- 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) + H2(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.