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Chapter 12: Problem 66

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

Expert verified
a. The number of moles of \(\mathrm{SO}_{3}\) will increase due to the increased concentration of \(\mathrm{O}_{2}\). b. The number of moles of \(\mathrm{SO}_{3}\) will increase due to the decrease in volume and increase in pressure. c. There will be no change in the number of moles of \(\mathrm{SO}_{3}\), as the addition of argon gas does not affect the partial pressures of reactants and products. d. The number of moles of \(\mathrm{SO}_{3}\) will increase when the temperature is decreased, as the reaction is endothermic. e. The number of moles of \(\mathrm{SO}_{3}\) will decrease when gaseous sulfur dioxide is removed.

Step by step solution

01

a. Oxygen gas is added.

When oxygen gas is added to the system, the concentration of \(\mathrm{O}_{2}\) increases. According to Le Châtelier's principle, the equilibrium will shift to the side where the oxygen is being consumed, i.e., to the left. As the reaction shifts to the left, the number of moles of \(\mathrm{SO}_{3}\) will increase.
02

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

When the volume of the container is decreased, the pressure increases. According to Le Châtelier's principle, the equilibrium will shift to the side with fewer moles of gas to counteract the increase in pressure. In this reaction, there are 3 moles of gas on the right side and 2 moles of gas on the left side. So, the equilibrium will shift to the left, where there are fewer moles of gas. Consequently, the number of moles of \(\mathrm{SO}_{3}\) will increase.
03

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

In this case, only the total pressure is increased, but the partial pressures of the reactants and products remain the same. Since argon gas is inert and does not take part in the reaction, the equilibrium will not be affected, and there will be no change in the number of moles of \(\mathrm{SO}_{3}\).
04

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

Since the reaction is endothermic, it absorbs heat to proceed. When the temperature is decreased, according to Le Châtelier's principle, the equilibrium will shift in the direction which absorbs heat, i.e., to the left. As a result, the number of moles of \(\mathrm{SO}_{3}\) will increase.
05

e. Gaseous sulfur dioxide is removed.

If gaseous sulfur dioxide (\(\mathrm{SO}_{2}\)) is removed, its concentration decreases. According to Le Châtelier's principle, the equilibrium will shift towards the side where the removed substance is being generated, i.e., to the right. This shift will cause the reaction to produce more \(\mathrm{SO}_{2}\) and consume \(\mathrm{SO}_{3}\). Consequently, the number of moles of \(\mathrm{SO}_{3}\) will decrease.

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Most popular questions from this chapter

At a particular temperature, \(K=3.75\) for the reaction $$\mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g)$$.If all four gases had initial concentrations of 0.800 \(M,\) calculate the equilibrium concentrations of the gases.

At a particular temperature, 12.0 moles of \(\mathrm{SO}_{3}\) is placed into a 3.0-L rigid container, and the \(\mathrm{SO}_{3}\) dissociates by the reaction $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$. At equilibrium, 3.0 moles of \(\mathrm{SO}_{2}\) is present. Calculate \(K\) for this reaction.

At \(25^{\circ} \mathrm{C}, K_{\mathrm{p}}=5.3 \times 10^{5}\) for the reaction $$\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)$$ When a certain partial pressure of \(\mathrm{NH}_{3}(g)\) is put into an otherwise empty rigid vessel at \(25^{\circ} \mathrm{C},\) equilibrium is reached when \(50.0 \%\) of the original ammonia has decomposed. What was the original partial pressure of ammonia before any decomposition occurred?

Which of the following statements is(are) true? Correct the false statement(s). a. When a reactant is added to a system at equilibrium at a given temperature, the reaction will shift right to reestablish equilibrium. b. When a product is added to a system at equilibrium at a given temperature, the value of \(K\) for the reaction will increase when equilibrium is reestablished. c. When temperature is increased for a reaction at equilibrium, the value of \(K\) for the reaction will increase. d. When the volume of a reaction container is increased for a system at equilibrium at a given temperature, the reaction will shift left to reestablish equilibrium. e. Addition of a catalyst (a substance that increases the speed of the reaction) has no effect on the equilibrium position.

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is \(\mathrm{HCO}_{3}^{-}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \quad K=5.6 \times 10^{-11}\) If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?
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