Chemistry | Chemical Equilibria

Welcome to Chemical Equilibria!

Hello future chemist! This chapter is one of the most important concepts in Physical Chemistry. We've previously learned about how fast reactions go (Kinetics/Rates). Now, we ask: How far do they go?

Most reactions don't simply run until one reactant is completely gone. Instead, they reach a state of balance. Understanding this balance—known as Equilibrium—allows industrial chemists to maximise the production of useful chemicals, like ammonia or sulfuric acid. Don't worry if this seems tricky at first; we will break it down into simple, manageable steps!

1. Reversible Reactions and Dynamic Equilibrium

1.1 Reversible Reactions

An Irreversible Reaction goes in one direction only. Think of burning wood; once it's ash, you can't easily turn it back into a log.

A Reversible Reaction can proceed in both the forward and reverse directions.
We use a special double arrow (\(\rightleftharpoons\)) to show this:
\(A + B \rightleftharpoons C + D\)

Example: The Haber process, where nitrogen and hydrogen react to form ammonia, but ammonia can also decompose back into nitrogen and hydrogen.

1.2 Defining Dynamic Equilibrium

When a reversible reaction is carried out in a closed system (meaning nothing can enter or leave, like reactants, products, or energy), it will eventually reach a state of equilibrium.

What does "Dynamic Equilibrium" actually mean?

1. The Rates Are Equal: The rate of the forward reaction (A + B \(\to\) C + D) is exactly equal to the rate of the reverse reaction (C + D \(\to\) A + B).
2. The Concentrations Are Constant: Because the chemicals are being used up and replaced at the same rate, the overall concentrations of reactants and products remain constant (they are not necessarily equal, just constant).

B The Key Word is "Dynamic"
This is not a static state where everything has stopped! It is dynamic—reactions are still occurring constantly at the microscopic level, but because they are perfectly balanced, nothing changes at the macroscopic (visible) level.

Analogy: The Escalator
Imagine an escalator going up. If 10 people step onto the escalator going up every minute (forward reaction rate) and 10 people step off the top onto the floor every minute (reverse reaction rate), the number of people *on* the escalator remains constant. It looks static from a distance, but the people are constantly moving! That’s dynamic equilibrium.

2. Factors Affecting Equilibrium: Le Chatelier's Principle (LCP)

Equilibrium is delicate. If we change the conditions (the stress), the system will try to undo the change. This behaviour is summed up by:

Le Chatelier's Principle: If a change is made to the conditions of a system at equilibrium, the system responds to counteract the change.

We say the equilibrium "shifts" to the left (favouring reactants) or shifts to the right (favouring products).

2.1 Effect of Concentration Changes

This is the most straightforward change.

• If you increase the concentration of a reactant: The system will shift to use up the extra reactant. The equilibrium shifts to the right (favouring products).
• If you increase the concentration of a product: The system will shift to remove the extra product. The equilibrium shifts to the left (favouring reactants).

Simple Trick: The equilibrium always shifts AWAY from what you add and TOWARDS what you take away.

Example: \(A + B \rightleftharpoons C\)
If we add more A, the system shifts right to make more C.
If we remove C (perhaps by distillation), the system shifts right to replace the lost C.

2.2 Effect of Temperature Changes

To apply LCP here, you must know whether the reaction is exothermic (releases heat, \(\Delta H\) is negative) or endothermic (absorbs heat, \(\Delta H\) is positive).

Strategy: Treat "Heat" as if it were a reactant or a product in the equation.

1. Exothermic Reaction (Heat is a Product):
   \(Reactants \rightleftharpoons Products + \text{Heat}\)
   • Increase Temperature (Add Heat): The system shifts away from the added heat. Shifts Left (favours reactants).
   • Decrease Temperature (Remove Heat): The system shifts towards the missing heat. Shifts Right (favours products).
2. Endothermic Reaction (Heat is a Reactant):
   \(\text{Heat} + Reactants \rightleftharpoons Products\)
   • Increase Temperature (Add Heat): Shifts Right (favours products).
   • Decrease Temperature (Remove Heat): Shifts Left (favours reactants).

Did you know? Industrial processes like the Haber Process use moderate temperatures (around 450 °C). Why not very low temperatures? Although lower temperatures shift the equilibrium to the right (as it is exothermic), very low temperatures also significantly slow down the rate of reaction! A compromise is needed.

2.3 Effect of Pressure Changes (Gaseous Systems Only)

Pressure changes only affect reactions involving gases. Pressure is caused by gas particles hitting the container walls.

Rule: Pressure is directly related to the number of gas moles.
The system shifts to reduce the pressure change.

• Increase Pressure: The system shifts to the side with the FEWER moles of gas to relieve the pressure.
• Decrease Pressure: The system shifts to the side with the MORE moles of gas to increase the pressure.

