Welcome to Chemical Equilibria! (AS Level Chemistry 9701, Topic 7.1)

Welcome to one of the most important topics in Physical Chemistry: Chemical Equilibria!
This chapter is all about reactions that don't go to completion. Instead, they reach a stable state where reactants and products coexist. Understanding this allows chemists to control and optimize industrial processes—which is huge!

Don't worry if this seems tricky at first. We will use simple analogies to break down the key ideas: reversible reactions, the concept of 'dynamic equilibrium', and the powerful rule chemists use to predict changes: Le Chatelier's Principle. Let's get started!

Part 1: Reversible Reactions and Dynamic Equilibrium

1.1 Reversible Reactions

In Kinetics (Topic 8), we studied reactions that proceed in one direction until one reactant runs out (irreversible reactions). However, many chemical reactions are reversible.

A reversible reaction is one where the products can react together to reform the original reactants.

  • We use a double arrow, \(\rightleftharpoons\), to indicate a reversible reaction, showing that the reaction proceeds both ways:
    $A + B \rightleftharpoons C + D$
  • The reaction going from left to right ($A + B \rightarrow C + D$) is the forward reaction.
  • The reaction going from right to left ($C + D \rightarrow A + B$) is the reverse reaction.

1.2 Establishing Dynamic Equilibrium

Imagine two identical buckets connected by a pipe. You pour water into the first bucket (Reactants). As water flows into the second bucket (Products), it starts to flow back into the first bucket too.

When a reversible reaction takes place in a container, it eventually reaches dynamic equilibrium.

Key Characteristics of Dynamic Equilibrium
  1. Rate of forward reaction equals the rate of reverse reaction. (This is the "Dynamic" part—the reaction hasn't stopped, it's constantly moving both ways.)
  2. Concentrations of reactants and products remain constant. (This is the "Equilibrium" part—the amounts of A, B, C, and D are stable.)

Analogy: Think of a crowded escalator. If 10 people step onto the escalator every minute (forward rate), and 10 people step off at the top every minute (reverse rate), the total number of people on the escalator (concentration) stays constant, even though there is constant movement.

1.3 The Closed System Requirement

In order to establish dynamic equilibrium, the reaction must take place in a closed system.

A closed system is one where nothing can enter or leave the reaction vessel (especially gaseous reactants or products).

  • If the system were open, a gas might escape. This loss would reduce the concentration of that product, preventing the reverse reaction from ever reaching a rate equal to the forward reaction.
Quick Review: The Three Must-Haves for Dynamic Equilibrium
1. It must be a Reversible Reaction (\(\rightleftharpoons\)).
2. The reaction must be in a Closed System.
3. The rates must be Equal (Rate$_{\text{forward}}$ = Rate$_{\text{reverse}}$).

Key Takeaway for Part 1: Equilibrium is a balance of movement, not a standstill. The rates are balanced, which means the amounts (concentrations) are stable, but the reaction hasn't stopped.

Part 2: Controlling Equilibrium – Le Chatelier's Principle

2.1 Defining Le Chatelier's Principle (LCP)

If you disturb a system that is already at dynamic equilibrium, how does it react? It follows the rule laid down by Henry Le Chatelier.

Le Chatelier's Principle states: If a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.

Analogy: Think of LCP as the system being stubborn. If you push it one way, it pushes back the other way.

2.2 Effect of Concentration Changes

If you change the concentration of a reactant or product, the system shifts to try and restore balance.

Consider the reaction: \(A + B \rightleftharpoons C\)

  • If you increase the concentration of A: The system attempts to reduce the amount of A you added. It does this by favouring the forward reaction, shifting the equilibrium position to the right. This produces more C.
  • If you decrease the concentration of C (e.g., by removing it as soon as it forms): The system attempts to increase C to replace what was lost. It does this by favouring the forward reaction, shifting the equilibrium position to the right.

2.3 Effect of Pressure Changes

Pressure changes only affect reactions involving gases, and they rely on the number of moles of gas on each side of the equation.

Consider: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
Left side: 4 moles of gas (1 + 3). Right side: 2 moles of gas.

  • If you increase the pressure: The system attempts to reduce the pressure. It achieves this by favouring the side with fewer moles of gas.
    In the example above, the equilibrium shifts to the right (towards 2 moles).
  • If you decrease the pressure: The system attempts to increase the pressure. It favours the side with more moles of gas.
    In the example above, the equilibrium shifts to the left (towards 4 moles).

Important Note: If the number of moles of gas is the same on both sides (e.g., \(\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons 2\text{HI}(g)\)), changing the pressure has no effect on the position of equilibrium.

2.4 Effect of Temperature Changes

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

Consider the reaction: \(A + B \rightleftharpoons C\). Assume the forward reaction is exothermic (\(\Delta H \text{ is negative}\)).

