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The volume of the reaction vessel containing an equilibrium mixture in the reaction \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\) is increased. When equilibrium is re-established, (a) the amount of \(\mathrm{Cl}_{2}\) will have increased; (b) the amount of \(\mathrm{SO}_{2}\) will have decreased; (c) the amounts of \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) will have remained the same; (d) the amount of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) will have increased.

Short Answer

Expert verified
(a) The amount of Cl2 will have increased. (b) Incorrect, the amount of SO2 will have increased. (c) Incorrect, the amounts of SO2 and Cl2 will have increased. (d) Incorrect, the amount of SO2Cl2 will have decreased.

Step by step solution

01

Identifying The Reaction Direction

In the given reaction, \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), the number of moles of gas on the products side (SO2 and Cl2) is more than the reactants side (SO2Cl2). So, when the volume of the reaction vessel increases, the reaction will proceed in the forward direction to increase the number of moles of gas and thus, counteract the change.
02

Changes in the amounts of the gases

Now, as the reaction proceeds in the forward direction, the amount of SO2Cl2 which is the reactant will decrease. As a result, the amount of both Cl2 and SO2 will increase as they are the products of the forward reaction. This is in accordance with Le Chatelier's principle as it occurs to counteract the increase in volume and decrease in the pressure of the system.
03

Conclusion for each statement

On analyzing the scenario, (a) the amount of Cl2 will have increased, which is correct. (b) The amount of SO2 will have decreased, which is incorrect as the amount of SO2 will have increased. (c) The amounts of SO2 and Cl2 will have remained the same, which is incorrect as both will have increased. (d) The amount of SO2Cl2 will have increased. This statement is also incorrect as the amount of SO2Cl2 will have decreased.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Le Chatelier's Principle
Le Chatelier's Principle is a handy tool in understanding how a system at equilibrium reacts to external changes. The principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change and restore balance.
In the context of chemical reactions, this often involves changes in concentration, temperature, or pressure. To simplify:
  • If you add a substance, the system shifts to use it.
  • If you remove a substance, the system shifts to replace it.
  • If pressure increases, the system shifts to reduce it by favoring the side with fewer gas moles.

In gas reactions, like our given example involving \\(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g})\), Le Chatelier's Principle predicts the direction of the shift when the volume changes.
Reaction Direction
Determining the direction of a chemical reaction when equilibrium is disturbed is essential. For the reaction between \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) and its products, \(\mathrm{SO}_{2}(\mathrm{g})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\), know that the reaction moves towards the side with more gas molecules if the volume of the container is increased.
Since gas occupies more volume, increasing the reaction vessel's size decreases pressure. Le Chatelier's Principle advises that the reaction will adjust by shifting towards the side with more moles of gas, thereby increasing the volume again.
This means the forward reaction is favored, converting more \(\mathrm{SO}_{2} \mathrm{Cl}_{2}(\mathrm{g})\) into \(\mathrm{SO}_{2}(\mathrm{g})\) and \(\mathrm{Cl}_{2}(\mathrm{g})\). In our example:
  • Moles increase from one to two as you shift to the right.
  • This adjustment isn't just theoretical; it's a direct response to the change in conditions.
Gas Reactions
Gas reactions provide a practical context for exploring equilibrium concepts, especially in changing conditions like volume and pressure. Gas reactions, like our given example \(\mathrm{SO}_{2}\mathrm{Cl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{SO}_{2}(\mathrm{g}) + \mathrm{Cl}_{2}(\mathrm{g})\), are significant due to their immediate reaction to volume changes.
In a gas reaction, when you change the volume of the reaction vessel, it directly affects the pressure. According to Le Chatelier's Principle, the reaction shifts in a direction to balance this pressure change.
Consider these points:
  • Volume Increase: If you increase the volume, the reaction shifts towards the side with more gas molecules, as this helps increase pressure again.
  • Volume Decrease: Conversely, if the volume is reduced, the reaction shifts toward fewer gas molecules.

Explorations of gas reactions can help one visualize the dynamic nature of equilibria and their dependence on the physical context of the reaction environment.

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Most popular questions from this chapter

Explain why the percent of molecules that dissociate into atoms in reactions of the type \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g})\) always increases with an increase in temperature.

Cadmium metal is added to \(0.350 \mathrm{L}\) of an aqueous solution in which \(\left[\mathrm{Cr}^{3+}\right]=1.00 \mathrm{M} .\) What are the concentrations of the different ionic species at equilibrium? What is the minimum mass of cadmium metal required to establish this equilibrium? $$\begin{array}{r} 2 \mathrm{Cr}^{3+}(\mathrm{aq})+\mathrm{Cd}(\mathrm{s}) \rightleftharpoons 2 \mathrm{Cr}^{2+}(\mathrm{aq})+\mathrm{Cd}^{2+}(\mathrm{aq}) \\ K_{\mathrm{c}}=0.288 \end{array}$$

A 1.100 L flask at \(25^{\circ} \mathrm{C}\) and 1.00 atm pressure contains \(\mathrm{CO}_{2}(\mathrm{g})\) in contact with \(100.0 \mathrm{mL}\) of a saturated aqueous solution in which \(\left[\mathrm{CO}_{2}(\mathrm{aq})\right]=3.29 \times 10^{-2} \mathrm{M}\) (a) What is the value of \(K_{c}\) at \(25^{\circ} \mathrm{C}\) for the equilibrium \(\mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{aq}) ?\) (b) If 0.01000 mol of radioactive \(^{14} \mathrm{CO}_{2}\) is added to the flask, how many moles of the \(^{14} \mathrm{CO}_{2}\) will be found in the gas phase and in the aqueous solution when equilibrium is re-established? [Hint: The radioactive \(^{14} \mathrm{CO}_{2}\) distributes itself between the two phases in exactly the same manner as the nonradioactive \(\left.^{12} \mathrm{CO}_{2} .\right]\)

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.

For the dissociation of \(\mathrm{I}_{2}(\mathrm{g})\) at about \(1200^{\circ} \mathrm{C}\) \(\mathrm{I}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{I}(\mathrm{g}), K_{\mathrm{c}}=1.1 \times 10^{-2} .\) What volume flask should we use if we want 0.37 mol I to be present for every \(1.00 \mathrm{mol} \mathrm{I}_{2}\) at equilibrium?

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