/*! 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 49 Write equilibrium-constant expre... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Write equilibrium-constant expressions \(K_{p}\) for each of the following reactions: a) \(\mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g)\) b) \(\mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{~S}(g)\) c) \(4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Cl}_{2}(g)\) d) \(\mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\)

Short Answer

Expert verified
a) \( K_{p, a} = \frac{(P_{\mathrm{HBr}})^2}{P_{\mathrm{H}_2} \cdot P_{\mathrm{Br}_2}} \); b) \( K_{p, b} = \frac{P_{\mathrm{CH}_{4}} \cdot (P_{\mathrm{H}_{2}S})^2}{P_{\mathrm{CS}_{2}} \cdot (P_{\mathrm{H}_2})^4} \); c) \( K_{p, c} = \frac{(P_{\mathrm{H}_2O})^2 \cdot (P_{\mathrm{Cl}_2})^2}{(P_{\mathrm{HCl}})^4 \cdot P_{\mathrm{O}_2}} \); d) \( K_{p, d} = \frac{P_{\mathrm{CH}_3OH}}{P_{\mathrm{CO}} \cdot (P_{\mathrm{H}_2})^2} \).

Step by step solution

01

Understand the Equilibrium Constant Expression

The equilibrium constant expression in terms of partial pressures, denoted as \( K_{p} \), is based on the partial pressures of gases in a balanced chemical equation. For a general reaction of the form \( aA + bB \rightleftharpoons cC + dD \), the expression is \( K_p = \frac{P_C^c \cdot P_D^d}{P_A^a \cdot P_B^b} \) where \( P \) refers to the partial pressure of each component.
02

Write the \( K_{p} \) Expression for Reaction a

For the reaction \( \mathrm{H}_{2}(g)+\mathrm{Br}_{2}(g) \rightleftharpoons 2 \mathrm{HBr}(g) \), apply the formula. The coefficients in the balanced equation become exponents: \[ K_{p, a} = \frac{(P_{\mathrm{HBr}})^2}{P_{\mathrm{H}_2} \cdot P_{\mathrm{Br}_2}} \]
03

Write the \( K_{p} \) Expression for Reaction b

For the reaction \( \mathrm{CS}_{2}(g)+4 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+2 \mathrm{H}_{2} \mathrm{S}(g) \), use the equilibrium constant expression format: \[ K_{p, b} = \frac{P_{\mathrm{CH}_{4}} \cdot (P_{\mathrm{H}_{2}S})^2}{P_{\mathrm{CS}_{2}} \cdot (P_{\mathrm{H}_2})^4} \]
04

Write the \( K_{p} \) Expression for Reaction c

For \( 4 \mathrm{HCl}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{H}_{2} \mathrm{O}(g)+2 \mathrm{Cl}_{2}(g) \), the \( K_p \) expression is: \[ K_{p, c} = \frac{(P_{\mathrm{H}_2O})^2 \cdot (P_{\mathrm{Cl}_2})^2}{(P_{\mathrm{HCl}})^4 \cdot P_{\mathrm{O}_2}} \]
05

Write the \( K_{p} \) Expression for Reaction d

For the reaction \( \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) \), write the expression: \[ K_{p, d} = \frac{P_{\mathrm{CH}_3OH}}{P_{\mathrm{CO}} \cdot (P_{\mathrm{H}_2})^2} \]

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.

Equilibrium Expressions
An equilibrium expression is essential when dealing with chemical reactions that reach a state of balance. In a balanced chemical equation, the rate at which reactants are converted into products is equal to the rate at which products revert back to reactants. The equilibrium constant expression is derived from this balance.
For reactions involving gases, we use the partial pressures of the reactants and products to express this equilibrium state mathematically using the equilibrium constant \( K_{p} \). It is given by the ratio of the product of the partial pressures of the products raised to their stoichiometric coefficients, divided by the product of the partial pressures of the reactants raised to their stoichiometric coefficients.

This enables chemists to predict whether a reaction favors the formation of products or the retention of reactants under specific conditions. A high \( K_{p} \) value indicates a greater concentration of products at equilibrium, while a low \( K_{p} \) signifies a greater concentration of reactants.
Partial Pressure
In chemistry, partial pressure is a critical concept that applies to gaseous reactions and equilibrium expressions. It refers to the pressure exerted by a single type of gas in a mixture of gases, as though it occupied the entire volume by itself. Each gas in a mixture contributes to the total pressure according to its mole fraction and its individual properties.
The basic formula for calculating partial pressure \( P_i \) of a gas \( i \) in a mixture is:\[ P_i = X_i \cdot P_{total} \]where:- \( X_i \) is the mole fraction of gas \( i \).- \( P_{total} \) is the total pressure of the gas mixture.

