What will happen to the number of moles of \(\mathrm{SO}_{3}\) in equilibrium with \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) in the reaction $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ in each of the following cases? a. Oxygen gas is added. b. The pressure is increased by decreasing the volume of the reaction container. c. In a rigid reaction container, the pressure is increased by adding argon gas. d. The temperature is decreased (the reaction is endothermic). e. Gaseous sulfur dioxide is removed.

Short Answer

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a. Number of moles of \(\mathrm{SO}_{3}\) will increase. b. Number of moles of \(\mathrm{SO}_{3}\) will increase. c. Number of moles of \(\mathrm{SO}_{3}\) remains unchanged. d. Number of moles of \(\mathrm{SO}_{3}\) will increase. e. Number of moles of \(\mathrm{SO}_{3}\) will decrease.

Step by step solution

01

a. Oxygen gas is added.

According to Le Chatelier's principle, when an equilibrium system is subjected to a change, the system will adjust the position of the equilibrium in a way that counteracts the change imposed. In this case, adding more oxygen gas will favor the reverse reaction, as the equilibrium will try to consume the excess \(\mathrm{O}_{2}\). This results in an increase in the number of moles of \(\mathrm{SO}_{3}\) in equilibrium.
02

b. The pressure is increased by decreasing the volume of the reaction container.

According to Le Chatelier's principle, increasing pressure favors the side with fewer moles of gas. In our reaction, \[2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g) + \mathrm{O}_{2}(g)\] the total number of moles on the reactant side is 2, while on the product side it is 3. Therefore, increasing the pressure by decreasing the volume will favor the side with fewer moles, i.e., the reactant side. Consequently, the number of moles of \(\mathrm{SO}_{3}\) in equilibrium will increase.
03

c. In a rigid reaction container, the pressure is increased by adding argon gas.

Adding an inert gas like argon doesn't affect equilibrium because it doesn't participate in the reaction. So, in this case, the number of moles of \(\mathrm{SO}_{3}\) in equilibrium remains unchanged.
04

d. The temperature is decreased (the reaction is endothermic).

Since the reaction is endothermic, it absorbs heat. According to Le Chatelier's principle, when the temperature is decreased, the equilibrium will shift towards the side that absorbs heat (the reactant side), in an attempt to counteract the change imposed. Thus, the number of moles of \(\mathrm{SO}_{3}\) in equilibrium will increase in this situation.
05

e. Gaseous sulfur dioxide is removed.

Removing gaseous sulfur dioxide will result in an imbalance, per Le Chatelier's principle. Therefore, the equilibrium will try to counteract this change by shifting towards the side where sulfur dioxide was removed (the product side) to restore equilibrium. Consequently, this will lead to a decrease in the number of moles of \(\mathrm{SO}_{3}\) in equilibrium.

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Most popular questions from this chapter

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