Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 18, Problem 40PS

Determine whether the reactions listed below are entropy-favored or disfavored under standard conditions. Predict how an increase in temperature will affect the value of ΔrG°.

  1. (a) I2(g) → 2 I(g)
  2. (b) 2 SO2(g) + O2(g) → 2 SO3(g)
  3. (c) SiCl4(g) + 2 H2O() → SiO2(s) + 4 HCl(g)
  4. (d) P4(s, white) + 6 H2(g) → 4 PH3(g)

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ΔrGo.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔGo=ΔHo-TΔSo

Here, ΔHo is the change in enthalpy and ΔSo is the change in entropy.

Entropy for any reaction is expressed as,

ΔrS°=nS°(products)nS°(reactants)

A reaction is said to be entropy-favored if the value of entropy change for reaction is positive.

Answer to Problem 40PS

The formation of I(g) is entropy favourable, the reaction will become product-favoured at higher temperatures.

Explanation of Solution

The value of ΔrGoΔHo and ΔSo is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

I2(g)2I(g)ΔfH°(kJ/mol)62.438106.838So(J/K×mol)260.69180.791

ΔrH°=fH°(products)fH°(reactants)=[(2 mol I(g)/mol-rxn)ΔfH°[I(g)]-(1 mol I2(g)/mol-rxn)ΔfH°[I2(g)] ] 

Substituting the respective values

ΔrH°=[(2 mol I(g)/mol-rxn)(106.838 kJ/mol)-(1 mol I2(g)/mol-rxn)(62.438 kJ/mol) ]=151.2 kJ/mol-rxn

Also,

ΔrS°nS°(products)-nS°(reactants)=[(2 mol I(g)/mol-rxn)S°[I(g)](1 mol I2(g)/mol-rxn)S°[I2(g)] ]  

Substituting the respective values

ΔrS°=[(2 mol I(g)/mol-rxn)(180.791 J/K×mol)-(1 mol I2(g)/mol-rxn)(260.69 J/K×mol)]=100.892 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo=151.2 kJ/mol-rxn-[(298K)(100.892 J/K×mol- rxn)](1 kJ1000 J)=121.14 kJ/mol-rxn

The formation of I(g) is entropy favourable as the value of entropy change is positive.

The reaction will become product-favored at higher temperature.

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ΔrGo.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔGo=ΔHo-TΔSo

Here, ΔHo is the change in enthalpy and ΔSo is the change in entropy.

Entropy for any reaction is expressed as,

ΔrS°=nS°(products)nS°(reactants)

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 40PS

The formation of SO3(g) is entropy unfavourable as the value of entropy change is negative. The reaction is spontaneous at lower temperatures.

Explanation of Solution

The value of ΔrGo, ΔHo and ΔSo is calculated below.

Given:

The Appendix L referred for the values of standard entropies and enthalpies.

2SO2(g)+O2(g)2SO3(g)ΔfH°(kJ/mol)-296.840-395.77So(J/K×mol)248.21205.07256.77

The standard enthalpy change is expressed as,

ΔrH°=fH°(products)fH°(reactants)=[(2 mol SO3(g)/mol-rxn)ΔfH°[SO3(g)]-[(2 mol SO2(g)/mol-rxn)ΔfH°[SO2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substituting the respective values

ΔrH°=[(2 mol SO3(g)/mol-rxn)(-395.77 kJ/mol)-[(2 mol SO2(g)/mol-rxn)(-296.84 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)] ]= -197.86 kJ/mol-rxn

Also,

ΔrS°nS°(products)-nS°(reactants)[(2 mol SO3(g)/mol-rxn)S°[SO3(g)]-[(2 mol SO2(g)/mol-rxn)S°[SO2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]] ]

Substituting the respective values

ΔrS°=[(2 mol SO3(g)/mol-rxn)(256.77 J/K×mol)-[(2 mol SO2(g)/mol-rxn)(248.21 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)] ]=-187.95 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo= -197.86 kJ/mol-rxn-[(298K)(-187.95 J/K×mol- rxn)](1 kJ1000 J)= -141.86 kJ/mol-rxn

The formation of SO3(g) is entropy unfavourable as the value of entropy change is negative. The reaction is spontaneous at lower temperatures.

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ΔrGo.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔGo=ΔHo-TΔSo

Here, ΔHo is the change in enthalpy and ΔSo is the change in entropy.

Entropy for any reaction is expressed as,

ΔrS°=nS°(products)nS°(reactants)

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 40PS

The reaction of SiCl4(g) and water is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at all temperatures.

Explanation of Solution

The Appendix L referred for values for the values of standard entropies and enthalpies.

