Question

Hydrogen for use in ammonia production is produced by the reaction

CH₄(g) + H₂O(g)$\underset{{\mathbf{750}}^{\mathbf{o}}\mathbf{C}}{\overset{\mathbf{N}\mathbf{i}\mathbf{ }\mathbf{}\mathbf{c}\mathbf{a}\mathbf{t}\mathbf{a}\mathbf{l}\mathbf{y}\mathbf{s}\mathbf{t}}{\mathbf{\rightleftharpoons }}}$CO(g) + 3H₂(g)

What will happen to a reaction mixture at equilibrium if

a. H₂O(g) is removed?

b. the temperature is increased (the reaction is endothermic)?

c. an inert gas is added to a rigid reaction container?

d. CO(g) is removed?

e. the volume of the container is tripled?

Short answer

a. The reaction mixture at equilibrium will shift to the left.

b. The reaction mixture at equilibrium will shift to the right.

c. Addition of an inert gas has no effect.

d. The reaction mixture at equilibrium will shift to the right.

e. The reaction mixture at equilibrium will shift to the right.

Step-by-step solution

Part a.

When water vapor is removed from the reaction mixture, the reaction will shift to left to produce more water vapor to reach the equilibrium again. Therefore, the reaction mixture at equilibrium will shift to the left.

Part b.

At equilibrium, the reaction mixture will shift to the left when the temperature is increased because the given reaction is endothermic so that it absorbs added heat to produce the more products. Hence, the reaction mixture at equilibrium will shift to the right.

Part c.

Addition of an inert gas has no effect because added inert gas is not a part of reaction (reactants or products). Hence, it has no effect on the equilibrium.

Part d.

Carbon monoxide gas is one of the products. So, the removal of CO(g) will shift the reaction to the right to generate more carbon monoxide in order to reestablish the equilibrium. Hence, the reaction mixture at equilibrium will shift to the right.

Part e.

When the volume of the reaction container is increased by three-fold, the reaction will move to the right that has more gaseous product molecules in order to fill the larger volume. Therefore, the reaction mixture at equilibrium will shift to the right.