Question

The reaction $2H_2(g) + CO(g) \rightleftharpoons CH_3OH(g) \quad \Delta H = -90.2 \text{ kJ}$ is at equilibrium. Predict how the concentration of $H_2$, CO and $CH_3OH$ will differ at a new equilibrium if each of the following changes is made: (1) more $H_2$ is added. (2) CO is removed. (3) $CH_3OH$ is added. (4) the pressure on the system is increased. (5) the temperature of the system is increased. (6) more catalyst is added.

Answer 1

1. More $H_2$ is added: $[H_2]$ increases, $[CO]$ decreases, $[CH_3OH]$ increases.
2. CO is removed: $[H_2]$ increases, $[CO]$ decreases, $[CH_3OH]$ decreases.
3. $CH_3OH$ is added: $[H_2]$ increases, $[CO]$ increases, $[CH_3OH]$ increases.
4. The pressure on the system is increased: $[H_2]$ decreases, $[CO]$ decreases, $[CH_3OH]$ increases.
5. The temperature of the system is increased: $[H_2]$ increases, $[CO]$ increases, $[CH_3OH]$ decreases.
6. More catalyst is added: $[H_2]$ remains the same, $[CO]$ remains the same, $[CH_3OH]$ remains the same.

Answer 2

Le Chatelier's Principle states that if a change of condition is applied to a system in equilibrium, the system will shift in a direction that relieves the stress.

1. Adding $H_2$ (reactant) shifts equilibrium to the right.
2. Removing $CO$ (reactant) shifts equilibrium to the left.
3. Adding $CH_3OH$ (product) shifts equilibrium to the left.
4. Increasing pressure favors the side with fewer moles of gas. Reactants have 3 moles, products have 1 mole. Pressure increase shifts equilibrium to the right.
5. The reaction is exothermic ($\Delta H < 0$). Increasing temperature shifts equilibrium in the endothermic direction (left).
6. A catalyst does not affect equilibrium position.