Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
9th Edition
ISBN: 9781133949640
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 18, Problem 23PS

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) N2(g) + 2 O2(g) → 2 NO2(g)
  2. (b) 2 C(s) + O2(g) → 2 CO(g)
  3. (c) CaO(s) + CO2(g) → CaCO3(s)
  4. (d) 2 NaCl(s) → 2 Na(s) + Cl2(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 ΔrGo. 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 23PS

The formation of NO2(g) is entropy unfavourable. There is no temperature at which reaction will become product-favoured at equilibrium.

Explanation of Solution

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

Given:

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

N2(g)+2O2(g)2NO2(g)ΔfH°(kJ/mol)0033.1So(J/K×mol)191.56205.07240.04

  ΔrH°=fH°(products)fH°(reactants)=[(2 mol NO2(g)/mol-rxn)ΔfH°[NO2(g)]-[(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]+(2 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substituting the respective values

  ΔrH°=[(2 mol NO2(g)/mol-rxn)(33.1 kJ/mol)-[(1 mol N2(g)/mol-rxn)(0 kJ/mol)+(2 mol O2(g)/mol-rxn)(0 kJ/mol)] ]=66.2 kJ/mol-rxn

  ΔrS°nS°(products)-nS°(reactants)=[(2 mol NO2(g)/mol-rxn)S°[NO2(g)]-[(1 mol N2(g)/mol-rxn)S°[N2(g)]+(2 mol O2(g)/mol-rxn)S°[O2(g)]] ]  

Substituting the respective values

  ΔrS°=[(2 mol NO2(g)/mol-rxn)(240.04 J/K×mol)-[(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)+(2 mol O2(g)/mol-rxn)(205.07 J/K×mol)] ]= -121.62 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

  ΔGo= 66.2 kJ/mol-rxn-[(298K)(-121.62 J/K×mol- rxn)](1 kJ1000 J)= 102.5 kJ/mol-rxn

The formation of NO2 is entropy unfavourable as the value of entropy change is negative.

The Table 18.1 was referred,

Both the enthalpy and entropy change are unfavorable. There is no temperature at which the reaction will become product-favored at equilibrium. It is a Type 4 reaction and as the temperature is increased, the reaction becomes more reactant-favored.

(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 ΔrGo. 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 23PS

The formation of CO(g) is entropy favourable. As temperature increases the reaction will become more product-favoured at equilibrium.

Explanation of Solution

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

Given:

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

2C(s)+O2(g)2CO(g)ΔfH°(kJ/mol)00-110.525So(J/K×mol)5.6205.07197.674

  ΔrH°fH°(products)fH°(reactants)[(2 mol CO(g)/mol-rxn)ΔfH°[CO(g)]-[(2 mol C(s)/mol-rxn)ΔfH°[C(s)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substituting the values,

  ΔrH°[(2 mol CO(g)/mol-rxn)(-110.525 kJ/mol)-[(2 mol C(s)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)] ]= -221.05 kJ/mol-rxn

  ΔrS°nS°(products)-nS°(reactants)=[(2 mol CO(g)/mol-rxn)S°[CO(g)][(2 mol C(s)/mol-rxn)S°[C(s)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]] ]  

Substituting the values,

  ΔrS°=[(2 mol CO(g)/mol-rxn)(197.674 J/K×mol)-[(2 mol C(s)/mol-rxn)(5.6 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)] ]=179.1 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo= -221.05 kJ/mol-rxn-[(298K)(179.1 J/K×mol- rxn)](1 kJ1000 J)= -274.45 kJ/mol-rxn

The formation of CO is entropy favourable as the value of entropy change is positive.

The Table 18.1 was referred,

Both the enthalpy and entropy change are favorable. As temperature increases the reaction will become more product-favored at equilibrium. It is a Type 1 reaction and as the temperature is increased, the reaction becomes more product-favored.

