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
Question
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Chapter 3, Problem 3D.1AE

(i)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at 298K has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at 298K

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

(i)

Expert Solution
Check Mark

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at 298K is 636.62kJmol1_.  The standard reaction Gibbs energy for the given reaction at 298K is 575.25kJmol-1_.

Explanation of Solution

The given reaction is,

    2CH3CHO(g)+O2(g)2CH3COOH(l)

The standard reaction enthalpy of the reaction is calculated by the following formula,

    ΔHΘreaction=ΔfHΘproducts-ΔfHΘreactants

    ΔHΘreaction=((2×ΔfHΘ(CH3COOH(l)))-(2×ΔfHΘ(CH3CHO(g))+ΔfHΘ(O2(g))))

The values of standard enthalpy of formations are:

  • ΔfHΘ(CH3COOH(l))=484.5kJmol1
  • ΔfHΘ(CH3CHO(g))=166.19kJmol1
  • ΔfHΘ(O2(g))=0kJmol1

Substitute the corresponding values in the above equation.

    ΔHΘreaction=((2×484.5kJmol1)((2×166.19kJmol1)+(0kJmol1)))=(969kJmol1)(332.38kJmol1)=636.62kJmol1_

Hence, the standard reaction enthalpy for the given reaction at 298K is 636.62kJmol1_.

The reaction Gibbs energy (ΔGΘreaction) is given as,

    ΔGΘreaction=ΔHΘreactionTΔSΘreaction                                                               (1)

Where,

  • T is the temperature of the reaction
  • ΔSΘreaction is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, ΔSΘreaction is 205.938JK-1mol-1.

The temperature of the reaction, T is 298K.

Substitute the corresponding values in equation (1).

    ΔGΘreaction=636.62×103Jmol1(298K×205.938JK-1mol-1)=575250.476Jmol1=575250.476Jmol1×1kJ1000J=575.25kJmol-1_

Hence, the standard reaction Gibbs energy for the given reaction at 298K is 575.25kJmol-1_.

(ii)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at 298K has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at 298K.

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

(ii)

Expert Solution
Check Mark

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at 298K is 53.4kJmol_.  The standard reaction Gibbs energy for the given reaction at 298K is 25.79328kJmol-1_.

Explanation of Solution

The given reaction is,

    2AgCl(s)+Br2(l)2AgBr(s)+Cl2(g)

The standard reaction enthalpy of the reaction is calculated by the following formula,

    ΔHΘreaction=ΔfHΘproducts-ΔfHΘreactants=((2×ΔfHΘ(AgBr(s))+ΔfHΘ(Cl2(g)))-(2×ΔfHΘ(AgCl(s))+ΔfHΘ(Br2(l))))

The values of standard enthalpy of formations are:

  • ΔfHΘ(AgBr)=100.37kJmol-1
  • ΔfHΘ(Cl2(g))=0kJmol-1
  • ΔfHΘ(Br2(l))=0kJmol-1
  • ΔfHΘ(AgCl(s))=127.07kJmol-1

Substitute the corresponding values in the above equation ΔHΘreaction=((2×(100.37kJmol-1)+0)-(2×(127.07kJmol-1)+0))=53.4kJmol_-1

Hence, the standard reaction enthalpy for the given reaction at 298K 53.4kJmol_.

The reaction Gibbs energy (ΔGΘreaction) is given as,

    ΔGΘreaction=ΔHΘreactionTΔSΘreaction                                                               (1)

Where,

  • T is the temperature of the reaction
  • ΔSΘreaction is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, ΔSΘreaction is 92.64JK-1mol-1.

The temperature of the reaction, T is 298K.

Substitute the corresponding values in equation (1).

    ΔGΘreaction=53.4×103Jmol1(298K×92.64JK-1mol-1)=25793.28Jmol1=25793.28Jmol1×1kJ1000J=25.79328kJmol-1_

Hence, the standard reaction Gibbs energy for the given reaction at 298K is 25.79328kJmol-1_.

(iii)

Interpretation Introduction

Interpretation: The standard reaction enthalpy of the given reaction at 298K has to be predicted.  It has to be used to predict the standard reaction Gibbs energy at 298K.

Concept introduction: The standard reaction enthalpy of a given reaction represents the exothermic or endothermic property of a reaction.  If the value of standard reaction enthalpy is negative, the reaction is exothermic and releases heat.  If the value of standard reaction enthalpy is positive, the reaction is endothermic and requires heat.  The standard Gibbs energy is a criterion for determining the spontaneity of a reaction.

(iii)

Expert Solution
Check Mark

Answer to Problem 3D.1AE

The standard reaction enthalpy for the given reaction at 298K is 224.3kJmol-1_.  The standard reaction Gibbs energy for the given reaction at 298K is 178.67918kJmol-1_.

Explanation of Solution

The given reaction is,

    Hg(l)+Cl2(g)HgCl2(s)

The standard reaction enthalpy of the reaction is calculated by the following formula.

    ΔHΘreaction=ΔfHΘproducts-ΔfHΘreactantsΔHΘreaction=((ΔfHΘ(HgCl2(s)))(ΔfHΘ(Hg(l))+ΔfHΘ(Cl2(g))))

The values of standard enthalpies of formations are:

  • ΔfHΘ(HgCl2(s))=224.3kJmol-1
  • ΔfHΘ(Cl2(g))=0kJmol-1
  • ΔfHΘ(Hg(l))=0kJmol-1

Substitute the corresponding values in the above equation.    ΔHΘreaction=((224.3kJmol-1)(0+0))=224.3kJmol-1_

Hence, the standard entropy of the reaction at 298K is 224.3kJmol-1_.

