Le Chatelier’s Principle

Many reactions will reach equilibrium. Theoretically all reactions will eventually reach equilibrium. If a reaction is in equilibrium, no net changes are observed. Equilibrium is dynamic, but the ratio of the product to reactant concentrations remain constant. So, how would a chemical manufacturer make money if this is the case? Well, according to Le Chatelier’s Principle:

If a system at equilibrium is disturbed, the system will shift in such a way as to counteract that disturbance.


 
A disturbance can be a change in concentration, a temperature change, or a change in pressure by increasing or decreasing the volume of the system at equilibrium. In simple terms, if something is added to the system, it will shift in such a way to remove what has been added. If something is removed from the system, it will shift to replace what has been removed. Recall the value of the equilibrium constant is constant at a constant temperature. A change in temperature is the only parameter that will change the equilibrium constant.

Le Chatelier’s Principle allows us to determine what happens when an equilibrium is disturbed, but it does not explain why.

Chemical manufacturers are able to manipulate systems at equilibrium to produce a higher yield of product. In the following sections, we will discuss these methods.

Effects of Changes in Concentration

Consider the following reaction at equilibrium:

3 H2 (g) + N2 (g) ⇄ 2 NH3 (g)

 
If more H2 or N2 is added, the system will shift to the right to counteract the change. If H2 or N2 is removed, the system will shift to the left to counteract the change. To see why this occurs, let’s look at the reaction quotient below:

\(\displaystyle Q_c=\frac{[NH_3]^2}{[H_2]^3[N_2]}\)

 
A change in concentration, at a constant temperature, does not affect the equilibrium constant. If we add more H2, the system will shift to the right until equilibrium is established again. The concentration of N2 will decrease as the concentration of NH3 will increase until equilibrium is reached. There will be no change in the equilibrium constant as long as the temperature remains constant. In order for the equilibrium constant to remain unchanged, the concentrations of the reactants and products do have to change.

If we were to decrease the concentration of H2 at constant temperature, the system would shift to the left. The concentrations of H2 and N2 would increase while the concentration of NH3 would decrease. The equilibrium constant would remain unchanged.

If we decreased the concentration of NH3, the system would shift to the right. The concentration of NH3 would increase while the concentrations of H2 and N2 would decrease until equilibrium is again reached. Some manufacturers continuously remove product so the reaction never reaches equilibrium to increase the product yield.

Effects of Changes in Pressure by Increasing or Decreasing the Volume

A change in pressure by increasing or decreasing the volume only applies to systems that have at least one gas. Recall, the only parameter that will change the equilibrium constant is a change in temperature. A change in pressure will not affect the equilibrium constant.

The products and reactants in the reaction below are gases. We need to determine if there is a change in the number of moles of gas by counting the number of moles of gas on both sides of the equation.

2 A (g) + B (g) ⇄C (g) + D (g)

There are 3 moles of gas on the reactant side and 2 moles of gas on the product side. Next, we write Qp.

\(\displaystyle Q_p=\frac{P_CP_D}{P_A^2P_B}\)

If the volume was decreased, the pressure would increase. The reaction would shift to the side with fewer moles of gas, to the right, in order for Kp to remain constant. If the pressure were decreased by increasing the volume, the system would shift to the side with the greater number of moles of gas, to the left. To see why this occurs, recall that the partial pressure of a gas is equal to its mole fraction multiplied by the total pressure:

PN2 = XN2PT

 
Substituting mole fractions and total pressure into the Kp expression:

\(\displaystyle K_p=\frac{(X_CP_T)(X_DP_T)}{(X_AP_T)^2(X_BP_T)}=\frac{X_CX_D}{X_A^2X_BP_T}\)

 
The pressure terms canceled out leaving only one PT in the denominator. If the pressure is increased, the mole fractions of C and D must also increase while the mole fractions of A and B must decrease to keep Kp constant. If the pressure is decreased, the mole fractions of C and D must decrease while the mole fractions of A and B will decrease to keep Kp constant.

For a reaction that contains gases at equilibrium, if the pressure is increased, the reaction will shift to the side that has fewer moles of gas. If the pressure is decreased, it will shift to the side with the greater number of moles of gas. This is only true if there is a difference in the number of moles of gas on both sides of the equation.

If there is no change in the number of moles of gas in the chemical equation, changing pressure by increasing or decreasing the volume will not cause a shift to either side. As an example, consider the following chemical reaction:

I2 (g) + H2 (g) ⇄ 2 HI (g)

 
There are 2 moles of gas on each side of the equation. Lets put the partial pressure terms in the Kp expression.

\(\displaystyle K_p=\frac{P_{HI}^2}{P_{I_2}P_{H_2}}\)

 
Next, we substitute with partial pressures.

\(\displaystyle K_p=\frac{(X_{HI}P_T)^2}{X_{I_2}P_TX_{H_2}P_T}=\frac{X_{HI}^2}{X_{I_2}X_{H_2}}\)

 
In this case there are no pressure terms left in Kp. A change in pressure will not affect Kp or the position of equilibrium.

The addition of an inert gas that is not included in the equilibrium constant expression will not affect either the equilibrium position or the equilibrium constant.

Effects of Changes in Temperature

Temperature is the only parameter that will change the equilibrium constant. It will also affect the equilibrium position.

Consider the following reaction:

N2 (g) + O2 (g) ⇄ 2 NO (g)   Δ H = +181 kJ

 
We need to know if the reaction is endothermic or exothermic. This reaction is endothermic. If the temperature is increased, the reaction will shift to the endothermic side. In this case it will shift to form more NO. If the temperature were decreased, it would shift to the left.

Even though this is technically incorrect, you can think of heat as a product or reactant. For an endothermic reaction heat would be a reactant.

heat + N2 (g) + O2 (g) ⇄ 2 NO (g)   ΔH = +181 kJ

 
If we remove heat by lowering the temperature, the reaction would shift to the left. The following reaction is exothermic.

2 NO (g) ⇄ N2 (g) + O2 (g)    ΔH = -181 kJ

 
We can rewrite it as:

2 NO (g) ⇄ N2 (g) + O2 (g) + heat

 
If the temperature is increased the reaction will shift to the left. If the temperature is decreased the reaction will shift to the right.

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