Standard molar enthalpy of combustion has to be calculated.

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Answer 1

Question

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

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Answer 2

Question

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

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Answer 3

Question

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

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Answer 4

Question

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

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Answer 5

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

Answer 6

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

Answer 7

Chapter 2, Problem 2.13P

(a)

Interpretation Introduction

Interpretation:

Standard molar enthalpy of combustion has to be calculated.

Answer 8

(a)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUo=ΔcUon=qvm/M=−CvΔTm/MΔcHo= (641 JK−1)(7.793 K)(0.3212 g)/(180.2 gmol−1)ΔcHo= −2802×103 Jmol−1=−2800 kJmol−1.

Standard molar enthalpy of combustion is −2800 kJmol−1.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

Answer 13

(b)

Interpretation Introduction

Interpretation:

Standard internal energy of combustion has to be calculated.

Answer 14

(b)

Expert Solution

Explanation of Solution

The heat released in a bomb calorimeter is equivalent to the charge in internal energy because the combustion occurs at constant volume.

ΔU= qv.

Assuming the volume of condensed phases is negligible in comparison to that of gases behave perfectly, then

ΔcUo= ΔcHo−ΔνgasRT.

For a combustion reaction of glucose,

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Here, Δνgas is zero, so

ΔcHo=ΔcUoΔcUo= −2800 kJmol−1.

Standard internal energy of combustion is −2800 kJmol−1.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.

Answer 19

(c)

Interpretation Introduction

Interpretation:

Standard enthalpy for the formation of glucose has to be calculated.

Answer 20

(c)

Expert Solution

Explanation of Solution

For the formation of glucose,

6C(s)+6H2(g)+ 3O2(g)→C6H12O6(s)

As the sum of the formation reaction for CO2(g) and H2O(l),

6C(s)+6O2(g)→6CO2(g) ΔH(1)6H2(g)+ 3O2(g)→ 6H2Og ΔH(2)

For which the enthalpy changes are known, minus the combustion reaction

C6H12O6(s)+6O2(g)→ 6CO2(g)+6H2O(l)

Thus,

ΔfHo(C6H12O6(s))=ΔH(1)+ΔH(2)−ΔH(3) = {6×ΔfHo(CO2)+6ΔfHo(H2O)} −ΔcHo(C6H12O6(s)) =(6×−393.5 kJmol−1)+(6×−285.8 kJmol−1)−(−2800 kJmol−1) = −1.27×103kJmol−1.