Welcome to Chemical Equilibrium!
Welcome to one of the most exciting and essential topics in AS Chemistry! In your previous chemistry studies, you might have thought that chemical reactions only go in one direction until the reactants run out. However, many chemical reactions are two-way streets: reactants turn into products, and products turn back into reactants.
Understanding equilibrium is vital not only for scoring top marks in your CCEA AS 2 exam, but also for understanding how the global chemical industry produces millions of tonnes of fertilisers, medicines, and plastics efficiently and safely.
Don't worry if this chapter seems tricky at first! We will break down every concept step-by-step, using simple analogies and clear rules so you can master equilibrium with total confidence.
1. Reversible Reactions and Dynamic Equilibrium
What is a Reversible Reaction?
A reversible reaction is one where the forward reaction (reactants forming products) and the reverse reaction (products forming reactants) happen at the same time.
We represent reversible reactions using the dynamic equilibrium arrow: \(\rightleftharpoons\)
General equation: \(aA + bB \rightleftharpoons cC + dD\)
What is Dynamic Equilibrium?
Imagine walking up a "down" escalator. If your walking speed upwards matches the downward speed of the escalator exactly, you remain in the exact same spot! To an outside observer, you look completely still, but continuous activity is happening under your feet.
This is exactly what happens in a chemical system at dynamic equilibrium.
For a system to achieve dynamic equilibrium, two strict conditions must be met:
- Closed System: No matter or energy can enter or leave the reaction vessel (e.g., a sealed flask with a stopper).
- Equal Rates: The rate of the forward reaction equals the rate of the reverse/backward reaction.
Key Features of a Dynamic Equilibrium
- Rates: The rate of the forward reaction equals the rate of the reverse reaction (\(\text{rate}_{\text{forward}} = \text{rate}_{\text{reverse}}\)).
- Concentrations: The concentrations of all reactants and products remain constant (they do not change over time, but they are usually not equal to each other).
- Macroscopic properties: Macroscopic properties (such as colour, pressure, density, and pH) remain constant.
- Continuity: The reaction has not stopped; both forward and reverse processes are ongoing at the molecular level.
Did You Know?
A common misconception is thinking that dynamic equilibrium means there are equal amounts of reactants and products. In reality, concentrations are simply constant, not necessarily equal. You might have \(95\%\) products and \(5\%\) reactants, or vice-versa!
Key Takeaway for Section 1
Dynamic equilibrium occurs in a closed system when the rate of the forward reaction equals the rate of the reverse reaction, causing the concentrations of reactants and products to remain constant.
2. Le Chatelier's Principle
Once a system reaches equilibrium, what happens if we change the temperature, pressure, or concentration? The French chemist Henri Le Chatelier figured this out!
Definition: Le Chatelier's Principle
"If a system at dynamic equilibrium is subjected to a change in conditions, the position of equilibrium will shift in the direction that opposes the change."
Analogy: Think of equilibrium as a stubborn teenager. Whatever you do, it does the exact opposite! If you heat it up, it tries to cool down. If you squeeze it, it tries to relieve the pressure.
A. Effect of Changing Concentration
- Increasing reactant concentration: The system opposes this by using up the added reactant. The equilibrium shifts to the right (producing more products).
- Decreasing reactant concentration: The system opposes this by making more reactant. The equilibrium shifts to the left.
- Removing product: The system opposes this by making more product to replace what was lost. The equilibrium shifts to the right.
B. Effect of Changing Pressure (Gaseous Equilibria Only)
Important Prerequisite: Pressure only affects reactions where at least one reactant or product is a gas (\(g\)). Pressure is determined by counting the total moles of gas on each side of the equation.
- Increasing pressure: The system shifts to the side with fewer moles of gas to reduce the pressure.
- Decreasing pressure: The system shifts to the side with more moles of gas to increase the pressure.
- Equal moles of gas on both sides: Changing pressure has no effect on the position of equilibrium.
