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 6, Problem 6B.5P
Interpretation Introduction

Interpretation:

The values of K,ΔrHΘ,ΔrSΘ and ΔrGΘ at 1443K for the dissociation of CO2(g) into CO(g) and O2(g) has to be calculated.

Concept introduction:

The Gibbs free energy of the reaction is the key factor of the identification of the direction of spontaneity of the reaction.  For the reaction to be spontaneous, the Gibbs free energy of the reaction must be negative.

Expert Solution & Answer
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Answer to Problem 6B.5P

The values of K,ΔrHΘ,ΔrSΘ and ΔrGΘ at 1443K for the dissociation of CO2(g) into CO(g) and O2(g) are calculated as 2.795×106_, 289kJmol1_, 94.24Jmol1K1_ and 153.4kJmol1_ respectively.

Explanation of Solution

The given reaction is shown below.

    CO2(g)CO(g)+12O2(g)

Initially total number of moles is considered as one.  After dissociation the total number of moles become 1+α2.

The ICE table for the given reaction is shown below.

                                 CO2(g)       CO(g)          +    12O2(g)Initialamounts        (1α)                α               α2Molefractions         1α1+α2                 α  1+α2                α2  1+α2Partialpressure       (1α)p1+α2          αp  1+α2           α2p  1+α2

The equilibrium constant K can be calculated using the equation shown below.    K=(pCO(g)p°)(pO2(g)p°)(pCO2(g)p°)                                                                          (1)

Where,

  • pCO(g) is the partial pressure of carbon monoxide.
  • pO2(g) is the partial pressure of oxygen.
  • pCO2 is the partial pressure of carbon dioxide.
  • p° is the standard pressure.

Substitute the values of partial pressure of carbon monoxide, oxygen and carbon dioxide in equation (1) as shown below.

    K=(pCO(g)p°)(pO2(g)p°)(pCO2(g)p°)=((α  1+α2)pp°)((α2  1+α2)pp°)((1α  1+α2)pp°)1/2=(α  1+α2)(α2  1+α2)(pp°)1/2(1α  1+α2)

The value of dissociation constant (α) is less than one at all temperatures, therefore, the above equation can be written as shown below.

    K=(α  1)(α2  1)(pp°)1/2(1  1)=α(α2)1/2=α3/22                                                                              (2)

The value of α at 1443K is given as 2.50×104.  Substitute the value of α in above equation as shown below.

    K=α3/22=(2.50×104)3/22=2.795×106_

The value of ΔrGΘ for the given reaction is calculated by the following formula.

    ΔrGΘ=RTlnK                                                                                           (3)

Where,

  • ΔrGΘ is the Gibbs free energy of the given reaction.
  • R is the gas constant.
  • K is the equilibrium constant.
  • T is the temperature.

The value of gas constant is 8.314JK-1mol-1.

The value of equilibrium constant for the given reaction is calculated as 2.795×106.

It is given that the temperature is 1443K.

Substitute the values of temperature, equilibrium constant and gas constant in equation (3).

    ΔrGΘ=8.314JK-1mol-1×1443K×ln(2.795×106)=153415.08Jmol-1=153.4kJmol1_

The relation between standard enthalpy and equilibrium constant is given by van’t Hoff equation as shown below.

    ln(K2K1)=ΔrHΘR[1T21T1]                                                                          (4)

Where,

  • K2,K1 are the equilibrium constants.
  • T1,T2 are the initial and final temperatures.
  • R is gas constant.
  • ΔrHΘ is the standard enthalpy.

The value of α at 1395K is given as 1.44×104.

The value of equilibrium constant at 1395K can be calculated using equation (2)  as shown below.

    K1=α3/22=(1.44×104)3/22=1.22×106

Substitute T1=1395K, T2=1443K, K1=1.22×106 and K2=2.795×106 in equation (4) as shown below.

    ln(K2K1)=ΔrHΘR[1T21T1]ln(2.795×1061.22×106)=ΔrHΘ8.314Jmol1K1[11443K11395K]0.8289=ΔrHΘ8.314Jmol1K1×[(1395K1443K)1443K×1395K]ΔrHΘ=(0.8289×8.314Jmol1K1×[1443K×1395K(1395K1443K)])

The above equation on solving gives the value of standard enthalpy as ΔrHΘ=289kJmol1_.

The ΔrSΘ is calculated by the formula given below.

    ΔrSΘ=ΔrHΘΔrGΘT

Substitute the values of ΔrHΘ, ΔrGΘ and T in above equation as shown below.

    ΔrSΘ=ΔrHΘΔrGΘT=289kJmol1153kJmol11443K=0.0942kJmol1K1=94.24Jmol1K1_

Therefore, the values of K,ΔrHΘ,ΔrSΘ and ΔrGΘ at 1443K for the dissociation of CO2(g) into CO(g) and O2(g) are calculated as 2.795×106_, 289kJmol1_, 94.24Jmol1K1_ and 153.4kJmol1_ respectively.

