Chemical reactions - Reversible reactions and equilibrium (3)
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1.
Describe the reaction that occurs in the Contact Process to convert sulfur dioxide to sulfur trioxide. State the symbol equation for this reaction. What effect does increasing the temperature have on the equilibrium position of this reaction?
In the Contact Process, sulfur dioxide reacts with oxygen in the presence of a catalyst to produce sulfur trioxide. The reaction is reversible.
The symbol equation for the reaction is:
- 2 SO2(g) + O2(g) ⇌ 2 SO3(g)
Increasing the temperature will shift the equilibrium to the right (endothermic reaction). This means that more sulfur trioxide will be produced. However, very high temperatures can lead to the decomposition of sulfur trioxide back into sulfur dioxide and oxygen, so the temperature is carefully controlled to optimize the yield of sulfur trioxide.
2.
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.
3.
Consider the following reversible reaction:
$N_2(g) + 3H_2(g) ⇌ 2NH_3(g)$
State, with a clear explanation, how changing the pressure of the gaseous reactants and products would affect the position of the equilibrium. Explain your reasoning using Le Chatelier's principle.
Changing the pressure of the gaseous reactants and products will affect the position of the equilibrium according to Le Chatelier's principle.
- Increasing the pressure: Increasing the pressure favors the side of the reaction with fewer moles of gas. In this case, the left side (forward reaction) has 4 moles of gas (1 mole N2 + 3 moles H2), while the right side (reverse reaction) has 2 moles of gas (2 moles NH3). Therefore, increasing the pressure will shift the equilibrium to the right, favoring the formation of ammonia (NH3). This is because the system will try to reduce the pressure by producing fewer gas molecules.
- Decreasing the pressure: Decreasing the pressure favors the side of the reaction with more moles of gas. As stated above, the forward reaction has more moles of gas. Therefore, decreasing the pressure will shift the equilibrium to the left, favoring the decomposition of ammonia back into nitrogen and hydrogen. This is because the system will try to increase the pressure by producing more gas molecules.