The reaction Gibbs energy (ΔGΘreaction) is given as,

    ΔGΘreaction=ΔHΘreactionTΔSΘreaction                                                             (1)

Where,

  • T is the temperature of the reaction
  • ΔSΘreaction is the standard reaction entropy of the reaction.

The standard reaction entropy of the given reaction, ΔSΘreaction is -153.09JK-1mol-1.

The temperature of the reaction, T is 298K.

Substitute the corresponding values in equation (1).

    ΔGΘreaction=224.3×103Jmol1(298K×-153.09JK-1mol-1)=178679.18Jmol1=178679.18Jmol1×1kJ1000J=178.67918kJmol-1_

Hence, the standard reaction Gibbs energy for the given reaction at 298K is 178.67918kJmol-1_.

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Chapter 3 Solutions

Atkins' Physical Chemistry

Ch. 3 - Prob. 3A.1AECh. 3 - Prob. 3A.1BECh. 3 - Prob. 3A.2AECh. 3 - Prob. 3A.2BECh. 3 - Prob. 3A.3AECh. 3 - Prob. 3A.3BECh. 3 - Prob. 3A.4AECh. 3 - Prob. 3A.4BECh. 3 - Prob. 3A.5AECh. 3 - Prob. 3A.5BECh. 3 - Prob. 3A.6AECh. 3 - Prob. 3A.6BECh. 3 - Prob. 3A.1PCh. 3 - Prob. 3A.2PCh. 3 - Prob. 3A.3PCh. 3 - Prob. 3A.4PCh. 3 - Prob. 3A.5PCh. 3 - Prob. 3A.6PCh. 3 - Prob. 3A.7PCh. 3 - Prob. 3B.1DQCh. 3 - Prob. 3B.1AECh. 3 - Prob. 3B.1BECh. 3 - Prob. 3B.2AECh. 3 - Prob. 3B.2BECh. 3 - Prob. 3B.3AECh. 3 - Prob. 3B.3BECh. 3 - Prob. 3B.4AECh. 3 - Prob. 3B.4BECh. 3 - Prob. 3B.5AECh. 3 - Prob. 3B.5BECh. 3 - Prob. 3B.6AECh. 3 - Prob. 3B.6BECh. 3 - Prob. 3B.7AECh. 3 - Prob. 3B.7BECh. 3 - Prob. 3B.1PCh. 3 - Prob. 3B.2PCh. 3 - Prob. 3B.3PCh. 3 - Prob. 3B.4PCh. 3 - Prob. 3B.5PCh. 3 - Prob. 3B.6PCh. 3 - Prob. 3B.7PCh. 3 - Prob. 3B.8PCh. 3 - Prob. 3B.9PCh. 3 - Prob. 3B.10PCh. 3 - Prob. 3B.12PCh. 3 - Prob. 3C.1DQCh. 3 - Prob. 3C.1AECh. 3 - Prob. 3C.1BECh. 3 - Prob. 3C.2AECh. 3 - Prob. 3C.2BECh. 3 - Prob. 3C.3AECh. 3 - Prob. 3C.3BECh. 3 - Prob. 3C.1PCh. 3 - Prob. 3C.2PCh. 3 - Prob. 3C.4PCh. 3 - Prob. 3C.5PCh. 3 - Prob. 3C.6PCh. 3 - Prob. 3C.7PCh. 3 - Prob. 3C.8PCh. 3 - Prob. 3C.9PCh. 3 - Prob. 3C.10PCh. 3 - Prob. 3D.1DQCh. 3 - Prob. 3D.2DQCh. 3 - Prob. 3D.1AECh. 3 - Prob. 3D.1BECh. 3 - Prob. 3D.2AECh. 3 - Prob. 3D.2BECh. 3 - Prob. 3D.3AECh. 3 - Prob. 3D.3BECh. 3 - Prob. 3D.4AECh. 3 - Prob. 3D.4BECh. 3 - Prob. 3D.5AECh. 3 - Prob. 3D.5BECh. 3 - Prob. 3D.1PCh. 3 - Prob. 3D.2PCh. 3 - Prob. 3D.3PCh. 3 - Prob. 3D.4PCh. 3 - Prob. 3D.5PCh. 3 - Prob. 3D.6PCh. 3 - Prob. 3E.1DQCh. 3 - Prob. 3E.2DQCh. 3 - Prob. 3E.1AECh. 3 - Prob. 3E.1BECh. 3 - Prob. 3E.2AECh. 3 - Prob. 3E.2BECh. 3 - Prob. 3E.3AECh. 3 - Prob. 3E.3BECh. 3 - Prob. 3E.4AECh. 3 - Prob. 3E.4BECh. 3 - Prob. 3E.5AECh. 3 - Prob. 3E.5BECh. 3 - Prob. 3E.6AECh. 3 - Prob. 3E.6BECh. 3 - Prob. 3E.1PCh. 3 - Prob. 3E.2PCh. 3 - Prob. 3E.3PCh. 3 - Prob. 3E.4PCh. 3 - Prob. 3E.5PCh. 3 - Prob. 3E.6PCh. 3 - Prob. 3E.8PCh. 3 - Prob. 3.1IACh. 3 - Prob. 3.2IA
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