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Chapter 6 Solutions

Atkins' Physical Chemistry

Ch. 6 - Prob. 6A.2BECh. 6 - Prob. 6A.3AECh. 6 - Prob. 6A.3BECh. 6 - Prob. 6A.4AECh. 6 - Prob. 6A.4BECh. 6 - Prob. 6A.5AECh. 6 - Prob. 6A.5BECh. 6 - Prob. 6A.6AECh. 6 - Prob. 6A.6BECh. 6 - Prob. 6A.7AECh. 6 - Prob. 6A.7BECh. 6 - Prob. 6A.8AECh. 6 - Prob. 6A.8BECh. 6 - Prob. 6A.9AECh. 6 - Prob. 6A.9BECh. 6 - Prob. 6A.10AECh. 6 - Prob. 6A.10BECh. 6 - Prob. 6A.11AECh. 6 - Prob. 6A.11BECh. 6 - Prob. 6A.12AECh. 6 - Prob. 6A.12BECh. 6 - Prob. 6A.13AECh. 6 - Prob. 6A.13BECh. 6 - Prob. 6A.14AECh. 6 - Prob. 6A.14BECh. 6 - Prob. 6A.1PCh. 6 - Prob. 6A.2PCh. 6 - Prob. 6A.3PCh. 6 - Prob. 6A.4PCh. 6 - Prob. 6A.5PCh. 6 - Prob. 6A.6PCh. 6 - Prob. 6B.1DQCh. 6 - Prob. 6B.2DQCh. 6 - Prob. 6B.3DQCh. 6 - Prob. 6B.1AECh. 6 - Prob. 6B.1BECh. 6 - Prob. 6B.2AECh. 6 - Prob. 6B.2BECh. 6 - Prob. 6B.3AECh. 6 - Prob. 6B.3BECh. 6 - Prob. 6B.4AECh. 6 - Prob. 6B.4BECh. 6 - Prob. 6B.5AECh. 6 - Prob. 6B.5BECh. 6 - Prob. 6B.6AECh. 6 - Prob. 6B.6BECh. 6 - Prob. 6B.7AECh. 6 - Prob. 6B.7BECh. 6 - Prob. 6B.8AECh. 6 - Prob. 6B.8BECh. 6 - Prob. 6B.1PCh. 6 - Prob. 6B.2PCh. 6 - Prob. 6B.3PCh. 6 - Prob. 6B.4PCh. 6 - Prob. 6B.5PCh. 6 - Prob. 6B.6PCh. 6 - Prob. 6B.7PCh. 6 - Prob. 6B.8PCh. 6 - Prob. 6B.9PCh. 6 - Prob. 6B.10PCh. 6 - Prob. 6B.11PCh. 6 - Prob. 6B.12PCh. 6 - Prob. 6C.1DQCh. 6 - Prob. 6C.2DQCh. 6 - Prob. 6C.3DQCh. 6 - Prob. 6C.4DQCh. 6 - Prob. 6C.5DQCh. 6 - Prob. 6C.1AECh. 6 - Prob. 6C.1BECh. 6 - Prob. 6C.2AECh. 6 - Prob. 6C.2BECh. 6 - Prob. 6C.3AECh. 6 - Prob. 6C.3BECh. 6 - Prob. 6C.4AECh. 6 - Prob. 6C.4BECh. 6 - Prob. 6C.5AECh. 6 - Prob. 6C.5BECh. 6 - Prob. 6C.1PCh. 6 - Prob. 6C.2PCh. 6 - Prob. 6C.3PCh. 6 - Prob. 6C.4PCh. 6 - Prob. 6D.1DQCh. 6 - Prob. 6D.2DQCh. 6 - Prob. 6D.1AECh. 6 - Prob. 6D.1BECh. 6 - Prob. 6D.2AECh. 6 - Prob. 6D.2BECh. 6 - Prob. 6D.3AECh. 6 - Prob. 6D.3BECh. 6 - Prob. 6D.4AECh. 6 - Prob. 6D.4BECh. 6 - Prob. 6D.1PCh. 6 - Prob. 6D.2PCh. 6 - Prob. 6D.3PCh. 6 - Prob. 6D.4PCh. 6 - Prob. 6D.5PCh. 6 - Prob. 6D.6PCh. 6 - Prob. 6.1IACh. 6 - Prob. 6.2IACh. 6 - Prob. 6.3IACh. 6 - Prob. 6.4IACh. 6 - Prob. 6.7IACh. 6 - Prob. 6.8IACh. 6 - Prob. 6.10IACh. 6 - Prob. 6.12IA
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