A muscle can be thought of as a fuel cell, producing work from the metabolism of glucose:

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2⟶6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$

(a) Use the data at the back of this book to determine the values of $\Delta H$and $\Delta G$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?

The chemical reaction for the fuel cell of the muscle is

${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6+6{\mathrm{O}}_2\to 6{\mathrm{CO}}_2+6{\mathrm{H}}_2\mathrm{O}$

Formula used:

Write the expression for the enthalpy change for the reaction.

$\Delta H=6H_{{\mathrm{Co}}_2}+6H_{{\mathrm{H}}_2\mathrm{O}}-{\mathrm{H}}_{{\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6}-6{\mathrm{H}}_{{\mathrm{O}}_2\dots \dots \dots }\text{(1)}$

Here, ${\mathrm{H}}_{{\mathrm{CO}}_2}$is the enthalpy for ${\mathrm{CO}}_2,{\mathrm{H}}_{{\mathrm{H}}_2\mathrm{O}}$is the enthalpy of ${\mathrm{H}}_2\mathrm{O},{\mathrm{H}}_{{\mathrm{C}}_4{\mathrm{H}}_2{\mathrm{O}}_4}$is the enthalpy for ${\mathrm{C}}_6{\mathrm{H}}_{12}{\mathrm{O}}_6$and ${\mathrm{Ho}}_2$is the enthalpy of ${\mathrm{O}}_2$.

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

Calculation:

Refer table at the back of the book.

Substitute$-285.83\mathrm{kJ}$for ${\mathrm{HCO}}_4,-393.51\mathrm{kJ}$for ${\mathrm{H}}_{{\mathrm{H}}_2\mathrm{O}},0$for ${\mathrm{H}}_{{\mathrm{C}}_n{\mathrm{H}}_2{\mathrm{O}}_6}$and $-1273\mathrm{kJ}$for ${\mathrm{Ho}}_2$in expression (1).

$$\begin{gathered} \Delta H=-6(285.83\mathrm{kJ}+393.51\mathrm{kJ})-6\left(0\right)+1273\mathrm{kJ} \\ =-2803.04\mathrm{kJ} \end{gathered}$$

Substitute$-237.13\mathrm{kJ}$for $G_{{\mathrm{CO}}_2},-394.36\mathrm{kJ}$for $G_{{\mathrm{H}}_2\mathrm{O}},0$for ${\mathrm{G}}_{{\mathrm{C}}_6{\mathrm{H}}_2{\mathrm{O}}_3}$and $-910\mathrm{kJ}$for $G_{{\mathrm{O}}_2}$in expression (2).

$$\begin{gathered} \Delta G=-6(237.13\mathrm{kJ}+394.36\mathrm{kJ})-6\left(0\right)+910\mathrm{kJ} \\ =-2878.94\mathrm{kJ} \end{gathered}$$

Thus, the value of $\Delta H$ is $-2803.04\mathrm{kJ}$ and the value of $\Delta G$ is $-2878.94\mathrm{kJ}$.

The chemical reaction for the fuel cell of the muscle is

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{+6O}}_2{\text{→6CO}}_2{\text{+6H}}_2\text{O}$.

Workdone for ideal operation,

$$\begin{gathered} \text{W=}\left|∆G\right| \\ \text{where} \\ \text{W=workdone} \\ \left|∆\text{G}\right|=\text{Gibbsenergy} \end{gathered}$$

As, $∆\text{G=2878.94KJ}$.

So,$\text{W=2878.94KJ}$

The maximum amount of workdone is 2878.94 KJ.

The chemical reaction for the fuel cell of the muscle is

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{+6O}}_2{\text{→6CO}}_2{\text{+6H}}_2\text{O}$.

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

The expression for the heat absorbed.

$\text{Q=W-∆H}$

where

$\text{Q=heatabsorbed}$.

Here

$$\begin{gathered} \text{W=2878.94KJ} \\ ∆\text{H=2803.04KJ} \end{gathered}$$

So,$$\begin{gathered} \text{Q=2878.94KJ-2803.04KJ} \\ =75.9\text{KJ} \end{gathered}$$

The amount of heat absorbed is 75.9 KJ.

The chemical reaction for the furl cell of the muscle is

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{+6O}}_2{\text{→6CO}}_2{\text{+6H}}_2\text{O}$.

The expression for the entropy change for the reaction.

$$\begin{gathered} ∆S=6S_{C{O}_2}+6S_{H_{2}O}-S_{C_6{H}_{12}{O}_6}-6S_{O_2}·······\left(3\right) \\ where \\ {S}_{C{O}_2}=EntropyforC{O}_2 \\ {S}_{H_{2}O}=Entropyof{H}_{2}O \\ {S}_{C_6{H}_{12}{O}_6}=Entropyof{C}_6{H}_{12}{O}_6 \\ {S}_{O_2}=Entropyof{O}_2 \end{gathered}$$

Entropy in terms of heat absorbed

$$\begin{gathered} ∆S=\frac{Q}{T}········\left(4\right) \\ where \\ Q=heatabsorbed \\ T=absolutetemperature \end{gathered}$$

From the table

$$\begin{gathered} S_{C{O}_2}=69.91J·K^{-1} \\ {S}_{H_{2}O}=213.74J·K^{-1} \\ {S}_{C_6{H}_{12}{O}_6}=205.14J·K^{-1} \\ {S}_{O_2}=212J·K^{-1} \end{gathered}$$

So,

$$\begin{gathered} ∆S=\left(6\left(69.91+213.74\right)=6\left(205.14\right)-212\right)J·K^{-1} \\ =259.06J·K^{-1} \end{gathered}$$

Hence,

$$\begin{gathered} Q=\left(259.06J·K^{-1}\right)\left(298K\right) \\ =77.2KJ \end{gathered}$$

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

The chemical reaction for the fuel cell of the muscle is

${\text{C}}_6{\text{H}}_{12}{\text{O}}_6{\text{+6O}}_2{\text{→6CO}}_2{\text{+6H}}_2\text{O}$.

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.

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