The common sugar glucose C6H12O6 is oxidized essentially continuously in the human body to produce the energy needed for various metabolic processes. The reaction is C6H12O6+6O2 → 6CO2+6H2O This reaction is done biochemically in the body and is referred to as glycolysis. The specific path is

Question

The common sugar glucose [latex]C _{6} H _{12} O _{6}[/latex] is oxidized essentially continuously in the human body to produce the energy needed for various metabolic processes. The reaction is

[latex]C _{6} H _{12} O _{6}+6 O _{2} \rightarrow 6 CO _{2}+6 H _{2} O[/latex]

This reaction is done biochemically in the body and is referred to as glycolysis. The specific path is referred to as the Krebs cycle, and part of the energy released in the reaction is stored by producing two adenosine triphosphate (ATP) molecules from 2 adenosine diphosphate (ADT) molecules. The dissociation of ATP to ADT is then used as an energy source elsewhere in the body to drive biochemical processes.

The following data are available at 25ºC:

a. Complete the table above.

b. Calculate the availability of each component and the maximum flow shaft work that can be obtained from this reaction with respect to ambient conditions of 25ºC.

c. Calculate the maximum flowshaftwork from this reactionwith respect to body temperature of 37ºC.

Accepted Answer

a. The enthalpy, Gibbs free energy, and entropy of formation are at 25ºC. Therefore the missing items in the table are computed using [latex]\Delta_{ f } \underline{G}=\Delta_{ f } \underline{H}-T \Delta_{ f } \underline{S}=\Delta_{ f } \underline{H}-298.15[/latex].

[latex]\Delta_{ f } \underline{S} .[/latex] Also, here at 25ºC the availability is equal to the Gibbs energy of formation. The results are:

	[latex]\Delta_{ f } \underline{H}( kJ / mol )[/latex]	[latex]\Delta_{ f } \underline{G}( kJ / mol )[/latex]	[latex]\Delta_{ f } \underline{S}( J / mol \cdot K )[/latex]	CP (J/mol · K)	[latex]\underline{ B }( kJ / mol )[/latex]
Glucose	−1271	−1333.4	209.2	218.6	−1333.4
[latex]O _{2}[/latex]	0	0	0	29.4	0
[latex]CO _{2}[/latex]	−393.5	−394.4	3.0	37.0	−394.4
[latex]H _{2} O[/latex]	−285.8	−237.1	−163.3	73.4	−237.1

b. At 25ºC the flow availability of each compound is equal to its Gibbs free energy of formation. Therefore, the maximum shaft work available from this reactio; is

[latex]\begin{aligned}W_{s, \max } &=6 \underline{ B }_{ CO _{2}}+6 \underline{ B }_{ H _{2} O }-\underline{ B }_{ glucose }-6 \underline{ B }_{ O _{2}} \&=6 imes(-394.4)+6 imes(-237.1)-(-1333.4)-6 imes 0 \&=-2455.6 kJ / mol ext { glucose }\end{aligned}[/latex]

c. As the first step in the calculation, we have to correct the enthalpy and the entropy to 37ºC. This is accomplished using the following equations

[latex]\begin{array}{l}\underline{H}\left(37^{\circ} C ight)=\underline{H}\left(25^{\circ} C ight)+\int_{298.15}^{310.15} C_{ P } d T=\underline{H}\left(25^{\circ} C ight)+12 imes C_{ P } \\underline{S}\left(37^{\circ} C ight)=\underline{S}\left(25^{\circ} C ight)+\int_{298.15}^{310.15} \frac{C_{ P }}{T} d T=\underline{S}\left(25^{\circ} C ight)+C_{ P } imes \ln \left(\frac{310.15}{298.15} ight) \\quad=\underline{S}\left(25^{\circ} C ight)+0.03946 imes C_{ P }\end{array}[/latex]

[latex] ext { Also, here the flow availability is } \underline{ B }\left(37^{\circ} C ight)=\underline{H}\left(37^{\circ} C ight)-310.15 imes \underline{S}\left(37^{\circ} C ight) ext {. }[/latex]

The results at 37ºC are given in the following table.

	[latex]\underline{H}( kJ / mol )[/latex]	[latex]\underline{S}( J / mol \cdot K )[/latex]	[latex]\underline{ B }( kJ / mol )[/latex]
Glucose	−1268.4	217.9	−1336.0
[latex]O _{2}[/latex]	−0.350	1.12	0.01
[latex]CO _{2}[/latex]	−393.1	4.48	−394.5
[latex]H _{2} O[/latex]	−284.8	−160.44	−235.2

and [latex]W_{S, \max }[/latex] = −2441.6 kJ/mol glucose

This maximum work at 310.15 K is only slightly less than at 298.15 K in part b.

Comments

In humans approximately 32% of the energy released on the glycolysis reaction is used in the ADT to ATP reaction, the rest is released as heat. Therefore, the glycolysis reaction is only 32% efficient.

Question 14.5.3: The common sugar glucose C6H12O6 is oxidized essentially con...

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