13.6 The Le Chatelier’s Principle

The Le Chatelier’s Principle states that if a dynamic equilibrium is disturbed by external conditions changes such as temperature, volume, pressure and concentration, then chemical equilibrium is adjusting itself by shifting to the side of the chemical equilibrium with the least stress to offset the introduced change.

The videos below illustrate Le Chatelier’s Principle in details:

https://www.youtube.com/watch?v=wjcPEHhiik8

An animation and simulation video are given below as well: An animation video of Le Chatelier’s Principle:

A simulation video of Le Chatelier’s Principle:

Examples:

Effect of change in concentration:

Consider the following chemical equilibrium and predict the shift of the chemical equilibrium to the right (to products’ side) or to left (to reactants’ side) or no shift:

Heat + CH₄ (g)  + 2 H₂S (g)  CS₂ (g)  + 4 H₂ (g)

1. Adding CH₄:

Shift: to the right

• Adding H2S Shift: to the right
• Adding CS₂:

Shift: to the left

• Adding H₂:

Shift: to the left

• Removing CH₄:

Shift: to the left

• Removing H₂S:

Shift: to the left

• Removing CS₂:

Shift: to the right

• Removing H₂:

Shift: to the right

Effect of change in Temperature:

Heat + CH₄ (g)  + 2 H₂S (g)  CS₂ (g)  + 4 H₂ (g)

1. Adding Heat:

Shift: to the right

• Cooling the chemical equilibrium: Cooling means removing the heat Shift: to the left

Effect of change in Pressure:

1. Increasing the Pressure: the chemical equilibrium shifts to the side with LEAST number of moles Number of moles at reactants’ side (reactants’ side) = 1 + 2 = 3 moles

Number of moles at products’ side (products’ side) = 1 + 4 = 5 moles

The reactants have the least number of moles, therefore increasing the pressure will cause the system to shift to the side with least number of moles, namely to the left (reactants’ side).

• Decreasing the Pressure: the chemical equilibrium shifts to the side with MOST number of moles Number of moles at reactants’ side (reactants’ side) = 1 + 2 = 3 moles

Number of moles at products’ side (products’ side) = 1 + 4 = 5 moles

The products have the greatest number of moles, therefore decreasing the pressure will cause the system to shift to the side with the greatest number of moles, namely to the right (products’ side).

Effect of change in Volume:

1. Increasing the Volume: the chemical equilibrium shifts to the side with MOST number of moles Number of moles at reactants’ side (reactants’ side) = 1 + 2 = 3 moles

Number of moles at products’ side (products’ side) = 1 + 4 = 5 moles

The products have the greatest number of moles, therefore increasing the volume will cause the system to shift to the side with greatest number of moles, namely to the right (products’ side).

• Decreasing the Pressure: the chemical equilibrium shifts to the side with LEAST number of moles Number of moles at reactants’ side (reactants’ side) = 1 + 2 = 3 moles

Number of moles at products’ side (products’ side) = 1 + 4 = 5 moles

The reactants have the least number of moles, therefore decreasing the volume will cause the system to shift to the side with the least number of moles, namely to the left (reactants’ side).

Effect of adding a Catalyst:

No effect at all. The catalyst is only speeding the rate of reaction by decreasing the activation energy. The catalyst does not have any effect of the chemical equilibrium as whole.

Further examples:

Consider the reaction below and predict the chemical equilibrium shift:

CO (g) + 2 H₂ (g) CH₃OH              ∆ H = – 75.2 KJ

To the left = to the reactants’ side —à it means the reverse rate of reaction rr is greater than the forward rate of reaction rf

To the right = to the products’ side —à it means the forward rate of reaction rf is greater than the reverse rate of reaction rr

Number of moles of reactants = 3 moles Number of moles or products = 1 mole

1. CO gas is added into the container:

To the right

• The temperature is increased:

To the left

• The total pressure of the system is increased:

To the right

• H₂ is removed from the system

To the left

• A catalyst is added:

No effect

• The total volume of the container is increased

To the left

• The total pressure of the container is decreased

To the left

1. Inspection of the Effect of the changes on Chemical Equilibrium

1. Concentration Change

The following example will go over the Haber – Bosch Procedure and chemical equilibrium to produce ammonia gas from hydrogen gas and nitrogen gas:

Forward Reaction: N₂ (g)  +  3 H₂ (g)     2 NH₃ (g)     + Heat

The forward rate of chemical equilibrium = rf = K_f [ N₂ ] ^m x [ H₂ ] ⁿ

2 NH3 (g)     + Heat      N2 (g)  +  3 H2 (g)

The reverse rate of chemical equilibrium = rr = K_r [NH₃ ] ^x

At equilibrium both rates (the forward and the reverse) are equal:

r_f = r_r

K_f [ N₂ ] ^m x [ H₂ ] ⁿ = K_f [NH₃ ] ^x

When the chemical equilibrium is stressed by adding one of the reactants either N2 or H2. This is causing the concentration of the reactants to increase and therefore the rate of the forward reaction will increase and it will be greater than the reverse rate of reaction:

r_f > r_r

The stressed chemical equilibrium will shift toward the product side (to the right) to offset this incoming stress and more NH3 will be produced.

