   Chapter 17, Problem 74E

Chapter
Section
Textbook Problem

# 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)

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

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

(b)

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.

(c)

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.

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