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Chapter 13: Problem 76

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 \(SO_{3}\) will increase since the equilibrium will shift to the left to counteract the addition of \(O_{2}\). b. The number of moles of \(SO_{3}\) will increase as the equilibrium shifts to the left to decrease the pressure. c. The number of moles of \(SO_{3}\) will remain the same since the addition of argon gas doesn't affect the reacting species' equilibrium concentrations. d. The number of moles of \(SO_{3}\) will increase due to the shift in equilibrium towards the endothermic direction to increase temperature. e. The number of moles of \(SO_{3}\) will decrease as the equilibrium shifts to the right to produce more \(SO_{2}\) to counteract its removal.

Step by step solution

01

Analyze the effect of adding Oxygen gas

When Oxygen gas (\(O_2\)) is added to the reaction system, the concentration of \(O_2\) increases. According to the Le Chatelier's principle, the system will try to decrease the concentration of \(O_2\) to counteract the change.
02

Determine the shift in reaction equilibrium

To decrease the concentration of \(O_2\), the system will shift its equilibrium position towards the left, in the direction that consumes \(O_2\). Consequently, the number of moles of \(SO_3\) will increase. #b. The pressure is increased by decreasing the volume of the reaction container.#
03

Analyze the effect of increasing pressure

When the pressure of the reaction container is increased by decreasing its volume, the system will try to decrease the pressure according to Le Chatelier's principle.
04

Determine the shift in reaction equilibrium

To decrease the pressure, the system will shift its equilibrium position towards the side with fewer moles of gas. In this case, there are 3 moles of gas on the right side and 2 moles of gas on the left side of the reaction, so the equilibrium will shift to the left. This will cause an increase in the number of moles of \(SO_3\). #c. In a rigid reaction container, the pressure is increased by adding argon gas.#
05

Analyze the effect of adding argon gas

Adding argon gas in a rigid container will increase the overall pressure. However, since argon gas is inert and not involved in the reaction, it won't affect the equilibrium concentrations of the reacting species.
06

Determine the shift in reaction equilibrium

Since argon gas doesn't affect the equilibrium concentrations of the reacting species, there will be no shift in the equilibrium position. Hence, the number of moles of \(SO_3\) will remain the same. #d. The temperature is decreased (the reaction is endothermic).#
07

Analyze the effect of decreasing temperature

As the temperature is decreased, according to Le Chatelier's principle, the system will try to increase its temperature by shifting the equilibrium position towards the direction that absorbs heat, which is the endothermic reaction direction.
08

Determine the shift in reaction equilibrium

Since the reaction is endothermic, decreasing the temperature will cause the system to shift its equilibrium position towards the left to produce more heat. As a result, the number of moles of \(SO_3\) will increase. #e. Gaseous sulfur dioxide is removed.#
09

Analyze the effect of removing sulfur dioxide

When gaseous sulfur dioxide (\(SO_2\)) is removed from the system, its concentration decreases. According to Le Chatelier's principle, the system will try to increase the concentration of \(SO_2\) to counteract the change.
10

Determine the shift in reaction equilibrium

To increase the concentration of \(SO_2\), the system will shift its equilibrium position towards the right, in the direction that produces \(SO_2\). Consequently, the number of moles of \(SO_3\) will decrease.

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Most popular questions from this chapter

In a solution with carbon tetrachloride as the solvent, the compound \(\mathrm{VCl}_{4}\) undergoes dimerization: $$2 \mathrm{VCl}_{4} \rightleftharpoons \mathrm{V}_{2} \mathrm{Cl}_{8}$$ When 6.6834 g \(\mathrm{VCl}_{4}\) is dissolved in 100.0 \(\mathrm{g}\) carbon tetrachloride, the freezing point is lowered by \(5.97^{\circ} \mathrm{C}\) . Calculate the value of the equilibrium constant for the dimerization of \(\mathrm{VCl}_{4}\) at this temperature. (The density of the equilibrium mixture is \(1.696 \mathrm{g} / \mathrm{cm}^{3},\) and \(K_{\mathrm{f}}=29.8^{\circ} \mathrm{C} \mathrm{kg} / \mathrm{mol}\) for \(\mathrm{CCl}_{4} . )\)

Methanol, a common laboratory solvent, poses a threat of blindness or death if consumed in sufficient amounts. Once in the body, the substance is oxidized to produce formaldehyde (embalming fluid) and eventually formic acid. Both of these substances are also toxic in varying levels. The equilibrium between methanol and formaldehyde can be described as follows: $$\mathrm{CH}_{3} \mathrm{OH}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{CO}(a q)+\mathrm{H}_{2}(a q)$$ Assuming the value of \(K\) for this reaction is \(3.7 \times 10^{-10},\) what are the equilibrium concentrations of each species if you start with a 1.24\(M\) solution of methanol? What will happen to the concentration of methanol as the formaldehyde is further converted to formic acid?

At high temperatures, elemental nitrogen and oxygen react with each other to form nitrogen monoxide: $$\mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)$$ Suppose the system is analyzed at a particular temperature, and the equilibrium concentrations are found to be \(\left[\mathrm{N}_{2}\right]=\) \(0.041 M,\left[\mathrm{O}_{2}\right]=0.0078 M,\) and $[\mathrm{NO}]=4.7 \times 10^{-4} M .\( Calculate the value of \)K$ for the reaction.

At a particular temperature, a \(3.0-\mathrm{L}\) flask contains 2.4 moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\) , and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$

Write the equilibrium expression (K) for each of the following gas-phase reactions. a. \(N_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)\) b. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)\) c. $\operatorname{SiH}_{4}(g)+2 \mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{SiCl}_{4}(g)+2 \mathrm{H}_{2}(g)$ d. $2 \mathrm{PBr}_{3}(g)+3 \mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{PCl}_{3}(g)+3 \mathrm{Br}_{2}(g)$
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