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Chapter 13: 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. b. The number of moles of \(\mathrm{SO}_{3}\) will increase. c. There will be no change in the number of moles of \(\mathrm{SO}_{3}\). d. The number of moles of \(\mathrm{SO}_{3}\) will decrease. e. The number of moles of \(\mathrm{SO}_{3}\) will decrease.

Step by step solution

01

a. Oxygen gas is added.

Adding more oxygen gas to the system will increase the concentration of \(\mathrm{O}_{2}\). To counteract this, the reaction will shift to the left to consume the excess \(\mathrm{O}_{2}\). This results in an increase in the concentration of \(\mathrm{SO}_{3}\).
02

b. The pressure is increased by decreasing the volume.

When the volume of the container is decreased, the pressure inside increases. According to Le Chatelier's Principle, the system will respond by shifting towards the side with fewer moles of gas. In this case, the left side of the reaction (2 moles of \(\mathrm{SO}_{3}\)) has fewer moles than the right side (3 moles - 2 moles of \(\mathrm{SO}_{2}\) and 1 mole of \(\mathrm{O}_{2}\)). Therefore, the reaction will shift to the left, and the concentration of \(\mathrm{SO}_{3}\) will increase.
03

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

Since argon is an inert gas and does not participate in the reaction, adding it does not affect the equilibrium directly. However, when the pressure is increased due to the addition of argon, the partial pressures of the components will not change. Therefore, there will be no change in the number of moles of \(\mathrm{SO}_{3}\).
04

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

In an endothermic reaction, heat can be considered a reactant. Decreasing the temperature is equivalent to removing heat from the system. According to Le Chatelier's Principle, the reaction will shift towards the side that absorbs the heat, in this case, the right side. The shift to the right will result in a decrease in the concentration of \(\mathrm{SO}_{3}\).
05

e. Gaseous sulfur dioxide is removed.

When \(\mathrm{SO}_{2}\) is removed, its concentration decreases. To counteract the change, the reaction will shift towards the right side to produce more \(\mathrm{SO}_{2}\). This means that more \(\mathrm{SO}_{3}\) will be consumed, and its concentration will decrease.

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Most popular questions from this chapter

A 1.00-L flask was filled with \(2.00\) mol gaseous \(\mathrm{SO}_{2}\) and \(2.00 \mathrm{~mol}\) gaseous \(\mathrm{NO}_{2}\) and heated. After equilibrium was reached, it was found that \(1.30\) mol gaseous NO was present. Assume that the reaction $$ \mathrm{SO}_{2}(g)+\mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g)+\mathrm{NO}(g) $$ occurs under these conditions. Calculate the value of the equilibrium constant, \(K\), for this reaction.

For the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) $$ \(K_{\mathrm{p}}=3.5 \times 10^{4}\) at \(1495 \mathrm{~K}\). What is the value of \(K_{\mathrm{p}}\) for the following reactions at \(1495 \mathrm{~K}\) ? a. \(\mathrm{HBr}(g) \rightleftharpoons \frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)\) b. \(2 \mathrm{HBr}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g)\) c. \(\frac{1}{2} \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{Br}_{2}(g) \rightleftharpoons \mathrm{HBr}(g)\)

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

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 \(0.16 \mathrm{~mol}\) of \(\mathrm{HCO}_{3}^{-}\) is placed into \(1.00 \mathrm{~L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}{ }^{2-}\) ?

The formation of glucose from water and carbon dioxide is one of the more important chemical reactions in the world. Plants perform this reaction through the process of photosynthesis, creating the base of the food chain: $$ 6 \mathrm{H}_{2} \mathrm{O}(g)+6 \mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) $$ At a particular temperature, the following equilibrium concentrations were found: \(\left[\mathrm{H}_{2} \mathrm{O}(g)\right]=7.91 \times 10^{-2} M,\left[\mathrm{CO}_{2}(g)\right]=\) \(9.3 \times 10^{-1} M\), and \(\left[\mathrm{O}_{2}(g)\right]=2.4 \times 10^{-3} M .\) Calculate the value of \(K\) for the reaction at this temperature.
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