Hydrogen for use in ammonia production is produced by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \stackrel{\mathrm{Nicatalyst}}{750^{\circ} \mathrm{C}} \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) $$ What will happen to a reaction mixture at equilibrium if a. \(\mathrm{H}_{2} \mathrm{O}(g)\) is removed? b. the temperature is increased (the reaction is endothermic)? c. an inert gas is added to a rigid reaction container? d. \(\mathrm{CO}(g)\) is removed? e. the volume of the container is tripled?

Short Answer

Expert verified
a. If \(H_2O(g)\) is removed, the reaction will shift to the left, increasing concentrations of \(CH_4(g)\) and \(H_2O(g)\), and decreasing concentrations of \(CO(g)\) and \(H_2(g)\). b. If the temperature is increased, the reaction will shift to the right, increasing concentrations of \(CO(g)\) and \(H_2(g)\) and decreasing concentrations of \(CH_4(g)\) and \(H_2O(g)\). c. If an inert gas is added, there will be no change to the equilibrium position or the concentrations of the reacting species. d. If \(CO(g)\) is removed, the reaction will shift to the right, decreasing concentrations of \(CH_4(g)\) and \(H_2O(g)\), and increasing concentrations of \(CO(g)\) and \(H_2(g)\). e. If the volume of the container is tripled, the reaction will shift to the right, increasing concentrations of \(CO(g)\) and \(H_2(g)\), and decreasing concentrations of \(CH_4(g)\) and \(H_2O(g)\).

Step by step solution

01

Scenario A: Removal of \(H_2O(g)\)

When \(H_2O(g)\) is removed from the reaction mixture, the concentration of \(H_2O(g)\) decreases. According to Le Chatelier's principle, the system will adjust its equilibrium position to counteract the disturbance. In this case, the reaction will shift to the left to produce more \(H_2O(g)\) and consume \(CO(g)\) and \(H_2(g)\). Therefore, the concentration of \(CH_4(g)\) and \(H_2O(g)\) will increase, while the concentration of \(CO(g)\) and \(H_2(g)\) will decrease.
02

Scenario B: Increase in temperature (endothermic reaction)

As the reaction is endothermic, increasing the temperature will cause the system to absorb more heat. According to Le Chatelier's principle, the system will adjust its equilibrium position to counteract the disturbance. In this case, the reaction will shift to the right as this is the direction of the endothermic reaction. The equilibrium concentrations of \(CO(g)\) and \(H_2(g)\) will increase, while the equilibrium concentrations of \(CH_4(g)\) and \(H_2O(g)\) will decrease.
03

Scenario C: Addition of an inert gas to a rigid reaction container

When an inert gas is added to a rigid reaction container, the total pressure of the system increases, but the partial pressures of the reacting species remain unchanged. The inert gas does not participate in the reaction and, therefore, does not affect the equilibrium. In this case, there will be no change to the equilibrium position or the concentrations of the reacting species.
04

Scenario D: Removal of \(CO(g)\)

When \(CO(g)\) is removed from the reaction mixture, the concentration of \(CO(g)\) decreases. According to Le Chatelier's principle, the system will adjust its equilibrium position to counteract the disturbance. In this case, the reaction will shift to the right to produce more \(CO(g)\) and consume \(CH_4(g)\) and \(H_2O(g)\). Therefore, the concentration of \(CH_4(g)\) and \(H_2O(g)\) will decrease, while the concentration of \(CO(g)\) and \(H_2(g)\) will increase.
05

Scenario E: Tripling the volume of the container

When the volume of the container is tripled, the pressure of the system decreases. According to Le Chatelier's principle, the system will adjust its equilibrium position to counteract the disturbance. As the forward reaction produces more moles of gas (4 moles) than the reverse reaction (2 moles), the system will shift to the side where there is a greater number of moles to restore pressure. In this case, the reaction will shift to the right. The equilibrium concentrations of \(CO(g)\) and \(H_2(g)\) will increase, while the equilibrium concentrations of \(CH_4(g)\) and \(H_2O(g)\) will decrease.

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