8.6 Factors Affecting Equilibrium (The Le Chatelier's Principle)

The position of a chemical equilibrium is influenced by various external factors. Understanding these factors is crucial, especially in industrial chemistry, to optimize conditions for maximizing the yield of desired products. Le Chatelier's principle provides a qualitative way to predict these effects.

At equilibrium, the rates of the forward and reverse reactions are equal. Changes in conditions alter the frequency and energy of molecular collisions, which in turn changes the reaction rates and causes the equilibrium position to shift until a new equilibrium is established.

8.6.1 The Effect of a Change in Concentration

When the concentration of a reactant or product in an equilibrium mixture is changed, the system is no longer at equilibrium. The system will then shift to restore equilibrium.

Consider the gas-phase equilibrium:

${\mathrm{CO}}_{2(g)}+{\mathrm{H}}_{2(g)}\rightleftharpoons {\mathrm{CO}}_{(g)}+{\mathrm{H}}_2{\mathrm{O}}_{(g)}$

• Adding a Substance: If more ${\mathrm{CO}}_2$ is added, the concentration of reactants increases. This increases the rate of the forward reaction. The equilibrium shifts to the right (towards the products) to consume the added ${\mathrm{CO}}_2$. This results in higher concentrations of $\mathrm{CO}$ and ${\mathrm{H}}_2\mathrm{O}$ at the new equilibrium.
• Removing a Substance: If a product, like ${\mathrm{H}}_2\mathrm{O}$, is removed from the system, the rate of the reverse reaction decreases. The equilibrium shifts to the right to produce more ${\mathrm{H}}_2\mathrm{O}$ and restore the balance.

General Rules for Concentration Changes:

Microscopic Explanation: The rate of reaction depends on the number of effective collisions. Increasing the concentration of a reactant increases the number of collisions for the forward reaction, causing the equilibrium to shift right until the forward and reverse reaction rates become equal again.

8.6.2 The Effect of a Change in Pressure

Changes in pressure primarily affect gaseous equilibria where the number of moles of gas on the reactant and product sides are different.

• Increasing Pressure: When pressure is increased, the system tries to reduce the pressure by shifting to the side with fewer moles of gas. This reduces the total number of gas molecules and thus the volume.

  • Example: $2{\mathrm{SO}}_{2(g)}+{\mathrm{O}}_{2(g)}\rightleftharpoons 2{\mathrm{SO}}_{3(g)}$
  • On the left, there are $2+1=3$ moles of gas. On the right, there are $2$ moles of gas.
  • Increasing the pressure will shift the equilibrium to the right to produce more ${\mathrm{SO}}_3$.

• Decreasing Pressure: When pressure is decreased, the system shifts to the side with more moles of gas to counteract the drop in pressure.

• No Effect: If the number of moles of gas is the same on both sides of the equation, a change in pressure will have no effect on the position of the equilibrium.

  • Example: ${\mathrm{H}}_{2(g)}+{\mathrm{I}}_{2(g)}\rightleftharpoons 2{\mathrm{HI}}_{(g)}$
  • There are $2$ moles of gas on both the left and right sides.

• Inert Gas at Constant Volume: Adding an inert (noble) gas at constant volume increases total pressure but does not change the partial pressures of the reacting gases. Therefore, the equilibrium position is unaffected.

8.6.3 The Effect of a Change in Temperature

The effect of temperature depends on whether the reaction is exothermic ($\Delta H$ is negative) or endothermic ($\Delta H$ is positive). A change in temperature is the only factor that changes the value of the equilibrium constant, $K_c$.

• Exothermic Reactions (Heat is a product):

  • Example: $2{\mathrm{SO}}_{2(g)}+{\mathrm{O}}_{2(g)}\rightleftharpoons 2{\mathrm{SO}}_{3(g)}\Delta H^{\circ }=-198\text{kJ}$
  • We can write this as: $2{\mathrm{SO}}_{2(g)}+{\mathrm{O}}_{2(g)}\rightleftharpoons 2{\mathrm{SO}}_{3(g)}+\text{heat}$
  • Increasing temperature (adding heat) shifts the equilibrium to the left (reactant side) to absorb the excess heat. The value of $K_c$ decreases.
  • Decreasing temperature (removing heat) shifts the equilibrium to the right (product side) to produce more heat. The value of $K_c$ increases.

• Endothermic Reactions (Heat is a reactant):

  • Example: ${\mathrm{N}}_2{\mathrm{O}}_{4(g)}\rightleftharpoons 2{\mathrm{NO}}_{2(g)}\Delta H^{\circ }=+57.2\text{kJ}$
  • We can write this as: ${\mathrm{N}}_2{\mathrm{O}}_{4(g)}+\text{heat}\rightleftharpoons 2{\mathrm{NO}}_{2(g)}$
  • Increasing temperature shifts the equilibrium to the right (product side) to consume the added heat. The value of $K_c$ increases.
  • Decreasing temperature shifts the equilibrium to the left (reactant side) to release heat. The value of $K_c$ decreases.

Summary Table — Effect of Temperature on $K_c$:

8.6.4 The Effect of a Catalyst

A catalyst provides an alternative reaction pathway with a lower activation energy. It increases the rates of both the forward and reverse reactions equally.

• A catalyst does not shift the equilibrium position.
• A catalyst does not change the value of $K_c$.
• A catalyst only helps the system reach equilibrium faster.

Summary of Factors Affecting Equilibrium