Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

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.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

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.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The Deacon process for producing chlorine gas from hydrogen chloride is used in situations where \(\mathrm{HCl}\) is available as a by-product from other chemical processes. $$\begin{aligned} 4 \mathrm{HCl}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})+2 \mathrm{Cl}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-114 \mathrm{kJ} \end{aligned}$$ A mixture of \(\mathrm{HCl}, \mathrm{O}_{2}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{Cl}_{2}\) is brought to equilibrium at \(400^{\circ} \mathrm{C}\). What is the effect on the equilibrium amount of \(\mathrm{Cl}_{2}(\mathrm{g})\) if (a) additional \(\mathrm{O}_{2}(\mathrm{g})\) is added to the mixture at constant volume? (b) \(\mathrm{HCl}(\mathrm{g})\) is removed from the mixture at constant volume? (c) the mixture is transferred to a vessel of twice the volume? (d) a catalyst is added to the reaction mixture? (e) the temperature is raised to \(500^{\circ} \mathrm{C} ?\)

At \(2000 \mathrm{K}, K_{c}=0.154\) for the reaction \(2 \mathrm{CH}_{4}(\mathrm{g}) \rightleftharpoons\) \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})+3 \mathrm{H}_{2}(\mathrm{g}) .\) If a \(1.00 \mathrm{L}\) equilibrium mixture at \(2000 \mathrm{K}\) contains \(0.10 \mathrm{mol}\) each of \(\mathrm{CH}_{4}(\mathrm{g})\) and \(\mathrm{H}_{2}(\mathrm{g})\) (a) what is the mole fraction of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) present? (b) Is the conversion of \(\mathrm{CH}_{4}(\mathrm{g})\) to \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) favored at high or low pressures? (c) If the equilibrium mixture at \(2000 \mathrm{K}\) is transferred from a 1.00 L flask to a 2.00 L flask, will the number of moles of \(\mathrm{C}_{2} \mathrm{H}_{2}(\mathrm{g})\) increase, decrease, or remain unchanged?

A sample of pure \(\mathrm{PCl}_{5}(\mathrm{g})\) is introduced into an evacuated flask and allowed to dissociate. $$ \mathrm{PCl}_{5}(\mathrm{g}) \rightleftharpoons \mathrm{PCl}_{3}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) $$ If the fraction of \(\mathrm{PCl}_{5}\) molecules that dissociate is denoted by \(\alpha,\) and if the total gas pressure is \(P\) show that $$ K_{\mathrm{p}}=\frac{\alpha^{2} P}{1-\alpha^{2}} $$

Equilibrium is established in the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\) \(\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \quad\) at \(\quad \mathrm{a} \quad\) temperature \(\quad\) where \(K_{\mathrm{c}}=100 .\) If the number of moles of \(\mathrm{SO}_{3}(\mathrm{g})\) in the equilibrium mixture is the same as the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}),\) (a) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is also equal to the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}) ;\) (b) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is half the number of moles of \(\mathrm{SO}_{2} ;\) (c) \(\left[\mathrm{O}_{2}\right]\) may have any of several values; (d) \(\left[\mathrm{O}_{2}\right]=0.010 \mathrm{M}\)

Refer to Example \(15-4 . \mathrm{H}_{2} \mathrm{S}(\mathrm{g})\) at \(747.6 \mathrm{mmHg}\) pressure and a \(1.85 \mathrm{g}\) sample of \(\mathrm{I}_{2}(\mathrm{s})\) are introduced into a \(725 \mathrm{mL}\) flask at \(60^{\circ} \mathrm{C} .\) What will be the total pressure in the flask at equilibrium? $$\begin{aligned} \mathrm{H}_{2} \mathrm{S}(\mathrm{g})+\mathrm{I}_{2}(\mathrm{s}) \rightleftharpoons 2 \mathrm{HI}(\mathrm{g})+\mathrm{S}(\mathrm{s}) & \\ K_{\mathrm{p}}=& 1.34 \times 10^{-5} \mathrm{at} 60^{\circ} \mathrm{C} \end{aligned}$$

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free