Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 77

A mixture containing 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO}_{2}\) was allowed to react in a flask at a certain temperature according to the equation $$ \mathrm{NO}(g)+\mathrm{CO}_{2}(g) \rightleftharpoons \mathrm{NO}_{2}(g)+\mathrm{CO}(g) $$ At equilibrium, 0.11 mole of \(\mathrm{CO}_{2}\) was present. Calculate the equilibrium constant \(K_{\mathrm{c}}\) of this reaction.

Short Answer

Expert verified
The equilibrium constant \(K_c\) for the reaction is approximately 17.2.

Step by step solution

01

Reactant and Product Mole Calculation

We first need to determine the change in moles for each reactant and product. Initially, we have 3.9 moles of \(\mathrm{NO}\) and 0.88 mole of \(\mathrm{CO_2}\). At equilibrium, we have 0.11 moles of \(\mathrm{CO_2}\). From the reaction equation, we know for every mole of \(\mathrm{NO}\) used, one mole of \(\mathrm{CO_2}\) is also used. So, the change in moles for \(\mathrm{NO}\) is the same as the change in moles for \(\mathrm{CO_2}\).
02

Change in Moles Calculation

The change in moles for \(\mathrm{CO_2}\) is the initial moles minus the moles at equilibrium (0.88 - 0.11 = 0.77 moles). So, the change in moles for \(\mathrm{NO}\) is also 0.77 moles. Consequently, the moles of \(\mathrm{NO}\) at equilibrium become 3.9 - 0.77 = 3.13 moles. Because for each mole of reactant consumed, one mole of product is created (according to the equation), the moles of \(\mathrm{NO_2}\) and \(\mathrm{CO}\) at equilibrium would be 0.77 moles.
03

Finding Equilibrium Constant \(K_c\)

The equilibrium constant \(K_c\) for the reaction is given by the ratio of the concentrations (in moles) of the products to the reactants each raised to the power of their stoichiometric coefficients. Algebraically, this becomes \[K_c = \frac{[\mathrm{NO_2}][\mathrm{CO}]}{[\mathrm{NO}][\mathrm{CO_2}]}\]. Substituting the calculated equilibrium moles of the reactants and products into the equation, we find \(K_c = \frac{(0.77)(0.77)}{(3.13)(0.11)}\).
04

Calculating the Numerical Value of \(K_c\)

After performing the mentioned arithmetic operation, the value of \(K_c\) will be obtained.

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.

Start your free trial

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

The vapor pressure of mercury is \(0.0020 \mathrm{mmHg}\) at \(26^{\circ} \mathrm{C}\). (a) Calculate \(K_{\mathrm{c}}\) and \(K_{P}\) for the process \(\mathrm{Hg}(l) \rightleftharpoons \mathrm{Hg}(g) .\) (b) A chemist breaks a thermometer and spills mercury onto the floor of a laboratory measuring \(6.1 \mathrm{~m}\) long, \(5.3 \mathrm{~m}\) wide, and \(3.1 \mathrm{~m}\) high. Calculate the mass of mercury (in grams) vaporized at equilibrium and the concentration of mercury vapor in \(\mathrm{mg} / \mathrm{m}^{3}\). Does this concentration exceed the safety limit of \(0.050 \mathrm{mg} / \mathrm{m}^{3} ?\) (Ignore the volume of furniture and other objects in the laboratory.)

The decomposition of ammonium hydrogen sulfide $$ \mathrm{NH}_{4} \mathrm{HS}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g) $$ is an endothermic process. A \(6.1589-\mathrm{g}\) sample of the solid is placed in an evacuated 4.000 - \(L\) vessel at exactly \(24^{\circ} \mathrm{C}\). After equilibrium has been established, the total pressure inside is \(0.709 \mathrm{~atm}\). Some solid \(\mathrm{NH}_{4} \mathrm{HS}\) remains in the vessel. (a) What is the \(K_{P}\) for the reaction? (b) What percentage of the solid has decomposed? (c) If the volume of the vessel were doubled at constant temperature, what would happen to the amount of solid in the vessel?

The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$ 2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) $$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C}\). What is the equilibrium constant for the reaction $$ \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g) $$ at the same temperature?

Baking soda (sodium bicarbonate) undergoes thermal decomposition as $$ 2 \mathrm{NaHCO}_{3}(s) \rightleftharpoons \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ Would we obtain more \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) by adding extra baking soda to the reaction mixture in (a) a closed vessel or (b) an open vessel?

Consider this reaction at equilibrium in a closed container: $$ \mathrm{CaCO}_{3}(s) \rightleftharpoons \mathrm{CaO}(s)+\mathrm{CO}_{2}(g) $$ What would happen if (a) the volume is increased, (b) some \(\mathrm{CaO}\) is added to the mixture, (c) some \(\mathrm{CaCO}_{3}\) is removed, (d) some \(\mathrm{CO}_{2}\) is added to the mixture, (e) a few drops of an \(\mathrm{NaOH}\) solution are added to the mixture, (f) a few drops of an \(\mathrm{HCl}\) solution are added to the mixture (ignore the reaction between \(\mathrm{CO}_{2}\) and water \(),(\mathrm{g})\) the temperature is increased?
See all solutions

Recommended explanations on Chemistry Textbooks

Nuclear Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Reactions

Read Explanation
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