Le Chatelier's Principle

Summary

• In 1884, the French Chemist Henri Le Chatelier suggested that equilibrium systems tend to compensate for the effects of perturbing influences.
• When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimise, or counteract, the effect of the disturbance.
  • If the concentration of a reactant is increased, the equilibrium position shifts to use up the added reactants by producing more products.
  • If the pressure on an equilibrium system is increased, then the equilibrium position shifts to reduce the pressure.
  • If the volume of a gaseous equilibrium system is reduced (equivalent to an increase in pressure) then the equilibrium position shifts to increase the volume (equivalent to a decrease in pressure)
  • If the temperature of an endothermic equilibrium system is increased, the equilibrium position shifts to use up the heat by producing more products.

Examples

• Changes in Concentration.
  Consider the folowing system at equilibrium
  

  Fe3+(aq) + SCN-(aq) (colourless)

  FeSCN2+(aq) (red)

  • Increasing the concentration of either Fe³⁺(aq) or SCN^-(aq) will result in the equilibrium position moving to the right, using up the some of the additional reactants and producing more FeSCN²⁺(aq). The solution will become a darker red colour.
  • Removing some of the Fe³⁺(aq) or SCN^-(aq) will result in the equilibrium position moving to the left to produce more Fe³⁺(aq) and SCN^-(aq). The solution will become less red as FeSCN²⁺(aq) is consumed.
  • Increasing the concentration of FeSCN²⁺(aq) will result in the equilibrium position moving to the left to use up some of the additional FeSCN²⁺(aq) and producing more Fe³⁺(aq) and SCN^-(aq). The solution will become less red.
  • Removing some of the FeSCN²⁺(aq) will result in the equilibrium position moving to the right to produce more FeSCN²⁺(aq) which will lead the solution to become a darker red colour.

• Changes in Pressure of Gaseous Equilibrium Systems
  Gas pressure is related to the number of gas particles in the system, more gas particles means more gas pressure.
  

  Consider the following gaseous equilibrium system:

  2NO2(g) (red-brown)

  N2O4(g) (colourless)

  • Increasing the pressure on this equilibrium system will result in the equilibrium position shifting to reduce the pressure, that is, to the side that has the least number of gas particles.
    There are 2 gas particles on the left hand side of the reaction and 1 gas particle on the right hand side of the reaction.
    Increasing the pressure on this system results in the equilibrium position moving to the right, consuming NO₂(g) and producing more N₂O₄(g). The system will become a lighter red-brown colour.
  • Reducing the pressure on this equiibrium system will result in the equilibrium position moving to the left, that is, to the side that has the most gas particles, in order to increase the pressure. The red-brown colour of the system becomes darker.

  Consider the following equilibrium system:

  C(s) + H2O(g)

  CO(g) + H2(g)

  • Increasing the pressure on this equilibrium system results in the equilibrium position shifting to the left, CO(g) and H₂(g) are consumed to produce more C(s) and H₂O(g).
    There is only 1 gas particle on the left of the equation (H₂O) and 2 gas particles on the right hand side of the equation (CO(g) + H_2(g)), so the equilibrium position moves to the side that has the least number of gas particles to reduce the pressure, that is, to the left.
  • Reducing the pressure on this equilibrium system results in the equilibrium position shifting to the side that has the most gas particles in order to increase the pressure, that is, to the right. C(s) and H₂O(g) are consumed to produce more CO(g) and H₂(g).
    

• Changes in Volume of Gaseous Equilibrium Systems
  Gas volume is related to gas pressure, a gas at reduced volume has a higher pressure, a gas at increased volume has a lower pressure. Consider the following gaseous equilibrium system:

  2NO2(g) (colourless)

  N2O4(g) (red-brown)

  • Increasing the volume of this equilibrium system is equivalent to reducing the pressure, so the equilibrium position moves to the side of the reaction that has the most gas particles in order to increase the pressure. The equilibrium position shifts to the left producing more NO₂(g) by consuming N₂O₄(g). The system becomes less red-brown in colour.
  • Reducing the volume of this equilibrium system is equivalent to increasing the pressure, so the equilibrium position moves to the side of the reaction that has the least gas particles in order to reduce the pressure. The equilibrium position shifts to the right producing more N₂O₄(g) by consuming more NO₂(g). The system becomes a darker red-brown.

• Changes in Temperature
  In an endothermic reaction, energy can be considered as a reactant of the reaction.
  In an exothermic reaction, energy can be considered as a product of the reaction.
  

  • Endothermic Equilibrium Systems
    Consider the following reaction at equilibrium:

    H2(g) (colourless)

    +

    I2(s) (purple)

    2HI(g) (colourless)

    H = + 52 kJ mol-1

    This reaction can also be written with the energy term incorporated into the equation on the side with the reactants:

    H2(g) (colourless)

    +

    I2(s) (purple)

    + 52 kJ

    2HI(g) (colourless)

    • Increasing the temperature of the equilibrium system will shift the equilibrium position to the side that does not include the energy term in order to reduce the temperature, that is to the right. The increased heat will be consumed to produce more HI(g) product, since H₂(g) and I₂(s) will be consumed, there will be less purple solid in the reaction vessel.
    • Reducing the temperature of the equilibrium system will shift the equilibrium to the left in order to produce more heat. HI(g) will be consumed in order to produce more H₂(g) and I₂(s). The amount of purple solid will increase.
    • Exothermic Equilibrium Systems
      Consider the following reaction at equilibrium:

      Ag+(aq) + Cl-(aq) (colourless)

      AgCl(s) (white)

      H = -112 kJ mol-1

      This reaction can also be written with the energy term incorporated into the equation on the side with the the products:

      Ag+(aq) + Cl-(aq) (colourless)

      AgCl(s) (white)

      + 112 kJ

      • Increasing the temperature of this equilibrium system shifts the equilibrium position to the left, consuming some of the energy and products to produce more reactants. There will be less white AgCl(s) in the reaction vessel.
      • Reducing the temperature of this equilibrium system shifts the equilibrium position to the right, producing more heat and more white AgCl(s).