Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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Chapter 17, Problem 80E

The Ostwald process for the commercial production of nitric acid involves three steps:

4 NH 3 ( g ) + 5 O 2 ( g ) 825 C Pt 4 NO ( g ) + 6 H 2 O ( g ) 2 NO ( g ) + O 2 ( g ) 2 NO 2 ( g ) 3 NO 2 ( g ) + H 2 O ( l ) 2 HNO 3 ( l ) + NO ( g )

a. Calculate ∆H°, ∆S°,∆G° and K (at 298 K) for each of the three steps in the Ostwald process (see Appendix 4).

b. Calculate the equilibrium constant for the first step at 825°C, assuming ∆H° and ∆S° do not depend on temperature.

c. Is there a thermodynamic reason for the high temperature in the first step, assuming standard conditions?

(a)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔHο,ΔSο,ΔGο and K at 298K for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at 825°C is to be calculated. The independency of ΔHο and ΔSο on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term ΔGο is a thermodynamic function. The superscript on this function represents its standard form. The term Δ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

Explanation

Given

The reactions are given as,

4NH3(g)+5O2(g)825°CPt4NO(g)+H2O(g)4NO(g)+O2(g)2NO2(g)3NO2(g)+H2O(l)2HNO3(l)+NO(g)

The standard values of ΔHfο,ΔSfο and ΔGfο given in appendix 4 are as follows,

SubstanceΔHfο(kJ/mol)ΔSfο(J/Kmol)ΔGfο(kJ/mol)NH3(g)4617193O2(g)00205NO(g)9087211H2O(g)242229189NO2(g)3452240H2O(l)28623770HNO3(l)17481156

The value of standard enthalpy change ΔHο of the given reaction is calculated by the formula,

ΔHο=npΔHfο(products)nrΔHfο(reactants)

Where,

  • ΔHfο(reactants) are the standard enthalpy of formation for the reactants.
  • ΔHfο(products) are the standard enthalpy of formation for the products.
  • np is the number of products molecule.
  • nr is the number of reactants molecule.
  • is the symbol of summation.

For the first reaction the representation in the above form is written as,

ΔHο=npΔHfο(products)nrΔHfο(reactants)ΔHο=[4ΔHfοNO(g)+6ΔHfοH2O(g)][4ΔHfοNH3(g)+5ΔHfοO2(g)]

Substitute the value of standard enthalpy of formations in the above equation.

ΔHο=[4ΔHfοNO(g)+6ΔHfοH2O(g)][4ΔHfοNH3(g)+5ΔHfοO2(g)]ΔHο=[[4mol(90kJ/mol)+6mol(243kJ/mol)][4mol×(46kJ/mol)+5mol×(0kJ/mol)]]ΔHο=(1092kJ)+(184kJ)ΔHο=908kJ_

For the second reaction the value of standard enthalpy of formation is calculated as,

ΔHο=[2ΔHfοNO2(g)][2ΔHfοNO(g)+ΔHfοO2(g)]ΔHο=[2mol(34kJ/mol)][(2mol(90kJ/mol))+1mol(0kJ/mol)]ΔHο=(68kJ)(180kJ)ΔHο=112kJ_

For the third reaction the value of standard enthalpy of formation is calculated as,

ΔHο=[2ΔHfοHNO3(g)+ΔHfοNO(g)][3ΔHfοNO2(g)+5ΔHfοHO2(l)]ΔHο=[[2mol(174kJ/mol)+1mol(90kJ/mol)][3mol×(34kJ/mol)+1mol×(286kJ/mol)]]ΔHο=(258kJ)+(184kJ)ΔHο=74kJ_

The calculation of standard entropy change for both the reactions is given below.

The value of standard entropy change ΔSο of the given reaction is calculated by the formula,

ΔSο=npΔSfο(products)nrΔSfο(reactants)

Where,

  • ΔSfο(reactants) are the standard entropy of formation for the reactants.
  • ΔSfο(products) are the standard entropy of formation for the products.
  • np is the number of products molecule.
  • nr is the number of reactants molecule.
  • is the symbol of summation.

For the first reaction the representation in the above form is written as,

ΔSο=npΔSfο(products)nrΔSfο(reactants)ΔSο=[4ΔSfοNO(g)+6ΔSfοH2O(g)][4ΔSfοNH3(g)+5ΔSfοO2(g)]

Substitute the values of standard entropy of formations in the above equation.

ΔSο=[4ΔSfοNO(g)+6ΔSfοH2O(g)][4ΔSfοNH3(g)+5ΔSfοO2(g)]ΔSο=[[4mol(211J/Kmol)+6mol(189J/Kmol)][4mol×(193J/Kmol)+5mol×(205J/Kmol)]]ΔSο=(1978J/Kmol)(1797J/Kmol)ΔSο=181J/K_

For the second reaction the representation in the above form is written as,

ΔSο=npΔSfο(products)nrΔSfο(reactants)ΔSο=[2ΔSfοNO2(g)][2ΔSfοNO(g)+ΔSfοO2(g)]

Substitute the values of standard entropy of formations in the above equation.

ΔSο=[2ΔSfοNO2(g)][2ΔSfοNO(g)+ΔSfοO2(g)]ΔSο=[2mol(240J/Kmol)][2mol(211J/Kmol)+1mol(205J/Kmol)]ΔSο=[(480J/K)][(422J/K)+(205J/K)]ΔSο=147J/K_

For the third reaction the representation in the above form is written as,

ΔSο=npΔSfο(products)nrΔSfο(reactants)ΔSο=[2ΔSfοHNO3(l)+ΔSfοNO(g)][3ΔSfοNO2(g)+ΔSfοHO2(l)]

Substitute the values of standard entropy of formations in the above equation.

