CHANGING PRESSURE

Using Le Chatelier's Principle with a change of pressure

Lets discuss about changing pressure in reactions involving gases
What would happen if the condition of an equilibrium is changed by increasing the pressure?
According to Le Chatelier, the position of equilibrium will shift to a certain direction to counteract the change. That means that the position of equilibrium will shift so that the pressure is reduced again (If there is an action, there is a reaction).
The more molecules in the container, the higher the pressure will be. The system can reduce the pressure by reacting in such a way to produce fewer molecules.
For example : N2(g)+ 3H2(g) = 2NH3(g)
In this reaction, there are 4 molecules on the left-hand side of the equation, but only 2 on the right. By forming more NH3, the system causes the pressure to reduce.
Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.
In summary, when P >, Volume <, equilibrium shift to the fewer molecules. N2 and H2 molecules are closely, the efective collision increases, the rate of forward reaction increases. The equilibrium will achieve after the rate of forward reaction = the rate of reverse reaction. The value of equilibrium constant is the same. What would happen if the condition of equilibrium is changed by decreasing the pressure? The equilibrium will shift to the left, because P <, Volume >, equilibrium will shift to the more molecules, e.i. N2 and H2 molecules.
Another example : H2(g) + I2(g) = 2HI(g).
In the above equilibrium, the number of molecules in the both sides are the same. What happens if P > ?
Increasing the pressure has no effect on the position of the equilibrium, because there are the same numbers of molecules on both sides.