Chemistry for Engineering Students
Chemistry for Engineering Students
3rd Edition
ISBN: 9781285199023
Author: Lawrence S. Brown, Tom Holme
Publisher: Cengage Learning
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Textbook Question
Chapter 10, Problem 10.69PAE

Using tabulated thermodynamic data, calculate Δ G ° for these reactions.

(a) Fe ( s ) + 2HCl ( g ) FeCl 2 ( s ) + H 2 ( g )

(b) 3NO 2 ( g ) + H 2 O ( l ) 2HNO 3 ( l ) + NO ( g )

(c) 2K ( s ) + Cl 2 ( g ) 2KCl ( s )

(d) Cl 2 ( g ) + 2NO ( g ) 2NOCl ( g )

(e) SiCl 4 ( g ) Si ( s ) + 2Cl 2 ( g )

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

Using tabulated thermodynamic data, value of ΔG for the following reactions needs to be calculated.

Fe(s) + 2HCl (g) FeCl2 (s) + H2(g)

3NO2(g) + H2O(l)2HNO3(l) + NO(g)

2K(s) + Cl2(g) 2KCl(s)

Cl2(g) + 2NO(g) 2NOCl(g)

SiCl4(g)Si (s) + 2 Cl2(g)

Concept introduction:

The standard change of Gibbs free energy is defined as the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state.

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0ΔGreactants0

Here, ΔGproducts0 is change in standard Gibbs free energy of product and ΔGreactants0 is change in standard Gibbs free energy of reactant.

Answer to Problem 10.69PAE

Solution:

  1. -111.702 kJ
  2. 8.329 kJ
  3. -818.28 kJ
  4. -40.94 kJ
  5. 616 kJ

(a)

Explanation of Solution

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0  ΔGreactants0

Thu, for the following reaction:

Fe(s) + 2HCl (g) FeCl2 (s) + H2(g)

The change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔG FeCl2(s)0+ ΔGH2(g)0 ΔGFe(s)0 2ΔGHCl(g)0       =  [( 1 mol)( 302.3 kJ mol )+( 1 mol)( 0 kJ mol )( 1 mol)( 0 kJ mol )( 2 mol)( 95.299 kJ mol )]      =  (1 mol)(302.3 kJ mol)+190.598 kJ     = 111.702 kJ

(b)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0  ΔGreactants0

Thu, for the following reaction:

3NO2(g) + H2O(l)2HNO3(l) + NO(g)

The change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = 2ΔG HNO3(l)0+ ΔGNO(g)0 ΔGH2O(l)0 3ΔG NO2(g)0       =  [( 2 mol)( 80.71 kJ mol )+( 1 mol)( 86.55 kJ mol )( 1 mol)( 237.129 kJ mol )( 3 mol)( +51.31 kJ mol )]       = 161.42 kJ+86.55 kJ+237.129 kJ153.93 kJ     = 8.329 kJ

(c)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0  ΔGreactants0

Thu, for the following reaction:

2K(s) + Cl2(g) 2KCl(s)

The change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = 2ΔGKCl(s)0 2ΔGK(s)0 ΔG Cl2(g)0      =  (2 mol)(409.14 kJ mol)(2 mol)(0  kJ mol) (1 mol)(0 kJ mol)     = 818.28 kJ

(d)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0  ΔGreactants0

Thu, for the following reaction:

Cl2(g) + 2NO(g) 2NOCl(g)

The change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = 2ΔGNOCl(g)0 2ΔGNO(g)0 ΔG Cl2(g)0       =(2 mol)(66.08 kJ mol)(2 mol)(86.55 kJ mol)(2 mol)(0 kJ mol)      = 132.16 kJ173.1 kJ     = 40.94 kJ

(e)

For a reaction where reactant gives product, change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGproducts0  ΔGreactants0

Thu, for the following reaction:

SiCl4(g)Si (s) + 2 Cl2(g)

The change in standard Gibbs free energy of reaction is as calculated as follows:

ΔG(reaction)0 = ΔGSi(s)0+2ΔG Cl2(g)0 ΔG SiCl4(g)0=(1 mol)(0 kJ mol)+(2 mol)(0 kJ mol)(1 mol)(616.98  kJ mol)      = 616.98  kJ

Conclusion

The standard change of Gibbs free energy is defined as the formation of 1 mole of a substance in its standard state from its constituent elements in their standard state. Based on the thermodynamic table we have determined the below values: -

  • -111.702 kJ
  • 8.329 kJ
  • -818.28 kJ
  • -40.94 kJ
  • 616kJ

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Chapter 10 Solutions

Chemistry for Engineering Students

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