Chemistry (9620) | Chemical equilibria, Le Chatelier’s principle and Kc

Hello, Future Chemist! Welcome to one of the most fundamental and important topics in Physical Chemistry: Chemical Equilibria. While kinetics tells us how fast a reaction happens, this chapter tells us how far it goes – which is crucial for maximizing product yield in industry! Don't worry if this seems tricky at first; we will break down the rules into simple, predictable steps.

3.1.7 Chemical Equilibria and Le Chatelier’s Principle

What is Dynamic Equilibrium?

Most chemical reactions you have studied so far seem to go one way, from reactants to products. These are irreversible reactions. However, many reactions are reversible, meaning the products can react together to reform the original reactants. We show this using a double arrow (\(\rightleftharpoons\)).

In a reversible reaction, the state of chemical equilibrium is reached when:

• The rate of the forward reaction (Reactants \(\rightarrow\) Products) is exactly equal to the rate of the reverse reaction (Products \(\rightarrow\) Reactants).
• The concentrations of reactants and products remain constant over time.

It is important to remember that equilibrium is dynamic. This doesn't mean the reaction has stopped! It means both the forward and reverse reactions are still happening, but they are balancing each other out. Imagine two identical conveyor belts moving opposite ways—items are constantly moving, but the total number of items on each side stays the same.

Key Takeaway: At equilibrium, rates are equal, and concentrations are constant.

3.1.7.1 Le Chatelier’s Principle (LCP)

What happens if we disturb a system that is happily sitting at equilibrium? That is where Le Chatelier’s Principle comes in. This principle allows us to predict how an equilibrium mixture will respond to external changes.

Definition of Le Chatelier’s Principle

LCP states that if a change is made to the conditions (temperature, pressure, or concentration) of a system at equilibrium, the system will move to counteract the change, restoring a new equilibrium.

Analogy: Think of LCP as Chemistry’s way of saying, "Whatever you do to me, I’m going to try my best to undo it."

Applying LCP to Homogeneous Systems

LCP is used to predict the effect on the position of equilibrium (meaning, whether the reaction shifts to the left or the right).

1. Effect of Changing Concentration

• If you increase the concentration of a reactant: The system attempts to use up the excess reactant. The equilibrium shifts to the right (forward reaction favoured) to produce more product.
• If you decrease the concentration of a reactant: The system attempts to replace the missing reactant. The equilibrium shifts to the left (reverse reaction favoured).

Example: \( \text{A} + \text{B} \rightleftharpoons \text{C} \). If we add more A, the reaction shifts right to consume A and make more C.

2. Effect of Changing Temperature

To predict temperature effects, you need to know the enthalpy change (\(\Delta H\)) for the reaction.

• If \(\Delta H\) is negative (exothermic), heat is a product.
• If \(\Delta H\) is positive (endothermic), heat is a reactant.

Rule of Thumb: Treat Heat like a chemical reactant or product.

Change	Exothermic Reaction (\(\Delta H\)-)	Endothermic Reaction (\(\Delta H\)+)
Increase Temperature (Add Heat)	Shifts Left (Reverse/Endothermic direction) to use up the added heat.	Shifts Right (Forward/Endothermic direction) to use up the added heat.
Decrease Temperature (Remove Heat)	Shifts Right (Forward/Exothermic direction) to replace the missing heat.	Shifts Left (Reverse/Exothermic direction) to replace the missing heat.

3. Effect of Changing Pressure (Gases only)

Pressure changes only significantly affect equilibria involving gases where there is a change in the total number of moles of gas.

Rule of Thumb: The system shifts to the side with the fewest moles of gas to counteract pressure increases.

• If you increase the pressure: The system relieves the stress by shifting to the side with the fewer moles of gas. (This decreases the volume occupied, counteracting the pressure increase).
• If you decrease the pressure: The system increases the pressure by shifting to the side with the greater moles of gas.

Example: \( \text{N}_2(\text{g}) + 3\text{H}_2(\text{g}) \rightleftharpoons 2\text{NH}_3(\text{g}) \). (4 moles of gas on the left, 2 moles of gas on the right). Increasing pressure shifts Right (to the 2 moles side).

4. Effect of Adding a Catalyst

A catalyst increases the rate of reaction by providing an alternative pathway with a lower activation energy (\(E_a\)).

• A catalyst speeds up the forward reaction rate and the reverse reaction rate by the same amount.
• Therefore, a catalyst causes the system to reach equilibrium faster.
• A catalyst does not affect the position of equilibrium.

