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 84E

Consider the following reaction at 298 K:

2 SO 2 ( g ) + O 2 ( g ) 2 SO 3 ( g )

An equilibrium mixture contains O2(g) and SO3(g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO2 in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions?

Expert Solution & Answer
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Interpretation Introduction

Interpretation: The equilibrium partial pressure of SO2 in the given mixture is to be determined. Standard conditions are to be assumed and the most favored dependency on the high or low temperature is to be decided.

Concept introduction: The equilibrium partial pressure is defined as the total pressure experienced by the individual gases present in the equilibrium mixture. The equilibrium constant in terms of partial pressure is denoted by Kp .

Answer to Problem 84E

Answer

The equilibrium partial pressure of SO2 in the given mixture is 2.98atm_ .

At the low temperature the reaction is most favored.

Explanation of Solution

Explanation

To determine: The equilibrium partial pressure of SO2 in the given mixture; whether this reaction be most favored at a high or low temperature by assuming standard conditions.

Given

The reaction is given as,

2SO2(g)+O2(g)2SO3(g)

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

ΔHfο(kJ/mol)ΔSfο(J/Kmol)ΔGfο(kJ/mol)SO3396257371O20205300SO22972480

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 given reaction the representation in the above form is written as,

ΔHο=npΔHfο(products)nrΔHfο(reactants)ΔHο=[2ΔHfοSO3(g)][2ΔHfοSO2(g)+ΔHfοO2(g)]

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

ΔHο=[2ΔHfοSO3(g)][2ΔHfοSO2(g)+ΔHfοO2(g)]ΔHο=[[2mol(396kJ/mol)][2mol×(297kJ/mol)+1mol×(0kJ/mol)]]ΔHο=(792kJ)(594kJ)ΔHο=198kJ

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 given reaction the representation in the above form is written as,

ΔSο=npΔSfο(products)nrΔSfο(reactants)ΔSο=[2ΔSfοSO3(g)][2ΔSfοSO2(g)+ΔSfοO2(g)]

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

ΔSο=[2ΔSfοSO3(g)][2ΔSfοSO2(g)+ΔSfοO2(g)]ΔSο=[[2mol(257J/Kmol)][2mol×(248J/Kmol)+1mol×(205J/Kmol)]]ΔSο=(514J/Kmol)(701J/Kmol)ΔSο=187J/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 given reaction the representation in the above form is written as,

ΔGο=npΔGfο(products)nrΔGfο(reactants)ΔGο=[2ΔGfοSO3(g)][2ΔGfοSO2(g)+ΔGfοO2(g)]

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

ΔGο=[2ΔGfοSO3(g)][2ΔGfοSO2(g)+ΔGfοO2(g)]ΔGο=[[2mol(371kJ/mol)][2mol×(300kJ/mol)+1mol×(0kJ/mol)]]ΔGο=(742kJ)+(600kJ)ΔGο=142kJ

The free energy change of the given reaction in terms of equilibrium constant is given as,

ΔGο=ΔG+RTlnK (1)

The Gibb’s energy change in terms of entropy and enthalpy is given as,

ΔGο=ΔHοTΔSο (2)

Compare the equation (1) and (2) We get,

ΔGο+RTlnK=ΔHοTΔSο

Substitute the value is the above formula.

ΔHοTΔSο=ΔGο+RTlnK[(198×103J)(298K)(187J/K)]=[(142×103J)+(8.314J/Kmol)(298K)lnK][(142274)+(142000)][(8.314)(298)]=[lnK]lnK=(0.1105)

The equilibrium constant in terms of partial pressure for the given reaction is written as,

K=(PSO3(g)2)(PSO2(g)2PO2(g))

Where,

  • PSO3(g)2 is the partial pressure of SO3 gas.
  • PSO2(g)2 is the partial pressure of SO2 gas.
  • PO2(g) is the partial pressure of O2(g) gas.

Substitute the value of K in the above formula.

lnK=(0.1105)ln(PSO3(g)2)(PSO2(g)2PO2(g))=(0.1105)(2.0)2(PSO2(g)2)(0.50)=e(0.1105)PSO2(g)2=(4.0)(0.8953)(0.50)

Simplify the above equation.

PSO2(g)=(8.9365)1/2PSO2(g)=2.98atm_

This reaction is most favored at low temperature. The negative value of enthalpy and entropy clearly indicates that only at low temperature this reaction is spontaneous. Therefore, by assuming standard conditions only at low temperature reaction is most favored.

Conclusion

Conclusion

The equilibrium partial pressure of SO2 in the given mixture is 2.98atm_ .

The reaction is most favored at a low temperature.

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