To determine the impact of increasing pressure on the equilibrium of the chemical reaction
[tex]\[ \text{CH}_4(g) + \text{H}_2\text{O}(g) \Leftrightarrow \text{CO}(g) + 3\text{H}_2(g) \][/tex]
we need to apply Le Chatelier's principle. Le Chatelier's principle states that when a system at equilibrium is subjected to a change in pressure, the system will adjust itself to counteract the change and reestablish equilibrium. Specifically, if the pressure is increased, the equilibrium shifts towards the side with fewer moles of gas, thereby reducing the pressure.
Let's analyze the number of moles of gas on each side of the reaction:
- Reactants side (left side):
- CH[tex]\(_4\)[/tex](g): 1 mole
- H[tex]\(_2\)[/tex]O(g): 1 mole
- Total moles of gas on the reactants side: [tex]\(1 + 1 = 2\)[/tex] moles
- Products side (right side):
- CO(g): 1 mole
- 3H[tex]\(_2\)[/tex](g): 3 moles
- Total moles of gas on the products side: [tex]\(1 + 3 = 4\)[/tex] moles
When the pressure increases, the equilibrium will shift towards the side with fewer moles of gas to reduce the pressure. By comparing the total moles of gas on both sides:
- Reactants side: 2 moles
- Products side: 4 moles
Since there are fewer moles of gas on the reactants side (2 moles) compared to the products side (4 moles), the equilibrium will shift towards the reactants side (left side).
Therefore, the equilibrium will shift to the left to favor the reverse reaction.
The correct answer is:
The equilibrium will shift to the left to favor the reverse reaction.
