Le Chateliers principle and equilibrium constants 

In a nutshell

Dynamic equilibrium is when the concentrations of the reactants and products stay constant in an equilibrium reaction. Le Chateliers principle is defined as any change in temperature, concentration or pressure, will cause the equilibrium to move to help counteract this change.

Equilibrium

Dynamic equilibrium is when the concentrations of the reactants and products stay constant in an equilibrium reaction. Dynamic equilibrium only occurs in a closed system. A closed system is a system when nothing can come in or out. 

In a dynamic equilibrium reaction, the rate of the forward and backward reactions are the same. When a condition changes, the equilibrium position changes. 

If the condition causes more product formation, the equilibrium will shift to the left. If the condition causes less product formation, the equilibrium will shift to the right. If the condition causes more reactant formation, the equilibrium will shift to the right. If the condition causes less reactant formation, the equilibrium will shift to the left. 

Le Chateliers principle is defined as any change in temperature, concentration or pressure, will cause the equilibrium to move to help counteract this change. If the amount of product increases, the equilibrium constant will also increase. If the amount of reactant increases, the equilibrium constant will decrease.

Temperature 

Temperature changes causes the equilibrium to shift. Increasing the temperature favours endothermic reaction. Decreasing the temperature favour exothermic reaction. Temperature changes affect the equilibrium constant because the amount of reactant/product changes. 

Determining if the equilibrium constant will increase or decrease during a reaction is described below.

procedure

Example

The temperature of the following equilibrium is increased. Will the value of the gaseous equilibrium constant, $$K_p$$​, increase or decrease?

​$$2NO(g)+O_2(g)⇌2NO_2(g)$$

$$Enthalpy\>change=\triangle H=-110\>kJ\>mol^{-1}$$

Increasing the temperature favours the backwards endothermic reaction. 

Write the $$K_p$$ expression:

$$K_p=\dfrac{(p(NO_2))^2}{(p(NO))^2\>p(O_2)}$$​​

The amount of $$NO_2$$ decreases. Therefore, the value of $$K_p$$ $$\underline{decreases}$$.

Concentration

Concentration changes in an equilibrium reaction does not change the value of the equilibrium constant. Even if the concentration is changed, the equilibrium will shift to counteract this change. This results in the ratio of the equilibrium concentrations of the products and reactants to remain the same.

Overall, changes in concentrations do result in changes of the amount of reactants/products. However, it doesn't affect the value of the equilibrium constant as the ratio remains the same. The equilibrium position shifts to keep the ratio of the product/reactant equilibrium concentrations the same.

Example

The concentration of hydrochloric acid is increased. Determine if the value of $$K_c$$ will increase or decrease.

​

​$$HCl(l) + NaOH(l) ⇌ NaCl(l) + H_2O(l)$$

Increasing the concentration of hydrochloric acid will shift the equilibrium to the right to counteract this increase in concentration. This removes the extra concentration of hydrochloric acid by increasing the concentration of sodium chloride and water. Therefore, $$K_c$$ remains constant. 

Pressure

Pressure changes in an equilibrium reaction does not change the value of the equilibrium constant. This is because increasing or decreasing the pressure causes all the partial pressures to increase or decrease by the same amount. Therefore, the ratio of the product and reactant pressures remain the same. 

Pressure changes causes the equilibrium position to shift to the side with the fewest gaseous moles to counteract the change.

Example

The pressure of the gas mixture in this equilibrium reaction is increased. Determine if the value of $$K_p$$ will increase or decrease.  ​

​$$SO_2Br_2 (g)⇌ SO_2(g)+Br_2(g)$$

Increasing the pressure causes the equilibrium to shift to the left as there's only one gaseous mole. The equilibrium constant, $$K_p$$, is not affected.