Example: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
Left side = \(1 + 3 = 4\text{ moles of gas}\)
Right side = \(2\text{ moles of gas}\)
Increasing pressure shifts the equilibrium to the right (fewer gas moles).
C. Effect of Changing Temperature
To predict temperature effects, always check the enthalpy change sign (\(\Delta H\)) for the forward reaction:
- If \(\Delta H\) is negative (\(-\)), the forward reaction is exothermic (gives out heat).
- If \(\Delta H\) is positive (\(+\)), the forward reaction is endothermic (absorbs heat).
Rules for temperature changes:
- Increasing temperature: The system opposes the increase by absorbing heat. Equilibrium shifts in the endothermic direction.
- Decreasing temperature: The system opposes the decrease by releasing heat. Equilibrium shifts in the exothermic direction.
Example: \(N_2O_4(g) \rightleftharpoons 2NO_2(g)\) \(\quad \Delta H = +58\text{ kJ mol}^{-1}\) (Endothermic forward)
- Increasing temperature shifts equilibrium to the right (endothermic), producing more brown \(NO_2(g)\).
- Decreasing temperature shifts equilibrium to the left (exothermic), producing more colourless \(N_2O_4(g)\).
D. Effect of Adding a Catalyst
A catalyst increases the rate of both the forward and reverse reactions equally by providing an alternative pathway with a lower activation energy (\(E_a\)).
Therefore, a catalyst:
- Has no effect on the position of equilibrium.
- Has no effect on the yield of product.
- Simply allows equilibrium to be reached faster.
Common Mistakes to Avoid
- Mistake 1: Stating that a catalyst increases the yield. (Correction: It only increases the rate of reaching equilibrium).
- Mistake 2: Counting solid or liquid moles when evaluating pressure changes. (Correction: Only count gas moles!).
- Mistake 3: Forgetting to state which reaction direction is endothermic or exothermic in exam explanations.
Key Takeaway for Section 2
Equilibrium shifts to counteract applied changes: higher pressure favours the side with fewer gas moles; higher temperature favours the endothermic direction; adding a catalyst changes only the rate, not the equilibrium position.
3. The Equilibrium Constant (\(K_c\))
While Le Chatelier's principle tells us qualitative directions (left or right), the equilibrium constant (\(K_c\)) gives us exact mathematical values for the position of equilibrium at a specific temperature.
Writing the \(K_c\) Expression
For a general homogeneous reaction in the liquid or gaseous state:
\(aA + bB \rightleftharpoons cC + dD\)
The equilibrium expression is written as:
\(K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}\)
Where:
- \([A], [B], [C], [D]\) represent equilibrium concentrations in \(\text{mol dm}^{-3}\).
- Superscript powers \(a, b, c, d\) are the stoichiometric balancing numbers from the balanced equation.
- Convention: Products always go on the top (numerator), reactants go on the bottom (denominator).
Step-by-Step Guide to Calculating \(K_c\) (The "RICE" Method)
When solving exam problems, use a structured RICE table:
- R (Reaction): Write the balanced chemical equation.
- I (Initial moles): Write the starting moles given in the question.
- C (Change in moles): Use mole ratios to determine how moles change as equilibrium is established.
- E (Equilibrium moles): Calculate final moles at equilibrium (\(\text{Initial} \pm \text{Change}\)).
- Divide by Volume: Divide equilibrium moles by volume (\(V\) in \(\text{dm}^3\)) to get concentrations (\(\text{mol dm}^{-3}\)).
- Substitute into \(K_c\): Plug numbers into the expression and solve.
Worked Example: Determining Units of \(K_c\)
Units for \(K_c\) are not universal; they vary depending on the reaction stoichometry!
Example: Find the units of \(K_c\) for \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\)
1. Write out the expression with concentration units:
\(K_c = \frac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}] \times [\text{mol dm}^{-3}]^3} = \frac{[\text{mol dm}^{-3}]^2}{[\text{mol dm}^{-3}]^4}\)
2. Cancel terms mathematically:
\(K_c = \frac{1}{[\text{mol dm}^{-3}]^2} = [\text{mol dm}^{-3}]^{-2} = \text{mol}^{-2}\text{dm}^6\)
What Factors Affect the Value of \(K_c\)?
