Le Châtelier's Principle

In order to try to figure out how to optimize the production of ammonia from hydrogen and nitrogen, Haber studied the equilibrium concentrations of ammonia in his famous process:

N₂(g) + 3H₂(g) <=> 2NH₃(g)

• He noted the equilibrium concentration of ammonia at different temperatures (while keeping pressure constant)
• He also noted the equilibrium concentration of ammonia at different pressures (while keeping the temperature constant)

• Haber observed that the equilibrium concentration of ammonia:
• decreased with increasing temperature,
• and increased with increasing pressure

The underlying basis behind both of these phenomena was described by Henri-Louis Le Châtelier; Le Châtelier's principle:

At equilibrium the rate of the forward reaction is equal to the rate of the reverse reaction. Le Châtelier's principle states that if the concentration of one of the components of the reaction (either product or reactant) is changed, the system will respond in such a way as to counteract the effect

• If a substance (either reactant or product) is removed from a system, the equilibrium will shift so as to produce more of that component (and once again achieve equilibrium)
• If a substance (either reactant or product) is added to a system, the equilibrium will shift so as to consume more of that component (and once again achieve equilibrium)

As an example, consider the Haber reaction:

N₂(g) + 3H₂(g) <=> 2NH₃(g)

What would happen if we started with a Haber reaction at equilibrium, and then suddenly added some H₂(g) to the reaction mix?

• The following is a graphical representation of how the concentrations of the individual components, and the overall system, would react in response to adding H₂(g):

What has happened to the equilibrium of the system in response to the added H₂(g)?

• Prior to the addition of H₂(g) the system is in equilibrium. This can be seen because the concentrations of the different components do not change with time (the forward and reverse rates must be equal)
• When the H₂(g) is added, we see that the system responds by consuming N₂(g) and producing NH₃(g).

Forward rate = k₁ [N₂] [H₂]³

• The production of NH₃(g) requires both N₂(g) and H₂(g) as reactants. Therefore, the production of NH₃(g) not only consumes N₂(g) but also H₂(g)

N₂ + 3H₂ à 2NH₃

• After some time, the system reaches a new state of equilibrium. It will not be identical to the original state, however. Although the system has responded to resist the effects of the added H₂(g), the new equilibrium state contains a slightly higher concentration of NH₃(g), and slightly lower concentration of N₂(g) (as well as a slightly higher concentration of H₂(g).

What would happen if we repeated the experiment, but added NH₃(g) instead of H₂(g)?

• The system would respond by decomposing some of the added NH₃(g) and a new equilibrium condition would be established (with slightly higher concentrations of H₂(g) and N₂(g) - as well as slightly higher equilibrium concentrations of NH₃(g).

Reverse rate = k_-1 [NH₃]²

• At the new equilibrium the concentrations of reactants and product is slightly different from before, but Kc has the same value

A chemical system in equilibrium can respond to the effects of pressure also. According to Le Châtelier's Rule, if the pressure is increased on a system, it will respond by trying to reduce the pressure. How does it do this?

• We are primarily concerned with homogeneous gaseous reactions
• The stoichiometry of the reaction may lead to a greater number of molecules on one side of the equation.
• For example, in the Haber reaction, N₂(g) + 3H₂(g) <=> 2NH₃(g) there are twice as many moles of reactants as products
• If the Haber reaction were in equilibrium, and the pressure was increased, the reaction would respond to oppose the increase in pressure. It could accomplish this by shifting the equilibrium to the right (producing NH₃(g))
• This would reduce the overall number of moles in the reaction, and therefore, lower the pressure

An interesting point about pressure effects is that they do not cause a change in the value of the equilibrium constant, K (as long as T is held constant). Their affects are upon concentration of reactants and products

• For the Haber process at 472°C, the value of K is 0.105. An example of one such system in equilibrium at this temp might include [H₂]=0.121M, [N₂]=0.0402M and [NH₃]=0.00272M
• K for the Haber process is defined as [NH₃]²/([N₂]*[H₂]³)
• If the volume of the sample is suddenly decreased by half, from the ideal gas equation the concentrations of the components in the sample have doubled (and the pressure has doubled):

PV = nRT

P = (n/V)RT

• The concentration of the components in this reaction at this new pressure would therefore be [H₂]=0.242M, [N₂]=0.0804M and [NH₃]=0.00544M
• The reaction quotient, Q, at this new condition would therefore be: (0.00544)²/(0.0804*0.242³) = 0.0262. This is less than K=0.105, thus, in response, the reaction equilibrium shifts to the right, consuming reactants and producing NH₃(g) product.
• This shift to the right reduces the number of molecules, therefore, the pressure is reduced.

The intrinsic value of K does not change when we increase concentrations or pressures of components in a reaction. However, almost every equilibrium constant (K) changes in response to changes in temperature.

Once again, we can apply Le Châtelier's rule in order to predict the effects of temperature changes upon chemical reactions

• Most chemical reactions have some heat change associated with the reaction. (Note, the energy of a reaction can be used to either do work - i.e. accelerate an object against some force - or to change temperature. We will consider reaction conditions under which no work is done, and therefore all energy changes associated with reactions will be manifested by temperature changes)
• Exothermic reactions are associated with heat release when the reaction proceeds in the forward direction
• Endothermic reactions are associated with heat release when the reaction proceeds in the reverse direction (i.e. heat is absorbed in the forward direction)

• These two types of reactions and their associated heat changes can be written as:

Exothermic: Reactants ó Products + Heat

Endothermic: Reactants + Heat ó Products

• If temperature is increased, the equilibrium will shift so as to minimize the effect of the added heat

• When heat is added to exothermic reactions at equilibrium, products will be consumed to produce reactants (shift to the LEFT)
• When heat is added to endothermic reactions at equilibrium, reactants will be consumed to produce products (shift to the RIGHT)

Based on this behavior, what is the effect of T upon K?

• Assume K = 1.0 for an exothermic reaction at equilibrium.
• Added heat causes the reaction to shift to the left.

Reactants <= Products + Heat

• Thus, 1.0 must represent a reaction quotient, Q, that is too large in comparison to the new value of K.
• Thus, the effect of increasing temperature on an exothermic reaction is to lower the value of K.
• Conversely, the effect of increasing temperature on an endothermic reaction is to increase the value of K

• A catalyst lowers the activation energy barrier, E_a
• Although the activation energy barrier is a different magnitude for the forward and reverse reactions, the change in the activation energy (DE_a) is the same for both the forward and reverse reactions
• Therefore, a catalyst changes the rate at which equilibrium is achieved, but does not change the composition of the equilibrium mixture (i.e. does not alter the equilibrium constant, K)