Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 27

An equilibrium mixture at 1000 K contains an equilibrium mixter \(0.276\ \mathrm{mol}\ \mathrm{H}_{2}, 0.276 \mathrm{mol}\ \mathrm{CO}_{2}, 0.224\ \mathrm{mol}\ \mathrm{CO},\) and \(0.224\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{O}\) $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Show that for this reaction, \(K_{\mathrm{c}}\) is independent of the reaction volume, \(V\) (b) Determine the value of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\)

Short Answer

Expert verified
Part (a) demonstrates that the equilibrium constant \(K_c\) is indeed independent of volume. By quantifying the given equilibrium concentrations, the constants \(K_c\) and \(K_p\) for Part (b) are determined to both be 0.6596.

Step by step solution

01

Step 1- Demonstrating that \(K_c\) is independent of volume

First, rewrite \[K_c = \frac{[Products]}{[Reactants]}\] as \[K_c=\frac{([CO][H_{2}O])}{([CO_{2}][H_{2}])}\]Molar concentration is moles/volume. So, substituting concentration in terms of moles and volume into the \(K_c\) expression:\[K_c=\frac{\frac{(mol\ CO)(mol\ H_{2}O)}{V^2}}{\frac{(mol\ CO_{2})(mol\ H_{2})}{V^2}} = \frac{(mol\ CO)(mol\ H_{2}O)}{(mol\ CO_{2})(mol\ H_{2})} = K'_c\]As it can be seen, the volumes cancel each other out, concluding that \(K_c\) is independent of volume.
02

Step 2- Computing \(K_c\)

Substitute the given equilibrium concentrations into \(K_c\):\[K_c=\frac{(0.224)(0.224)}{(0.276)(0.276)} = 0.6596\]
03

Step 3- Computing \(K_p\)

To find \(K_p\), apply the equation \[K_p = K_c(RT)^{\Delta n}\] where \(R\) is the ideal gas constant (0.08206 L atm / K mol), \(T\) is the temperature in Kelvin (1000 K), and \(\Delta n\) is the change in moles of gas (final - initial). The reaction is \(CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2} O(g)\). Therefore, \(\Delta n = (1 + 1) - (1 + 1) = 0\). So, \(K_p = K_c(RT)^{\Delta n} = 0.6596(0.08206 \times 1000)^{0} = 0.6596\]

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

An equilibrium mixture of \(\mathrm{SO}_{2}, \mathrm{SO}_{3},\) and \(\mathrm{O}_{2}\) gases is maintained in a \(2.05 \mathrm{L}\) flask at a temperature at which \(K_{\mathrm{c}}=35.5\) for the reaction $$2 \mathrm{SO}_{2}(\mathrm{g})+\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g})$$ (a) If the numbers of moles of \(\mathrm{SO}_{2}\) and \(\mathrm{SO}_{3}\) in the flask are equal, how many moles of \(\mathrm{O}_{2}\) are present? (b) If the number of moles of \(\mathrm{SO}_{3}\) in the flask is twice the number of moles of \(\mathrm{SO}_{2}\), how many moles of \(\mathrm{O}_{2}\) are present?

Recall the formation of methanol from synthesis gas, the reversible reaction at the heart of a process with great potential for the future production of automotive fuels (page 663 ). $$\begin{aligned} \mathrm{CO}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{g}) & \\ K_{\mathrm{c}}=& 14.5 \mathrm{at} 483 \mathrm{K} \end{aligned}$$ A particular synthesis gas consisting of 35.0 mole percent \(\mathrm{CO}(g)\) and 65.0 mole percent \(\mathrm{H}_{2}(\mathrm{g})\) at a total pressure of 100.0 atm at \(483 \mathrm{K}\) is allowed to come to equilibrium. Determine the partial pressure of \(\mathrm{CH}_{3} \mathrm{OH}(\mathrm{g})\) in the equilibrium mixture.

\(1.00 \mathrm{g}\) each of \(\mathrm{CO}, \mathrm{H}_{2} \mathrm{O},\) and \(\mathrm{H}_{2}\) are sealed in a \(1.41 \mathrm{L}\) vessel and brought to equilibrium at 600 K. How many grams of \(\mathrm{CO}_{2}\) will be present in the equilibrium mixture? $$\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g}) \rightleftharpoons \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \quad K_{\mathrm{c}}=23.2$$

Equilibrium is established in the reaction \(2 \mathrm{SO}_{2}(\mathrm{g})+\) \(\mathrm{O}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{SO}_{3}(\mathrm{g}) \quad\) at \(\quad \mathrm{a} \quad\) temperature \(\quad\) where \(K_{\mathrm{c}}=100 .\) If the number of moles of \(\mathrm{SO}_{3}(\mathrm{g})\) in the equilibrium mixture is the same as the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}),\) (a) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is also equal to the number of moles of \(\mathrm{SO}_{2}(\mathrm{g}) ;\) (b) the number of moles of \(\mathrm{O}_{2}(\mathrm{g})\) is half the number of moles of \(\mathrm{SO}_{2} ;\) (c) \(\left[\mathrm{O}_{2}\right]\) may have any of several values; (d) \(\left[\mathrm{O}_{2}\right]=0.010 \mathrm{M}\)

What is the apparent molar mass of the gaseous mixture that results when \(\mathrm{COCl}_{2}(\mathrm{g})\) is allowed to dissociate at \(395^{\circ} \mathrm{C}\) and a total pressure of 3.00 atm? $$\begin{aligned} \operatorname{COCl}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+& \mathrm{Cl}_{2}(\mathrm{g}) \\ & K_{\mathrm{p}}=4.44 \times 10^{-2} \mathrm{at} 395^{\circ} \mathrm{C} \end{aligned}$$ Think of the apparent molar mass as the molar mass of a hypothetical single gas that is equivalent to the gaseous mixture.
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

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