   Chapter 17, Problem 64E

Chapter
Section
Textbook Problem

# Consider two reactions for the production of ethanol: C 2 H 4 ( g ) + H 2 O ( g ) → CH 3 CH 2 OH ( l ) C 2 H 6 ( g ) + H 2 O ( g ) → CH 3 CH 2 OH ( l ) + H 2 ( g ) Which would be the more thermodynamically feasible at standard conditions? Why?

Interpretation Introduction

Interpretation: The thermodynamic feasibility of the reaction at standard conditions is to be stated and the reason for this is to be explained.

Concept introduction: The feasibility of the chemical reaction at standard conditions is decided by the standard values of enthalpy change. For the given reactions, the standard enthalpy change is negative. The difference in the reactants and products molecules also decides the feasibility of the reaction.

Explanation

Explanation

Given

The reactions are given as,

C2H4(g)+H2O(g)CH3CH2OH(l)C2H6(g)+H2O(g)CH3CH2OH(l)+H2(g)

The standard values of enthalpy and entropy is given as,

ΔHfο(kJ/mol)ΔSfο(J/Kmol)CH3CH2OH(l)276.98161.0C2H4(g)52.3219.5H2O(g)241.8188.7H2(g)0131.0C2H6(g)84.7229.5

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 free energy of formation for the reactants.
• ΔHfο(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,

ΔHο=npΔHfο(products)nrΔHfο(reactants)ΔHο=ΔHfοCH3CH2OH(l)ΔHfοC2H4(g)ΔHfοH2O(g)

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

ΔHο=ΔHfοCH3CH2OH(l)ΔHfοC2H4(g)ΔHfοH2O(g)ΔHο=1mol(276.98kJ/mol)1mol(52.3kJ/mol)1mol×(241.8kJ/mol)ΔHο=(276.98kJ)+(52.3kJ)+(241.8kJ)ΔHο=87.48kJ

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

ΔHο=npΔHfο(products)nrΔHfο(reactants)ΔHο=ΔHfοCH3CH2OH(l)+ΔHfοH2(g)ΔHfοC2H6(g)ΔHfοH2O(g)

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

ΔHο=ΔHfοCH3CH2OH(l)+ΔHfοH2(g)ΔHfοC2H6(g)ΔHfοH2O(g)ΔHο=1mol(276.98kJ/mol)+1mol(0kJ/mol)1mol×(84.7kJ/mol)1mol×(241.8kJ/mol)ΔHο=(276.98kJ)+(84.7kJ)+(241.8kJ)ΔHο=+49.5kJ

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

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