Step-by-Step Example:
\(N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)}\)
1. Count moles on the left (Reactants): \(1 + 3 = \mathbf{4}\) moles of gas.
2. Count moles on the right (Products): \(\mathbf{2}\) moles of gas.
3. If we increase pressure, the system shifts towards fewer moles: shifts Right (towards the 2 moles of \(NH_3\)).

B Common Mistake Alert: If the total number of moles of gas is the same on both sides (e.g., \(H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}\) - 2 moles on both sides), then changing the pressure has no effect on the position of equilibrium.

2.4 Effect of Catalysts

A catalyst speeds up the reaction by lowering the activation energy.

B Crucial Fact: A catalyst speeds up the forward reaction AND the reverse reaction by exactly the same amount.
Therefore, adding a catalyst has NO effect on the position of equilibrium. It only helps the system reach equilibrium faster.

Key Takeaway: LCP tells us the direction of the shift—to the left or to the right—to restore balance. Pressure, concentration, and temperature cause a shift. Only concentration and pressure shifts are reversible processes that restore the original \(K_c\).

3. The Equilibrium Constant, \(K_c\)

While LCP tells us *how* the equilibrium shifts, the Equilibrium Constant (\(K_c\)) gives us a quantitative measure of *how far* the reaction goes at a specific temperature.

3.1 Defining the \(K_c\) Expression

For a general reaction:
\(aA + bB \rightleftharpoons cC + dD\)

The equilibrium constant expression is defined as the ratio of the concentration of products to the concentration of reactants, each raised to the power of their stoichiometric coefficient:

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

Where \([ ]\) indicates the concentration of the substance at equilibrium, measured in \(\text{mol dm}^{-3}\).

Rules for Writing \(K_c\) Expressions

1. Products over Reactants: Always put product concentrations in the numerator (top) and reactant concentrations in the denominator (bottom).
2. Stoichiometry is the Power: The coefficient from the balanced equation becomes the exponent (power).
3. Exclusion Rule: Concentrations of pure solids (s) and pure liquids (l) are considered constant and are therefore omitted from the \(K_c\) expression. Only substances that change concentration significantly—gases (g) and aqueous solutions (aq)—are included.

Example: The synthesis of methanol
\(CO_{(g)} + 2H_{2(g)} \rightleftharpoons CH_3OH_{(g)}\)
\[K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\]

3.2 Calculating Units for \(K_c\)

Since concentration is measured in \(\text{mol dm}^{-3}\), \(K_c\) often has units.

Step 1: Substitute the units \(\text{mol dm}^{-3}\) into the \(K_c\) expression.
Step 2: Cancel out the terms.

Using the methanol example again:
\[K_c = \frac{[CH_3OH]}{[CO][H_2]^2}\]
Units \(= \frac{(\text{mol dm}^{-3})}{(\text{mol dm}^{-3}) \times (\text{mol dm}^{-3})^2}\]
Units \(= \frac{1}{(\text{mol dm}^{-3})^2}\]
Units \(= (\text{mol dm}^{-3})^{-2}\]
Units \(= \mathbf{\text{mol}^{-2} \text{ dm}^{6}}\]

3.3 Interpreting the Magnitude of \(K_c\)

The size of the \(K_c\) value tells us where the equilibrium lies:

• \(K_c > 1\): The numerator (Products) is larger than the denominator (Reactants). Equilibrium lies to the right, meaning products are favoured.
• \(K_c < 1\): The denominator (Reactants) is larger than the numerator (Products). Equilibrium lies to the left, meaning reactants are favoured.
• \(K_c \approx 1\): Significant amounts of both reactants and products are present.

3.4 The Unique Role of Temperature on \(K_c\)

This is the most critical distinction in this chapter:

B The Value of \(K_c\) is constant for a given reaction ONLY at a specified temperature.

• Changes in concentration or pressure cause the system to shift (LCP), but the ratio of \(\text{Products}/\text{Reactants}\) quickly adjusts back to the original \(K_c\) value. They do NOT change \(K_c\).
• Only changing the temperature changes the value of \(K_c\).
  • If the equilibrium shifts to the right (more products) due to temperature change, \(K_c\) increases.
  • If the equilibrium shifts to the left (more reactants) due to temperature change, \(K_c\) decreases.

Memory Aid: $K_c$ is a Temperature Queen: She only cares about Temperature (T).

Key Takeaway: \(K_c\) is a mathematical tool that precisely defines the position of equilibrium. It must be derived using equilibrium concentrations only, and its value is temperature-dependent.

You’ve covered the core of chemical equilibria! You now understand the balance (Dynamic Equilibrium), how to disturb that balance (Le Chatelier's Principle), and how to measure the balance (\(K_c\)). Practice writing those \(K_c\) expressions and calculating the units—it’s where many marks are gained! Keep up the great work!