(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 ΔrGo. 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 23PS

The formation of CaCO3(s) is entropy unfavourable. As temperature increases the reaction will become less product-favoured at equilibrium.

Explanation of Solution

The value of ΔrGoΔHo and ΔSo is calculated below.

Given:

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

CaO(s)+CO2(g)CaCO3(s)ΔfH°(kJ/mol)-635.09-393.509-1207.6So(J/K×mol)38.2213.7491.7

ΔrH°=fH°(products)fH°(reactants)=[(1mol CaCO3(s)/mol-rxn)ΔfH°[CaCO3(s)][(1 mol CaO(s)/mol-rxn)ΔfH°[CaO(s)]+(1 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]] ] 

Substituting the respective values

ΔrH°=[(1mol CaCO3(s)/mol-rxn)(1207.6 kJ/mol)[(1 mol CaO(s)/mol-rxn)(635.09 kJ/mol)(1 mol CO2(g)/mol-rxn)(393.509 kJ/mol)] ] =179 kJ/mol-rxn

ΔrS°nS°(products)-nS°(reactants)=[(1mol CaCO3(s)/mol-rxn)S°[CaCO3(s)]-[(2 mol CaO(s)/mol-rxn)S°[CaO(s)]+(1 mol CO2(g)/mol-rxn)S°[CO2(g)]] ]  

Substituting the respective values

ΔrS°=[(1 mol CaCO3(s)/mol-rxn)(91.7 J/K×mol)-[(1 mol CaO(s)/mol-rxn)(38.2 J/K×mol)+(1 mol CO2(g)/mol-rxn)(213.74 J/K×mol)] ]=-160.2 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo=-179 kJ/mol-rxn-[(298K)(-160.2 J/K×mol-rxn)](1 kJ1000 J)=-131.23 kJ/mol-rxn

The formation of CaCO3 is entropy unfavourable as the value of entropy change is negative. The reaction is enthalpy favored. As the temperature increases, the reaction thus becomes less product-favored.

(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 ΔrGo. 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 23PS

The decomposition of NaCl(s) is entropy favourable. As temperature increases the reaction will become more product-favored at equilibrium.

Explanation of Solution

The value of ΔrGoΔHo and ΔSo is calculated below.

Given:

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

2NaCl(s)2Na(s)+Cl2(g)ΔfH°(kJ/mol)-411.1200So(J/K×mol)72.1151.21223.08

ΔrH°=fH°(products)fH°(reactants)=[[(2 mol Na(s)/mol-rxn)ΔfH°[Na(s)]+(1mol Cl2(g)/mol-rxn)ΔfH°[Cl2(g)]]-(2 mol NaCl(s)/mol-rxn)ΔfH°[NaCl(s)] ]

Substituting the respective values

ΔrH°=[[(2 mol Na(s)/mol-rxn)(-0 kJ/mol)+(1mol Cl2(g)/mol-rxn)(-0 kJ/mol)]-(2 mol NaCl(s)/mol-rxn)(-411.12 kJ/mol) ]=822.2 kJ/mol-rxn

ΔrS°nS°(products)-nS°(reactants)=[[(2 mol Na(s)/mol-rxn)S°[Na(s)]+(1mol Cl2(g)/mol-rxn)S°[Cl2(g)]]-(2 mol NaCl(s)/mol-rxn)S°[NaCl(s)] ]  

Substituting the respective values

ΔrS°=[[(2 mol Na(s)/mol-rxn)(51.21 J/K×mol)+(1mol Cl2(g)/mol-rxn)(223.08 J/K×mol)]-(2 mol NaCl(s)/mol-rxn)(72.11 J/K×mol) ]=181.28 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

ΔGo= 822.2 kJ/mol-rxn-[(298K)(181.28 J/K×mol- rxn)](1 kJ1000 J)= 768.19 kJ/mol-rxn

The decomposition of NaCl is entropy favourable as the value of entropy change is positive. The reaction is enthalpy unfavored. As the temperature increases, the reaction thus becomes more product-favored.

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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