Equilibrium I and II

• All reversible reactions reach a position of dynamic equilibrium, with the following features.

  1. rate of forwards reaction = rate of backwards reaction.
  2. concentration of reactants and products remain constant.

  for the forwards reaction, initially, reactants are used up quickly, but then this slows as their conc drops.

  for the backwards reaction, initially reactants are reformed slowly, but this process speeds up as the concentration of products increases.
• How a dynamic equilibrium is formed (only occurs in a closed system):

  1. high concentration of reactants and zero products.
  2. concentration of reactants decreases and concentration of products increases as reaction progresses.
  3. rate of forwards reaction slows and rate of backwards reaction increases.
  4. speed of forwards and backwards reactions are equal.
  5. dynamic equilibrium is formed.
• position of equilibrium is used to describe which side of the reaction the equilibrium lies:

  • towards the left - there are more reactants
  • towards the right - there are more products
• if one direction of an equilibrium is exothermic, the opposite direction will be endothermic and vice versa.
• 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 to oppose this and move in the endothermic direction to reduce the temperature. This will give a greater yield of the products of the endothermic direction.
• if temperature is decreased, the equilibrium will shift to oppose this and move in the exothermic direction to increase the temperature. This will give a greater yield of the products of the exothermic direction.
• if pressure is decreased, equilibrium will shift to oppose this and move to the side with more moles of gas to increase the pressure. This gives a greater yield of the side with more moles of gas.
• if pressure is increased, equilibrium will shift to oppose this and move to the side with fewer moles of gas to increase the pressure. This gives a greater yield of the side with fewer moles of gas.
• increasing the concentration of reactants causes the equilibrium to shift to oppose this and move to the right-hand side (products side) to remove/use up the reactants and decrease the increased concentration on the left-hand side.
• decreasing the concentration of a species on the right-hand side (products) causes the equilibrium to shift to oppose this and move to the right-hand side to replace and increase the decreased concentration on the right-hand side.
• in industry, conditions used in reversible reactions are often compromised.

  
  

  e.g. a low temperature may give a high yield of product, but will also cause a slow rate of reaction- compromise is thus needed for a reasonable yield and rate.

  
  

  e.g. a high pressure may give a high yield of product and fast rate, but high pressures are expensive to produce - compromise is thus needed for a reasonable yield/rate and price.
• catalysts have no net effect on the position of equilibrium. This is because they speed up the rates of forward and backward reactions equally, increasing the rate at which the equilibrium is reached without affecting the position.

  
  

  catalysts hence have no effect on the value of Kc.
• generalised equilibrium constant (Kc)

  m A + n B = p C + q D

  $Kc=$$[C]^p[D]^q/[A]^m[B]n$

  
  

  Kc is calculated from equilibrium concentrations at constant temperature

  (only gaseous and aqueous species are included)

  (solids and pure liquids are not included in the Kc expression)
• units:

  
• calculating concentration at equilibrium:

  1. initial moles
  2. change in moles
  3. moles at equilbrium
  4. concentration at equilibrium

  equilibrium moles of reactant = initial moles - moles reacted

  equilibrium moles of product - initial moles + moles formed
• n= c x v

  moles = concentration x volume
• Kc calculation example:

  
• a large Kc/Kp value means the position of equilibrium lies to the right and a small Kc value means the position of equilibrium lies to the left.
• Kc is not changed by changes in concentration or by adding a catalyst. It's only changed by changing temperature.

  
  

  when the concentration of a reactant or product is changed, the system responds according to Le Châtelier's principle, shifting to restore the balance to ensure that Kc remains constant.
• the impact of Kc depends on the direction the change of position of equilibrium as a result of the temperature change.
• partial pressure - the pressure that a gas would exert if it occupied the volume occupied by the mixture
• mole fraction = moles of gas/total gaseous moles present in mixture
• total pressure - the sum of all the partial pressures of gases present in a mixture
• partial pressure (gas) = total pressure (system) x mole fraction (gas)
• generalised equation for Kp:

  m A(g) + n B(g) = p C(g) + q D(s)

  $Kp=$$p(C)^p/p(B)^np(A)^m$

  the equilibrium constant is given by the partial pressures of the gases present + their balancing numbers (only includes gases in a heterogenous equilibrium)

  
  

  When you have two products or two reactants in a 1:1 ratio, you know that they have the same value for pressure/concentration.

  In other words you can't calculate the moles of a reactant from the moles of a product.
• Kp calculation example

  
• Units of Kp:

  
• Kp question example:

  
• effect of temperature on Kp:

  changing the temperature will change equilibrium pressures, which will hence change the value for Kp too.

  if temperature change causes equilibrium to shift right, Kp will increase.

  if temperature change causes equilibrium to shift left, Kp will decrease.
• effect of concentration on Kp:

  the value of Kp is unaffected by any changes in pressure — this is because if the value for the total pressure changes the system responds (Le Châtelier's principle) shifting the equilibrium position to counteract this change and keep Kp constant.

  e.g. an increase in pressure with result in a higher partial pressure on the side of the equation with fewer gaseous moles and vice versa for a decrease in pressure.
• why concentration and pressure changes don't affect the the value for Kc or Kp, but temperature does:

  • changes in temperature fundamentally alter the energy balance of a reaction.
  • changes in concentration and pressure temporarily change equilibriums, but the system then reverses these changes to keep Kc and Kp constant.
• For Kc to be exactly 1, reactant and product concentrations need to be the same.