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 increases the fraction of products in an equilibrium mixture. (b) The balanced chemical equation has more molecules on the product side, as increasing the volume of the reaction vessel increases the fraction of products.

Step by step solution

01

When the temperature of a reaction at equilibrium is increased, the reaction will shift in the direction of the reaction that can consume the added energy (heat). If the reaction is exothermic, heat is released, and if it is endothermic, heat is absorbed. As increasing the temperature increases the fraction of products, the reaction must be shifting towards the products, indicating that it is an endothermic reaction. So, the answer to question (a) is endothermic. #Step 2: Determine if there are more molecules on the reactant or product side of the balanced chemical equation#

When the volume of a reaction mixture at equilibrium is increased, the pressure decreases, and the system will shift to the side with more moles of gas. Increasing the volume of the reaction vessel results in an increase in the fraction of products. Therefore, the reaction must be shifting towards the products side, which means the product side has more molecules than the reactant side. Thus, the answer to question (b) is that 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

Mercury(I) oxide decomposes into elemental mercury and elemental oxygen: $2 \mathrm{Hg}_{2} \mathrm{O}(s) \rightleftharpoons 4 \mathrm{Hg}(l)+\mathrm{O}_{2}(g)$. (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Suppose you run this reaction in a solvent that dissolves elemental mercury and elemental oxygen. Rewrite the equilibriumconstant expression in terms of molarities for the reaction, using (solv) to indicate solvation.

Consider the reaction $$\mathrm{CaSO}_{4}(s) \rightleftharpoons \mathrm{Ca}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)$$ At \(25^{\circ} \mathrm{C}\), the equilibrium constant is $K_{c}=2.4 \times 10^{-5}\( for this reaction. (a) If excess \)\operatorname{CaSO}_{4}(s)$ is mixed with water at \(25^{\circ} \mathrm{C}\) to produce a saturated solution of \(\mathrm{CaSO}_{4},\) what are the equilibrium concentrations of \(\mathrm{Ca}^{2+}\) and \(\mathrm{SO}_{4}^{2-}\) ? (b) If the resulting solution has a volume of \(1.4 \mathrm{~L},\) what is the minimum mass of \(\operatorname{CaSO}_{4}(s)\) needed to achieve equilibrium?

Two different proteins \(\mathrm{X}\) and \(\mathrm{Y}\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\). The proteins bind in a 1: 1 ratio to form XY. A solution that is initially \(1.00 \mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, \(0.20 \mathrm{mM}\) of free \(\mathrm{X}\) and \(0.20 \mathrm{~m} M\) of free Y remain. What is \(K_{c}\) for the reaction?

Ethene \(\left(\mathrm{C}_{2} \mathrm{H}_{4}\right)\) reacts with halogens \(\left(\mathrm{X}_{2}\right)\) by the following reaction: $$\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{X}_{2}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4} \mathrm{X}_{2}(g)$$ The following figures represent the concentrations at equilibrium at the same temperature when \(\mathrm{X}_{2}\) is \(\mathrm{Cl}_{2}\) (green), \(\mathrm{Br}_{2}\) (brown), and \(\mathrm{I}_{2}\) (purple). List the equilibria from smallest to largest equilibrium constant. [Section 15.3\(]\)

A 5.37 -g sample of \(\mathrm{SO}_{3}\) is placed in a 5.00-L container and heated to \(1000 \mathrm{~K}\). The \(\mathrm{SO}_{3}\) 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 157 \(\mathrm{kPa}\). Find the values of \(K_{p}\) and \(K_{c}\) for this reaction at \(1100 \mathrm{~K}\).

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