LeChatelier’s Principle and Chemical Equilibrium: Effects of Disturbances on Equilibrium Systems

LeChatelier’s Principle and Chemical Equilibrium

Introduction to Chemical Equilibrium and LeChatelier’s Principle

Chemical equilibrium is a dynamic state in which the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. LeChatelier’s Principle provides a framework for predicting how a system at equilibrium responds to external disturbances.

• Dynamic Equilibrium: Even though concentrations remain constant, reactions continue to occur in both directions.

• LeChatelier’s Principle: If a system at equilibrium is disturbed, the system will shift in a direction that counteracts the disturbance and restores equilibrium.

Types of Disturbances Affecting Equilibrium

Concentration Changes

Changing the concentration of reactants or products will shift the equilibrium to minimize the effect of the change.

• Adding a species: The equilibrium shifts away from the added species.

• Removing a species: The equilibrium shifts toward the removed species.

• Law of Mass Action: Only species that appear in the equilibrium constant expression ($K$) are affected by these changes. Solids and pure liquids are not included in $K$.

Example: For the reaction $\mathrm{NH_4OCO NH_2(s) \rightleftharpoons 2NH_3(g) + CO_2(g)}$, the reaction quotient is:

$Q = \frac{[NH_3]^2[CO_2]}{1}$

Adding $NH_3$ or $CO_2$ shifts the equilibrium to the left; removing them shifts it to the right.

Volume and Pressure Changes (for Gaseous Systems)

Changes in pressure or volume affect equilibria involving gases. The system shifts to favor the side with fewer or more moles of gas, depending on the change.

• Increase in pressure (decrease in volume): Shifts equilibrium toward the side with fewer moles of gas.

• Decrease in pressure (increase in volume): Shifts equilibrium toward the side with more moles of gas.

• Rationale: The system seeks to minimize the change in the number of gas particle collisions.

Example: For $SO_2(g) + NO_2(g) \rightleftharpoons SO_3(g) + NO(g)$, decreasing the volume (increasing pressure) shifts equilibrium toward the side with fewer moles of gas.

Temperature Changes

The effect of temperature depends on whether the reaction is endothermic or exothermic. Temperature changes alter both the position of equilibrium and the value of the equilibrium constant ($K$).

• Endothermic reaction: Heat is a reactant. Increasing temperature shifts equilibrium to the right (toward products).

• Exothermic reaction: Heat is a product. Increasing temperature shifts equilibrium to the left (toward reactants).

Example Equations:

• Endothermic: $\text{heat} + A \rightleftharpoons B$

• Exothermic: $C \rightleftharpoons D + \text{heat}$

Effect on $K$: Increasing temperature increases $K$ for endothermic reactions and decreases $K$ for exothermic reactions.

$K = \frac{k_f}{k_r} = \frac{A_f e^{-E_{af}/RT}}{A_r e^{-E_{ar}/RT}}$

Effect of Catalysts

Catalysts speed up the attainment of equilibrium by lowering the activation energy for both forward and reverse reactions equally. They do not affect the position of equilibrium or the value of $K$.

• Key Point: Catalysts do not change the composition of the equilibrium mixture.

Worked Example: Applying LeChatelier’s Principle

Consider the reaction: $\text{PCl}_5(g) \rightleftharpoons \text{PCl}_3(g) + \text{Cl}_2(g)$, $\Delta H = +87.9\,\text{kJ}$ (endothermic)

Disturbance	Direction of Shift	Reason
Remove Cl2	Left	Shifts toward removed species to produce more Cl2
Decrease temperature	Left	Heat is a reactant; removing heat favors the reverse reaction
Increase volume	Right	More moles of gas on the right; system shifts to increase pressure
Add PCl5	Right	Shifts away from added species to produce more products

ICE Table Shortcut for Equilibrium Calculations

ICE tables (Initial, Change, Equilibrium) are used to calculate equilibrium concentrations. For reactions with small equilibrium constants, the change in concentration ($x$) may be negligible compared to the initial concentration, allowing for a simplification.

• Example: For $CH_3OH(g) \rightleftharpoons CO(g) + 2H_2(g)$, with $K = 4.42 \times 10^{-7}$ and initial $[CH_3OH] = 0.500$ M:

Set up the ICE table and solve for $x$:

$K = \frac{[CO][H_2]^2}{[CH_3OH]}$

Assume $x$ is small: $0.500 - x \approx 0.500$

Solve for $x$:

$4.42 \times 10^{-7} = \frac{x (2x)^2}{0.500}$

$x = 3.91 \times 10^{-3}$ M$

Additional info: This approximation is valid when $K$ is very small and the initial concentration is much larger than $x$.