For the following endothermic reaction at equilibrium: $$ 2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) $$ which of the following changes will increase the value of \(K ?\) a. increasing the temperature b. decreasing the temperature c. removing \(\mathrm{SO}_{3}(g)\) (constant \(T\) ) d. decreasing the volume (constant \(T\) ) e. adding \(\operatorname{Ne}(g)\) (constant \(T\) ) f. adding \(\mathrm{SO}_{2}(g)\) (constant \(T\) ) g. adding a catalyst (constant \(T\) )

Short Answer

Expert verified
In conclusion, the only change that will increase the value of K is "a. Increasing the temperature".

Step by step solution

01

Write down the given equilibrium reaction

: The given reaction is: \(\mathrm{2SO_3(g)} \rightleftharpoons \mathrm{2SO_2(g) + O_2(g)}\) This is an endothermic reaction, meaning it absorbs heat.
02

Effect of changing temperature

: For an endothermic reaction, increasing the temperature will shift the equilibrium towards the products. Thus, increasing the temperature will result in a higher value of \(K\). a. Increasing the temperature: Yes, it will increase \(K\). b. Decreasing the temperature: No, it will not increase \(K\), as it will shift the equilibrium towards the reactants.
03

Effect of changing concentrations

: According to Le Chatelier's Principle, a change in the concentration of a species in the reaction will shift the equilibrium to counteract the change. c. Removing \(\mathrm{SO_3(g)}\): No, removing reactants shifts the equilibrium towards reactants and not the products. d. Decreasing the volume: No, as decreasing the volume of the gas will favor the side with fewer moles, which is the reactants. e. Adding \(\operatorname{Ne}(g)\): No, it will not affect the reaction since it's inert. f. Adding \(\mathrm{SO_2(g)}\): No, adding products will shift the equilibrium towards reactants.
04

Effect of a catalyst

: g. Adding a catalyst: No, the catalyst doesn't affect the equilibrium constant. It only speeds up the rate at which equilibrium is reached. In conclusion, the only change that will increase the value of K is "a. Increasing the temperature".

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
Understanding Le Chatelier's Principle is crucial in predicting how a change in conditions can affect a chemical equilibrium. This principle states that if an external condition is imposed upon a system at equilibrium, the system will adjust to minimize the effect of that change. Whether it involves changes in concentration, pressure, volume, or temperature, the system will shift its equilibrium position to counteract the change.

For example, if there is an increase in the concentration of the reactants, the system will shift towards the products to lower the reactant concentration. This is because the principle seeks to establish a new equilibrium state that restores balance. Similarly, when the pressure increases by reducing the volume of a gaseous system, the equilibrium will shift towards the side with fewer moles of gas, as this will help alleviate the increased pressure.

Le Chatelier's Principle is vital in industrial processes where conditions are constantly adjusted to optimize the yield of desired products. By understanding this principle, chemists can make informed decisions on how to manipulate reactions to their advantage.
Endothermic Reaction
An endothermic reaction is a chemical process that absorbs heat from its surroundings. This is in contrast to exothermic reactions, which release heat. In an endothermic reaction, the energy required to break bonds in the reactants is greater than the energy released when new bonds form in the products. As a result, the reaction vessel or area often feels cooler as the system absorbs heat.

In the context of Le Chatelier's Principle, for an endothermic process, the addition of heat can be thought of as adding a reactant. Therefore, increasing temperature pushes the equilibrium toward the products. This increase in product concentration at higher temperatures is reflected by a larger equilibrium constant, represented as K. Conversely, lowering the temperature favors the formation of reactants.

Many everyday phenomena, like photosynthesis and the dissolution of salts in water, involve endothermic reactions. Recognizing an endothermic process helps in determining the direction of equilibrium shifts when temperature changes occur.
Equilibrium Constant
The equilibrium constant, denoted as K, quantifies the balance between reactants and products in a chemical reaction at equilibrium. It is defined by the concentration of the products raised to the power of their coefficients, divided by the concentration of the reactants raised to the power of their coefficients. The value of K is determined at a specific temperature; a change in temperature results in a new equilibrium constant.

When K is large, it indicates that at equilibrium, the concentration of products is high relative to the concentration of reactants. Conversely, a small K value suggests a reaction mixture with more reactants than products at equilibrium. It is important to note that K does not change with the addition of a catalyst as the catalyst only speeds up the reaction rate, without affecting the position of the equilibrium. However, changing the temperature will result in a different K value.

Understanding the equilibrium constant provides invaluable insight into the extent of a reaction and the concentrations of chemicals involved at equilibrium. Therefore, knowledge of K aids in predicting the effects of various changes to the system, in accordance with Le Chatelier's Principle.

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

An initial mixture of nitrogen gas and hydrogen gas is reacted in a rigid container at a certain temperature by the reaction $$ 3 \mathrm{H}_{2}(g)+\mathrm{N}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) $$ At equilibrium, the concentrations are \(\left[\mathrm{H}_{2}\right]=5.0 \mathrm{M},\left[\mathrm{N}_{2}\right]=\) \(8.0 M\), and \(\left[\mathrm{NH}_{3}\right]=4.0 M .\) What were the concentrations of nitrogen gas and hydrogen gas that were reacted initially?

Old-fashioned "smelling salts" consist of ammonium carbonate, \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3} .\) The reaction for the decomposition of ammonium carbonate $$ \left(\mathrm{NH}_{4}\right)_{2} \mathrm{CO}_{3}(s) \rightleftharpoons 2 \mathrm{NH}_{3}(g)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ is endothermic. Would the smell of ammonia increase or decrease as the temperature is increased?

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.

At \(25^{\circ} \mathrm{C}, K=0.090\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ Calculate the concentrations of all species at equilibrium for each of the following cases. a. \(1.0 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}\) and \(2.0 \mathrm{~g} \mathrm{Cl}_{2} \mathrm{O}\) are mixed in a \(1.0-\mathrm{L}\) flask. b. \(1.0\) mole of pure \(\mathrm{HOCl}\) is placed in a \(2.0-\mathrm{L}\) flask.

A sample of gaseous nitrosyl bromide (NOBr) was placed in a container fitted with a frictionless, massless piston, where it decomposed at \(25^{\circ} \mathrm{C}\) according to the following equation: $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ The initial density of the system was recorded as \(4.495 \mathrm{~g} / \mathrm{L}\). After equilibrium was reached, the density was noted to be \(4.086 \mathrm{~g} / \mathrm{L}\) a. Determine the value of the equilibrium constant \(K\) for the reaction. b. If \(\operatorname{Ar}(g)\) is added to the system at equilibrium at constant temperature, what will happen to the equilibrium position? What happens to the value of \(K\) ? Explain each answer.

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