This is a major CCEA exam favourite:
- Temperature: ONLY temperature changes the numerical value of \(K_c\).
- For an exothermic reaction: Increasing temperature decreases \(K_c\) (equilibrium shifts left).
- For an endothermic reaction: Increasing temperature increases \(K_c\) (equilibrium shifts right).
- Concentration changes: Do NOT change \(K_c\).
- Pressure changes: Do NOT change \(K_c\).
- Catalysts: Do NOT change \(K_c\).
Key Takeaway for Section 3
\(K_c\) is calculated as \(\frac{[\text{products}]}{[\text{reactants}]}\) at equilibrium. Only temperature changes the numerical value of \(K_c\); changes in concentration, pressure, or adding a catalyst leave \(K_c\) unchanged.
4. Industrial Applications: Compromise Conditions
In large-scale industrial chemistry, companies want to make products as quickly and as cheaply as possible while obtaining a high yield.
Case Study 1: The Haber Process
Equation: \(N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\) \(\quad \Delta H = -92\text{ kJ mol}^{-1}\)
1. Temperature Analysis:
- Equilibrium yield perspective: Since the forward reaction is exothermic, a low temperature produces the highest equilibrium yield of \(NH_3\).
- Rate perspective: A low temperature gives a very slow rate of reaction because fewer particles have energy \(\ge E_a\).
- The Compromise: A temperature of approximately \(400 - 450^\circ\text{C}\) is used. It gives a reasonably fast rate with an acceptable yield.
2. Pressure Analysis:
- Equilibrium yield perspective: \(4\text{ moles of gas} \rightarrow 2\text{ moles of gas}\). A high pressure favours the forward reaction, increasing yield.
- Rate perspective: High pressure increases particle collision frequency, speeding up the rate.
- The Compromise / Safety & Cost: Extremely high pressure requires thick, reinforced pipes, expensive pumping equipment, and poses severe safety risks.
- A pressure of approximately \(200\text{ atm}\) (\(20\text{ MPa}\)) is chosen as an economic compromise.
3. Catalyst:
- An iron (\(Fe\)) catalyst is used to allow the reaction to proceed at a commercially viable rate at this compromise temperature.
Case Study 2: The Contact Process (Stage 2)
Equation: \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\) \(\quad \Delta H = -196\text{ kJ mol}^{-1}\)
- Temperature: \(\approx 450^\circ\text{C}\) (A compromise between high yield and practical reaction rate).
- Pressure: \(\approx 1 - 2\text{ atm}\). Why not higher? Even at just \(1 - 2\text{ atm}\), the equilibrium yield of \(SO_3\) is already over \(98\%\)! Applying higher pressure would add unnecessary equipment and running costs for negligible extra yield.
- Catalyst: Vanadium(V) oxide (\(V_2O_5\)).
Summary of Industrial Conditions Table
- Haber Process (\(NH_3\)): \(\approx 400 - 450^\circ\text{C}\), \(\approx 200\text{ atm}\), \(\text{Iron }(Fe)\) catalyst.
- Contact Process (\(SO_3\)): \(\approx 450^\circ\text{C}\), \(\approx 1 - 2\text{ atm}\), \(\text{Vanadium(V) oxide }(V_2O_5)\) catalyst.
Key Takeaway for Section 4
Industrial conditions are an economic compromise balancing reaction rate (kinetics), equilibrium yield (thermodynamics), and operating costs/safety.
Quick Review Checklist
Before sitting your exam, make sure you can:
- Define dynamic equilibrium and list its key characteristics.
- State Le Chatelier's Principle accurately word-for-word.
- Predict shifts caused by changing temperature, pressure, and concentration.
- Write expressions for \(K_c\) and calculate equilibrium concentrations and units.
- Explain why only temperature changes \(K_c\).
- Explain the choice of compromise conditions for the Haber Process and Contact Process.