Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 53

At \(900^{\circ} \mathrm{C}, K_{p}=51.2\) for the equilibrium $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g)$$ If the pressure of \(\mathrm{NO}(g)\) is half the pressure of \(\mathrm{NOBr}(g)\), what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Short Answer

Expert verified
The equilibrium pressure of \(Br_2(g)\) is 204.8 atm.

Step by step solution

01

Write the expression for Kp

Since the equilibrium constant is given in terms of pressure, we will write the expression for Kp in terms of the partial pressures of the reactants and products. For the reaction $$2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g),$$ the expression for Kp is given by: \[K_p = \frac{(\mathrm{P_{NO}})^2 \times (\mathrm{P_{Br_2}})}{(\mathrm{P_{NOBr}})^2}\]
02

Set up the relationship between the pressures

We are given that the pressure of NO(g) is half the pressure of NOBr(g). Let's denote the equilibrium pressure of NOBr(g) as x. Therefore, the pressure of NO(g) will be equal to 0.5x. Now, we can rewrite the pressures in our equation in terms of x.
03

Substitute the pressures into the Kp expression

Using the expression for Kp, we substitute the pressures of NO(g) and NOBr(g) as functions of x and plug in the given value for Kp: \[51.2 = \frac{(0.5x)^2 \times (\mathrm{P_{Br_2}})}{(x)^2}\]
04

Solve for the Br2 pressure

Now, we solve the equation for the pressure of Br2(g) using algebraic steps: \[51.2 = \frac{0.25x^2 \times (\mathrm{P_{Br_2}})}{(x)^2}\] \[(51.2)(x^2) = 0.25x^2 \times (\mathrm{P_{Br_2}})\] Divide both sides by (0.25x^2): \[\frac{(51.2)(x^2)}{0.25x^2} = \mathrm{P_{Br_2}}\] Simplify: \[204.8 = \mathrm{P_{Br_2}}\] So, the equilibrium pressure of Br2(g) is 204.8 atm.

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

At \(1200 \mathrm{~K}\), the approximate temperature of automobile exhaust gases (Figure 15.15 ), \(K_{p}\) for the reaction $$2 \mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g)+\mathrm{O}_{2}(g)$$ is about \(1 \times 10^{-11}\). Assuming that the exhaust gas (total pressure \(101.3 \mathrm{kPa}\) ) contains \(0.2 \% \mathrm{CO}, 12 \% \mathrm{CO}_{2},\) and \(3 \% \mathrm{O}_{2}\) by volume, is the system at equilibrium with respect to the \(\mathrm{CO}_{2}\) reaction? Based on your conclusion, would the CO concentration in the exhaust be decreased or increased by a catalyst that speeds up the \(\mathrm{CO}_{2}\) reaction? Recall that at a fixed pressure and temperature, volume \(\%=\mathrm{mol} \%\).

(a) If \(Q_{c}>K_{c}\), how must the reaction proceed to reach equilibrium? (b) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of \(Q_{c}\) at this point in the reaction?

A mixture of \(1.374 \mathrm{~g}\) of \(\mathrm{H}_{2}\) and \(70.31 \mathrm{~g}\) of \(\mathrm{Br}_{2}\) is heated in a 2.00-L vessel at \(700 \mathrm{~K}\). These substances react according to $$\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)$$ At equilibrium, the vessel is found to contain \(0.566 \mathrm{~g}\) of \(\mathrm{H}_{2}\). (a) Calculate the equilibrium concentrations of $\mathrm{H}_{2}, \mathrm{Br}_{2},\( and \)\mathrm{HBr} .$ (b) Calculate \(K_{c}\)

At \(700 \mathrm{~K}\), the equilibrium constant for the reaction $$\mathrm{CCl}_{4}(g) \rightleftharpoons \mathrm{C}(s)+2 \mathrm{Cl}_{2}(g)$$ is \(K_{p}=77\). A flask is charged with \(202.7 \mathrm{kPa}\) of \(\mathrm{CCl}_{4}\), which then reaches equilibrium at \(700 \mathrm{~K}\). (a) What fraction of the \(\mathrm{CCl}_{4}\) is converted into \(\mathrm{C}\) and \(\mathrm{Cl}_{2} ?\) (b) What are the partial pressures of \(\mathrm{CCl}_{4}\) and \(\mathrm{Cl}_{2}\) at equilibrium?

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

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemistry Branches

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