State that some chemical reactions are reversible as shown by the symbol $ ightleftharpoons$

Chemical Reactions – Reversible Reactions and Equilibrium (IGCSE 0620)

1. What is a reversible reaction?

• A chemical change that can proceed forward (reactants → products) and reverse (products → reactants).
• It is written with a double‑arrow: \( \rightleftharpoons \) to show that the products can react to reform the reactants.

2. General representation

$$\text{Reactants} \; \rightleftharpoons \; \text{Products}$$

Example – synthesis of ammonia (Haber process):

$$\mathrm{N_2(g) + 3\,H_2(g) \; \rightleftharpoons \; 2\,NH_3(g)}$$

3. Dynamic equilibrium

• In a closed system the forward and reverse reactions continue to occur.
• When the two rates become equal, the concentrations of all species stop changing – this is dynamic equilibrium.
• At equilibrium:
  • Rate forward = Rate reverse
  • Concentrations are constant (they need not be equal).

3.1 Graphical illustration (Supplement 6.3.2)

How to use the graph – Students should be able to state at which point the curve becomes straight (no further change) and explain why this represents dynamic equilibrium.

4. Reversible reactions involving hydrated salts (Core 6.3.1)

Many ionic salts exist as hydrates. Heating removes water as vapour; adding liquid water reforms the hydrate. Both directions are shown with the double‑arrow.

• Heating drives the reaction to the **anhydrous side** (left of the double‑arrow) because water is removed as vapour.
• Adding water drives the reaction to the **hydrated side** (right of the double‑arrow) as the system absorbs the added water.

5. Le Chatelier’s principle – how conditions shift equilibrium (Supplement 6.3.2)

When a system at equilibrium is disturbed, it responds so as to minimise the disturbance.

5.1 Example of the pressure rule (required for the syllabus)

For the Haber process:

$$\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}$$

4 mol of gas on the left → 2 mol on the right. Increasing pressure therefore shifts the equilibrium **right** (towards fewer gas molecules).

6. Interpreting equilibrium data (Quantitative/graphical – Supplement 6.3.2)

Students are often asked to read a simple concentration‑vs‑time diagram and answer questions such as:

1. At what time does equilibrium occur? (the point where the curve becomes straight)
2. Which direction is dominant before equilibrium? (compare the slopes of reactant and product curves)
3. How would the curve change if the temperature were increased for an exothermic forward reaction? (the equilibrium point would move left; the product curve would fall, the reactant curve would rise).

7. Industrial examples (Core 6.3.2)

7.1 Haber process – ammonia synthesis

$$\mathrm{N_2(g) + 3\,H_2(g) \rightleftharpoons 2\,NH_3(g)}\qquad \Delta H = -92\ \text{kJ mol}^{-1}$$

• Forward reaction is exothermic → raising temperature shifts the equilibrium **left**.
• 4 mol (gas) → 2 mol (gas). High pressure (e.g., 150 atm) shifts the equilibrium **right**.
• Iron catalyst speeds up the approach to equilibrium but does **not** change its position.

7.2 Contact process – sulphuric acid production

Key step:

$$\mathrm{2\,SO_2(g) + O_2(g) \rightleftharpoons 2\,SO_3(g)}\qquad \Delta H = -198\ \text{kJ mol}^{-1}$$

• Exothermic forward reaction → increasing temperature shifts the equilibrium **left**.
• 3 mol (gas) → 2 mol (gas). Raising pressure therefore shifts the equilibrium **right**.
• Vanadium(V) oxide (V₂O₅) acts as a catalyst; it does not affect the equilibrium position.

8. Summary table – factors that affect equilibrium position

9. Quick experimental design idea (to show a catalyst does not shift equilibrium)

Place two identical sealed tubes containing a coloured equilibrium system (e.g., CuSO₄·5H₂O ⇌ CuSO₄ + water vapour). Add a solid catalyst to one tube only. Record the time taken for the colour to stop changing. The tube with the catalyst reaches equilibrium faster, but the final colour intensity (i.e., the equilibrium position) is the same in both tubes.

10. Key points to remember

1. Reversible reactions are written with the double‑arrow \( \rightleftharpoons \).
2. Dynamic equilibrium means equal forward and reverse rates; concentrations become constant.
3. Le Chatelier’s principle predicts the *direction* of shift when concentration, temperature, pressure or a catalyst is changed. The temperature effect depends on whether the forward reaction is endothermic (ΔH > 0) or exothermic (ΔH < 0).
4. For hydrated salts, heating drives the reaction to the anhydrous side; adding water drives it to the hydrated side (CuSO₄·5H₂O and CoCl₂·6H₂O are the required examples).
5. Industrial processes (Haber and Contact) illustrate how manufacturers manipulate temperature, pressure and catalysts to obtain maximum yield while respecting Le Chatelier’s principle.
6. Students must be able to read a concentration‑vs‑time graph, identify the point of equilibrium, and explain how a change in conditions would alter the curve.