Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 12: Problem 30

The following equilibrium pressures were observed at a certain temperature for the reaction $$\begin{array}{c}\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \\\P_{\mathrm{NH}}=3.1 \times 10^{-2} \mathrm{atm} \\\P_{\mathrm{N}_{2}}=8.5 \times 10^{-1} \mathrm{atm} \\\P_{\mathrm{H}_{2}}=3.1 \times 10^{-3} \mathrm{atm}\end{array}$$.Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature. If \(P_{\mathrm{N}_{2}}=0.525\) atm, \(P_{\mathrm{NH},}=0.0167\) atm, and \(P_{\mathrm{H}_{2}}=0.00761\) atm, does this represent a system at equilibrium?

Short Answer

Expert verified
The equilibrium constant \(K_\mathrm{p}\) for the given reaction at the given temperature is approximately 54.28. When given new pressure values (\(P_{N_2} = 0.525 \mathrm{atm}\), \(P_{NH_3} = 0.0167 \mathrm{atm}\), and \(P_{H_2} = 0.00761 \mathrm{atm}\)), we find that the reaction quotient, \(Q_\mathrm{p} \approx 58.25\). Since \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.

Step by step solution

01

Write the expression for the equilibrium constant Kp

For the given reaction: \[N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)\] The expression for the equilibrium constant \(K_\mathrm{p}\) can be written as: \[K_\mathrm{p} = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}\]
02

Substitute the given pressures into the Kp expression

We are given: \(P_{NH_3} = 3.1 \times 10^{-2} \mathrm{atm}\), \(P_{N_2} = 8.5 \times 10^{-1} \mathrm{atm}\), and \(P_{H_2} = 3.1 \times 10^{-3} \mathrm{atm}\). Substitute these pressure values into the expression for Kp: \[K_\mathrm{p} = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1}) \times (3.1 \times 10^{-3})^3}\]
03

Calculate Kp

Now, calculate the value of Kp: \[K_\mathrm{p} = \frac{(3.1 \times 10^{-2})^2}{(8.5 \times 10^{-1}) \times (3.1 \times 10^{-3})^3} \approx 54.28\] So, the equilibrium constant \(K_\mathrm{p}\) at the given temperature is approximately 54.28.
04

Analyze the new pressures

We are given new pressure values: \(P_{N_2} = 0.525 \mathrm{atm}\), \(P_{NH_3} = 0.0167 \mathrm{atm}\), and \(P_{H_2} = 0.00761 \mathrm{atm}\). We need to check if these pressures represent a system at equilibrium. To do this, we can calculate \(Q_\mathrm{p}\) using the same expression as for \(K_\mathrm{p}\), and compare the two values. \[Q_\mathrm{p} = \frac{P_{NH_3}^2}{P_{N_2} \times P_{H_2}^3}\]
05

Calculate Qp

Substitute the new pressure values into the expression for Qp: \[Q_\mathrm{p} = \frac{(0.0167)^2}{(0.525) \times (0.00761)^3} = 58.25\] Now compare \(Q_\mathrm{p}\) to \(K_\mathrm{p}\): - If \(Q_\mathrm{p} < K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the right. - If \(Q_\mathrm{p} = K_\mathrm{p}\), the reaction is at equilibrium. - If \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.
06

Compare Qp to Kp

Comparing our calculated \(Q_\mathrm{p}\) and \(K_\mathrm{p}\) values: \(Q_\mathrm{p} = 58.25\), \(K_\mathrm{p} \approx 54.28\). Since \(Q_\mathrm{p} > K_\mathrm{p}\), the reaction is not at equilibrium and will proceed to the left.

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 a particular temperature, 8.1 moles of \(\mathrm{NO}_{2}\) gas is placed in a 3.0 -L container. Over time the \(\mathrm{NO}_{2}\) decomposes to NO and \(\mathrm{O}_{2}:\) $$2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g)$$ At equilibrium the concentration of \(\mathrm{NO}(g)\) was found to be 1.4 mol/L. Calculate the value of \(K\) for this reaction.

The synthesis of ammonia gas from nitrogen gas and hydrogen gas represents a classic case in which a knowledge of kinetics and equilibrium was used to make a desired chemical reaction economically feasible. Explain how each of the following conditions helps to maximize the yield of ammonia. a. running the reaction at an elevated temperature b. removing the ammonia from the reaction mixture as it forms c. using a catalyst d. running the reaction at high pressure

For the reaction $$\mathrm{NH}_{3}(g)+\mathrm{H}_{2} \mathrm{S}(g) \rightleftharpoons \mathrm{NH}_{4} \mathrm{HS}(s)$$,\(K=400 .\) at \(35.0^{\circ} \mathrm{C} .\) If 2.00 moles each of \(\mathrm{NH}_{3}, \mathrm{H}_{2} \mathrm{S},\) and NH,HS are placed in a 5.00 -L vessel, what mass of \(\mathrm{NH}_{4} \mathrm{HS}\) will be present at equilibrium? What is the pressure of \(\mathrm{H}_{2} \mathrm{S}\) at equilibrium?

Nitric oxide and bromine at initial partial pressures of 98.4 and 41.3 torr, respectively, were allowed to react at \(300 .\) K. At equilibrium the total pressure was 110.5 torr. The reaction is $$2 \mathrm{NO}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{NOBr}(g)$$ a. Calculate the value of \(K_{\mathrm{p}}\). b. What would be the partial pressures of all species if NO and \(\mathrm{Br}_{2},\) both at an initial partial pressure of 0.30 atm, were allowed to come to equilibrium at this temperature?

At \(35^{\circ} \mathrm{C}, K=1.6 \times 10^{-5}\) for the reaction $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$.If 2.0 moles of \(\mathrm{NO}\) and 1.0 mole of \(\mathrm{Cl}_{2}\) are placed into a \(1.0-\mathrm{L}\) flask, calculate the equilibrium concentrations of all species.
See all solutions

Recommended explanations on Chemistry Textbooks

Chemistry Branches

Read Explanation

Chemical Analysis

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Ionic and Molecular Compounds

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