/*! 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 34 Methanol, \(\mathrm{CH}_{3} \mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Methanol, \(\mathrm{CH}_{3} \mathrm{OH},\) formerly known as wood alcohol, is manufactured commercially by the following reaction: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ A 1.500-L vessel was filled with \(0.1500 \mathrm{~mol} \mathrm{CO}\) and 0.3000 mol \(\mathrm{H}_{2}\). When this mixture came to equilibrium at \(500 \mathrm{~K}\), the vessel contained \(0.1187 \mathrm{~mol} \mathrm{CO}\). How many moles of each substance were in the vessel?

Short Answer

Expert verified
At equilibrium: \(0.1187 \, \text{mol CO}\), \(0.2374 \, \text{mol H}_2\), \(0.0313 \, \text{mol CH}_3\mathrm{OH}\).

Step by step solution

01

Initial information and equation setup

We are given the initial moles of \( \mathrm{CO} \) and \( \mathrm{H}_2 \), as well as the reaction. We need to determine the changes in their amounts when equilibrium is established. Start by writing the balanced chemical equation: \( \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \). Initial moles in the vessel are \( n(\mathrm{CO}) = 0.1500 \, \text{mol} \) and \( n(\mathrm{H}_2) = 0.3000 \, \text{mol} \).
02

Using stoichiometry to determine change in moles

At equilibrium, the moles of \( \mathrm{CO} \) have decreased to \( 0.1187 \, \text{mol} \). Thus, the change in \( \mathrm{CO} \) is \( \Delta n(\mathrm{CO}) = 0.1500 - 0.1187 = 0.0313 \, \text{mol} \). Due to the stoichiometry of the reaction, 2 moles of \( \mathrm{H}_2 \) are required to react with each mole of \( \mathrm{CO} \). Therefore, \( \Delta n(\mathrm{H}_2) = 2 \times \Delta n(\mathrm{CO}) = 0.0626 \, \text{mol} \).
03

Calculating equilibrium moles of hydrogen

Subtract the change in moles from the initial moles of \( \mathrm{H}_2 \) to find the equilibrium moles: \( n_{eq}(\mathrm{H}_2) = 0.3000 - 0.0626 = 0.2374 \, \text{mol} \).
04

Determining moles of methanol at equilibrium

For every mole of \( \mathrm{CO} \) that reacts, one mole of \( \mathrm{CH}_3\mathrm{OH} \) is produced, due to the 1:1 ratio in the reaction. Therefore, \( n_{eq}(\mathrm{CH}_3\mathrm{OH}) = \Delta n(\mathrm{CO}) = 0.0313 \, \text{mol} \). This is the amount of \( \mathrm{CH}_3\mathrm{OH} \) produced at equilibrium.

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Stoichiometry
Stoichiometry is a crucial concept in chemistry that involves calculating the quantities of reactants and products in a chemical reaction. It is based on the balanced chemical equation for the reaction, which indicates the ratio in which molecules interact. Stoichiometry allows chemists to predict how much of a substance is consumed or produced in a reaction, given the amounts of the other substances involved.

In the methanol production problem, we used stoichiometry to assess how much of the initial substances would react. The balanced equation is:
  • \( \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \)
From this equation, it is clear that one mole of carbon monoxide (CO) reacts with two moles of hydrogen (H₂) to produce one mole of methanol (CH₃OH).

By applying stoichiometric principles, one can determine the change in mole quantities as the reaction proceeds. When we know how much one component changes, stoichiometry helps us deduce the changes in others, maintaining a balance between reactants and products. This balance is dictated by the coefficients in the chemical equation.
Methanol Production
Methanol, also known as wood alcohol, is an important chemical used in various industries. It is often produced through a chemical reaction involving carbon monoxide and hydrogen gases. This reaction occurs under specific conditions of temperature and pressure to favor the formation of methanol.

