An increase in pressure affects a chemical equilibrium involving gases by shifting the equilibrium position toward the side with fewer moles of gas. This is explained by Le Chatelier's principle, which states that the system will adjust to counteract the change in pressure. For the general case:
- If the reaction has fewer gas molecules on the product side, increasing pressure shifts equilibrium toward the products.
- If there are fewer gas molecules on the reactant side, increasing pressure shifts equilibrium toward the reactants.
- If both sides have the same number of gas molecules, pressure changes have no effect on the equilibrium position
Applying this to the specific reaction:
C(s)+H2O(g)⇌CO(g)+H2(g)\text{C(s)}+\text{H}_2\text{O(g)}\rightleftharpoons \text{CO(g)}+\text{H}_2\text{(g)}C(s)+H2O(g)⇌CO(g)+H2(g)
- Gaseous reactants: 1 mole (H₂O)
- Gaseous products: 2 moles (CO + H₂)
- Δn (change in moles of gas) = 2 - 1 = +1
Since the product side has more moles of gas, increasing pressure will shift the equilibrium to the side with fewer moles of gas, which is the reactant side (left), favoring the formation of C(s) and H₂O(g)
. In summary:
- Increasing pressure shifts the equilibrium toward the side with fewer gas molecules.
- For the given reaction, this means the equilibrium shifts to the left (toward reactants) when pressure is increased.
This shift reduces the total number of gas molecules, partially offsetting the pressure increase
