For the given reaction, the value of Δ r H ° for the given oxidation reaction of one mole of glucose is − 2803 kJ/mol-rxn has to be verified. C 6 H 12 O 6 ( s ) + 6 O 2 ( g ) → 6 CO 2 ( g ) + 6 H 2 O ( 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, Δ r H ° = ∑ n Δ f H ° ( products ) − ∑ n Δ f H ° ( reactants ) (1) Here, Δ f H ° is the standard enthalpy of formation and n is the number of moles of reactant and product in the balanced chemical reaction.

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Answer 1

Question

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

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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Answer 2

Question

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

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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Answer 3

Question

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

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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Answer 4

Question

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

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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Answer 5

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

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 $−2803 kJ/mol-rxn$ .

Answer 6

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

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 $−2803 kJ/mol-rxn$ .

Answer 7

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.

Answer 8

Expert Solution & Answer

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 $ΔfH°$ .

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