Edexcel Chemistry - Topic 11: Equilibrium II

• The Equilibrium Constant for a reaction can be calculated using the equation below.
  Finding Kc by experiment
• Mix measured quantities of reactants and/or products - then allow to reach equilibrium
• Analyse the mixture to find the equilibrium concentration of one of the chemicals at equilibrium
• Use the equation to work out the values for the equilibrium concentrations of all substances
• Substitute these values into the expression for Kc

• If a reaction contains gases an alternative equilibrium expression can be used:
  Kp = (pC)^c x (pD)^d
           (pA)^a x (pB)^b
• The partial pressure of a gas (p) is the pressure that the gas would have if it alone occupied the volume occupied by the whole mixture
• Kp expression only involves gases in heterogenous equilibrium. Any substance in another state is left out of the expression

• Temperature affects both equilibrium position and Kc and Kp values

N2(g) + 3H2(g) ⇌ 2NH3(g) (ΔH = -ve)

• If temperature is increased, the reaction shifts in the endothermic direction. The position of the equilibrium shifts left and the value of Kc gets smaller as there are fewer products
• If temperature is decreased, the reaction shifts in the exothermic direction. The position of the equilibrium shifts right and the value of Kc gets bigger as there are more products
• As temperature increases, a significantly bigger proportion of particles have energy greater than the activation energy, so the frequency of successful collisions increases

• Changing concentrations/pressures will shift the position of the equilibrium but won't affect the Kc or Kp values

H2(g) + Cl2(g) ⇌ 2HCl(g)

• Increasing the concentration of H2 will move the equilibrium position to the right lowering the concentrations of H2 and Cl2 and increasing the concentration of HCl
• The new concentrations will restore the equilibrium to the same Kc
• At higher concentrations/pressures there are more particles per unit volume so the particles collide with a greater frequency and there will be a higher frequency of effective collisions
• If pressure is increased, the reaction will shift to oppose the change and will move in the direction with the fewer moles of gas. The Kp value will remain the same
• The increased pressure, increases the pressure terms on the bottom of the Kp expression more than the top so the system isn't at equilibrium so it shifts to oppose the change. The top of the Kp expression increases and the bottom decreases until the original Kp value is restored

N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH = -ve

T = 450C   P = 200-1000atm  Catalyst: Iron

• Low temperatures give a good yield but at a slow rate: compromise temperature is used
• High pressure gives a good yield and a high rate: too high a pressure would lead to high energy costs and apparatus costs
• Catalyst is only used to speed up the rate of reaction. Catalysts have no effect on position of the equilibrium or on Kc or Kp values

S(s) + O2(g) --> SO2(g)
SO2(g) + 0.5 O2(g) ⇌ SO3(g)   ΔH = -ve

T = 450C   P = 10atm   Catalyst = V2O5

• Low temperatures give good yield but at a slow rate: compromise temperature used
• High pressures gives better yield and a high rate: too high a pressure would lead to high energy and apparatus costs

• A catalyst has no effect on the position of the equilibrium or values of Kc or Kp - they only increase the rate at which equilibrium is achieved
• Recycling unreacted reagents back into the reactor can improve the overall yields and atom economy 
• Industrial Processes cannot be in equilibrium as the products are removed as they are formed to improve the conversion of the reactants (they are not closed systems)
• Companies use lower pressures which form a compromise with temperatures to reduce the cost of maintaining high temperatures and pressures