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

A gaseous reaction mixture contains \(0.30 \mathrm{~atm} \mathrm{SO}_{2}\), \(0.16 \mathrm{~atm} \mathrm{Cl}_{2},\) and \(0.50 \mathrm{~atm} \mathrm{SO}_{2} \mathrm{Cl}_{2}\) in a \(2.0-\mathrm{L}\) container. \(K=0.011\) for the equilibrium system $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Is the system at equilibrium? Explain. (b) If it is not at equilibrium, in which direction will the system move to reach equilibrium?

Short Answer

Expert verified
(b) If not, in which direction will the reaction proceed to reach equilibrium? a) To determine if the system is at equilibrium, compare the calculated reaction quotient (Q) with the given equilibrium constant (K). If Q = K, the system is at equilibrium. If Q ≠ K, the system is not at equilibrium. b) If the system is not at equilibrium, the direction in which the reaction will proceed depends on the comparison between Q and K. If Q > K, the reaction will move towards the reactants (left). If Q < K, the reaction will move towards the products (right). The reason for the chosen direction is to establish the equilibrium condition where Q = K.

Step by step solution

01

(Step 1: Calculate the reaction quotient (Q))

(Find the reaction quotient Q by dividing the product of the partial pressures of the products by the partial pressure of the reactant:) $$ Q=\frac{[SO_2][Cl_2]}{[SO_2Cl_2]} $$ Substitute the given partial pressures for each component and calculate Q.
02

(Step 2: Compare Q with K)

(Determine if the system is at equilibrium by comparing the calculated Q with the given equilibrium constant K. If Q = K, the system is at equilibrium. If Q > K, the reaction will move towards the reactants (left). If Q < K, the reaction will move towards the products (right).)
03

(Step 3: Determine the direction of the reaction)

(Based on the comparison made in Step 2, decide whether the reaction will move towards the reactants or the products to reach equilibrium. Explain the reason for the chosen direction.)
04

(Step 4: Answer both parts (a) and (b) of the question)

(a) Determine if the system is at equilibrium from the comparison made in Step 2 and provide an explanation. (b) If the system is not at equilibrium, state which direction the reaction will move to reach equilibrium and the reason for the chosen direction.)

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

Calculate \(K\) for the formation of methyl alcohol at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ given that at equilibrium, the partial pressures of the gases $$ \text { are } P_{\mathrm{CO}}=0.814 \mathrm{~atm}, \quad P_{\mathrm{H}_{2}}=0.274 \mathrm{~atm}, \text { and } P_{\mathrm{CH}_{3} \mathrm{OH}}= $$ \(0.0512 \mathrm{~atm} .\)

At a certain temperature, nitrogen and oxygen gases combine to form nitrogen oxide gas. $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g) $$ When equilibrium is established, the partial pressures of the gases $$ \text { are: } P_{\mathrm{N}_{2}}=1.200 \mathrm{~atm}, P_{\mathrm{O}_{2}}=0.800 \mathrm{~atm}, P_{\mathrm{NO}}=0.0220 \mathrm{~atm} . $$ (a) Calculate \(K\) at the temperature of the reaction. (b) After equilibrium is reached, more oxygen is added to make its partial pressure 1.200 atm. Calculate the partial pressure of all gases when equilibrium is reestablished.

At \(800 \mathrm{~K}\), hydrogen iodide can decompose into hydrogen and iodine gases. $$ 2 \mathrm{HI}(g) \rightleftharpoons \mathrm{I}_{2}(g)+\mathrm{H}_{2}(g) $$ At this temperature, \(K=0.0169 .\) What are the partial pressures at equilibrium of the hydrogen and iodine if initially a sealed flask at \(800 \mathrm{~K}\) contains only HI at a pressure of 0.200 atm?

At a certain temperature, \(K\) is 4.9 for the formation of one mole of bromine chloride gas (BrCl) from its elements. A mixture at equilibrium at this temperature contains all three gases. The partial pressures at equilibrium of bromine and chlorine gas is 0.19 atm. What is the partial pressure of bromine chloride in this mixture at equilibrium?

Mustard gas, used in chemical warfare in World War I, has been found to be an effective agent in the chemotherapy of Hodgkin's disease. It can be produced according to the following reaction: $$ \mathrm{SCl}_{2}(g)+2 \mathrm{C}_{2} \mathrm{H}_{4}(g) \rightleftharpoons \mathrm{S}\left(\mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{Cl}\right)_{2}(g) $$ An evacuated \(5.0-\mathrm{L}\) flask at \(20.0^{\circ} \mathrm{C}\) is filled with \(0.258 \mathrm{~mol}\) \(\mathrm{SCl}_{2}\) and \(0.592 \mathrm{~mol} \mathrm{C}_{2} \mathrm{H}_{4}\). After equilibrium is established, 0.0349 mol mustard gas is present. (a) What is the partial pressure of each gas at equilibrium? (b) What is \(K\) at \(20.0^{\circ} \mathrm{C} ?\)

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