Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 14: Problem 76

A mixture of 0.47 mole of \(\mathrm{H}_{2}\) and 3.59 moles of \(\mathrm{HCl}\) is heated to \(2800^{\circ} \mathrm{C}\). Calculate the equilibrium partial pressures of \(\mathrm{H}_{2}, \mathrm{Cl}_{2},\) and \(\mathrm{HCl}\) if the total pressure is 2.00 atm. For the reaction $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$ \(K_{P}\) is 193 at \(2800^{\circ} \mathrm{C}\)

Short Answer

Expert verified
The solution to the problem requires the application of the Le Chatelier's principle, the use the ideal gas law to convert between moles and pressure and the equilibrium constant. At the end of the calculations, values are obtained for the equilibrium partial pressures of \(\mathrm{H}_{2}\), \(\mathrm{Cl}_{2}\), and \(\mathrm{HCl}\).

Step by step solution

01

- Defining Initial and Change in Moles

From the reaction's stoichiometry, we know the initial amount of HCl is 3.59 moles and we have 0.47 moles of \(H_{2}\). We do not know the initial amount of \(Cl_{2}\), but we know the total pressure is 2 atm. We define \(x\) as the change in moles of H2 and Cl2, and 2x as the change in molar quantity of HCl. At equilibrium, the moles of \(H_{2}\) and \(Cl_{2}\) will both be \(0.47 - x\) while for HCl it'll be \(3.59 - 2x\). Putting these into an ICE table, we get: \nH2, \tInitial \(0.47 - x\) \tChange \(x\) \tEquilibrium \(0.47 - x\) \nCl2, \tInitial \(0.47 - x\) \tChange \(x\) \tEquilibrium \(0.47 - x\) \nHCl, \tInitial \(3.59 - 2x\) \tChange -\(2x\) \tEquilibrium \(3.59 - 2x\)
02

- Calculating the Partial Pressures

Next, convert the molar quantities at equilibrium into partial pressures using the ideal gas law: \[P = \frac{n \cdot R \cdot T}{V}\]\nSince the total pressure is given, and assuming the volume of the container and the temperature are constant, the partial pressures for each component will be equal to their mole fraction times the total pressure. Therefore, at equilibrium the partial pressures are:\n \(P_{H_{2}} = \frac{(0.47 - x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\)\n \[P_{Cl_{2}} = \frac{(0.47 - x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\]\n \[P_{HCl} = \frac{(3.59 -2x)}{(0.47 -x) + (0.47 -x ) + (3.59-2x)} \cdot 2\]
03

- Solving for x using the Equilibrium Constant

Now we can substitute the expressions for the partial pressures into the expression for \(K_P\):\n\(193 = K_P = \frac{P_{HCl}^{2}}{P_{H_{2}} \cdot P_{Cl_{2}}}\)\nSubstituting the expression for the partial pressures from step 2 and solving for \(x\) will give the changes in molar quantities at equilibrium. From which we calculate the equilibrium partial pressures.

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

The equilibrium constant \(\left(K_{\mathrm{c}}\right)\) for the reaction $$2 \mathrm{HCl}(g) \rightleftharpoons \mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g)$$ is \(4.17 \times 10^{-34}\) at \(25^{\circ} \mathrm{C} .\) What is the equilibrium constant for the reaction $$\mathrm{H}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{HCl}(g)$$ at the same temperature?

The forward and reverse rate constants for the reaction \(\mathrm{A}(g)+\mathrm{B}(g) \rightleftharpoons \mathrm{C}(g)\) are \(3.6 \times 10^{-3} / M \cdot \mathrm{s}\) and \(8.7 \times 10^{-4} \mathrm{~s}^{-1},\) respectively, at \(323 \mathrm{~K}\). Calculate the equilibrium pressures of all the species starting at \(P_{\mathrm{A}}=1.6 \mathrm{~atm}\) and \(P_{\mathrm{B}}=0.44 \mathrm{~atm}\).

What effect does an increase in pressure have on each of the following systems at equilibrium? The temperature is kept constant and, in each case, the reactants are in a cylinder fitted with a movable piston. (a) \(\mathrm{A}(s) \rightleftharpoons 2 \mathrm{~B}(s)\) (b) \(2 \mathrm{~A}(l) \rightleftharpoons \mathrm{B}(l)\) (c) \(\mathrm{A}(s) \rightleftharpoons \mathrm{B}(g)\) (d) \(\mathrm{A}(g) \rightleftharpoons \mathrm{B}(g)\) (e) \(\mathrm{A}(g) \rightleftharpoons 2 \mathrm{~B}(g)\)

The equilibrium constant \(K_{\mathrm{c}}\) for the reaction $$\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{HI}(g)$$ is 54.3 at \(430^{\circ} \mathrm{C}\). At the start of the reaction there are 0.714 mole of \(\mathrm{H}_{2}, 0.984\) mole of \(\mathrm{I}_{2}\), and 0.886 mole of HI in a 2.40 -L reaction chamber. Calculate the concentrations of the gases at equilibrium.

Consider the dissociation of iodine: $$\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)$$ A 1.00-g sample of \(I_{2}\) is heated to \(1200^{\circ} \mathrm{C}\) in a \(500-\mathrm{mL}\) flask. At equilibrium the total pressure is 1.51 atm. Calculate \(K_{P}\) for the reaction. [Hint: Use the result in \(14.117(\mathrm{a}) .\) The degree of dissociation \(\alpha\) can be obtained by first calculating the ratio of observed pressure over calculated pressure, assuming no dissociation.]
See all solutions

Recommended explanations on Chemistry Textbooks

The Earths Atmosphere

Read Explanation

Physical Chemistry

Read Explanation

Making Measurements

Read Explanation

Chemical Reactions

Read Explanation

Chemistry Branches

Read Explanation

Inorganic 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