Question

1. explain how each of the following affects the amount of h₂ present in an equilibrium mixture in the reaction

3 fe(s) + 4 h₂o(g) ⇌ fe₃o₄(s) + 4 h₂(g) δh° = -150 kj
a. raising the temperature of the mixture.
b. introducing more h₂o(g) to the mixture.
c. removing some (not all) of the fe₃o₄.
d. doubling the volume of the reaction vessel.
e. adding an appropriate catalyst.

Explanation:

a. The reaction is exothermic ($\Delta H^{\circ}=- 150\ kJ$). According to Le - Chatelier's principle, raising the temperature will shift the equilibrium in the endothermic direction (reverse direction), so the amount of $H_2$ will decrease.
b. Introducing more $H_2O(g)$ will increase the concentration of a reactant. The equilibrium will shift to the forward direction to consume the added $H_2O$, and the amount of $H_2$ will increase.
c. Removing some $Fe_3O_4$ (a solid) does not affect the equilibrium position because the concentration of a solid is considered constant in an equilibrium expression. So the amount of $H_2$ remains unchanged.
d. Doubling the volume of the reaction vessel will decrease the pressure. Since there are equal moles of gas on both sides of the equation ($4$ moles of $H_2O(g)$ and $4$ moles of $H_2(g)$), the equilibrium will not shift and the amount of $H_2$ remains unchanged.
e. Adding a catalyst will increase the rate of both the forward and reverse reactions equally. It does not affect the equilibrium position, so the amount of $H_2$ remains unchanged.

Answer:

a. The amount of $H_2$ decreases.
b. The amount of $H_2$ increases.
c. The amount of $H_2$ remains unchanged.
d. The amount of $H_2$ remains unchanged.
e. The amount of $H_2$ remains unchanged.