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 90SCQ

Consider the formation of NO(g) from its elements.

N2(g) + O2(g) ⇄ 2 NO(g)

  1. (a) Calculate Kp at 25 °C. Is the reaction product-favored at equilibrium at this temperature?
  2. (b) Assuming ΔrH° and ΔrS° are nearly constant with temperature, calculate ΔrG° at 700 °C. Estimate Kp from the new value of ΔrG° at 700 °C. Is the reaction product-favored at equilibrium at 700 °C?
  3. (c) Using Kp at 700 °C, calculate the equilibrium partial pressures of the three gases if you mix 1.00 bar each of N2 and O2.

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The equilibrium constant for formation of NO(g) from its elements should be calculated and identified that whether the reaction is product favoured at equilibrium at given temperature.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

  ΔrG°fG°(products)fG°(reactants)   

ΔGo is related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

It is also 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.

Answer to Problem 90SCQ

The value of Kp at 25 ° is 3.5×1031.

The reaction is reactant-favoured at equilibrium.

Explanation of Solution

The value of Kp at 25 °C  is calculated below.

Given:

The Appendix L referred for values of standard free energy values.

The given reaction is,

  N2(g) + O2(g)2NO(g)

The ΔrGo for NO(g) is 86.58 kJ/mol.

The ΔrGo for O2(g) is 0 kJ/mol.

The ΔrGo for N2(g) is 0 kJ/mol.

  ΔrG°fG°(products)fG°(reactants)=[(2 mol NO(g)/mol-rxn)ΔfG°[NO(g)]-[(1 mol N2(g)/mol-rxn)ΔfG°[N2(g)]+(1 mol O2(g)/mol-rxn)ΔfG°[O2(g)]] ] 

Substituting the respective values,

ΔrG°=[(2 mol NO(g)/mol-rxn)(86.58 kJ/mol)-[(1 mol N2(g)/mol-rxn)(0)+(1 mol O2(g)/mol-rxn)(0)] ] = +173.16 kJ/mol-rxn

The value of ΔrGo is positive, thus the reaction is reactant-favored at equilibrium.

ΔGo is related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

Substituting the respective values of ΔrGo, T and R.

  Kp= e - +173.16 kJ/mol- rxn(0.008314 kJ/K×mol)(298.15 K)= 3.5×10-31

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The ΔrGo value for formation of NO(g) from its elements should be calculated at 700oC and Kp for respective ΔrGo at 700oC should be estimated also it should be identified that the reaction is product favoured at equilibrium at this given temperature.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

  ΔrG°fG°(products)fG°(reactants)   

ΔGo is related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

It is also 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.

Answer to Problem 90SCQ

The value of ΔrGo at 700 ° is 179.59 kJ/mol-rxn.

The value of Kp at 700 ° is 4×109.

Explanation of Solution

The value of ΔrGo and Kp at 700 ° is calculated below.

Given:

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

N2(g)+O2(g)2NO(g)ΔfH°(kJ/mol)0090.29So(J/K×mol)191.56205.07210.76

  ΔrH°fH°(products)fH°(reactants)[(2 mol NO(g)/mol-rxn)ΔfH°[NO(g)]-[(1 mol N2(g)/mol-rxn)ΔfH°[N2(g)]+(1 mol O2(g)/mol-rxn)ΔfH°[O2(g)]] ] 

Substitute the values,

  ΔrH°[(2 mol NO(g)/mol-rxn)(90.29 kJ/mol)-[(1 mol N2(g)/mol-rxn)(0 kJ/mol)+(1 mol O2(g)/mol-rxn)(0 kJ/mol)] ] = 180.58 kJ/mol-rxn

  ΔrS°nS°(products)-nS°(reactants)=[(2 mol NO(g)/mol-rxn)S°[NO(g)]-[(1 mol N2(g)/mol-rxn)S°[N2(g)]+(1 mol O2(g)/mol-rxn)S°[O2(g)]] ]

Substitute the values,

  ΔrS°=[(2 mol NO(g)/mol-rxn)(210.76 J/K×mol)-[(1 mol N2(g)/mol-rxn)(191.56 J/K×mol)+(1 mol O2(g)/mol-rxn)(205.07 J/K×mol)] ]= 24.89 J/K×mol-rxn

Now, ΔGo= ΔHo- TΔSo

Substitute the value of ΔHo and ΔSo.

  ΔGo= 180.58 kJ/mol-rxn-[(973.15K)(24.89 J/K×mol-rxn)](1 kJ1000 J)= 156.36 kJ/mol-rxn

The value of ΔrGo is positive, thus the reaction is reactant-favored at equilibrium.

ΔG is related to the equilibrium constant Kp by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

Substitute the value of ΔrGo, T and R.

  Kp= e+156.36 kJ/mol- rxn(0.008314 kJ/K×mol)(973.15 K)= 4.0×10-9

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation:

The equilibrium partial pressures of the given three gases should be calculated if they were mixed 1bar of each N2andO2.

Concept introduction:

The Gibbs free energy or the free energy change is a thermodynamic quantity represented by ΔrGo. It can be calculated in a similar manner as entropy and enthalpy.  The expression for the free energy change is:

  ΔrG°fG°(products)fG°(reactants)   

ΔGo is related to the equilibrium constant K by the equation,

  ΔrGo= -RTlnKp

The rearranged expression is,

  Kp= eΔrGoRT

It is also 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.

Answer to Problem 90SCQ

The equilibrium partial pressure of N2(g) is 1 bar.

The equilibrium partial pressure of O2(g) is 1 bar.

The equilibrium partial pressure of NO(g) is 3×105.

Explanation of Solution

The equilibrium partial pressure of N2(g), O2(g) and NO(g) is calculated below.

Given:

The initial pressures of N2(g) and O2(g) is 1 bar. The partial pressure of NO(g) is supposed to be x.

N2(g)+O2(g)2NO(g)Initial110Equilibrium1-x1-x2x

Now,

  Kp= 4.0×10-9

The equilibrium constant is related to the equilibrium partial pressure by the expression,

  Kpp2NO(pN2)(pO2)

Substitute the values,

  4.0×10-9 = 4x2(1-x)(1-x)x = 3×10-5

Thus, the value of x is very small and can be neglected in comparison to 1 bar.

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Chapter 18 Solutions

Chemistry & Chemical Reactivity

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