Atkins' Physical Chemistry
Atkins' Physical Chemistry
11th Edition
ISBN: 9780198769866
Author: ATKINS, P. W. (peter William), De Paula, Julio, Keeler, JAMES
Publisher: Oxford University Press
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Chapter 2, Problem 2C.5AE

(i)

Interpretation Introduction

Interpretation: The values of ΔrH° and ΔrU° has to be calculated for the given reaction.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.

(i)

Expert Solution
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Answer to Problem 2C.5AE

The value of ΔrH° for the reaction (3) is 114.4kJmol1_.  The value of ΔrU° for the reaction (3) is 111.922kJmol1_.

Explanation of Solution

The given reactions are:

    H2(g)+Cl2(g)2HCl(g)                                                                        (1)

    2H2(g)+O2(g)2H2O(g)                                                                      (2)

    4HCl(g)+O2(g)2Cl2(g)+2H2O(g)                                                   (3)

The given value of ΔrH° for the reaction (1) is 184.62kJmol1 and the given value of ΔrH° for the reaction (2) is 483.64kJmol1.

To calculate the ΔrH° value for the reaction (3), multiply reaction (1) by 2 and reverse it as shown below.

    2×(H2(g)+Cl2(g)2HCl(g))

    4HCl(g)2H2(g)+2Cl2(g)                                                                     (4)

The value of ΔrH° for the reaction (4) will be,

  2×(184.62kJmol1)=369.24kJmol1

Add reaction (4) to reaction (2).

    4HCl(g)2H2(g)+2Cl2(g)2H2(g)+O2(g)2H2O(g)__________________________________4HCl(g)+O2(g)2Cl2(g)+2H2O(g)

To calculate the ΔrH° value for the reaction (3), use the enthalpy of reaction (4) and (2).

    ΔrH°=369.24kJmol1+(483.64)kJmol1=-114.4 kJmol-1_

Hence, the value of ΔrH° for the reaction (3) is -114.4 kJmol-1_.

The change in number of moles of the reaction (3) is calculated by the formula,

  Numberofmoles,Δng=MolesofgaseousproductMolesofgaseousreactant

Hence, the number of moles for the reaction (3) is,

  Δng=(2+2)(4+1)=1

The expression to calculate the value of ΔrU° for the reaction (3) is,

    ΔrU°=ΔrH°ΔngRT

Substitute the values of ΔrH°, R, Δng and T=298K in the above equation to calculate the value of ΔrU°.

  ΔrU°=114.4kJmol1(1×8.314×103kJK1mol1×298K)=114.4kJmol1(2477.572×103kJmol1)=-111.922kJmol-1_

Hence, the value of ΔrU° for the reaction (3) is -111.922kJmol-1_.

(ii)

Interpretation Introduction

Interpretation: The value of ΔfH° for both HCl(g) and H2O(g) has to be calculated.

Concept introduction: Enthalpy is the internal heat content of the system.  Under constant pressure, a system can do expansion work under the effect of energy supplied as heat known as enthalpy change of the system.  Therefore, the measure of amount of heat supplied to the system gives the enthalpy change of the system.  The total amount of energy possessed within the given system is known as internal energy of the system. It is denoted by U.

(ii)

Expert Solution
Check Mark

Answer to Problem 2C.5AE

The value of ΔfH° for HCl(g) is 92.31kJmol_1.  The value of ΔfH° for H2O(g) is 241.82kJmol_1.

Explanation of Solution

The given reactions are:

    H2(g)+Cl2(g)2HCl(g)                                                                        (1)

    2H2(g)+O2(g)2H2O(g)                                                                      (2)

    4HCl(g)+O2(g)2Cl2(g)+2H2O(g)                                                   (3)

The given value of ΔrH° for the reaction (1) is 184.62kJmol1 and the given value of ΔrH° for the reaction (2) is 483.64kJmol1.

The value of ΔfH° for HCl(g) is calculate by the expression,

  ΔfH°=ΔrH°2

The value of ΔrH° for the reaction (1) is used to calculate the value of ΔfH° for HCl(g).  The value of ΔrH° for the reaction (1) is 184.62kJmol1.

From the above expression, the value of ΔfH° for HCl(g) is,

    ΔfH°=184.62kJmol12=92.31kJmol_1

Hence, the value of ΔfH° for HCl(g) is 92.31kJmol_1.

The value of ΔfH° for H2O(g) is calculate by the expression,

  ΔfH°=ΔrH°2

The value of ΔrH° for the reaction (1) is used to calculate the value of ΔfH° for H2O(g).  The value of ΔrH° for the reaction (2) is 483.64kJmol1.

From the above expression, the value of ΔfH° for H2O(g) is,

    ΔfH°=483.64kJmol12=241.82kJmol_1

Hence, the value of ΔfH° for H2O(g) is 241.82kJmol_1.

