Effect of Change in Pressure on Equilibrium
In chemical equilibrium, an increase in pressure applied to a system at equilibrium, favors the reaction in the direction that produces lower number of moles of gases and a decrease in pressure favors the opposite reaction. Since, pressure and number of moles are related. The relation between equilibrium constants kp and kc is given by the following formula:
Kp = Kc (RT)Δn = Kc (p/v)Δn
where,
- kp and kc are equilibrium constant in partial pressure and number of moles
- R is gas constant
- T is temperature
- n is number of moles
High pressure is favorable for the reaction in which there is increase in volume. Change in pressure only affect the gaseous reaction.
- Increase in pressure shifts the equilibrium in the direction of lesser number of gaseous moles.
- Decrease in pressure shifts the equilibrium in the direction of greater number of gaseous moles.
Note: If there is no change in number of gas molecules in a reaction, then the pressure change doesn’t affect the equilibrium.
- n1 = Number of gaseous moles of reactants
- n2 = Number of gaseous moles of products
△ng = n2 – n1
- â–³ng = 0; No effect of change in pressure
- â–³ng = +ve; Increase in pressure will decrease product formation and vice versa.
- â–³ng = -ve; Increase in pressure will increase product formation.
Let us understand it with examples
When â–³ng is -ve
Increase in pressure will increase product formation and shift the equilibrium towards products and hence concentration of NH3 and SO3 will increase.
- High pressure favors the shift of equilibrium towards products.
N2(g) + 3H2(g) ⇄ 2NH3(g); n1 = 4, n2 = 2; △ng = -2
2SO2(g) + O2(g) ⇄ 2SO3(g); n1 = 3, n2 = 2; △ng = -1
When â–³ng is +ve
Decrease in pressure will increase reactants and shift the equilibrium towards reactants and hence concentration of PCl5 and N2O4 will increase.
- Low pressure favors the shift of equilibrium towards reactants.
PCl5(g) ⇄ PCl3(g) + Cl2(g); n1 = 1, n2 = 2; △ng = 1
N2O4(g) ⇄ 2NO2(g); n1 = 1, n2 = 2; △ng = 1
When â–³ng is Zero
There will be no effect of pressure if number of moles of gases is the same on both sides of reaction. Here, the concentration of both reactants and products remain same. Hence, â–³ng = 0
H2(g) + I2(g) ⇄ 2HI(g) ; n1 = 2, n2 = 2; △ng = 0
N2(g) + O2(g) ⇄ 2NO(g); n1 = 2, n2 = 2; △ng = 0
If the species are not in their gaseous state hence there will be no effect of pressure change.
2NaNO3(s) ⇄ 2NaNO3(s) + O2(g); n1 = 2, n2 = 3; △ng = 1
NH4HS(s) ⇄ NH3(g) + H2S(g) ; n1 = 1, n2 = 2; △ng = 1
Le Chateliers Principle
Le Chatelier’s Principle is a fundamental concept in chemistry that describes how a chemical system at equilibrium responds to changes in temperature, pressure, or concentration of reactants or products. This principle is named after the French chemist Henry Louis Le Chatelier, who formulated it in the late 19th century. Equilibrium refers to the state of balance which tells that there is equal weight on both sides of the scale. Chemical equilibrium is attained when the rate of forward reaction is equal to the rate of backward reaction.
In this article, we will have a thorough knowledge of equilibrium, Le Chatelier’s principle, and the effect of various factors on chemical equilibrium.
Table of Content
- What is Le Chatelier’s Principle of Equilibrium?
- Effect of Concentration Change on Equilibrium
- Effect of Change in Pressure on Equilibrium
- Effect of Volume Change on Product Formation
- Effect of Change in Temperature on Equilibrium
- Effect of Catalyst on Equilibrium
- Effect of Addition of an Inert gas on Equilibrium
- Application of Le Chatelier’s Principle