ΔSο=[2ΔSfοHNO3(l)+ΔSfοNO(g)][3ΔSfοNO2(g)+ΔSfοHO2(l)]ΔSο=[[2mol(156J/Kmol)+1mol(211J/Kmol)][3mol(240J/Kmol)+1mol(70J/Kmol)]]ΔSο=[(523J/K)][(790J/K)]ΔSο=267J/K_

The value of standard Gibb’s free energy ΔGο of the given reaction is calculated by the formula,

ΔGο=npΔGfο(products)nrΔGfο(reactants)

Where,

  • ΔGfο(reactants) are the standard free energy of formation for the reactants.
  • ΔGfο(products) are the standard free energy of formation for the products.
  • np is the number of products molecule.
  • nr is the number of reactants molecule.
  • is the symbol of summation.

For the first reaction the representation in the above form is written as,

ΔGο=npΔGfο(products)nrΔGfο(reactants)ΔGο=[4ΔGfοNO(g)+6ΔGfοH2O(g)][4ΔGfοNH3(g)+5ΔGfοO2(g)]

Substitute the values of Gibb’s energy of formations in the above equation.

ΔGο=[4ΔGfοNO(g)+6ΔGfοH2O(g)][4ΔGfοNH3(g)+5ΔGfοO2(g)]ΔGο=[[4mol(87kJ/mol)+6mol(229kJ/mol)][4mol×(17kJ/mol)+5mol×(0kJ/mol)]]ΔGο=(1026kJ)+(68kJ)ΔGο=958kJ_

For the second reaction the representation in the above form is written as,

ΔGο=npΔGfο(products)nrΔGfο(reactants)ΔGο=[2ΔGfοNO2(g)][2ΔGfοNO(g)+ΔGfοO2(g)]

Substitute the values of Gibb’s energy of formations in the above equation.

ΔGο=[2ΔGfοNO2(g)][2ΔGfοNO(g)+ΔGfοO2(g)]ΔGο=[2mol(52kJ/mol)][(2mol(87kJ/mol))+1mol(0kJ/mol)]ΔGο=(104kJ)(174kJ)ΔGο=70kJ_

For the third reaction the representation in the above form is written as,

ΔGο=npΔGfο(products)nrΔGfο(reactants)ΔGο=[2ΔGfοHNO3(l)+ΔGfοNO(g)][3ΔGfοNO2(g)+ΔGfοHO2(l)]

Substitute the values of Gibb’s energy of formations in the above equation.

ΔGο=[2ΔGfοHNO3(g)+ΔGfοNO(g)][3ΔGfοNO2(g)+5ΔGfοHO2(g)]ΔGο=[[2mol(81kJ/mol)+1mol(87kJ/mol)][3mol×(52kJ/mol)+1mol×(237kJ/mol)]]ΔGο=(75kJ)(81kJ)ΔGο=+6kJ_

The value of rate constant, K is calculated by the formula,

ΔGο=RTlnKlnK=ΔGοRTK=eΔGοRT

Where,

  • R is gas constant.
  • T is the temperature in Kelvin.
  • eΔGοRT is the exponential term have the power of ΔGο/RT .

For the first reaction substitute the values in the above formula.

K=eΔGοRTK=e[(958×103J/mol)8.314J/molK×298K]K=e(387)K=1.18×10168_

For the second reaction substitute the values in the above formula.

K=eΔGοRTK=e[(70×103J/mol)8.314J/molK×298K]K=e(30)K=1013_

For the third reaction substitute the values in the above formula.

K=eΔGοRTK=e[(6×103J/mol)8.314J/molK×298K]K=e-(2.421)K=0.088_

(b)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔHο,ΔSο,ΔGο and K at 298K for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at 825°C is to be calculated. The independency of ΔHο and ΔSο on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term ΔGο is a thermodynamic function. The superscript on this function represents its standard form. The term Δ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

Explanation

Given

The entropy and enthalpy are independent of the temperature.

Temperature is 825°C(825+273=1098K) .

The free energy change of the first reaction at the given temperature is calculated by the formula,

ΔGο=ΔHοTΔSο

Substitute the values in the above equation.

ΔGο=ΔHοTΔSο=(908000J)(1098K)(181J/K)=(1106735J)

The equilibrium constant for the first step is calculated as,

K=eΔGοRTK=e[(1106738J/mol)8.314J/molK×1098K]K=e(121.23)K=4.7×1052_

(c)

Expert Solution
Check Mark
Interpretation Introduction

Interpretation: The values of ΔHο,ΔSο,ΔGο and K at 298K for each of the three steps in the Ostwald process are to be calculated. The equilibrium constant for the first step at 825°C is to be calculated. The independency of ΔHο and ΔSο on the temperature is to be assumed. The reason for the high temperature in the first step is to be stated.

Concept introduction: The term ΔGο is a thermodynamic function. The superscript on this function represents its standard form. The term Δ represents the change. This function is known as the standard Gibb’s free energy change. It correlates the enthalpy and entropy of the system in a mathematical formula.

Explanation of Solution

Explanation

There is no thermodynamic reason observed in the Ostwald process for the high temperature in first step. Only kinetic reason is responsible. As at the low temperature the reaction is very slow and a side reaction is observed. The side reaction is given as,

4NH3+6NO5N2+6H2O .

This side reaction occurs at the low temperature in the first step and it decreases the percent yield of the product. Therefore, to increase the percent yield the high temperature is to be given.

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Chapter 17 Solutions

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