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Chapter 2 Solutions

Atkins' Physical Chemistry

Ch. 2 - Prob. 2A.4DQCh. 2 - Prob. 2A.5DQCh. 2 - Prob. 2A.1AECh. 2 - Prob. 2A.1BECh. 2 - Prob. 2A.2AECh. 2 - Prob. 2A.2BECh. 2 - Prob. 2A.3AECh. 2 - Prob. 2A.3BECh. 2 - Prob. 2A.4AECh. 2 - Prob. 2A.4BECh. 2 - Prob. 2A.5AECh. 2 - Prob. 2A.5BECh. 2 - Prob. 2A.6AECh. 2 - Prob. 2A.6BECh. 2 - Prob. 2A.1PCh. 2 - Prob. 2A.2PCh. 2 - Prob. 2A.3PCh. 2 - Prob. 2A.4PCh. 2 - Prob. 2A.5PCh. 2 - Prob. 2A.6PCh. 2 - Prob. 2A.7PCh. 2 - Prob. 2A.8PCh. 2 - Prob. 2A.9PCh. 2 - Prob. 2A.10PCh. 2 - Prob. 2B.1DQCh. 2 - Prob. 2B.2DQCh. 2 - Prob. 2B.1AECh. 2 - Prob. 2B.1BECh. 2 - Prob. 2B.2AECh. 2 - Prob. 2B.2BECh. 2 - Prob. 2B.3AECh. 2 - Prob. 2B.3BECh. 2 - Prob. 2B.4AECh. 2 - Prob. 2B.4BECh. 2 - Prob. 2B.1PCh. 2 - Prob. 2B.2PCh. 2 - Prob. 2B.3PCh. 2 - Prob. 2B.4PCh. 2 - Prob. 2B.5PCh. 2 - Prob. 2C.1DQCh. 2 - Prob. 2C.2DQCh. 2 - Prob. 2C.3DQCh. 2 - Prob. 2C.4DQCh. 2 - Prob. 2C.1AECh. 2 - Prob. 2C.1BECh. 2 - Prob. 2C.2AECh. 2 - Prob. 2C.2BECh. 2 - Prob. 2C.3AECh. 2 - Prob. 2C.3BECh. 2 - Prob. 2C.4AECh. 2 - Prob. 2C.4BECh. 2 - Prob. 2C.5AECh. 2 - Prob. 2C.5BECh. 2 - Prob. 2C.6AECh. 2 - Prob. 2C.6BECh. 2 - Prob. 2C.7AECh. 2 - Prob. 2C.7BECh. 2 - Prob. 2C.8AECh. 2 - Prob. 2C.8BECh. 2 - Prob. 2C.1PCh. 2 - Prob. 2C.2PCh. 2 - Prob. 2C.3PCh. 2 - Prob. 2C.4PCh. 2 - Prob. 2C.5PCh. 2 - Prob. 2C.6PCh. 2 - Prob. 2C.7PCh. 2 - Prob. 2C.8PCh. 2 - Prob. 2C.9PCh. 2 - Prob. 2C.10PCh. 2 - Prob. 2C.11PCh. 2 - Prob. 2D.1DQCh. 2 - Prob. 2D.2DQCh. 2 - Prob. 2D.1AECh. 2 - Prob. 2D.1BECh. 2 - Prob. 2D.2AECh. 2 - Prob. 2D.2BECh. 2 - Prob. 2D.3AECh. 2 - Prob. 2D.3BECh. 2 - Prob. 2D.4AECh. 2 - Prob. 2D.4BECh. 2 - Prob. 2D.5AECh. 2 - Prob. 2D.5BECh. 2 - Prob. 2D.1PCh. 2 - Prob. 2D.2PCh. 2 - Prob. 2D.3PCh. 2 - Prob. 2D.4PCh. 2 - Prob. 2D.5PCh. 2 - Prob. 2D.6PCh. 2 - Prob. 2D.7PCh. 2 - Prob. 2D.8PCh. 2 - Prob. 2D.9PCh. 2 - Prob. 2E.1DQCh. 2 - Prob. 2E.2DQCh. 2 - Prob. 2E.1AECh. 2 - Prob. 2E.1BECh. 2 - Prob. 2E.2AECh. 2 - Prob. 2E.2BECh. 2 - Prob. 2E.3AECh. 2 - Prob. 2E.3BECh. 2 - Prob. 2E.4AECh. 2 - Prob. 2E.4BECh. 2 - Prob. 2E.5AECh. 2 - Prob. 2E.5BECh. 2 - Prob. 2E.1PCh. 2 - Prob. 2E.2PCh. 2 - Prob. 2.1IACh. 2 - Prob. 2.3IACh. 2 - Prob. 2.4IACh. 2 - Prob. 2.5IACh. 2 - Prob. 2.6IACh. 2 - Prob. 2.7IA
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