilution, it shifts the equilibrium to the left, more blue.

      • Increasing the chloride ion concentration by adding concentrated hydrochloric acid or concentrated sodium chloride solution (or solid) shifts the equilibrium to the right, more green ==> yellowish brown.

      •   + 4Cl^-      + 6H₂O

      • -

• Example 1.4.7 - The high temperature formation of nitrogen(II) oxide.
  • N_2(g) + O_2(g) 2NO_(g) (ΔH = +181 kJ mol^–1)
  • Rule 1 – temperature and energy change (ΔH)
    • Increase in temperature favours the endothermic formation of NO.
    • This reaction does not happen at room temperature but is formed at the high temperatures in car engines.
    • Unfortunately when released through the car exhaust, it cools to normal temperatures when NO irreversibly reacts with oxygen in air to form nitrogen(IV) oxide, NO₂, which is acidic, a lung irritant and a reactive free radical molecule involved in the chemistry of photochemical smog not good!
  • Rule 2 – gas pressure (ΔV)
    • Since 2 mol gaseous reactants gives 2 mol gaseous products, pressure does not affect the position of the equilibrium.
  • Rule 3 – concentration
    • The concentration of nitrogen is high from air, but although the concentration of oxygen is low in the exhaust gases, there is sufficient present in the combustion process to ensure a small % of NO is formed.
  • Rule 4 – catalyst: Not applicable.
    • -
• Example 1.4.8 - The oxidation of sulfur dioxide to sulfur trioxide
  • The oxidation of sulfur dioxide to sulfur trioxide
  • e.g. in the Contact Process for manufacturing sulphuric acid, see Equilibria Part 3.3.
  • 2SO_2(g) + O_2(g) 2SO_3(g) (ΔH = –196 kJ mol^–1)
  • Rule 1 – temperature and energy change (ΔH)
    • The exothermic formation of sulphur trioxide is favoured by low temperature.
  • Rule 2 – gas pressure (ΔV)
    • Higher pressure favours a higher yield of sulphur trioxide as 3 gas moles ==> 2 gas moles, though 1–2 only atm is used in practice because the equilibrium is already so far to the right (about 99%).
  • Rule 3 – concentration
    • Air is used as the source of oxygen and despite its dilution with nitrogen the concentration, the oxygen concentration is high enough to move the equilibrium very much to the RHS.
  • Rule 4 – catalyst
    • A vanadium(V) oxide, V₂O₅, catalyst ensures the high yield of 99% SO₃ is attained fast, but no more!
    • -
• Example 1.4.9 - Reforming reaction to manufacture hydrogen
  • One way to produce hydrogen for the Haber synthesis of ammonia is to react methane gas with steam.
  • CH_4(g) + H₂O_(g) 3H_2(g) + CO_(g) (ΔH = +206 kJ mol^–1)
  • Rule 1 – temperature and energy change (ΔH)
    • Increase in temperature favours the endothermic formation of hydrogen (and carbon monoxide).
  • Rule 2 – gas pressure (ΔV)
    • For the desired forward reaction, 2 mol of reactant gases ==> 4 mol of product gases, so the increase in product volume is favoured by lower pressure.
  • Rule 3 – concentration
    • Theoretically increase in methane and steam concentrations will increase the hydrogen concentration, but this essentially means increasing pressure favouring the LHS, so you might not gain as much hydrogen as you like to!
  • Rule 4 – catalyst
    • A nickel catalyst is used, but cannot affect the yield, only the rate of reaction – speeds it up for more efficient production.
    • -

Part 1 sub–index: 1.1 Reversible reactions * 1.2 The equilibrium concept

1.3 Le Chatelier's Principle – equilibrium rules * 1.4 Applying the equilibrium rules

Chemical Equilibrium Notes Index

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