The equilibrium for methanol production is represented by the equation:
  • \( \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \)
This reaction is reversible, meaning it can proceed in both forward and backward directions depending on the surrounding conditions.

In an industrial setting, this reaction is carried out in a controlled environment to maximize methanol yield. Optimizing the pressure, temperature, and proportions of reactants are key. Understanding chemical equilibrium, such as achieving the right balance between reactants and products at a certain temperature like 500 K as mentioned in the exercise, is essential in maximizing methanol production efficiency.
Reaction Stoichiometry
Reaction stoichiometry specifically refers to the quantitative relationships between reactants and products in a chemical reaction. It uses the balanced chemical equation to derive how much of each substance is involved in the reaction, based on their molar ratios.

The methanol production problem highlights how the stoichiometry of the reaction is used to calculate the equilibrium state. For instance, we determined the change in moles of carbon monoxide (CO) from the initial amount to its equilibrium amount, finding that 0.0313 moles of CO reacted. Using the stoichiometry indicated by the reaction equation:
  • \( \mathrm{CO}(g) + 2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \)
We derived that for every mole of CO that reacts, two moles of Hâ‚‚ are consumed and likewise, one mole of methanol is produced.

This 1:2:1 ratio is crucial for solving such problems, as it helps determine the quantities of each substance involved when equilibrium is reached. By consistently applying reaction stoichiometry, one can predict the outcomes of chemical processes and adjust conditions as necessary for desired results.

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

Phosgene was used as a poisonous gas in World War I. At high temperatures it decomposes as follows: $$ \begin{aligned} \mathrm{COCl}_{2}(g) & \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \\\ \text { with } K_{c}=4.6 \times 10^{-3} \text {at } 800 \mathrm{~K} \end{aligned} $$ a) A sample of \(7.80 \mathrm{~g}\) of \(\mathrm{COCl}_{2}\) is placed in a \(1.00-\mathrm{L}\) reaction vessel and heated to \(800 \mathrm{~K}\). b. What are the equilibrium concentrations of all of the species? What fraction of \(\mathrm{COCl}_{2}\) has decomposed? c. If \(3 \mathrm{~g}\) of carbon monoxide is inserted into the reaction vessel, what qualitative effect would this have on the fraction of \(\mathrm{COCl}_{2}\) that has decomposed?

When a continuous stream of hydrogen gas, \(\mathrm{H}_{2}\), passes over hot magnetic iron oxide, \(\mathrm{Fe}_{3} \mathrm{O}_{4},\) metallic iron and water vapor form. When a continuous stream of water vapor passes over hot metallic iron, the oxide \(\mathrm{Fe}_{3} \mathrm{O}_{4}\) and \(\mathrm{H}_{2}\) form. Explain why the reaction goes in one direction in one case but in the reverse direction in the other.

The reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ has \(K_{p}\) equal to 6.55 at \(627^{\circ} \mathrm{C}\). What is the value of \(K_{c}\) at this temperature?

Write the expression for the equilibrium constant \(K_{c}\) for each of the following equations: $$ \begin{array}{l} \text { a } \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) \\ \text { b } \mathrm{FeO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Fe}(s)+\mathrm{CO}_{2}(g) \\ \text { c } \mathrm{Na}_{2} \mathrm{CO}_{3}(s)+\mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Na}_{2} \mathrm{SO}_{4}(s)+\mathrm{CO}_{2}(g) \\ \text { d } \mathrm{PbI}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{I}^{-(a q)} \end{array} $$

At \(850^{\circ} \mathrm{C}\) and 1.000 atm pressure, a gaseous mixture of carbon monoxide and carbon dioxide in equilibrium with solid carbon is \(90.55 \%\) CO by mass. $$ \mathrm{C}(s)+\mathrm{CO}_{2}(g) \rightleftharpoons 2 \mathrm{CO}(g) $$ Calculate \(K_{c}\) for this reaction at \(850^{\circ} \mathrm{C}\)

See all solutions

Recommended explanations on Chemistry Textbooks

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.