Industrial Application: The Compromise

In industry (like making ammonia or sulfur trioxide), chemists often have conflicting goals:

1. Achieve a high yield (get lots of product).
2. Achieve a fast rate (make the product quickly).
3. Keep the process economical (don't use too much energy or expensive equipment).

For example, if the desired reaction is exothermic and has fewer moles of gas on the product side:

• High Pressure gives high yield (LCP shifts right) but is expensive and dangerous.
• Low Temperature gives high yield (LCP shifts right) but results in a very slow reaction rate (Kinetics says slow T = slow rate).

Industrial chemists must use a compromise temperature and pressure. They choose conditions that give a reasonable yield quickly and economically, often using a catalyst to speed things up without sacrificing yield.

3.1.7.2 The Equilibrium Constant, K_c (Homogeneous Systems)

Le Chatelier's principle is qualitative (it tells us *which way* the equilibrium shifts). The equilibrium constant, \(K_c\), is quantitative; it gives us a numerical value for how far the reaction proceeds.

We only deal with homogeneous systems in AS Chemistry, meaning all reactants and products are in the same physical state (e.g., all gases, or all aqueous solutions).

Constructing the \(K_c\) Expression

For a general reversible reaction:

\( a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D} \)

The equilibrium constant expression is defined as the ratio of product concentrations to reactant concentrations, each raised to the power of their stoichiometric coefficients:

\( K_c = \frac{[\text{C}]^c [\text{D}]^d}{[\text{A}]^a [\text{B}]^b} \)

• \([\text{X}]\) represents the concentration of species X in mol dm\(^{-3}\) at equilibrium.
• The concentration of a pure solid or pure liquid is considered constant and is omitted from the \(K_c\) expression. Since we are focused on homogeneous systems, everything will be included (usually (g) or (aq)).

Did you know? The units of \(K_c\) are variable and depend on the exponents (the powers) in the expression. Always calculate the units using \((\text{mol dm}^{-3})^{\Delta n}\), where \(\Delta n\) is (moles of products) - (moles of reactants).

Example of \(K_c\) Construction

Consider the equilibrium between sulfur dioxide and oxygen:

\( 2\text{SO}_2(\text{g}) + \text{O}_2(\text{g}) \rightleftharpoons 2\text{SO}_3(\text{g}) \)

The \(K_c\) expression is:

\( K_c = \frac{[\text{SO}_3]^2}{[\text{SO}_2]^2 [\text{O}_2]} \)

Interpreting the Value of \(K_c\)

The numerical value of \(K_c\) tells us about the extent of the reaction (how far to the right it lies):

• If \(K_c\) is large (>> 1): The numerator (products) is much larger than the denominator (reactants). The equilibrium lies far to the right, favouring the products.
• If \(K_c\) is small (<< 1): The denominator (reactants) is much larger than the numerator (products). The equilibrium lies far to the left, favouring the reactants.

Calculations Involving \(K_c\)

You must be able to calculate the value of \(K_c\) using given equilibrium concentrations, or use a known \(K_c\) value to find an unknown equilibrium concentration.

Step-by-step for calculating \(K_c\):

1. Write the balanced chemical equation and the \(K_c\) expression.
2. Ensure all concentrations are in mol dm\(^{-3}\). If moles are given, you must divide by the volume of the reaction vessel (V) to get concentration (\(\text{concentration} = n/V\)).
3. Substitute the equilibrium concentrations into the \(K_c\) expression.
4. Calculate the final value and determine the units.

Common mistake alert: Students sometimes forget that the numbers used in the \(K_c\) calculation must be the equilibrium concentrations, not the initial concentrations.

Factors Affecting the Value of K_c

This is a critical distinction. While Le Chatelier's principle tells us that changes in concentration or pressure change the position of equilibrium, they do not change the numerical value of \(K_c\).

If you change the concentration or pressure, the system shifts temporarily, but the ratio of \(\frac{[\text{Products}]}{[\text{Reactants}]}\) (i.e., \(K_c\)) returns to the same constant value.

The only factor that changes the value of the equilibrium constant, \(K_c\), is temperature.

• If the reaction is Exothermic (\(\Delta H\)-): Increasing temperature shifts the equilibrium left (reactant side). Since the concentration of reactants increases and products decreases, the value of \(K_c\) will decrease.
• If the reaction is Endothermic (\(\Delta H\)+): Increasing temperature shifts the equilibrium right (product side). Since the concentration of products increases and reactants decreases, the value of \(K_c\) will increase.

A catalyst does not affect the value of \(K_c\). It only helps the system achieve that constant ratio faster.

Key Takeaway: \(K_c\) is a constant at a specific temperature. Only changing the temperature changes the value of \(K_c\).