Student Solutions Manual to accompany Atkins' Physical Chemistry 11th  edition
Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition
11th Edition
ISBN: 9780198807773
Author: ATKINS
Publisher: Oxford University Press
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Chapter 6, Problem 6A.4P
Interpretation Introduction

Interpretation:

Thermodynamically, the most stable solid at 190K has to be predicted.

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 6A.4P

The most stable solid at 190K is HNO3.3H2O(s).

Explanation of Solution

For the formation of H2O(s):

The formation of H2O(s) at 190K is shown by the following chemical reaction.

    H2O(g)H2O(s)ΔrGΘ=23.6kJmol-1

The equilibrium constant for the formation of H2O(s) at 190K is calculated by the following formula.

    K1=pH2O(s)pH2O(g)                                                                                                 (1)

Where,

  • pH2O(g) is the partial pressure of H2O(g) at 190K.
  • pH2O(s) is the partial pressure of H2O(s) at 190K.

The partial pressure of a substance in the solid state is 1bar.  Therefore, the partial pressure of H2O(s) is 1bar.

It is given that the partial pressure of H2O(g) is 0.13μbar.

The conversion of μbar to bar is done as,

    1μbar=1×106bar

Therefore, the conversion of 0.13μbar to bar is done as,

    0.13μbar=0.13×106bar=1.3×107bar

Therefore, the partial pressure of H2O(g) is 1.3×107bar.

Substitute the value of the partial pressure of H2O(g) and H2O(s) in equation (1).

    K1=11.3×107=7.69×106

Therefore, the equilibrium constant for the formation of H2O(s) at 190K is 7.69×106.

For the formation of HNO3.H2O(s):

The formation of HNO3.H2O(s) at 190K is shown by the following chemical reaction.

    H2O(g)+HNO3(g)HNO3.H2O(s)ΔrGΘ=57.2kJmol-1

The equilibrium constant for the formation of HNO3.H2O(s) at 190K is calculated by the following formula.

    K2=pHNO3.H2O(s)pH2O(g)pHNO3(g)                                                                                      (2)

Where,

  • pH2O(g) is the partial pressure of H2O(g) at 190K.
  • pHNO3(g) is the partial pressure of HNO3(g) at 190K.
  • pHNO3.H2O(s) is the partial pressure of HNO3.H2O(s) at 190K.

The partial pressure of a substance in the solid state is 1bar.  Therefore, the partial pressure of HNO3.H2O(s) is 1bar.

It is given that the partial pressure of HNO3(g) is 0.41nbar.

The conversion of nbar to bar is done as,

    1nbar=1×109bar

Therefore, the conversion of 0.41nbar to bar is done as,

    0.41nbar=0.41×109bar=4.1×1010bar

Therefore, the partial pressure of HNO3(g) is 4.1×1010bar.

The partial pressure of H2O(g) is 1.3×107bar.

Substitute the value of the partial pressure of HNO3.H2O(s), H2O(g) and HNO3(g) in equation (2).

    K2=1(1.3×107)(4.1×1010bar)=15.33×1017=1.87×1016

Therefore, the equilibrium constant for the formation of HNO3.H2O(s) at 190K is 1.87×1016.

For the formation of HNO3.2H2O(s):

The formation of HNO3.2H2O(s) at 190K is shown by the following chemical reaction.

    2H2O(g)+HNO3(g)HNO3.2H2O(s)ΔrGΘ=85.6kJmol-1

The equilibrium constant for the formation of HNO3.2H2O(s) at 190K is calculated by the following formula.

    K3=pHNO3.2H2O(s)(pH2O(g))2pHNO3(g)                                                                                 (3)

Where,

  • pH2O(g) is the partial pressure of H2O(g) at 190K.
  • pHNO3(g) is the partial pressure of HNO3(g) at 190K.
  • pHNO3.2H2O(s) is the partial pressure of HNO3.2H2O(s) at 190K.

The partial pressure of a substance in the solid state is 1bar.  Therefore, the partial pressure of HNO3.2H2O(s) is 1bar.

The partial pressure of HNO3(g) is 4.1×1010bar.

The partial pressure of H2O(g) is 1.3×107bar.

Substitute the value of the partial pressure of HNO3.2H2O(s), H2O(g) and HNO3(g) in equation (3).

    K3=1(1.3×107)2(4.1×1010)=11.69×1014×4.1×1010=16.92×1024=1.44×1023

Therefore, the equilibrium constant for the formation of HNO3.2H2O(s) at 190K is 1.44×1023.

For the formation of HNO3.3H2O(s):

The formation of HNO3.3H2O(s) at 190K is shown by the following chemical reaction.

    3H2O(g)+HNO3(g)HNO3.3H2O(s)ΔrGΘ=112.8kJmol-1

The equilibrium constant for the formation of HNO3.2H2O(s) at 190K is calculated by the following formula.

    K4=pHNO3.3H2O(s)(pH2O(g))3pHNO3(g)                                                                                 (4)

Where,

  • pH2O(g) is the partial pressure of H2O(g) at 190K.
  • pHNO3(g) is the partial pressure of HNO3(g) at 190K.
  • pHNO3.3H2O(s) is the partial pressure of HNO3.3H2O(s) at 190K.

The partial pressure of a substance in the solid state is 1bar.  Therefore, the partial pressure of HNO3.3H2O(s) is 1bar.

The partial pressure of HNO3(g) is 4.1×1010bar.

The partial pressure of H2O(g) is 1.3×107bar.

Substitute the value of the partial pressure of HNO3.3H2O(s), H2O(g) and HNO3(g) in equation (4).

    K4=1(1.3×107)3(4.1×1010)=12.197×1021×4.1×1010=19.00×1031=1.11×1030

Therefore, the equilibrium constant for the formation of HNO3.3H2O(s) at 190K is 1.11×1030.

The relation between the equilibrium constant and the standard Gibbs energy of the reaction is shown below.

    ΔrGΘ=RTlnK                                                                                         (5)

Where,

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

The equilibrium constant for the formation of H2O(s) at 190K is 7.69×106.

The equilibrium constant for the formation of HNO3.H2O(s) at 190K is 1.87×1016.

The equilibrium constant for the formation of HNO3.2H2O(s) at 190K is 1.44×1023.

The equilibrium constant for the formation of HNO3.3H2O(s) at 190K is 1.11×1030.

Equation (5) indicates that higher the value of equilibrium constant, higher will be the negative value of the standard Gibbs energy of the reaction.  Higher the negative value of the standard Gibbs energy of the reaction, higher will be the spontaneity of the reaction or the reaction will be thermodynamically stable.

The value of the equilibrium constant for the formation of HNO3.3H2O(s) is highest.  Therefore, the reaction for the formation of HNO3.3H2O(s) will give the highest negative value of the standard Gibbs energy or HNO3.3H2O(s) will be the most stable.

Hence, HNO3.3H2O(s) is the most stable solid at 190K.

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Chapter 6 Solutions

Student Solutions Manual to accompany Atkins' Physical Chemistry 11th edition

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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