  • Forward reaction: Exothermic (releases heat).
  • Reverse reaction: Endothermic (absorbs heat).
  • If you increase the temperature (add heat): The system tries to reduce the temperature by absorbing the extra heat. It favours the endothermic direction (the reverse reaction). The equilibrium shifts to the left.
  • If you decrease the temperature (remove heat): The system tries to replace the lost heat. It favours the exothermic direction (the forward reaction). The equilibrium shifts to the right.

2.5 Effect of a Catalyst

A catalyst speeds up a reaction by providing an alternative pathway with a lower activation energy.

  • A catalyst speeds up the forward reaction.
  • A catalyst speeds up the reverse reaction by the exact same amount.

Therefore, the presence of a catalyst does not affect the position of equilibrium. It simply allows the system to reach equilibrium much faster. Catalysts are incredibly important in industry for saving time and energy costs, even if they don't boost the final yield.

Common Mistake Alert!
Many students think adding a catalyst shifts the equilibrium. Remember: A catalyst helps you get to the destination faster, but it doesn't change where the destination (equilibrium position) is!

Key Takeaway for Part 2: Le Chatelier's Principle is the guide. The system moves to oppose the change you make, favouring one direction (forward or reverse) to relieve the stress.

Part 3: The Equilibrium Constant, K$_{c}$

3.1 Defining K$_{c}$

While LCP tells us which direction the equilibrium will shift, the equilibrium constant (\(K_c\)) tells us quantitatively where the equilibrium lies. It is a ratio that relates the concentration of products to the concentration of reactants at equilibrium.

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

The expression for the equilibrium constant in terms of concentration (\(K_c\)) is:
$$K_c = \frac{[C]^c [D]^d}{[A]^a [B]^b}$$

  • \([X]\) represents the equilibrium concentration of species X in \(\text{mol dm}^{-3}\).
  • The exponents (a, b, c, d) are the stoichiometric coefficients from the balanced equation.
  • We only include species that are variable in concentration. For heterogeneous equilibria, pure solids and pure liquids are omitted because their concentrations are constant. We only include (aq) and (g) species.

3.2 Interpreting the K$_{c}$ Value

The magnitude of \(K_c\) tells you about the position of equilibrium:

  • Large \(K_c\) (e.g., \(> 10^4\)): Products dominate. The equilibrium lies far to the right.
  • Small \(K_c\) (e.g., \(< 10^{-4}\)): Reactants dominate. The equilibrium lies far to the left.
  • \(K_c\) near 1: Significant amounts of both reactants and products are present.

3.3 Calculations Involving K$_{c}$

Calculating \(K_c\) involves finding the equilibrium concentrations of all species, usually done by tracking the initial concentrations and the change in concentration (using stoichiometry).

Crucial Calculation Note (Syllabus 7.1.7): Calculations using the \(K_c\) expression will not require the solving of quadratic equations. This means you will either be given the equilibrium concentrations directly, or the numbers will simplify nicely.

Key Takeaway for Part 3: \(K_c\) is a constant ratio of (Products)/(Reactants) at equilibrium. A large \(K_c\) means the reaction effectively goes to completion.

Part 4: Equilibrium Constant for Gases, K$_{p}$

4.1 Partial Pressure and Mole Fraction (Gases only)

When dealing only with gaseous reactions, it is often easier to use partial pressures instead of concentrations.

The total pressure of a mixture of gases is the sum of the pressures exerted by each individual gas. This individual pressure is called the partial pressure.

  • The mole fraction (\(x\)) of a gas is the fraction of the total moles in the mixture that belong to that specific gas.
    $$x_A = \frac{\text{moles of A}}{\text{total moles of gas}}$$
  • The partial pressure (\(P\)) of a gas is the mole fraction multiplied by the total pressure ($P_{total}$).
    $$P_A = x_A \times P_{total}$$

4.2 Defining K$_{p}$

The equilibrium constant in terms of partial pressure (\(K_p\)) is analogous to \(K_c\), but uses partial pressures instead of concentrations.

For the reaction: \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\)

The expression for \(K_p\) is:
$$K_p = \frac{(P_C)^c (P_D)^d}{(P_A)^a (P_B)^b}$$

Units of \(K_p\) (and \(K_c\)) depend on the stoichiometry. If the total moles of gas on the product side equals the total moles of gas on the reactant side, \(K_p\) will be dimensionless (have no units). Otherwise, units like \(\text{Pa}\), \(\text{kPa}\), or \(\text{atm}\) must be included and raised to the power of \((c+d) - (a+b)\).

4.3 Step-by-Step K$_{p}$ Calculation

To calculate \(K_p\) from initial moles and total pressure:

  1. Find Equilibrium Moles: Determine the moles of A, B, C, and D present at equilibrium.
  2. Find Total Moles: Sum the moles of all gases at equilibrium.
  3. Find Mole Fraction: Calculate the mole fraction (\(x\)) for each gas.
  4. Find Partial Pressure: Calculate the partial pressure (\(P\)) for each gas using \(P = x \times P_{total}\).
  5. Calculate K$_{p}$: Substitute the partial pressures into the \(K_p\) expression.

