/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 At \(373 \mathrm{~K}, K_{p}=0.41... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 15: Problem 53

At \(373 \mathrm{~K}, K_{p}=0.416\) for the equilibrium $$ 2 \mathrm{NOBr}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) $$ If the pressures of \(\operatorname{NOBr}(g)\) and \(\mathrm{NO}(g)\) are equal, what is the equilibrium pressure of \(\mathrm{Br}_{2}(g)\) ?

Short Answer

Expert verified
The equilibrium pressure of Br鈧�(g) can be expressed in terms of the pressure of NO(g) as: $$ P_{Br_{2}}=0.416 \cdot (P_{NO})^2 $$ Once the pressure of NO(g) is known, we can find the equilibrium pressure of Br鈧�(g).

Step by step solution

01

Write the equilibrium expression

For the given reaction, the equilibrium expression in terms of partial pressures (Kp) can be written as: $$ K_{p} = \frac{P_{NO}^2 \cdot P_{Br2}}{P_{NOBr}^2} $$ Where \(P_{NO}\), \(P_{Br2}\), and \(P_{NOBr}\) are the partial pressures of NO, Br2, and NOBr respectively at equilibrium.
02

Substitute the given value of Kp and equal partial pressures

We are given \(K_p = 0.416\) at 373K and the pressures of NOBr(g) and NO(g) are equal. Let the equal partial pressures be denoted as \(P\). The equilibrium expression becomes: $$ 0.416 = \frac{P^2 \cdot P_{Br2}}{P^2} $$
03

Solve for the equilibrium pressure of Br鈧�(g)

Let's solve the equation for \(P_{Br2}\): $$ P_{Br2} = 0.416 \cdot P^2 $$ Since the pressures of NOBr and NO are equal, we can substitute \(P\) from the given equation in terms of \(P_{NO}\) and \(P_{NOBr}\). We know that \(P_{NO} = P_{NOBr}\): $$ P_{Br2} = 0.416 \cdot (P_{NO})^2 $$ We cannot find the exact value of \(P_{Br2}\) since we do not have the value for \(P_{NO}\). However, we have expressed the equilibrium pressure of Br鈧�(g) in terms of the pressure of NO(g). Once the pressure of NO(g) is known, we can easily find the pressure of Br鈧�(g).

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 91影视!

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 \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons\) \(\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) ? (b) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Consider the reaction $$ 4 \mathrm{NH}_{3}(g)+5 \mathrm{O}_{2}(g) \underset{4 \mathrm{NO}(g)+6 \mathrm{H}_{2} \mathrm{O}(g), \Delta H=-904.4 \mathrm{~kJ}}{\rightleftharpoons} $$ Does each of the following increase, decrease, or leave unchanged the yield of NO at equilibrium? (a) increase \(\left[\mathrm{NH}_{3}\right] ;\) (b) increase \(\left[\mathrm{H}_{2} \mathrm{O}\right] ;(\) c \()\) decrease \(\left[\mathrm{O}_{2}\right]\); (d) decrease the volume of the container in which the reaction occurs: (e) add a catalyst: (f) increase temperature.

At \(1285^{\circ} \mathrm{C}\), the equilibrium constant for the reaction \(\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{Br}(g)\) is \(K_{c}=1.04 \times 10^{-3}\). A \(0.200-\mathrm{L}\) vessel containing an equilibrium mixture of the gases has \(0.245 \mathrm{~g}\) \(\mathrm{Br}_{2}(g)\) in it. What is the mass of \(\mathrm{Br}(g)\) in the vessel?

When \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a 2.00- \(\mathrm{L}\) flask at \(303 \mathrm{~K}\), \(56 \%\) of the \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) decomposes to \(\mathrm{SO}_{2}\) and \(\mathrm{Cl}_{2}\) : $$ \mathrm{SO}_{2} \mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\mathrm{Cl}_{2}(g) $$ (a) Calculate \(K_{c}\) for this reaction at this temperature. (b) Calculate \(K_{p}\) for this reaction at \(303 \mathrm{~K}\). (c) According to Le Ch芒telier's principle, would the percent of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes increase, decrease or stay the same if the mixture were transferred to a \(15.00\)-L vessel? (d) Use the equilibrium constant you calculated above to determine the percentage of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) that decomposes when \(2.00 \mathrm{~mol}\) of \(\mathrm{SO}_{2} \mathrm{Cl}_{2}\) is placed in a \(15.00\) - \(\mathrm{L}\) vessel at \(303 \mathrm{~K}\).

The water-gas shift reaction \(\mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons\) \(\mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\) is used industrially to produce hydrogen. The reaction enthalpy is \(\Delta H^{\circ}=-41 \mathrm{~kJ}\). (a) To increase the equilibrium yield of hydrogen would you use high or low temperature? (b) Could you increase the equilibrium yield of hydrogen by controlling the pressure of this reaction? If so would high or low pressure favor formation of \(\mathrm{H}_{2}(g) ?\)
See all solutions

Recommended explanations on Chemistry Textbooks

Physical Chemistry

Read Explanation

Chemistry Branches

Read Explanation

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

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.