Le Chatelier's Principle

We know that equilibrium is achieved in a reversible reaction when the rate of forward reaction becomes equal to the rate of backward reaction. But what happens when we disturb this equilibrium? This is where Le Chatelier's principle comes into play.

Le Chatelier's principle states that if a system in equilibrium is subjected to a change of concentration, temperature or pressure, the equilibrium shifts in a direction so as to undo the effect of the change imposed. $_\square$

As we can see from the definition, a change in concentration (of the reactants/products), temperature, or pressure can shift the equilibrium of a reaction. However, adding a catalyst makes the reaction faster, but does not affect equilibrium. Now we will discuss how some factors affect equilibrium.

When a substance is added at equilibrium state, the reaction occurs in the direction that decreases the concentration of that substance. When a substance is removed at equilibrium state, the reaction occurs in the direction that increases the concentration of that substance.

As we add or remove reactant (or product), the ratio of equilibrium concentration becomes $Q,$ which is called the reaction quotient.

• If $Q<K$, equilibrium shifts in forward direction;

• If $Q>K$, equilibrium shifts in backward direction;

where $K$ is the equilibrium constant of the reaction.

This can be simply understood from the definition. Let us take an example:

$\ce{N2}(g) +\ce{3H2}(g) \rightleftharpoons \ce{2NH3}(g).$

Now if we remove some amount of reactant ($\ce{N2}$ or $\ce{H2}$ or both), then we have disturbed the equilibrium and the concentration of the reactants gets decreased. To achieve equilibrium, the products will react to form reactants. Hence the reaction moves backwards.

Mathematically, let the equilibrium concentrations of $\ce{N2}, \ce{H2}$ and $\ce{NH3}$ be $c_1,c_2,$ and $c_3,$ respectively.

Then the equilibrium constant would be

$K= \frac{{c_3}^2}{(c_1)(c_2)^3}.$

Now let $c'(<c_1)$ be the concentration of $\ce{N2}$ after removal. Just after the removal, we have the reaction quotient

$Q=\frac{{c_3}^2}{(c')(c_2)^3}.$

Now clearly we have $Q>K,$ and hence the reaction will proceed backwards and reactants will be formed.

Pressure can shift the chemical equilibrium of a reaction that involves gaseous molecules. If pressure is increased, then the equilibrium shifts in the direction that decreases the number of gas molecules. If pressure is decreased, then the equilibrium shifts in the direction that increases the number of gas molecules. For example, let's take a look at the following reaction:

$\text{C}(s)+\text{H}_2\text{O}(g)\leftrightharpoons\text{CO}(g)+\text{H}_2(g).$

Note that the forward reaction increases the number of gas molecules, and the reverse reaction decreases that. Thus if pressure is increased, then the reverse reaction will occur. In contrast if the pressure is decreased, then the equilibrium will shift forward.

However, the pressure we apply must have effect on the partial pressure of the reactants and/or products. If we increase the pressure by adding an irrelevant gas (e.g. Helium) under constant volume, then the equilibrium will not shift.

If a system in equilibrium consists of gases, then the concentration of all the components can be altered by changing the pressure.

• When there is an increase in pressure, the equilibrium will shift towards the side of the reaction with fewer moles of gas.

• When there is a decrease in pressure, the equilibrium will shift towards the side of the reaction with higher moles of gas.

We know in a gaseous system pressure is directly proportional to moles and hence the above two points.

Pressure is inversely related to volume. Therefore, the effects of changes in pressure are opposite of the effects of changes in volume. Additionally, this does not apply to a change in the pressure in the system due to the addition of an inert gas.

$N_2(g) +3H_2(g) \rightleftharpoons 2NH_3(g).$

If we increase pressure, then the reaction will proceed forward (less number of moles) and if we decrease pressure then the reaction will proceed backwards (high moles).

Consider the following two cases:

• For solids whose volume increases on melting, e.g. $\ce{Fe}, \ce{Cu}, \ce{Ag}, \ce{Au},$ etc., $\text{Solid (Lower volume) } \rightleftharpoons \text{Liquid (Higher volume)}.$ In this case, the process of melting becomes difficult at high pressure, and thus the melting point becomes high.

