   # Consider the following reaction at 298 K: 2 SO 2 ( g ) + O 2 ( g ) → 2 SO 3 ( g ) An equilibrium mixture contains O 2 (g) and SO 3 (g) at partial pressures of 0.50 atm and 2.0 atm, respectively. Using data from Appendix 4, determine the equilibrium partial pressure of SO 2 in the mixture. Will this reaction be most favored at a high or a low temperature, assuming standard conditions? ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 16, Problem 78E
Textbook Problem
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## 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?

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 .

### Explanation of Solution

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

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