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

For a certain gas-phase reaction, the fraction of products in an equilibrium mixture is increased by either increasing the temperature or by increasing the volume of the reaction vessel. (a) Is the reaction exothermic or endothermic? (b) Does the balanced chemical equation have more molecules on the reactant side or product side?

Short Answer

Expert verified
(a) The reaction is endothermic, as increasing the temperature favors product formation. (b) The balanced chemical equation has more molecules on the product side, as increasing the volume favors product formation.

Step by step solution

01

Examine the effect of temperature change

The reaction has an increased fraction of products when the temperature is increased. According to Le Chatelier's principle, increasing the temperature of an endothermic reaction will favor the formation of products, as heat is absorbed in the process. On the other hand, increasing the temperature of an exothermic reaction will favor the formation of reactants, as heat is released in the process. Therefore, the given reaction must be endothermic, as increasing the temperature favors product formation. Answer (a): The reaction is endothermic.
02

Examine the effect of volume change

The reaction also has an increased fraction of products when the volume of the reaction vessel is increased. According to Le Chatelier's principle, increasing the volume of a reaction vessel will favor the side of the reaction with more gas molecules, as the pressure is decreased. In order to determine which side of the balanced chemical equation has more molecules, we can apply this information to the given reaction. Since increasing the volume favors product formation, the product side must have more gas molecules than the reactant side. Answer (b): The balanced chemical equation has more molecules on the product side.

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!

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

A \(0.831-\) g sample of \(\mathrm{SO}_{3}\) is placed in a 1.00 -L container and heated to 1100 \(\mathrm{K}\) . The SO \(_{3}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\) : $$2 \mathrm{SO}_{3}(g) \rightleftharpoons 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g)$$ At equilibrium, the total pressure in the container is 1.300 atm. Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at 1100 \(\mathrm{K}\) .

When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? $$\begin{array}{ll}{\text { (a) } \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)} & {K_{c}=1.5 \times 10^{-10}} \\ {\text { (b) } 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)} & {K_{p}=2.5 \times 10^{9}}\end{array}$$

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.

The following equilibria were attained at \(823 \mathrm{K} :\) $$\operatorname{CoO}(s)+\mathrm{H}_{2}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{H}_{2} \mathrm{O}(g) \quad K_{c}=67$$ $$\mathrm{CoO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Co}(s)+\mathrm{CO}_{2}(g) \quad K_{c}=490$$ Based on these equilibria, calculate the equilibrium constant for \(\mathrm{H}_{2}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g)\) at 823 \(\mathrm{K}\) .

The equilibrium constant \(K_{c}\) for \(C(s)+\mathrm{CO}_{2}(g) \rightleftharpoons\) 2 \(\mathrm{CO}(g)\) is 1.9 at 1000 \(\mathrm{K}\) and 0.133 at 298 \(\mathrm{K}\) . (a) If excess\(\mathrm{C}\) is allowed to react with 25.0 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) in a 3.00 -L vessel at \(1000 \mathrm{K},\) how many grams of CO are produced? (b) How many grams of \(\mathrm{C}\) are consumed? (c) If a smaller vessel is used for the reaction, will the yield of CO be greater or smaller? (d) Is the reaction endothermic or exothermic?

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