Chemistry & Chemical Reactivity
Chemistry & Chemical Reactivity
10th Edition
ISBN: 9781337399074
Author: John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher: Cengage Learning
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Chapter 24, Problem 23PS
Interpretation Introduction

Interpretation:

For the given reaction, the value of ΔrH° for the given oxidation reaction of one mole of glucose is 2803 kJ/mol-rxn has to be verified.

  C6H12O6(s)+6O2(g)6CO2(g)+6H2O(l)

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.

Expert Solution & Answer
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Explanation of Solution

The value of ΔrH° for the given oxidation reaction of one mole of glucose 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 O2(g) is 0 kJ/mol.

The standard enthalpy of formation of H2O(l) is 285.8 kJ/mol.

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

The given balanced chemical equation is:

  C6H12O6(s)+6O2(g)6CO2(g)+6H2O(l)

The ΔrH° can be calculated by the following expression,

ΔrH°=nΔfH°(products)nΔfH°(reactants)=[[(6 mol CO2(g)/mol-rxn)ΔfH°[CO2(g)]+(6 mol H2O(l)/mol-rxn)ΔfH°[H2O(l)]][(1 mol C6H12O6(s)/mol-rxn)ΔfH°[C6H12O6(s)]+(6 mol O2(g)/mol-rxn)ΔfH°[O2(g)]]]

Substitute the value of ΔfH°.

ΔrH°=[[(6 mol CO2(g)/mol-rxn)(393.5 kJ/mol)+(6 mol H2O(l)/mol-rxn)(285.8 kJ/mol)][(1 mol C6H12O6(s)/mol-rxn)(1273.3 kJ/mol)+(6 mol O2(g)/mol-rxn)(0)]]=2803 kJ/mol-rxn

The value of ΔrH° for the given oxidation reaction of one mole of glucose is 2803 kJ/mol-rxn.

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Chapter 24 Solutions

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