(a) Interpretation: The ∆ G ° ' for the reaction shown in Fig. P25.36. Concept Introduction: For a spontaneous reaction, the value for ∆ G ° ' is negative.

Question

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Answer 1

Question

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

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Answer 2

Question

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

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Answer 3

Question

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

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Answer 4

Question

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

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Answer 5

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Answer 6

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Conclusion

∆G°'=-14.0 kJ/mol

Answer 7

Chapter 25, Problem 25.36AP

Interpretation Introduction

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Answer 8

Expert Solution

Answer to Problem 25.36AP

∆G°'=-14.0 kJ/mol

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given reaction is:

fatty acyl-CoA+carnitine⇋ fatty acyl carnitine+ HSCoA

This reaction is the addition of two reactions that are shown below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+ ∆G°'=-31.4 kJ/mol (1)

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

Thus, the ∆G°' for the given reaction is calculated as below:

fatty acyl-CoA+H2O ⇋RCO2- + HSCoA +H+∆G°'=-31.4 kJ/mol (1)

+

RCO2-+carnitine+H+⇋ a fatty acyl carnitine+H2O ∆G°'=+17.4 kJ/mol (2)

_____________________________________________________________________

a fatty acyl-CoA+carnitine⇋a fatty acyl carnitine+ HSCoA∆G°'=-14.0 kJ/mol

Answer 13

Conclusion

∆G°'=-14.0 kJ/mol

Answer 14

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Conclusion

Keq=230

Answer 15

Interpretation Introduction

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Answer 16

Expert Solution

Answer to Problem 25.36AP

K=230 at 37℃ (310 K)

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Expert Solution

Answer 20

Explanation of Solution

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

By substituting the value of T=37℃ (310 K) and ∆G°'=-14.0 kJ/mol, the value for the equilibrium constant is calculated as below:

∆G°'=-2.3 RTlog⁡Keq

-14000Jmol=-2.3 ×8.314Jmol·K ×310 K×log⁡Keq

log⁡Keq=-14000 J/mol-2.3×8.314 J/mol·K×310 K

log⁡Keq=2.362

Keq=230

Thus, the equilibrium constant for the temperature T=37℃=310 K is 230.

Answer 21

Conclusion

Keq=230

Answer 22

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Answer 23

Interpretation Introduction

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Answer 24

Expert Solution

Answer to Problem 25.36AP

This reaction is favorable because the overall conversion is from a less stable thioester to a more stable oxoester.

Answer 25

Expert Solution

Answer 26

Expert Solution

Answer 27

Expert Solution

Answer 28

Explanation of Solution

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Answer 29

Conclusion

This reaction is favorable because the overall conversion is from a thioester to a more stable ester.

Answer 30

Ch. 25 - Prob. 25.1P Ch. 25 - Prob. 25.2P Ch. 25 - Prob. 25.3P Ch. 25 - Prob. 25.4P Ch. 25 - Prob. 25.5P Ch. 25 - Prob. 25.6P Ch. 25 - Prob. 25.7P Ch. 25 - Prob. 25.8P Ch. 25 - Prob. 25.9P Ch. 25 - Prob. 25.10P

(a) Interpretation: The ∆ G ° ' for the reaction shown in Fig. P25.36. Concept Introduction: For a spontaneous reaction, the value for ∆ G ° ' is negative.

Concept explainers

Videos

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

Answer to Problem 25.36AP

Explanation of Solution

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

Answer to Problem 25.36AP

Explanation of Solution

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.

Answer to Problem 25.36AP

Explanation of Solution

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Chapter 25 Solutions

(a) Interpretation: The ∆ G ° ' for the reaction shown in Fig. P25.36. Concept Introduction: For a spontaneous reaction, the value for ∆ G ° ' is negative.

Concept explainers

Videos

(a) Interpretation: The ∆G°' for the reaction shown in Fig. P25.36. Concept Introduction: For a spontaneous reaction, the value for ∆G°' is negative.

Answer to Problem 25.36AP

Explanation of Solution

(b) Interpretation: The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K). Concept Introduction: The relation between ∆G°' and the equilibrium constant is: ∆G°'=-2.3 RTlog⁡Keq

Answer to Problem 25.36AP

Explanation of Solution

(c) Interpretation: A structural reason for the reaction being so favorable, is to be suggested. Concept Introduction: Overall reaction is said to be favorable when the products are more stable than the reactants.

Answer to Problem 25.36AP

Explanation of Solution

Want to see more full solutions like this?

Chapter 25 Solutions

(a)

Interpretation:

The ∆G°' for the reaction shown in Fig. P25.36.

Concept Introduction:

For a spontaneous reaction, the value for ∆G°' is negative.

(b)

Interpretation:

The equilibrium constant (at pH=7) for this reaction at 37℃ (310 K).

Concept Introduction:

The relation between ∆G°' and the equilibrium constant is:

∆G°'=-2.3 RTlog⁡Keq

(c)

Interpretation:

A structural reason for the reaction being so favorable, is to be suggested.

Concept Introduction:

Overall reaction is said to be favorable when the products are more stable than the reactants.