Chemical reactions - Reversible reactions and equilibrium (3)
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1.
Describe the role of oxygen in the Contact Process. What would happen if the supply of oxygen was insufficient?
Oxygen (O2) is a reactant in the Contact Process. It reacts with sulfur dioxide (SO2) to produce sulfur trioxide (SO3). The overall reaction is:
SO2(g) + ½O2(g) → SO3(g)
If the supply of oxygen was insufficient, the reaction would not proceed efficiently. The amount of SO3 produced would be limited by the amount of oxygen available. This would result in a lower yield of sulfuric acid and a slower reaction rate. The process might even stall completely if there is no oxygen.
2.
The Haber process is used to produce ammonia (NH3) from nitrogen (N2) and hydrogen (H2). State the typical conditions used in the Haber process to maximise the yield of ammonia.
The typical conditions used in the Haber process are:
- Temperature: Approximately 450 °C. This temperature is high enough to allow the reaction to occur at a reasonable rate, but not so high that the equilibrium is shifted too far to the left.
- Pressure: Approximately 20,000 kPa (or 200 atm). High pressure favours the formation of ammonia because there are fewer moles of gas on the product side (left side of the equation).
- Catalyst: An iron catalyst is used. This speeds up the reaction rate without being consumed itself. Iron promotes the breaking and forming of the N-H bonds.
3.
Consider the reaction between copper(II) sulfate (CuSO₄) and water. Describe how the addition of water to anhydrous copper(II) sulfate can be influenced by changing the temperature. Explain your answer using Le Chatelier's principle.
The reaction between anhydrous copper(II) sulfate (CuSO₄) and water is an exothermic process, forming hydrated copper(II) sulfate. The equation is: CuSO₄(s) + 5H₂O(l) ⇌ [Cu(H₂O)₆]²⁺(aq) + 4H₂O(l).
Increasing the temperature will shift the equilibrium towards the formation of hydrated copper(II) sulfate. This is because the reaction is exothermic (ΔH is negative). Adding heat to the system is equivalent to adding a product, and the equilibrium will shift to favour the reactants (anhydrous CuSO₄ and water) to consume the excess heat and re-establish equilibrium. This results in more anhydrous copper(II) sulfate reacting with water to form hydrated copper(II) sulfate.
Decreasing the temperature will shift the equilibrium towards the formation of anhydrous copper(II) sulfate. This is because the reaction is exothermic. Removing heat from the system is equivalent to removing a product, and the equilibrium will shift to favour the products (hydrated copper(II) sulfate) to replace the lost heat. This results in some hydrated copper(II) sulfate losing water molecules and becoming anhydrous copper(II) sulfate.
Therefore, temperature is a key factor in controlling the hydration of copper(II) sulfate, and Le Chatelier's principle explains how changing the temperature can drive the reaction towards either the anhydrous or hydrated state.