Equilibria

• Equilibrium


  All reversible reactions reach a dynamic equilibrium state
• Many reactions are reversible
• Reversible reaction


  N2 + 3H2 ⇌ 2NH3
• Dynamic equilibrium


  • Forward and backward reactions are occurring at equal rates
  • The concentrations of reactants and products stay constant
• Position of equilibrium


  The composition of the equilibrium mixture
• Equilibrium favours reactants


  The equilibrium mixture will contain mostly reactants
• Le Chatelier's Principle


  If an external condition is changed the equilibrium will shift to oppose the change (and try to reverse it)
• If temperature is increased

  The equilibrium will shift in the endothermic direction to try to reduce the temperature
• If temperature is decreased


  The equilibrium will shift in the exothermic direction to try to increase the temperature
• Low temperatures may give a higher yield of product but will also result in slow rates of reaction. Often a compromise temperature is used that gives a reasonable yield and rate.
• Increasing pressure


  The equilibrium will shift towards the side with fewer moles of gas to oppose the change and reduce the pressure
• Decreasing pressure


  The equilibrium will shift towards the side with more moles of gas to oppose the change and increase the pressure
• If the number of moles of gas is the same on both sides of the equation then changing pressure will have no effect on the position of equilibrium.
• Increasing the concentration of OH- ions

  The equilibrium will shift in the forward direction to remove and decrease the concentration of OH- ions
• Adding H+ ions reacts with the OH- ions and reduces their concentration so the equilibrium shifts back to the left
• A catalyst has no effect on the position of equilibrium, but it will speed up the rate at which the equilibrium is achieved.
• Haber process


  • T= 450oC, P= 200 – 1000 atm, catalyst = iron
  • Low temp gives good yield but slow rate: compromise temp used
  • High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure
• Contact process


  • Stage 1: S (s) + O2 (g) → SO2 (g)

  Stage 2: SO2 (g) + ½ O2 (g) ⇌ SO3 (g) H = -98 kJ mol-1

  T= 450oC, P= 1 or 2 atm, catalyst = V2O5

  • Low temp gives good yield but slow rate: compromise moderate temp used
  • High pressure only gives slightly better yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure
• Hydration of ethene to produce ethanol

  • CH2=CH2 (g) + H2O (g) ⇌ CH3CH2OH(l) H = -ve

  T= 300oC, P= 70 atm, catalyst = conc H3PO4

  • Low temp gives good yield but slow rate: compromise temp used
  • High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure
  • High pressure also leads to unwanted polymerisation of ethene to poly(ethene)
• Production of methanol from CO

  • CO (g) + 2H2(g) ⇌ CH3OH (g) H = -ve exo

  T= 400oC, P= 50 atm, catalyst = chromium and zinc oxides

  • Low temp gives good yield but slow rate: compromise temp used
  • High pressure gives good yield and high rate: too high a pressure would lead to too high energy costs for pumps to produce the pressure
• Both methanol and ethanol can be used as fuels.
• If the carbon monoxide used to make methanol was extracted from the atmosphere then it could be classed as carbon neutral.
• Carbon neutral


  An activity that has no net annual carbon (greenhouse gas) emissions to the atmosphere
• Recycling unreacted reactants back into the reactor can improve the overall yields of all these processes.
• In all cases high pressure leads to too high energy costs for pumps to produce the pressure and too high equipment costs to have equipment that can withstand high pressures.
• In all cases catalysts speeds up the rate, allowing a lower temp to be used (and hence lower energy costs), but have no effect on position of equilibrium
• Equilibrium constant expressions


  • [NH3 (g)]^2 / [N2 (g)] [H2 (g)]^3

  [HCl (g)]^2 / [H2 (g)] [Cl2 (g)]
• Calculating Kc


  Work out the equilibrium moles and concentrations

  2. Put the concentrations into the Kc expression
• Calculating Kc examples


  • Example 3: Kc = [HCl (g)]^2 / [H2 (g)] [Cl2 (g)] = 0.196 (no unit)

  Example 4: Kc = [NH3 (g)]^2 / [N2 (g)] [H2 (g)]^3
• Preparing the equilibrium mixture

  1. Use burettes to prepare a mixture in boiling tube of carboxylic acid, alcohol, and dilute sulfuric acid

  2. Swirl and bung tube

  3. Leave the mixture to reach equilibrium for one week
• Titrating the equilibrium mixture

  1. Rinse a 250 cm3 volumetric flask with distilled water

  2. Use a funnel to transfer the contents of the boiling tube into the flask

  3. Use distilled water to make up the solution in the volumetric flask to exactly 250 cm3

  4. Use the pipette to transfer 25.0 cm3 of the diluted equilibrium mixture to a 250 cm3 conical flask

  5. Add 3 or 4 drops of phenolphthalein indicator to the conical flask

  6. Set up the burette with sodium hydroxide solution

  7. Add the sodium hydroxide solution from the burette until the mixture in the conical flask just turns pink

  8. Record the burette reading

  9. Repeat the titration until you obtain a minimum of two concordant titres
• The sodium hydroxide will react with the sulfuric acid catalyst and any unreacted carboxylic acid in the equilibrium mixture
• Esterification reaction


  Ethanol and ethanoic acid are mixed together with a sulphuric acid catalyst
• The pink colour of the phenolphthalein in the titration can fade after the end-point of the titration has been reached because the addition of sodium hydroxide may make the equilibrium shift towards the reactants
• Working out initial amount of moles of reactants

  1. Calculate the amount of moles of alcohol and carboxylic acid from the densities and volumes of liquids added

  2. Determine the initial amount of moles of acid catalyst by titrating a separate sample with sodium hydroxide
• Working out equilibrium amount of moles of acid present from the titre results

  1. Use the volume and concentration of sodium hydroxide used in the titration to calculate the total amount of H+ present

  2. Subtract the amount of H+ from the acid catalyst to get the amount of carboxylic acid at equilibrium
• Calculating the equilibrium constant

  1. Divide the equilibrium amounts by the total volume to get the equilibrium concentrations

  2. Put the equilibrium concentrations into the Kc expression
• To confirm one week was sufficient time for equilibrium, several mixtures could be made and left for different amounts of time. If the resulting Kc is the same value then it can be concluded the time is sufficient.
• Working out equilibrium amount of moles of other substances

  Calculate the equilibrium amount of ethanol, ethyl ethanoate and water using the initial amounts and the amount of carboxylic acid that reacted
• The amount of water at equilibrium would not really be 0 as there would be water present in the acid catalyst.