SiCl4(g)+2H2O(l)SiO2(s)+4HCl(g)ΔfH°(kJ/mol)-662.75-285.83-910.86-92.31So(J/K×mol)330.8669.9541.46186.2

The standard enthalpy change is expressed as,

ΔrH°=fH°(products)fH°(reactants)=[[(1 mol SiO2(s)/mol-rxn)ΔfH°[SiO2(s)]+(4 mol HCl(g)/mol-rxn)ΔfH°[HCl(g)]]-[(1 mol SiCl4(g)/mol-rxn)ΔfH°[SiCl4(g)]+(2 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]] ] 

Substituting the respective values

ΔrH°=[[(1 mol SiO2(s)/mol-rxn)(-910.86 kJ/mol)+(4 mol HCl(g)/mol-rxn)(-92.31 kJ/mol)]-[(1 mol SiCl4(g)/mol-rxn)(-662.75 kJ/mol)+(2 mol H2O(l)/mol-rxn)(-285.83 kJ/mol)] ]=-45.7 kJ/mol-rxn

Also,

ΔrS°nS°(products)-nS°(reactants)=[[(1 mol SiO2(s)/mol-rxn)S°[SiO2(s)]+(4 mol HCl(g)/mol-rxn)S°[HCl(g)]][(1 mol SiCl4(g)/mol-rxn)S°[SiCl4(g)]+(2 mol H2O(l)/mol-rxn)S°[H2O(l)]] ]   

Substituting the respective values

ΔrS°=[[(1 mol SiO2(s)/mol-rxn)(41.46 J/K×mol)+(4 mol HCl(g)/mol-rxn)(186.2 J/K×mol)]-[(1 mol SiCl4(g)/mol-rxn)(330.86 J/K×mol)+(2 mol H2O(l)/mol-rxn)(69.95 J/K×mol)] ]   =315.5 J/K×mol-rxn

Now, ΔGo=ΔHo-TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo=-45.7 kJ/mol-rxn-[(298K)(315.5 J/K×mol- rxn)](1 kJ1000 J)=-139.7 kJ/mol-rxn

The reaction of SiCl4(g) and water is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at all temperatures.

(d)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

It should be determined that whether the given reaction is entropy favorable and should be identified that how increase in temperature will affect the value of ΔrGo.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔGo. It is related to entropy and entropy by the following expression,

ΔGo=ΔHo-TΔSo

Here, ΔHo is the change in enthalpy and ΔSo is the change in entropy.

Entropy for any reaction is expressed as,

ΔrS°=nS°(products)nS°(reactants)

A reaction is said to be entropy-favoured if the value of entropy change for reaction is positive.

Answer to Problem 40PS

The reaction of P4(s) and hydrogen is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at higher temperatures.

Explanation of Solution

The value of ΔrGo, ΔHo and ΔSo is calculated below.

Given:

The Appendix L referred for values of standard entropies and enthalpies.

P4(s)+6H2(g)4PH3(g)ΔfH°(kJ/mol)005.47So(J/K×mol)41.1130.7210.24

The standard enthalpy change is expressed as,

 ΔrH°=fH°(products)fH°(reactants)=[(4 mol PH3(g)/mol-rxn)ΔfH°[PH3(g)][(6 mol H2(g)/mol-rxn)ΔfH°[H2(g)]+(1 mol P4(s)/mol-rxn)ΔfH°[P4(s)]] ] 

Substituting the respective values

ΔrH°=[(4 mol PH3(g)/mol-rxn)(5.47 kJ/mol)-[(6 mol H2(g)/mol-rxn)(0 kJ/mol)+(1 mol P4(s)/mol-rxn)(0 kJ/mol)] ] =21.88 kJ/mol-rxn

Also,

ΔrS°nS°(products)-nS°(reactants)=[(4 mol PH3(g)/mol-rxn)S°[PH3(g)]-[(6 mol H2(g)/mol-rxn)S°[H2(g)]+(1 mol P4(s)/mol-rxn)S°[P4(s)]] ]  

Substituting the respective values

ΔrS°=[(4 mol PH3(g)/mol-rxn)(210.24 J/K×mol)-[(6 mol H2(g)/mol-rxn)(130.7 J/K×mol)+(1 mol P4(s)/mol-rxn)(41.1 J/K×mol)] ]  =15.66 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo= 21.88 kJ/mol-rxn-[(298K)(15.66 J/K×mol- rxn)](1 kJ1000 J)= 17.21 kJ/mol-rxn

The reaction of P4(s) and hydrogen is entropy favorable as the value of entropy change is negative. The reaction is spontaneous at higher temperatures.

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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