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Author: Kotz
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Chapter 24, Problem 24PS

The chemical equation for the fermentation of glucose into ethanol is

C6H12O6(s) → 2 C2HsOH()+ 2 CO2(g)

Using ∆f values at 25 °C, calculate ∆r for this reaction. (See Question 23 for ∆f for glucose.)

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The value of ΔrH° for the given fermentation reaction of glucose into ethanol has to be calculated.

  C6H12O6(s)2C2H5OH(l)+2CO2(g)

Concept introduction:

The change in the enthalpy of a reaction when reactant is converted into product under standard conditions is called standard enthalpy of reaction.

The expression for standard enthalpy of reaction is,

ΔrH°=nΔfH°(products)nΔfH°(reactants) (1)

Here, ΔfH° is the standard enthalpy of formation and n is the number of moles of reactant and product in the balanced chemical reaction.

Explanation of Solution

The value of ΔrH° for the fermentation reaction of glucose into ethanol is calculated below.

Given:

Refer to Appendix L for the values of standard enthalpy of formation.

The standard enthalpy of formation of C6H12O6(s) is 1273.3 kJ/mol.

The standard enthalpy of formation of C2H5OH(l) is 277 kJ/mol.

The standard enthalpy of formation of CO2(g) is 393.5 kJ/mol.

The given balanced chemical equation is:

  C6H12O6(s)2C2H5OH(l)+2CO2(g)

The ΔrH° can be calculated by the following expression,

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(2 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]+(2 mol C2H5OH(l)/mol-rxn)ΔfH°[C2H5OH(l)]](1 mol C6H12O6(s)/mol-rxn)ΔfH°[C6H12O6(s)]]

Substitute the value of ΔfH°.

ΔrH°=[[(2 mol CO2(g)/mol-rxn)(393.5 kJ/mol)+(2 mol C2H5OH(l)/mol-rxn)277 kJ/mol](1 mol C6H12O6(s)/mol-rxn)1273.3 kJ/mol]=67.7 kJ/mol-rxn

The value of ΔrH° for the fermentation reaction of glucose into ethanol is 67.7 kJ/mol-rxn.

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Chapter 24 Solutions

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