Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 59

Methane, \(\mathrm{CH}_{4}\), reacts with \(I_{2}\) according to the reaction $\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{I}(g)+\mathrm{HI}(g)\(. At \)600 \mathrm{~K}, K_{p}$ for this reaction is \(1.95 \times 10^{-4}\). A reaction was set up at 600 \(\mathrm{K}\) with initial partial pressures of methane of \(13.3 \mathrm{kPa}\) and of $6.67 \mathrm{kPa}\( for \)\mathrm{I}_{2}$. Calculate the pressures, in \(\mathrm{kPa}\), of all reactants and products at equilibrium.

Short Answer

Expert verified
The equilibrium pressures of all reactants and products are approximately: \[\mathrm{CH}_4: 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: 0.131\,\mathrm{kPa}\]

Step by step solution

01

1. Write down the balanced equation and the equilibrium constant expression.

The balanced equation is: \[\mathrm{CH}_{4}(g)+\mathrm{I}_{2}(g) \rightleftharpoons \mathrm{CH}_{3}\mathrm{I}(g)+\mathrm{HI}(g)\] The equilibrium constant expression for this reaction is: \[K_p = \frac{[\mathrm{CH}_3\mathrm{I}][\mathrm{HI}]}{[\mathrm{CH}_4][\mathrm{I}_2]}\] Where \([\mathrm{CH}_4]\), \([\mathrm{I}_2]\), \([\mathrm{CH}_3\mathrm{I}]\) and \([\mathrm{HI}]\) represent the equilibrium pressures of each species.
02

2. Set up a table with initial, change, and equilibrium pressures.

We will set up a table with the initial pressures, the changes in pressures due to the reaction and the equilibrium pressures. ``` CH4 I2 CH3I HI Initial 13.3 6.67 0 0 Change -x -x +x +x Equilibrium 13.3-x 6.67-x x x ``` Where x is the amount of both reactants consumed at equilibrium.
03

3. Plug in the equilibrium concentrations into the equilibrium constant expression.

Now we substitute the equilibrium pressures into the equilibrium constant expression: \[1.95 \times 10^{-4} = \frac{x^2}{(13.3 - x)(6.67 - x)}\]
04

4. Solve for x

It is not a simple quadratic equation, but it can be solved using a numerical method or approximation. Additionally, since \(K_p\) is much smaller than 1, we can make an assumption to simplify the calculation. We can assume that the change in pressure x is relatively small compared to the initial pressures of methane and iodine. So we can rewrite the expression as: \[1.95 \times 10^{-4} = \frac{x^2}{(13.3)(6.67)}\] Now solve for x: \[x^2 = 1.95 \times 10^{-4} \times (13.3)(6.67)\] \[x^2 = 1.715 \times 10^{-2}\] \[x = \sqrt{1.715 \times 10^{-2}}\] \[x = 0.131\]
05

5. Find the equilibrium pressures for all species.

As we assumed that the change in pressure x is relatively small compared to the initial pressures of methane and iodine, the equilibrium pressures of all species are: \[\mathrm{CH}_4: 13.3 - x = 13.3 - 0.131 \approx 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.67 - x = 6.67 - 0.131 \approx 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: x \approx 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: x \approx 0.131\,\mathrm{kPa}\] So, the equilibrium pressures of all reactants and products are approximately: \[\mathrm{CH}_4: 13.169\,\mathrm{kPa}\] \[\mathrm{I}_2: 6.539\,\mathrm{kPa}\] \[\mathrm{CH}_3\mathrm{I}: 0.131\,\mathrm{kPa}\] \[\mathrm{HI}: 0.131\,\mathrm{kPa}\]

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

Methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: $\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) .$ An equilibrium mixture in a 10.00-L vessel is found to contain \(0.050 \mathrm{~mol}\) $\mathrm{CH}_{3} \mathrm{OH}, 0.850 \mathrm{~mol} \mathrm{CO},\( and \)0.750 \mathrm{~mol} \mathrm{H}_{2}\( at \)500 \mathrm{~K}\(. Calculate \)K_{c}$ at this temperature.

\(\mathrm{NiO}\) is to be reduced to nickel metal in an industrial process by use of the reaction $$\mathrm{NiO}(s)+\mathrm{CO}(g) \rightleftharpoons \mathrm{Ni}(s)+\mathrm{CO}_{2}(g)$$ At \(1600 \mathrm{~K}\), the equilibrium constant for the reaction is $K_{p}=6.0 \times 10^{2} .\( If a CO pressure of \)20 \mathrm{kPa}$ is to be employed in the furnace and total pressure never exceeds \(101.3 \mathrm{kPa}\), will reduction occur?

(a) Is the dissociation of fluorine molecules into atomic fluorine, \(\mathrm{F}_{2}(g) \rightleftharpoons 2 \mathrm{~F}(g)\) an exothermic or endothermic process? (b) If the temperature is raised by \(100 \mathrm{~K}\), does the equilibrium constant for this reaction increase or decrease? (c) If the temperature is raised by \(100 \mathrm{~K},\) does the forward rate constant \(k_{f}\) increase by a larger or smaller amount than the reverse rate constant \(k_{r} ?\)

Which of the following reactions lies to the right, favoring the formation of products, and which lies to the left, favoring formation of reactants? (a) $\mathrm{H}_{2} \mathrm{O}(g)+\mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g)\( at \)1300 \mathrm{~K} K_{c}=0.57$ (b) \(2 \mathrm{CO}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{C}(s)\) at \(900 \mathrm{~K} K_{p}=0.0572\)

The protein hemoglobin (Hb) transports \(\mathrm{O}_{2}\) in mammalian blood. Each Hb can bind \(4 \mathrm{O}_{2}\) molecules. The equilibrium constant for the \(\mathrm{O}_{2}\) binding reaction is higher in fetal hemoglobin than in adult hemoglobin. In discussing protein oxygen-binding capacity, biochemists use a measure called the \(P 50\) value, defined as the partial pressure of oxygen at which \(50 \%\) of the protein is saturated. Fetal hemoglobin has a \(\mathrm{P} 50\) value of \(2.53 \mathrm{kPa},\) and adult hemoglobin has a P50 value of \(3.57 \mathrm{kPa}\). Use these data to estimate how much larger \(K_{c}\) is for the aqueous reaction $4 \mathrm{O}_{2}(g)+\mathrm{Hb}(a q) \rightleftharpoons\left[\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)\right]\( in a fetus, compared to \)K_{c}$ for the same reaction in an adult.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Organic Chemistry

Read Explanation

The Earths Atmosphere

Read Explanation

Kinetics

Read Explanation

Making Measurements

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