Reversible reactions and equilibrium
Reversible Reactions: Some chemical reactions can proceed in both forward and reverse directions. This is indicated by the symbol \rightleftharpoons.
Dynamic Equilibrium: In a closed system, a reversible reaction reaches equilibrium when:
- The rate of the forward reaction equals the rate of the reverse reaction.
- The concentrations (or amounts) of reactants and products remain constant (but not necessarily equal).
Le Chatelier’s Principle: If conditions change, the position of equilibrium shifts to counteract the change.
- Temperature: Shifts towards the endothermic direction.
- Pressure: Shifts towards the side with fewer gas moles.
- Concentration: Shifts away from the added substance.
- Catalyst: No effect on position; only speeds up reaching equilibrium.
Example:
\text{Hydrated copper(II) sulfate} \rightleftharpoons \text{Anhydrous copper(II) sulfate} + \text{Water}
\text{CuSO}_4 \cdot 5\text{H}_2\text{O}(s) \rightleftharpoons \text{CuSO}_4(s) + 5\text{H}_2\text{O}(l)
- The rate of the forward reaction is equal to the rate of the reverse reaction.
- The concentrations of reactants and products are no longer changing.
Note: 'Closed system' means no matter can enter or leave. This is essential because if products escape, the reverse reaction cannot occur, and equilibrium is never established.
1. Copper(II) Sulfate:
- Forward (Dehydration): Heating blue hydrated \text{CuSO}_4 \cdot 5\text{H}_2\text{O} produces white anhydrous \text{CuSO}_4 and steam. This is endothermic.
- Reverse (Hydration): Adding water to white anhydrous \text{CuSO}_4 produces blue hydrated crystals and releases heat (exothermic).
2. Cobalt(II) Chloride:
- Forward: Heating pink hydrated \text{CoCl}_2 \cdot 6\text{H}_2\text{O} produces blue anhydrous \text{CoCl}_2.
- Reverse: Adding water to blue anhydrous \text{CoCl}_2 produces pink hydrated crystals.
| Contact Process (Sulfur Trioxide) |
|---|
| 2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g) |
| Sulfur dioxide from burning sulfur/roasting ores; Oxygen from air |
| 450^\circ\text{C} |
| 200 \text{ kPa} (2 \text{ atm}) |
| Vanadium(V) oxide (\text{V}_2\text{O}_5) |
Correction: At equilibrium, the rates are equal, but concentrations are constant (not necessarily equal). The position of equilibrium may favor products or reactants.
Mistake: Assuming a catalyst shifts the equilibrium position.
Correction: A catalyst increases the rate of both forward and reverse reactions equally. It helps reach equilibrium faster but does not change the yield or position.
Context: When asked to explain why specific conditions are used in the Haber or Contact process.
Reasoning: Examiners look for a 'compromise' explanation balancing rate (kinetics) and yield (equilibrium).
Correct Phrasing Example:
- Temperature (450^\circ\text{C}): This is a compromise. Lower temperatures would give a higher yield (since forward reaction is exothermic), but the rate would be too slow. Higher temperatures increase the rate but decrease the yield. 450^\circ\text{C} provides a reasonable rate and acceptable yield.
- Pressure:
- Haber: High pressure (200 \text{ atm}) increases yield (fewer moles on right) and rate. However, very high pressures are dangerous and expensive to maintain. 200 \text{ atm} is a safe economic compromise.
- Contact: Low pressure (2 \text{ atm}) is used because the yield is already >98% at atmospheric pressure. Increasing pressure offers negligible benefit but high cost.