Describe how changing the conditions can change the direction of a reversible reaction for: (a) the effect of heat on hydrated compounds (b) the addition of water to anhydrous compounds limited to copper(II) sulfate and cobalt(II) chloride

Chemical Reactions – Reversible Reactions and Equilibrium (IGCSE 0620)

Learning Objective

Key‑phrase (exact syllabus wording): “Changing the temperature or the concentration of water shifts the equilibrium to the right or left, respectively.”

Describe how changing the conditions can change the direction of a reversible reaction for:

• the effect of heat on hydrated compounds, and
• the addition of water to anhydrous compounds (limited to copper(II) sulfate and cobalt(II) chloride).

1. Reversible Reactions and Le Chatelier’s Principle

• A reversible reaction is written with a double arrow ⇌ and can proceed in both directions.
• At equilibrium the forward and reverse rates are equal; concentrations of reactants and products remain constant.
• Le Chatelier’s principle: If a stress (temperature, concentration, pressure, …) is applied to a system at equilibrium, the system shifts in the direction that counteracts the stress.

2. Quick Reference Box – CuSO₄ & CoCl₂

3. Effect of Heat on Hydrated Compounds

Dehydration of a hydrate is an endothermic process; heat is a reactant. Adding heat therefore drives the equilibrium to the right (towards the anhydrous salt + water vapour). Cooling removes this reactant and drives the equilibrium to the left (rehydration).

3.1 General form

\$\ce{M·nH2O(s) ⇌ M(s) + nH2O(g)}\$

where M·nH2O is the hydrate and M the anhydrous salt.

3.2 Observable changes

3.3 Example – Copper(II) sulfate pentahydrate

\$\ce{CuSO4·5H2O(s) ⇌ CuSO4(s) + 5H2O(g)}\$

• Heating above ≈ 150 °C supplies the required heat (reactant) → shift right → white CuSO₄.
• Cooling or placing the white solid in a humid environment supplies water vapour → shift left → blue CuSO₄·5H₂O.

3.4 Example – Cobalt(II) chloride hexahydrate

\$\ce{CoCl2·6H2O(s) ⇌ CoCl2(s) + 6H2O(g)}\$

• Heating above ≈ 150 °C drives the reaction right → blue anhydrous CoCl₂.
• Adding moisture drives it left → pink hexahydrate.

4. Effect of Adding Water to Anhydrous Compounds

In the hydrate‑formation equation water appears on the product side. Increasing the concentration of water therefore constitutes a stress that shifts the equilibrium to the right, producing the coloured hydrate. Removing water (by heating or using a desiccant) shifts the equilibrium to the left, giving the colourless/white anhydrous form.

4.1 Copper(II) sulfate

\$\ce{CuSO4(s) + 5H2O(l) ⇌ CuSO4·5H2O(s)}\$

• Adding a few drops of water: shift right → blue pentahydrate forms.
• Removing water (heating): shift left → white anhydrous CuSO₄.

4.2 Cobalt(II) chloride

\$\ce{CoCl2(s) + 6H2O(l) ⇌ CoCl2·6H2O(s)}\$

• Adding water: shift right → pink hexahydrate appears.
• Removing water (heating): shift left → blue anhydrous CoCl₂.

5. Summary of How Conditions Influence Direction

• Heat: Dehydration reactions are endothermic. Adding heat supplies a reactant, so the equilibrium shifts right (to the anhydrous salt). Cooling removes that reactant, shifting left (rehydration).
• Water concentration: Adding water increases the concentration of a product in the hydrate‑formation equation, so the equilibrium shifts right, producing the coloured hydrate. Removing water shifts left, giving the anhydrous form.
• Observable indicators: colour change, change in mass, evolution or absorption of water vapour.
• Magnitude of shift: Large temperature changes or a clear excess of water give an almost complete shift; small changes give only a partial shift (e.g., a pale‑blue mixture of CuSO₄·5H₂O and CuSO₄).

6. Extended Content – General Reversible Reactions (Syllabus 6.3.2)

Beyond hydrates, many industrial processes are reversible and are governed by Le Chatelier’s principle.

6.1 Example – Haber Process (N₂ + 3H₂ ⇌ 2NH₃)

• Temperature: Exothermic (heat released). Raising temperature shifts left; lowering temperature shifts right.
• Pressure: More moles of gas on the left (4) than on the right (2). Increasing pressure shifts right.
• Concentration: Adding N₂ or H₂ shifts right; removing NH₃ shifts right.

6.2 Example – Contact Process (2SO₂ + O₂ ⇌ 2SO₃)

• Exothermic; lower temperature favours SO₃ formation.
• Decrease volume (increase pressure) favours side with fewer gas moles (right side).
• Removing SO₃ (e.g., by absorption in water) drives the reaction to the right.

7. Practice Questions (AO1 & AO2)

1. When a sample of blue CuSO₄·5H₂O is heated strongly, it turns white. Explain the colour change using Le Chatelier’s principle and state whether the reaction is endothermic or exothermic.
2. A dry sample of CoCl₂ is placed in a test tube and a few drops of water are added. Predict the colour change, write the balanced reversible equation (including state symbols), and state the direction of the equilibrium shift.
3. In a sealed container, CuSO₄·5H₂O is heated until it becomes anhydrous. The container is then cooled to room temperature. Describe what will happen to the solid and why, referring to the equilibrium shift and observable changes.
4. Explain why a slight increase in temperature may only cause a pale‑blue mixture of CuSO₄·5H₂O and CuSO₄, whereas a large increase produces a completely white solid.
5. For the Haber process, predict the effect on ammonia yield of (a) raising the temperature, (b) increasing the pressure, and (c) removing ammonia as it forms.

8. Suggested Diagrams for Teachers

• Test‑tube sketch: blue CuSO₄·5H₂O → white CuSO₄ on heating, with arrows indicating water vapour loss.
• Two test‑tube sketch: (i) anhydrous CuSO₄ (white) turning blue after a drop of water; (ii) anhydrous CoCl₂ (blue) turning pink after water is added.
• Pressure‑volume diagram for the Haber process illustrating the shift toward fewer gas moles under increased pressure.