A system remains at equilibrium as long as the rate of the forward reaction equals the rate of the backward reaction. As soon as this balance is disrupted by any change in conditions such as temperature pressure and concentration, the position and the value of equilibrium will change. Le Chatelier’s principle states that a ststem at equilibrium when subjected to a change will respond in such a way as to minimize the effect of the change. This means that the system will respond in the opposite way. If we add something, the system will react to remove it. Similarly if we remove something, the system will react to replace it.
1) Change in concentration could shift the equilibrium point but not the value (magnitude of Kc). For example, if we increase the concentration of the reactants, the system will reduce this concentration by favouring the forward reaction (shifting the equilibrium to the right). Similarly, if we decrease the concentration of the products (e.g. removing it) the system will respond by increasing the concentration of product so shifting the equilibrium to the right. To shift the equilibrium to the left, we could either increase in concentration of product or a decrease in concentration of reactant.
2) Change in pressure only affects gases and if the reaction involves a change in the number of molecules. (equilibrium could shift but not the value of Kc) There is a direct relationship between the number of molecules and the pressure exterd by a gas in a fixed volume, so if we increase the pressure, the system will respond to decrease this pressure by favouring the side with the smaller number of molecules.
For example, consider the reaction used in the production of methanol:
CO(g) + 2H2(g) <-> CH3OH(g)
There are 3 molecules of gas on the left side and one on the right. So here high pressure will shift the equilibrium to the right, in favour of the smaller number of molecules, so increasing the production of methanol.
3) Change in temperature will change the value of Kc. However, in order to preduct how it will change we must look at the enthalpy change of the forward and backward reactions (exothermic releases heat, endothermic absorbs heat). Think that heat as a product in an exothermic reaction and as a reactant in an endothermic reaction. Note that the enthalpy changes of the forward and backward reactions are equal and opposite of each other.
Consider the reaction:
2NO2(g) <-> N2O4(g) ∆H = -24kJ mol-1
The negative sign of ∆H tells that the forward reaction is exothermic and so releases heat. If this reaction at equilibrium is subjected to a decrease in temperature, the system will respond by producing heat and favouring the exothermic reaction. This will shift the equilibrium to the right, as more product is formed, Kc will also increase. We can see that the reaction will give a higher yield of product at a lower temperature. However, at low temperature the reaction will proceed more slowly although a higher yield.
4) Addition of a catalyst will speed up the rate of a reaction by lowering its activation energy so increases the number of particles that have sufficient energy to react. However, by adding a catalyst this will speed up the forward as much as the backward so the value of Kc and the position of equilibrium remain unchanged however it will let it meet Kc earlier.
Industrial applications
In reactions involving the manufacture of a chemical, it is obviously a goal to obtain as high a yield of product as possible. Apply Le Chaterlier’s principle to the reactions involve enables us to maximize the yield by choosing conditions that will cause the equilibrium to shift to the right. However, the yield of a reaction in only part of the consideration as the rate is also clearly a great significant. For example, if we can get 95% yield of product, but takes several years t achieve this, it will not be worth it. They need to take consider the cost or production too because very high pressure is very expensive so they have to choose the best combinations.
The Haber process (Ammonia production): N2(g) + 3H2(g) <-> 2NH3(g) ∆H= -93kJ mol-1
Uses: fertilizers, plastics (nylon), refrigerants and powerful explosives
Note that there are 4 gas molecule on the left and 2 on the right. The forward reaction is exothermic and so releases heat. Using Le Chatelier’s principle, we can say that if we remove the product, increase the pressure and lower the temperature, this will all shift the equilibrium to the right and increase yield. However, it low temperature will slow down the reaction so we might want to add a catalyst.
Conditions: 450 Degrees Celsius, 200 atmospheres.
Contact process (Sulphuric acid production) :
i) Combustion of sulfur S(s) + O2(g) -> SO2(g)
ii) Oxidation of sulfur dioxide 2SO2(g) + O2(g) <-> 2SO3(g)
iii) Addition of water SO2(g) + H2SO4(l) -> H2S2O7 (l) + H2O(l) -> 2H2SO4(l)
Uses: fertilizers, detergents, dyes, explosives, drugs and plastics.
The overall rate of the process depends on step ii. So applying Le Chatelier’s principle to this, we can predict the condition that will most favour the formation of product. Forward reaction involves reduction I the number of molecules of gas (3 on left, 2 on right) so high pressure will favour product. Forward reaction is exothermic, so low temperature will increase the yield but decrease rate.
2SO2(g) + O2(g) <-> 2SO3(g) ∆H=-196kJ mol-1
Conditions: 450 Degrees Celsius, 2 Atmosphere, catalyst- Vanadium (V) oxide.
