To answer this question, we need to analyze each reaction and determine the impact of pressure changes on the equilibrium position according to Le Chatelier’s Principle, which states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.
1. Reaction 1:
[tex]\[ \text{CH}_4(g) + 2 \text{O}_2(g) \leftrightarrows \text{CO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]
- On the reactants side: 1 mole of CH₄ + 2 moles of O₂ = 3 moles of gas.
- On the products side: 1 mole of CO₂ + 2 moles of H₂O = 3 moles of gas.
Since both sides of the reaction have the same number of moles of gas, changing the pressure will not affect the equilibrium position. Hence, decreasing the pressure will not shift the equilibrium to the right for this reaction.
2. Reaction 2:
[tex]\[ \text{CaCO}_3(s) \leftrightarrows \text{CaO}(s) + \text{CO}_2(g) \][/tex]
- On the reactants side: 1 mole of solid (CaCO₃) which doesn't change with pressure.
- On the products side: 1 mole of solid (CaO) + 1 mole of gas (CO₂).
Solids are not affected by changes in pressure, so we only consider the gaseous component. Decreasing the pressure would favor the side with more moles of gas. Here, there's only one mole of gas on the products side, which would increase under lower pressure, shifting the equilibrium to the right.
3. Reaction 3:
[tex]\[ \text{Br}_2(g) + 3 \text{Cl}_2(g) \rightarrow 2 \text{BrCl}_3(g) \][/tex]
- This reaction is irreversible, meaning that the equilibrium cannot shift because there is no reverse reaction. Therefore, pressure changes will not affect the equilibrium position at all.
4. Reaction 4:
[tex]\[ 2 \text{H}_2\text{S}(g) + 3 \text{O}_2(g) \leftrightarrows 2 \text{SO}_2(g) + 2 \text{H}_2\text{O}(g) \][/tex]
- On the reactants side: 2 moles of H₂S + 3 moles of O₂ = 5 moles of gas.
- On the products side: 2 moles of SO₂ + 2 moles of H₂O = 4 moles of gas.
Decreasing the pressure would favor the side with fewer moles of gas. In this reaction, there are fewer moles of gas on the products side (4 moles) than on the reactants side (5 moles). Therefore, decreasing the pressure would shift the equilibrium to the right.
Based on the analysis above, the reactions where the equilibrium can be shifted to the right by decreasing the pressure are:
[tex]\[
\boxed{2 \text{ and } 4}
\][/tex]
