Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 15: Problem 24

Consider the following equilibrium: \(2 \mathrm{H}_{2}(g)+\mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{S}(g) \quad K_{c}=1.08 \times 10^{7} \mathrm{at} 700^{\circ} \mathrm{C}\) (a) Calculate \(K_{p}\) . (b) Does the equilibrium mixture contain mostly \(\mathrm{H}_{2}\) and \(\mathrm{S}_{2}\) or mostly \(\mathrm{H}_{2} \mathrm{S} ?(\mathbf{c})\) Calculate the value of \(K_{c}\) if you rewrote the equation \(\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{S}_{2}(g) \rightleftharpoons \mathrm{H}_{2} \mathrm{S}(g)\)

Short Answer

Expert verified
In summary, we have calculated the equilibrium constant \(K_p\) for the given reaction to be 0.042. The equilibrium mixture consists predominantly of hydrogen sulfide (H₂S) rather than hydrogen (H₂) and sulfur (S₂). Finally, the value of Kc for the rewritten equilibrium equation is approximately 3,286.

Step by step solution

01

Understand the relationship between Kc and Kp

The relationship between Kc and Kp is given by the equation: \[K_p = K_c(RT)^{\Delta n}\] where R is the ideal gas constant, T is the temperature in Kelvin, and Δn is the difference in the number of moles of gas between the products and the reactants. For the given equilibrium: \[2 \mathrm{H}_{2}(g) + \mathrm{S}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2}\mathrm{S}(g)\] Δn = (2 moles of H2S) - (2 moles of H2 + 1 mole of S2) = -1. Now we can proceed to calculate Kp.
02

Convert temperature to Kelvin and find Kp

First, convert the temperature from Celsius to Kelvin: T = 700 + 273.15 = 973.15 K Next, use the relationship between Kc and Kp to find Kp: \[K_p = K_c(RT)^{\Delta n} = (1.08 \times 10^7)(8.314 \times 973.15)^{-1} = 0.042\] So, \(K_p = 0.042\). For part (b), we will analyze the equilibrium mixture.
03

Determine the predominant components of the equilibrium mixture

Since \(K_c\) is greater than 1 (\(K_c = 1.08 \times 10^7\)), this means that the equilibrium favors the product, hydrogen sulfide (H2S). Therefore, the equilibrium mixture contains mostly H2S rather than H2 and S2. For part (c), we will calculate the value of Kc for the modified equilibrium equation.
04

Understand the relationship between the original and modified Kc values

The original equilibrium equation can be rewritten as twice the modified equation: \(\begin{array}{cccc} 2\mathrm{H}_{2}(g)&+&\mathrm{S}_{2}(g)&\rightleftharpoons2\mathrm{H}_{2}\mathrm{S}(g)\\ 2(\mathrm{H}_{2}(g)&+\frac{1}{2}\mathrm{S}_{2}(g)&\rightleftharpoons\mathrm{H}_{2}\mathrm{S}(g))& \end{array}\) We can denote the original Kc value as \(K_{c1}\) and the modified Kc value as \(K_{c2}\). Since the second reaction is equal to half of the first reaction, their equilibrium constants are related by the expression: \[K_{c1} = K_{c2}^2\] Now we can calculate the modified Kc value.
05

Calculate the modified Kc value

Use the relationship found in Step 4 to solve for the modified Kc value: \[K_{c2} = \sqrt{K_{c1}} = \sqrt{1.08 \times 10^7} = 3,286\] So, the value of Kc for the modified equilibrium equation is approximately 3,286.

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

Water molecules in the atmosphere can form hydrogen-bonded dimers, \(\left(\mathrm{H}_{2} \mathrm{O}\right)_{2}\) . The presence of these dimers is thought to be important in the nucleation of ice crystals in the atmosphere and in the formation of acid rain. (a) Using VSEPR theory, draw the structure of a water dimer, using dashed lines to indicate intermolecular interactions. (b) What kind of intermolecular forces are involved in water dimer formation? (c) The \(K_{p}\) for water dimer formation in the gas phase is 0.050 at 300 \(\mathrm{K}\) and 0.020 at 350 \(\mathrm{K}\) . Is water dimer formation endothermic or exothermic?

At \(800 \mathrm{K},\) the equilibrium constant for \(\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{I}(g)\) is \(K_{c}=3.1 \times 10^{-5} .\) If an equilibrium mixture in a 10.0 -L. vessel contains \(2.67 \times 10^{-2} \mathrm{g}\) of I(g), how many grams of \(\mathrm{I}_{2}\) are in the mixture?

Two different proteins \(X\) and \(Y\) are dissolved in aqueous solution at \(37^{\circ} \mathrm{C}\) . The proteins bind in a \(1 : 1\) ratio to form \(X Y . A\) solution that is initially 1.00 \(\mathrm{mM}\) in each protein is allowed to reach equilibrium. At equilibrium, 0.20 \(\mathrm{mM}\) of free \(\mathrm{X}\) and 0.20 \(\mathrm{mM}\) of free Y remain. What is \(K_{c}\) for the reaction?

Both the forward reaction and the reverse reaction in the following equilibrium are believed to be elementary steps: $$\mathrm{CO}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \operatorname{COCl}(g)+\mathrm{Cl}(g)$$ At \(25^{\circ} \mathrm{C},\) the rate constants for the forward and reverse reactions are \(1.4 \times 10^{-28} M^{-1} \mathrm{s}^{-1}\) and \(9.3 \times 10^{10} M^{-1} \mathrm{s}^{-1}\) respectively. (a) What is the value for the equilibrium constant at \(25^{\circ} \mathrm{C} ?\) (b) Are reactants or products more plentiful at equilibrium?

A flask is charged with 1.500 atm of \(\mathrm{N}_{2} \mathrm{O}_{4}(g)\) and 1.00 atm \(\mathrm{NO}_{2}(g)\) at \(25^{\circ} \mathrm{C},\) and the following equilibrium is achieved: $$\mathrm{N}_{2} \mathrm{O}_{4}(g) \rightleftharpoons 2 \mathrm{NO}_{2}(g)$$ After equilibrium is reached, the partial pressure of \(\mathrm{NO}_{2}\) is 0.512 atm. (a) What is the equilibrium partial pressure of \(\mathrm{N}_{2} \mathrm{O}_{4} ?\) (b) Calculate the value of \(K_{p}\) for the reaction. (c) Calculate \(K_{c}\) for the reaction.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Kinetics

Read Explanation

Chemistry Branches

Read Explanation

Organic Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Nuclear 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