Partial pressures are used in equilibrium expressions \( K_{p} \) to quantify how much of each gas participates in achieving equilibrium. Understanding how to calculate and use partial pressures helps to analyze and predict the outcome of reactions at a given temperature and pressure.
Chemical Reactions
Chemical reactions are processes where reactants convert into products through the breaking and forming of chemical bonds. Each reaction can be represented by a balanced chemical equation that accounts for all atoms or molecules involved.
These reactions can be reversible or irreversible. For reversible reactions, equilibrium can be established. Discovering this balance points us toward understanding how a system behaves over time. The stoichiometric coefficients in the equation become critical when setting up equilibrium expressions, as they determine the exponents for the partial pressures of each gas in the equilibrium constant expression.

The study of chemical reactions involves monitoring changes in concentration or pressure over time, enabling a deeper insight into the reaction dynamics and potential energy changes within the system.
Kp Expressions
\( K_{p} \) expressions are used specifically for reactions involving gases and utilize partial pressures in place of concentrations. These expressions are a tool for chemists to determine the position of equilibrium in gaseous reactions. They reveal how pressure changes can shift the equilibrium toward either the reactants or products.
For instance, consider a simple reaction: \( A(g) + B(g) \rightleftharpoons C(g) + D(g) \). The \( K_{p} \) expression for this reaction would be: \[ K_{p} = \frac{P_C \cdot P_D}{P_A \cdot P_B} \]
Here, each \( P \) stands for the partial pressure of the respective component. Proper application of \( K_{p} \) expressions allows predictions about how changes in pressure or volume may affect the system, adhering to Le Chatelier's Principle.

Practical applications of \( K_{p} \) involve setting up conditions that favor desired product formation in industrial chemical processes, proving \( K_{p} \) indispensable in both theoretical and practical chemistry pursuits.

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

Suppose \(1.000 \mathrm{~mol} \mathrm{CO}\) and \(3.000 \mathrm{~mol} \mathrm{H}_{2}\) are put in a 10.00 - \(\mathrm{L}\) vessel at \(1200 \mathrm{~K}\). The equilibrium constant \(K_{c}\) for $$ \mathrm{CO}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ equals 3.92. Find the equilibrium composition of the reaction mixture.

The equilibrium constant \(K_{c}\) for the reaction $$ \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons \mathrm{PCl}_{5}(g) $$ equals 4.1 at \(300^{\circ} \mathrm{C}\) a) A sample of \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) is placed in a 5.0 - \(\mathrm{L}\) reaction vessel and heated to \(300^{\circ} \mathrm{C}\). b) What are the equilibrium concentrations of all of the species? What fraction of \(\mathrm{PCl}_{5}\) has decomposed? c) If \(35.8 \mathrm{~g}\) of \(\mathrm{PCl}_{5}\) were placed in a 1.0 - \(\mathrm{L}\) vessel, what qualitative effect would this have on the fraction of \(\mathrm{PCl}_{5}\) that has decomposed (give a qualitative answer only; do not do the calculation)? Why?

Suppose sulfur dioxide reacts with oxygen at \(25^{\circ} \mathrm{C}\). $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ The equilibrium constant \(K_{c}\) equals \(8.0 \times 10^{35}\) at this temperature. From the magnitude of \(K_{c}\), do you think this reaction occurs to any great extent when equilibrium is reached at room temperature? If an equilibrium mixture is \(1.0 \mathrm{M}\) \(\mathrm{SO}_{3}\) and has equal concentrations of \(\mathrm{SO}_{2}\) and \(\mathrm{O}_{2}\), what is the concentration of \(\mathrm{SO}_{2}\) in the mixture? Does this result agree with what you expect from the magnitude of \(K_{c} ?\)

At some temperature, a \(100-\mathrm{L}\) reaction vessel contains a mixture that is initially \(1.00 \mathrm{~mol} \mathrm{CO}\) and \(2.00 \mathrm{~mol}\) \(\mathrm{H}_{2}\). The vessel also contains a catalyst so that the following equilibrium is attained: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ At equilibrium, the mixture contains \(0.100 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\) (methanol). In a later experiment in the same vessel, you start with \(1.00 \mathrm{~mol} \mathrm{CH}_{3} \mathrm{OH}\). How much methanol is there at equilibrium? Explain.

For the reaction $$ \mathrm{COCl}_{2}(g) \rightleftharpoons \mathrm{CO}(g)+\mathrm{Cl}_{2}(g) $$ show that $$ K_{c}=K_{p} /(R T) $$ Do not use the formula \(K_{p}=K_{c}(R T)^{\Delta n}\) given in the text. See Problem 14.99.

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.