   Chapter 24, Problem 23PS

Chapter
Section
Textbook Problem

The section about metabolism provided a value for ∆rH° for the oxidation of one mole of glucose (‒2803 kl/mol-rxn). Using ∆fH° values at 25°C, verify that this is the correct value for the equationC6H12O6(s) + 6 O2(g)→ 6 CO2(g) + 6 H2O(l)∆fH°[C6H12O6(s)] = ‒1273.3 kJ/mol

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.

Explanation

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)]][<

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