Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 31

At \(327^{\circ} \mathrm{C}\), the equilibrium concentrations are \(\left[\mathrm{CH}_{3} \mathrm{OH}\right]=0.15 \mathrm{M}\), \([\mathrm{CO}]=0.24 M\), and \(\left[\mathrm{H}_{2}\right]=1.1 M\) for the reaction $$ \mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) $$ Calculate \(K_{\mathrm{p}}\) at this temperature.

Short Answer

Expert verified
In this exercise, we have the reaction \(\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)\) with equilibrium concentrations: \(\left[\mathrm{CH}_{3} \mathrm{OH}\right] = 0.15 M\), \([\mathrm{CO}] = 0.24 M\), and \(\left[\mathrm{H}_{2}\right] = 1.1 M\). The equilibrium constant \(K_p\) can be calculated using the formula \(K_{p} = \frac{[\mathrm{CO}]^{1} [\mathrm{H}_{2}]^{2}}{[\mathrm{CH}_{3} \mathrm{OH}]^{1}}\). Substituting the equilibrium concentrations, we find \(K_p \approx 1.936\) at \(327^\circ \mathrm{C}\).

Step by step solution

01

Identify the stoichiometric coefficients

In the given reaction, the stoichiometric coefficients are: \(\mathrm{CH}_{3}\mathrm{OH}: 1\) \(\mathrm{CO}: 1\) \(\mathrm{H}_{2}: 2\)
02

Use the equilibrium constant formula

Using the \(K_p\) formula, we have: \[K_{p} = \frac{[\mathrm{CO}]^{1} [\mathrm{H}_{2}]^{2}}{[\mathrm{CH}_{3} \mathrm{OH}]^{1}}\]
03

Substitute the equilibrium concentrations

Now, substituting the equilibrium concentrations into the formula, we get: \(K_p = \frac{(0.24)^{1} (1.1)^{2}}{(0.15)^{1}}\)
04

Calculate the equilibrium constant

Perform the calculations to find the equilibrium constant: \(K_p = \frac{(0.24) (1.21)}{(0.15)}\) \(K_p = \frac{0.2904}{0.15}\) \(K_p \approx 1.936\) At \(327^\circ \mathrm{C}\), the equilibrium constant \(K_p\) for this reaction is approximately equal to 1.936.

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

For the reaction \(\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)\), consider two pos- sibilities: (a) you mix \(0.5\) mol of each reactant, allow the system to come to equilibrium, and then add another mole of \(\mathrm{H}_{2}\) and allow the system to reach equilibrium again, or (b) you mix \(1.5 \mathrm{~mol}\) \(\mathrm{H}_{2}\) and \(0.5 \mathrm{~mol} \mathrm{I}_{2}\) and allow the system to reach equilibrium. Will the final equilibrium mixture be different for the two procedures? Explain.

A sample of iron(II) sulfate was heated in an evacuated container to \(920 \mathrm{~K}\), where the following reactions occurred: $$ \begin{aligned} 2 \mathrm{FeSO}_{4}(s) & \rightleftharpoons \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+\mathrm{SO}_{3}(g)+\mathrm{SO}_{2}(g) \\ \mathrm{SO}_{3}(g) & \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \end{aligned} $$ After equilibrium was reached, the total pressure was \(0.836\) atm and the partial pressure of oxygen was \(0.0275\) atm. Calculate \(K_{\mathrm{p}}\) for each of these reactions.

At a particular temperature, \(8.0 \mathrm{~mol} \mathrm{NO}_{2}\) is placed into a \(1.0\) -L container and the \(\mathrm{NO}_{2}\) dissociates by the reaction $$ 2 \mathrm{NO}_{2}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) $$ At equilibrium the concentration of \(\mathrm{NO}(g)\) is \(2.0 \mathrm{M}\). Calculate \(K\) for this reaction.

The equilibrium constant is \(0.0900\) at \(25^{\circ} \mathrm{C}\) for the reaction $$ \mathrm{H}_{2} \mathrm{O}(g)+\mathrm{Cl}_{2} \mathrm{O}(g) \rightleftharpoons 2 \mathrm{HOCl}(g) $$ For which of the following sets of conditions is the system at equilibrium? For those which are not at equilibrium, in which direction will the system shift? a. A \(1.0\) - \(\mathrm{L}\) flask contains \(1.0 \mathrm{~mol} \mathrm{HOCl}, 0.10 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.10 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) b. A 2.0-L flask contains \(0.084 \mathrm{~mol} \mathrm{HOCl}, 0.080 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.98 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\) c. A 3.0-L flask contains \(0.25 \mathrm{~mol} \mathrm{HOCl}, 0.0010 \mathrm{~mol} \mathrm{Cl}_{2} \mathrm{O}\), and \(0.56 \mathrm{~mol} \mathrm{H}_{2} \mathrm{O}\)

The following equilibrium pressures at a certain temperature were observed for the reaction $$ \begin{aligned} 2 \mathrm{NO}_{2}(g) & \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \\\ P_{\mathrm{NO}_{2}} &=0.55 \mathrm{~atm} \\ P_{\mathrm{NO}} &=6.5 \times 10^{-5} \mathrm{~atm} \\ P_{\mathrm{O}_{2}} &=4.5 \times 10^{-5} \mathrm{~atm} \end{aligned} $$ Calculate the value for the equilibrium constant \(K_{\mathrm{p}}\) at this temperature.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Making Measurements

Read Explanation

The Earths Atmosphere

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

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