Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 13: Problem 110

A 4.72 -g sample of methanol (CH_ 3 \(\mathrm{OH}\) ) was placed in an otherwise empty \(1.00-\mathrm{L}\) flask and heated to \(250 .^{\circ} \mathrm{C}\) to vaporize the methanol. Over time, the methanol vapor decomposed by the following reaction: $$\mathrm{CH}_{3} \mathrm{OH}(g) \rightleftharpoons \mathrm{CO}(g)+2 \mathrm{H}_{2}(g)$$ After the system has reached equilibrium, a tiny hole is drilled in the side of the flask allowing gaseous compounds to effuse out of the flask. Measurements of the effusing gas show that it contains 33.0 times as much \(\mathrm{H}_{2}(g)\) as \(\mathrm{CH}_{3} \mathrm{OH}(g) .\) Calculate \(K\) for this reaction at \(250 . \mathrm{C} .\)

Short Answer

Expert verified
The equilibrium constant for the decomposition of methanol vapor at \(250\,\mathrm{C}\) is \(K \approx 7.94 \times 10^{-4}\).

Step by step solution

01

Calculate initial moles of methanol

First, we need to find the initial moles of methanol that was placed in the flask. We can use the mass of methanol given and its molar mass to determine this. The molar mass of methanol, \(\mathrm{CH}_3\mathrm{OH}\), is 12.01 (C) + 1.008 (H) + 1.008 (H) + 1.008 (H) + 16.00 (O) + 1.008 (H) = 32.042 g/mol. Now, we can find the moles of methanol: moles of \(\mathrm{CH}_3\mathrm{OH} = \frac{4.72 \text{g}}{32.042 \text{g/mol}} \approx 0.147\text{ moles}\).
02

Set up ICE table

To find the equilibrium concentrations, we will use the ICE (Initial, Change, Equilibrium) analysis: Initial: \(\mathrm{CH}_3\mathrm{OH}(g)\): \(0.147 \mathrm{mol/L}\) \(\mathrm{CO}(g) + 2 \mathrm{H}_2(g)\): \(0 \mathrm{mol/L}\) Change: \(\mathrm{CH}_3\mathrm{OH}(g)\): -\(x \mathrm{mol/L}\) \(\mathrm{CO}(g) + 2 \mathrm{H}_2(g)\): +\(x\) \(\mathrm{mol/L}, + 2x \,\mathrm{mol/L}\) Equilibrium: \(\mathrm{CH}_3\mathrm{OH}(g)\): \((0.147 - x)\,\mathrm{mol/L}\) \(\mathrm{CO}(g)\): \(x\,\mathrm{mol/L}\) \(\mathrm{H}_2(g)\): \(2x\,\mathrm{mol/L}\)
03

Determine equilibrium concentrations using given ratio

We're given that the ratio of \(\mathrm{H}_{2}(g)\) to \(\mathrm{CH}_{3} \mathrm{OH}(g)\) is 33.0 in the effusing gas. Using the equilibrium concentrations, we can write this equation: \(\frac{2x}{x} = \frac{\mathrm{H}_2}{\mathrm{CH}_3\mathrm{OH}} = 33\) Solving for \(x\), we get \(x = \frac{2}{33}\). Since \(x < 0.147\), this small initial approximation is valid. Now, we can find the equilibrium concentrations: \(\mathrm{CH}_3\mathrm{OH} = 0.147 - x \approx 0.141\,\mathrm{mol/L}\) \(\mathrm{CO} = x \approx \frac{2}{33}\,\mathrm{mol/L}\) \(\mathrm{H}_2 = 2x\approx \frac{4}{33}\,\mathrm{mol/L}\)
04

Find the equilibrium constant (K)

Now, we can use the calculated equilibrium concentrations to find the equilibrium constant: \(K = \frac{[\mathrm{CO}] [\mathrm{H}_2]^2}{[\mathrm{CH}_3\mathrm{OH}]} = \frac{\frac{2}{33} \times \left(\frac{4}{33}\right)^2}{0.141}\) Evaluating this expression, we get: \(K \approx 7.94 \times 10^{-4}\) So, at \(250\,\mathrm{C}\), the equilibrium constant for the decomposition of methanol vapor is \(K \approx 7.94 \times 10^{-4}\).

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

Consider the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ Suppose the system is at equilibrium, and then an additional mole of \(\mathrm{N}_{2} \mathrm{O}(g)\) is injected into the system at constant temperature. Once the reaction reestablishes equilibrium, has the amount of \(\mathrm{N}_{2} \mathrm{O}\) increased or decreased from its original equilibrium amount? Explain. What happens to the value of the equilibrium constant with this change?

The creation of shells by mollusk species is a fascinating process. By utilizing the \(\mathrm{Ca}^{2+}\) in their food and aqueous environment, as well as some complex equilibrium processes, a hard calcium carbonate shell can be produced. One important equilibrium reaction in this complex process is $$\mathrm{HCO}_{3}^{-}(a q) \leftrightharpoons \mathrm{H}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) K=5.6 \times 10^{-11}$$ If 0.16 mole of \(\mathrm{HCO}_{3}^{-}\) is placed into 1.00 \(\mathrm{L}\) of solution, what will be the equilibrium concentration of \(\mathrm{CO}_{3}^{2-2}\) ?

At a particular temperature, 8.0 moles of \(\mathrm{NO}_{2}\) 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 \(K_{\mathrm{p}}\) is \(2.4 \times 10^{3}\) at a certain temperature for the reaction $$2 \mathrm{NO}(g) \leftrightharpoons \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g)$$ For which of the following sets of conditions is the system at equilibrium? For those not at equilibrium, in which direction will the system shift? a. $P_{\mathrm{NO}}=0.012 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.11 \mathrm{atm}, P_{\mathrm{O}_{2}}=2.0 \mathrm{atm}$ b. $P_{\mathrm{NO}}=0.0078 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.36 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.67 \mathrm{atm}$ c. $P_{\mathrm{NO}}=0.0062 \mathrm{atm}, P_{\mathrm{N}_{2}}=0.51 \mathrm{atm}, P_{\mathrm{O}_{2}}=0.18 \mathrm{atm}$

At a particular temperature, a \(3.0-\mathrm{L}\) flask contains 2.4 moles of \(\mathrm{Cl}_{2}, 1.0\) mole of \(\mathrm{NOCl}\) , and \(4.5 \times 10^{-3}\) mole of NO. Calculate \(K\) at this temperature for the following reaction: $$2 \mathrm{NOCl}(g) \rightleftharpoons 2 \mathrm{NO}(g)+\mathrm{Cl}_{2}(g)$$
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Ionic and Molecular Compounds

Read Explanation

Chemical Reactions

Read Explanation

Making Measurements

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

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