For the following equilibria, use Le Châtelier's principle to predict the direction of the reaction when the pressure is increased by decreasing the volume of the equilibrium mixture.C(s) + H2O(g) ⇌ CO(g) + H2(g)

Verified step by step guidance

1. Identify the number of moles of gas on each side of the equilibrium equation. On the left side, there is 1 mole of H2O(g), and on the right side, there are 2 moles of gas (1 mole of CO(g) and 1 mole of H2(g)).
2. According to Le Châtelier's principle, if the pressure is increased by decreasing the volume, the equilibrium will shift in the direction that produces fewer moles of gas to counteract the change.
3. Compare the number of moles of gas on each side: the left side has 1 mole of gas, and the right side has 2 moles of gas.
4. Since the left side has fewer moles of gas, the equilibrium will shift to the left to reduce the pressure.
5. Conclude that the reaction will shift towards the reactants (C(s) + H2O(g)) when the pressure is increased by decreasing the volume.

Key Concepts

Le Châtelier's Principle

Le Châtelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the system will adjust itself to counteract that change and restore a new equilibrium. This principle is crucial for predicting how a system at equilibrium responds to changes in concentration, temperature, or pressure.

Equilibrium and Reaction Quotient

Chemical equilibrium occurs when the rates of the forward and reverse reactions are equal, resulting in constant concentrations of reactants and products. The reaction quotient (Q) helps determine the direction in which the reaction will shift to reach equilibrium, based on the concentrations of the reactants and products at any given moment.

Effect of Pressure on Gaseous Reactions

In gaseous reactions, changes in pressure affect the equilibrium position based on the number of moles of gas on each side of the reaction. According to Le Châtelier's Principle, increasing pressure by decreasing volume will favor the side of the reaction with fewer moles of gas, thus shifting the equilibrium to minimize the pressure change.