Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 52

For the equilibrium $$\mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g) $$ at \(400 \mathrm{~K}, K_{c}=7.0 .\) If $0.25 \mathrm{~mol}\( of \)\mathrm{Br}_{2}\( and \)0.55 \mathrm{~mol}$ of \(\mathrm{Cl}_{2}\) are introduced into a 3.0-L container at \(400 \mathrm{~K}\), what will be the equilibrium concentrations of $\mathrm{Br}_{2}, \mathrm{Cl}_{2}\(, and \)\mathrm{BrCl}$ ?

Short Answer

Expert verified
The equilibrium concentrations of the species are - \( [\mathrm{Br}_2]_{eq} = 0.034 \, M \) - \( [\mathrm{Cl}_2]_{eq} = 0.134 \, M \) - \( [\mathrm{BrCl}]_{eq} = 0.098 \, M \).

Step by step solution

01

Write down the balanced chemical equation and the given information

First, let's write the balanced chemical equation: \[ \mathrm{Br_{2}(g) + Cl_{2}(g) \rightleftharpoons 2 \operatorname{BrCl}(g)} \] Given information: - \( T = 400 \, K \) - \( K_c = 7.0 \) - \( [\mathrm{Br}_2]_0 = \frac{0.25 \, mol}{3.0 \, L} = 0.083 \, M \) - \( [\mathrm{Cl}_2]_0 = \frac{0.55 \, mol}{3.0 \, L} = 0.183 \, M \) - \( [\mathrm{BrCl}]_0 = 0 \, M \)
02

Write the equilibrium expression

The equilibrium expression for the given reaction can be written as: \[ K_c = \frac{[\mathrm{BrCl}]^2}{[\mathrm{Br}_2][\mathrm{Cl}_2]} \]
03

Set up the ICE table

Next, let's use the ICE (Initial, Change, Equilibrium) table to find the changes in concentrations for each species: \(\begin{array}{c|c|c|c} & [\mathrm{Br}_2] & [\mathrm{Cl}_2] & [\mathrm{BrCl}] \\ \hline \mathrm{Initial} & 0.083 & 0.183 & 0 \\ \mathrm{Change} & -x & -x & +2x \\ \mathrm{Equilibrium} & 0.083-x & 0.183-x & 2x \end{array}\) Where x is the change in molarity for Br₂ and Cl₂, and 2x represents the change in molarity for BrCl.
04

Substitute equilibrium concentrations into the equilibrium expression and solve for x

Substituting the equilibrium concentrations into the equilibrium expression: \[ 7.0 = \frac{(2x)^2}{(0.083 - x)(0.183 - x)} \] We now have a quadratic equation to solve for x. This quadratic equation can be a bit challenging to solve by hand, so using a solver might be best.
05

Solve for x and find the equilibrium concentrations

After using a quadratic equation solver, we get \(x = 0.049\). Now, we can find the equilibrium concentrations: - \( [\mathrm{Br}_2]_{eq} = 0.083 - 0.049 = 0.034 M \) - \( [\mathrm{Cl}_2]_{eq} = 0.183 - 0.049 = 0.134 M \) - \( [\mathrm{BrCl}]_{eq} = 2(0.049) = 0.098 M \) So, the equilibrium concentrations of the species are: - \( [\mathrm{Br}_2]_{eq} = 0.034 \, M \) - \( [\mathrm{Cl}_2]_{eq} = 0.134 \, M \) - \( [\mathrm{BrCl}]_{eq} = 0.098 \, M \)

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

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Consider the following equilibrium: $2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{~S}(g) \quad K_{c}=1.08 \times 10^{7}\( at \)700^{\circ} \mathrm{C}$ (a) Calculate \(K_{p \cdot}\) (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly $\mathrm{H}_{2} \mathrm{~S} ?(\mathbf{c})\( Calculatethevalue of \)K_{c}$ if you rewrote the equation $\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{~S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{~S}(g)$

The equilibrium constant for the reaction $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ is \(K_{c}=1.3 \times 10^{-2}\) at \(1000 \mathrm{~K}\). (a) At this temperature does the equilibrium favor \(\mathrm{NO}\) and \(\mathrm{Br}_{2}\), or does it favor NOBr? (b) Calculate \(K_{c}\) for $2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) .$ (c) Calculate \(K_{c}\) for $\mathrm{NOBr}(g) \rightleftharpoons \mathrm{NO}(g)+\frac{1}{2} \mathrm{Br}_{2}(g)$.

Solid \(\mathrm{NH}_{4} \mathrm{SH}\) is introduced into an evacuated flask at \(24^{\circ} \mathrm{C}\). The following reaction takes place: $$\mathrm{NH}_{4} \mathrm{SH}(s) \rightleftharpoons \mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{~S}(g)$$ At equilibrium, the total pressure (for \(\mathrm{NH}_{3}\) and $\mathrm{H}_{2} \mathrm{~S}\( taken together) is \)62.21 \mathrm{kPa}\(. What is \)K_{p}$ for this equilibrium at \(24^{\circ} \mathrm{C} ?\)

At \(120^{\circ} \mathrm{C}, K_{c}=0.090\) for the reaction $$\mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g)$$ In an equilibrium mixture of the three gases, the concentrations of \(\mathrm{SO}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{SO}_{2}\) are $0.100 \mathrm{M}\( and \)0.075 \mathrm{M}$, respectively. What is the partial pressure of \(\mathrm{Cl}_{2}\) in the equilibrium mixture?
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

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