• Solids whose volume decreases on melting, e.g. $\text{quartz, carborundum, ice, diamond},$ etc., $\text{Solid (Higher volume) } \rightleftharpoons \text{Liquid (Lower volume)}.$ In this case, the process of melting becomes favorable at high pressure, and thus the melting point is lowered.

When solid substances are dissolved in water, either heat is evolved (exothermic) or heat is absorbed (endothermic).

• For endothermic solubility process, solubility increases with increase in temperature .

• For exothermic solubility process, solubility decreases with increase in temperature .

When a gas dissolves in liquid, there is a decrease in volume. Thus the increase of pressure will favor the dissolution of gas in liquid.

For endothermic and exothermic reactions, the chemical equilibrium will shift according to temperature. If the temperature is risen, then the reaction will occur in the endothermic direction. If the temperature is dropped, the reaction occurs in the exothermic direction. Take a look at the following reaction:

$2\text{NO}_2(g)\leftrightharpoons\text{N}_2\text{O}_4(g),\qquad\Delta H=-54.8\text{ kJ}.$

Observe that the forward reaction is exothermic and the reverse reaction is endothermic. Thus, if we raise the temperature at equilibrium, then the equilibrium will shift backward. In this case the equilibrium constant $K$ becomes smaller. On the other hand if we lower the temperature, the equilibrium constant $K$ gets larger, and the equilibrium will shift forward.

Keep in mind that the only factor that alters the equilibrium constant $K$ is temperature.

• For reactions in which $n_p=n_r$ (number of moles of product = number of moles of reactant), there is no effect on adding an inert gas at constant volume or at constant pressure on the equilibrium.

• For reactions in which $n_p \ne n_r$, there is no effect on adding an inert gas at constant volume $BUT$ at constant pressure, the equilibrium shifts towards the side with higher number of moles.

Applying Le Chatelier's principle, the favorable conditions for dissociation of $\ce{NH_3}$ by Haber's process $\ce{N_2}(g)+3\ce{H_2}(g) \rightleftharpoons 2\ce{NH_3}, \quad \Delta H=-92.5kJ$ are

• high temperature

• low pressure

• removal of $\ce{N_2}$ and $\ce{H_2}$

• addition of inert gas at constant pressure.

Think why the above 4 points are true? (Remember we are talking about dissociation of $\ce{NH_3}.$)

$\text{N}_2(g)+2\text{O}_2(g)\leftrightharpoons2\text{NO}_2(g),\qquad\Delta H=+66\text{ kJ}.$

Find all of the options, if any, that will increase the yield of nitrogen dioxide.

$\qquad \text{(a)}$ Increasing the concentration of nitrogen.
$\qquad \text{(b)}$ Increasing the temperature.
$\qquad \text{(c)}$ Applying pressure.
$\qquad \text{(d)}$ Adding 3 moles of Neon under constant volume.
$\qquad \text{(e)}$ Adding a catalyst.

---

(a) If the concentration of nitrogen is increased, then the reaction will occur in the direction that decreases the concentration of nitrogen. Thus the equilibrium will shift forward, and increase the yield of $\text{NO}_2.$

(b) Since the forward reaction is endothermic, the equilibrium will shift forward when temperature is increased, increasing the yield of nitrogen dioxide. In this case, the equilibrium constant $K$ increases.

(c) If pressure is applied, then the equilibrium shifts in the direction that decreases the number of gas molecules, which is the forward direction in this case. Therefore increasing pressure will increase the yield of nitrogen dioxide.

(d) Neon is a noble gas, which has nothing to do with the above reaction. Adding 3 moles of Neon will change the total pressure, but will not affect the partial pressures of the reactants and products.

(e) Adding a catalyst will accelerate the speed of the reaction, but will not shift the equilibrium.

Therefore our answer is (a), (b), and (c). $_\square$