Book edition 1st

Author(s) Daniel V. Schroeder

Pages 356 pages

ISBN 9780201380279

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Short Answer

A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:
$C_{6} H_{12} O_{6} + 6 O_{2} ⟶ 6 {CO}_{2} + 6 H_{2} O$
(a) Use the data at the back of this book to determine the values of $Δ H$ and $Δ G$ for this reaction, for one mole of glucose. Assume that the reaction takes place at room temperature and atmospheric pressure.
(b) What is maximum amount of work that a muscle can perform , for each mole of glucose consumed, assuming ideal operation?
(c) Still assuming ideal operation, how much heat is absorbed or expelled by the chemicals during the metabolism of a mole of glucose?
(d) Use the concept of entropy to explain why the heat flows in the direction it does?
(e) How would your answers to parts (a) and (b) change, if the operation of the muscle is not ideal?

(a) The value of $Δ H$ is $- 2803.04 kJ$ and the value of $Δ G$ is $- 2878.94 k J$ .

(b) The maximum amount of workdone is 2878.94 KJ.

(d) Here, the heat is positive and entropy is positive. So, the heat flows into the system.

(e) Here less amount of energy will leave the system, but Gibbs free energy and enthalpy will be same.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Explanation

The chemical reaction for the fuel cell of the muscle is

$C_{6} H_{12} O_{6} + 6 O_{2} \to 6 {CO}_{2} + 6 H_{2} O$

Formula used:

Write the expression for the enthalpy change for the reaction.

$Δ H = 6 H_{{Co}_{2}} + 6 H_{H_{2} O} - H_{C_{6} H_{12} O_{6}} - 6 H_{O_{2} \dots \dots \dots} (1)$

Here, $H_{{CO}_{2}}$ is the enthalpy for ${CO}_{2}, H_{H_{2} O}$ is the enthalpy of $H_{2} O, H_{C_{4} H_{2} O_{4}}$ is the enthalpy for $C_{6} H_{12} O_{6}$ and ${Ho}_{2}$ is the enthalpy of $O_{2}$ .

Write the expression for the Gibbs energy change for the reaction.

Δ G = 6 G_{{CO}_{2}} + 6 G_{H_{2} O} - G_{C_{b} H_{12} O_{4}} - 6 G_{O_{2}} \dots \dots \dots (2)

Step 2: Calculation

Calculation:

Refer table at the back of the book.

Substitute $- 285.83 kJ$ for ${HCO}_{4}, - 393.51 kJ$ for $H_{H_{2} O}, 0$ for $H_{C_{n} H_{2} O_{6}}$ and $- 1273 kJ$ for ${Ho}_{2}$ in expression (1).

$Δ H = - 6 (285.83 kJ + 393.51 kJ) - 6 (0) + 1273 kJ = - 2803.04 kJ$

Substitute $- 237.13 kJ$ for $G_{{CO}_{2}}, - 394.36 kJ$ for $G_{H_{2} O}, 0$ for $G_{C_{6} H_{2} O_{3}}$ and $- 910 kJ$ for $G_{O_{2}}$ in expression (2).

$Δ G = - 6 (237.13 kJ + 394.36 kJ) - 6 (0) + 910 kJ = - 2878.94 kJ$

Thus, the value of $Δ H$ is $- 2803.04 kJ$ and the value of $Δ G$ is $- 2878.94 kJ$ .

Step 3. (b) GIven information

The chemical reaction for the fuel cell of the muscle is

$C_{6} H_{12} O_{6} {+6O}_{2} {→6CO}_{2} {+6H}_{2} O$ .

Workdone for ideal operation,

$W= |∆ G| where W=workdone |∆ G| = Gibbs energy$

Step 4. Calculation

As, $∆ G=2878.94 KJ$ .

So, $W=2878.94 KJ$

Step 5. Conclusion

The maximum amount of workdone is 2878.94 KJ.

Step 5. Given information

The chemical reaction for the fuel cell of the muscle is

$C_{6} H_{12} O_{6} {+6O}_{2} {→6CO}_{2} {+6H}_{2} O$ .

The enthalpy is less than the amount of the work extracted.

The expression for the heat absorbed.

$Q=W-∆H$

where

$Q=heat absorbed$ .

Step 7. Calculation

Here

$W=2878.94 KJ ∆ H=2803.04 KJ$

So, $Q=2878.94 KJ-2803.04 KJ = 75.9 KJ$

Step 8. Conclusion

The amount of heat absorbed is 75.9 KJ.

Step 9. Given information

The chemical reaction for the furl cell of the muscle is

$C_{6} H_{12} O_{6} {+6O}_{2} {→6CO}_{2} {+6H}_{2} O$ .

The expression for the entropy change for the reaction.

$∆ S = 6 S_{C O_{2}} + 6 S_{H_{2} O} - S_{C_{6} H_{12} O_{6}} - 6 S_{O_{2}} \cdot \cdot \cdot \cdot \cdot \cdot \cdot (3) w h e r e S_{C O_{2}} = E n t r o p y f o r C O_{2} S_{H_{2} O} = E n t r o p y o f H_{2} O S_{C_{6} H_{12} O_{6}} = E n t r o p y o f C_{6} H_{12} O_{6} S_{O_{2}} = E n t r o p y o f O_{2}$

Entropy in terms of heat absorbed

$∆ S = \frac{Q}{T} \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (4) w h e r e Q = h e a t a b s o r b e d T = a b s o l u t e t e m p e r a t u r e$

Step 10. Calculation

From the table

$S_{C O_{2}} = 69.91 J \cdot K^{- 1} S_{H_{2} O} = 213.74 J \cdot K^{- 1} S_{C_{6} H_{12} O_{6}} = 205.14 J \cdot K^{- 1} S_{O_{2}} = 212 J \cdot K^{- 1}$

So,

$∆ S = (6 (69.91 + 213.74) = 6 (205.14) - 212) J \cdot K^{- 1} = 259.06 J \cdot K^{- 1}$

Hence,

$Q = (259.06 J \cdot K^{- 1}) (298 K) = 77.2 K J$

Step 11. Conclusion

Thus, the heat is positive and entropy is also positive. So, the heat flows into the system.

Step 12. Given information

The chemical reaction for the fuel cell of the muscle is

$C_{6} H_{12} O_{6} {+6O}_{2} {→6CO}_{2} {+6H}_{2} O$ .

In the ideal reaction, the work equals the change in the Gibbs energy. But when the reaction is not ideal the amount of work is less than the Gibbs energy. Here, as the change of entropy is same, therefore less heat flow enters the system and less energy leaves the system. But, the Gibbs free energy and enthalpy are same whether or not the operation is ideal or not.

Step 13. Conclusion

Thus, less amount of energy will leave the system, but Gibbs free energy and enthalpy will be same.

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Step by Step Solution

Step 1: Explanation

Step 2: Calculation

Step 3. (b) GIven information

Step 4. Calculation

Step 5. Conclusion

Step 5. Given information

Step 7. Calculation

Step 8. Conclusion

Step 9. Given information

Step 10. Calculation

Step 11. Conclusion

Step 12. Given information

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