With same reasoning if one of reactants is removed, or some of product is added, then the shift of the chemical equilibrium will be to the left. In both cases: The rate of the forward reaction will be less than the reverse rate of the reaction.

r_f < r_r

The above analysis is based of kinetic inspection of both rates, forward and reverse.

One can come to the same conclusion, if one uses the reaction quotient Q C of the chemical equilibrium and compare it to the chemical equilibrium constant K C.

At chemical equilibrium: Q _C = K _C. =   [ NH₃ ] ² / { [ N₂ ] x [ H₂ ] ³ }

Now, if one of the reactants is added then the denominator value is increased which leads to decrease of Q C. This effect is observed if the product is removed which leads to a decrease of numerator and also leads to a decrease of Q C.

Then in case of adding reactants or removing product(s), the chemical equilibrium will shift to the right to offset these incoming stresses.

In these cases:     QC < KC

With similar reasonings, one can conclude:

If one of the reactants is removed then the denominator value is decreased which leads to increase of Q C. This effect is observed if the product is added which leads to an increase of numerator and also leads to an increase of Q C.

Then in case of removing reactants or adding product(s), the chemical equilibrium will shift to the left to offset these incoming stresses.

In these cases:     QC > KC

2.      The Gas Phase Chemical Equilibrium

The gas phase chemical equilibrium utilizes the partial pressure of both the reactants and the products. The ideal gas equation is used:

P V = n R T

P = (n / V) R T P = M R T

M = (n / V) = molarity = P / ( R T )

From the equation above any changes in partial pressures of the reactants and the products will lead in changes in the partial pressures of the gases of the reactants and the products will lead to changes in the concentrations of the reactants and the products. This is always true for any changes of the volume of the chemical equilibrium system which will be affected by any changes of the partial pleasures of the gases of the reactants and the products (Boyle’s Law).

Number of moles of the reactants and the products should be considered. At the equilibrium:

Q _P = ( P_NH3 ) ² / { ( P_N2 ) x ( P_H2 ) ³ } = K _P

Let us consider that the volume equilibrium is decreased by a factor of 2, then the partial pressures of the gases will increase by factor of 2 as well. ((Boyle’s Law).

Qp ^’ = ( 2 P   NH3   ) ² / { ( 2 P N2   ) x ( 2 PH2    ) ³ }

Qp ^’ = ( 4 PNH3      ) ² / { 16 ( P N2   ) x ( PH ) ³ } = 1/4 Q     < Kp

K _P > Q _P [The equilibrium constant is greater than the reaction quotient]

Decreasing the volume which is in turn meaning increasing the pressure of the equilibrium system: N2 (g) + 3 H2 (g) ßà 2 NH3 (g)                    + Heat

This leads the reaction system to shift to the right (towards the product(s)). This means the chemical equilibrium will shift towards the side with least number of moles when the pressure of the system is increased.

3.      The Effect of the Catalyst

The catalyst does not contribute to any changes in the chemical equilibrium. It does affect the activation energy. The addition of a catalyst will lower the activation energy of both the forward and the reverse rate of reaction. Both rates’ activation energies are reduced but their values are still identical and do not change as it is shown in the figure below.

Reference: https://slideplayer.com/slide/6380838/

4.      Effect of Temperature

The forward rate of reaction of the Habor – Bosch ammonia synthesis will equal the reverse rate of reaction at equilibrium

r_f = K_f [ N₂ ] ^m x [ H₂ ] ⁿ   = r_r = K_r [NH₃ ] ^x

K _C = K_f / K_r = [NH₃ ] ^x / { [ N₂ ] ^m x [ H₂ ] ⁿ }

The equilibrium constant K C is connected the rate constants of both the forward and the reverse rate of the reaction. The chemical equilibrium is exothermic:

N2 (g)  + 3 H2 (g)     2 NH3 (g)     + Heat

According to Arrhenius, the rate constant is changing with changing the temperature. Therefore, increasing the temperature will increasing the NH3 amount and the chemical equilibrium will shift left towards the reactants’ side. Likewise, removing the heat by cooling the equilibrium systems will increase the reactants amounts and the chemical equilibrium will shift to the right to the product’s side.