Remember: \(K_p\) expressions only include gaseous species!

Key Takeaway for Part 4: \(K_p\) is used for gas phase equilibrium. To find partial pressure, you must first calculate the mole fraction of each gas.

Part 5: Factors Affecting the VALUE of the Equilibrium Constant (K)

5.1 The One Factor That Changes K

This is a critical distinction that often confuses students. We learned that concentration, pressure, and catalysts shift the position of equilibrium (which species are favored). However, only one factor changes the numerical value of \(K_c\) or \(K_p\).

The numerical value of the equilibrium constant (K$_{c}$ or K$_{p}$) changes ONLY with a change in TEMPERATURE.

  • Changing concentration or pressure will shift the position of equilibrium, but the ratio of products to reactants (K) remains constant (as long as the temperature is fixed).
  • A catalyst speeds up the reaction but never changes the value of K.

5.2 How Temperature Affects K

The way temperature changes \(K\) links directly back to Le Chatelier's Principle:

Rule 1: If the forward reaction is EXOTHERMIC (\(\Delta H < 0\))

  • If you increase temperature, the equilibrium shifts left (towards reactants).
  • Since \(K = \frac{[Products]}{[Reactants]}\), if the products decrease and reactants increase, \(K\) decreases.

Rule 2: If the forward reaction is ENDOTHERMIC (\(\Delta H > 0\))

  • If you increase temperature, the equilibrium shifts right (towards products).
  • Since \(K = \frac{[Products]}{[Reactants]}\), if the products increase and reactants decrease, \(K\) increases.
Memory Aid: T is for Temperature, T is for TRUE Change
Only Temperature changes the numerical value of K (the constant). All other factors (C, P, Catalyst) only change the concentration/ratio until K is restored.

Key Takeaway for Part 5: If you want to change the equilibrium constant (K), you must change the temperature. For exothermic reactions, high T means lower K. For endothermic reactions, high T means higher K.

Part 6: Equilibrium in Industry – Haber and Contact Processes

Understanding equilibrium isn't just theory; it's essential for industry to maximize yield and profit. Industrial chemists must find a compromise between high yield (from LCP) and a fast rate (from kinetics/catalysis).

6.1 The Haber Process (Ammonia Production)

The Haber process produces ammonia (\(\text{NH}_3\)), vital for fertilizers, from nitrogen and hydrogen.

$$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g) \quad \Delta H = -92 \text{ kJ mol}^{-1}$$

The forward reaction is exothermic and involves a decrease in the number of moles of gas (4 moles $\rightarrow$ 2 moles).

Optimizing Conditions via Le Chatelier’s Principle (LCP)
  • Effect of Temperature: Since the forward reaction is exothermic, LCP predicts that a low temperature favours the formation of products (shifts right).
    Industrial Compromise: While low temperature gives high yield, it results in a very slow reaction rate. Therefore, a compromise temperature of around 400–450 °C is used to achieve a fast rate and an acceptable yield.
  • Effect of Pressure: Since the product side has fewer moles of gas (2 moles), LCP predicts that a high pressure favours the formation of products (shifts right).
    Industrial Compromise: Very high pressure is expensive and dangerous. A pressure of about 150–250 atmospheres (15–25 MPa) is used to achieve a good yield without excessive cost.
  • Catalyst: An iron catalyst is used to increase the rate of reaction, ensuring that the equilibrium position is reached quickly, but it does not affect the final yield.

6.2 The Contact Process (Sulfuric Acid Production)

The key equilibrium step in the Contact Process is the oxidation of sulfur dioxide to sulfur trioxide, which is then converted into sulfuric acid.

$$2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) \quad \Delta H = -197 \text{ kJ mol}^{-1}$$

The forward reaction is exothermic and involves a decrease in the number of moles of gas (3 moles $\rightarrow$ 2 moles).

Optimizing Conditions via Le Chatelier’s Principle (LCP)
  • Effect of Temperature: Since the reaction is exothermic, LCP favours low temperature for high yield.
    Industrial Compromise: A temperature of about 400–450 °C is used, similar to Haber, to maintain an industrially viable rate, despite the slight decrease in equilibrium yield.
  • Effect of Pressure: Since the product side has fewer moles (2 moles), LCP favours high pressure for high yield.
    Industrial Compromise: Since the yield is already over 99% at moderate pressures, a pressure of only 1–2 atmospheres is used, saving energy and equipment costs.
  • Catalyst: A Vanadium(V) oxide (\(\text{V}_2\text{O}_5\)) catalyst is used to achieve a rapid rate.

Key Takeaway for Part 6: Industrial processes always involve a compromise: high yield (favoured by low T/high P for exothermic reactions) versus high rate (favoured by high T/catalysts). The final conditions are chosen to achieve the best profit, not necessarily the highest possible yield.