Introduction to General, Organic and Biochemistry
Introduction to General, Organic and Biochemistry
11th Edition
ISBN: 9781285869759
Author: Frederick A. Bettelheim, William H. Brown, Mary K. Campbell, Shawn O. Farrell, Omar Torres
Publisher: Cengage Learning
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Chapter 7, Problem 7.25P
Interpretation Introduction

(a)

Interpretation:

Equilibrium expression for the reaction 2H2O2(g)◰  2H2O(g)+O2(g) must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (Keq) which is defined as.

Keq=[Products][Reactants].

Expert Solution
Check Mark

Answer to Problem 7.25P

The equilibrium expression for the reaction 2H2O2(g)◰  2H2O(g)+O2(g) can be written as.

Keq=[H2O(g)]2[O2(g)][H2O2(g)]2.

Explanation of Solution

2H2O2(g)◰  2H2O(g)+O2(g)

In this reversible reaction products are H2O(g) and O2(g). Reactant is H2O2(g)

Now we know that Keq=[Products][Reactants]

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, Keq=[H2O(g)]2[O2(g)][H2O2(g)]2.

Interpretation Introduction

(b)

Interpretation:

Equilibrium expression for the reaction 2N2O5(g)◰  2NO2(g)+O2(g) must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (Keq) which is defined as.

Keq=[Products][Reactants].

Expert Solution
Check Mark

Answer to Problem 7.25P

The equilibrium expression for the reaction 2N2O5(g)◰  2NO2(g)+O2(g) can be written as.

Keq=[NO2(g)]2[O2(g)][N2O5(g)]2.

Explanation of Solution

2N2O5(g)◰  2NO2(g)+O2(g)

In this reversible reaction products are NO2(g) and O2(g). Reactant is N2O5(g)

Now we know that Keq=[Products][Reactants]

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, Keq=[NO2(g)]2[O2(g)][N2O5(g)]2.

Interpretation Introduction

(c)

Interpretation:

Equilibrium expression for the reaction 6H2O(g)+6CO2(g)◰  C6H12O6(s)+6O2(g) must be written.

Concept Introduction:

These equilibrium concentrations are used for the calculation of equilibrium constant (Keq) which is defined as.

Keq=[Products][Reactants]

Active mass of solid substances is taken as unity and thus not used in any equilibrium expression.

Expert Solution
Check Mark

Answer to Problem 7.25P

The equilibrium expression for the reaction 6H2O(g)+6CO2(g)◰  C6H12O6(s)+6O2(g) can be written as.

Keq=[O2(g)]6[H2O(g)]6[CO2(g)]6.

Explanation of Solution

6H2O(g)+6CO2(g)◰  C6H12O6(s)+6O2(g)

In this reversible reaction products are C6H12O6(s) and O2(g).Reactant are H2O(g)and CO2(g).But C6H12O6 is solid and so must be excluded from equilibrium expression.

Now we know that Keq=[Products][Reactants]

Coefficients of reactants and products are raised to power of the concentrations of reactants and products.

Thus, Keq=[O2(g)]6[H2O(g)]6[CO2(g)]6.

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Chapter 7 Solutions

Introduction to General, Organic and Biochemistry

Ch. 7 - 7-11 Consider the following reaction: Suppose we...Ch. 7 - 7-12 Two kinds of gas molecules are reacted at a...Ch. 7 - 7-13 Why are reactions between ions in aqueous...Ch. 7 - Prob. 7.14PCh. 7 - 7-15 A certain reaction is exothermic by 9...Ch. 7 - 7-16 A quart of milk quickly spoils if left at...Ch. 7 - 7-17 If a certain reaction takes 16 h to go to...Ch. 7 - Prob. 7.18PCh. 7 - Prob. 7.19PCh. 7 - Prob. 7.20PCh. 7 - Prob. 7.21PCh. 7 - 7-22 If you add a piece of marble, CaCO3 to a 6 M...Ch. 7 - Prob. 7.23PCh. 7 - Prob. 7.24PCh. 7 - Prob. 7.25PCh. 7 - 7-26 Write the chemical equations corresponding to...Ch. 7 - Prob. 7.27PCh. 7 - 7-28 When the following reaction reached...Ch. 7 - 7-29 The following reaction was allowed to reach...Ch. 7 - Prob. 7.30PCh. 7 - 7-31 Here are equilibrium constants for several...Ch. 7 - 7-32 A particular reaction has an equilibrium...Ch. 7 - Prob. 7.33PCh. 7 - Prob. 7.34PCh. 7 - 7-35 A reaction has a high rate constant but a...Ch. 7 - 7-36 Complete the following table showing the...Ch. 7 - Prob. 7.37PCh. 7 - Prob. 7.38PCh. 7 - Prob. 7.39PCh. 7 - 7-40 Is there any change in conditions that change...Ch. 7 - 7-41 The equilibrium constant at 1127°C for the...Ch. 7 - Prob. 7.42PCh. 7 - 7-43 (Chemical Connections 7A and 7B) Why is a...Ch. 7 - Prob. 7.44PCh. 7 - 7-45 (Chemical Connections 7C) A painkiller—for...Ch. 7 - 7-46 (Chemical Connections 7D) What reaction takes...Ch. 7 - Prob. 7.47PCh. 7 - Prob. 7.48PCh. 7 - Prob. 7.49PCh. 7 - 7-50 Draw an energy diagram for an exothermic...Ch. 7 - Prob. 7.51PCh. 7 - Prob. 7.52PCh. 7 - Prob. 7.53PCh. 7 - Prob. 7.54PCh. 7 - Prob. 7.55PCh. 7 - Prob. 7.56PCh. 7 - 7-57 Write the reaction to which the following...Ch. 7 - Prob. 7.58PCh. 7 - Prob. 7.59PCh. 7 - Prob. 7.60PCh. 7 - Prob. 7.61PCh. 7 - Prob. 7.62PCh. 7 - Prob. 7.63PCh. 7 - 7-64 As we shall see in Chapter 20, there are two...Ch. 7 - Prob. 7.65PCh. 7 - Prob. 7.66PCh. 7 - Prob. 7.67PCh. 7 - Prob. 7.68PCh. 7 - 7-69 Pure carbon exists is several forms, two of...Ch. 7 - Prob. 7.70PCh. 7 - 7-71 You have a beaker that contains solid silver...Ch. 7 - Prob. 7.72PCh. 7 - Prob. 7.73PCh. 7 - Prob. 7.74PCh. 7 - Prob. 7.75PCh. 7 - Prob. 7.76PCh. 7 - Prob. 7.77PCh. 7 - Prob. 7.78PCh. 7 - Prob. 7.79PCh. 7 - Prob. 7.80PCh. 7 - Prob. 7.81PCh. 7 - 7-82 An equilibrium mixture of O2, SO2, and SO3...Ch. 7 - Prob. 7.83PCh. 7 - Prob. 7.84P
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