Thermodynamics

Group Work.6

- A Zen Moment

In silence contemplate the leaf for 2 minutes.

During the first minute, allow your mind to go where it wants.

During the second minute, become aware—think—about what the leaf is.

Topic: What is keeping you alive?

A. You also carry out combustion reactions to stay alive. The foods you eat are metabolized for energy.

(For a detailed discussion of this, take Biochemistry!) A simple example of one such process is glucose metabolism, part of your respiration process. In this reaction, glucose is "combusted" with oxygen to make CO2 and H2O, not unlike the candle burning process. Glucose is C6H12O6.

- Write the balanced combustion reaction for a glucose metabolism.

C6H12O6(s) + 6O₂(g) => 6CO₂(g) + 6H₂O(g) or (l)

- Calculate DH, DS and DG for this reaction. Is this reaction spontaneous? Is this reaction exothermic?

[answers when water is a gas]

DH = (6 mol)(-394 kJmol^-1) + (6 mol)(-242 kJmol^-1) - (-1274 kJmol^-1) - (6 mol)(0 kJmol^-1)

= -2540 kJ (H₂O(g)) Exothermic

\

DS = (6 mol)(214 Jmol^-1) + (6 mol)(189 Jmol^-1) - (212 Jmol^-1) - (6 mol)(205 Jmol^-1)

= 976 J (H₂O(g)) Driven by increasing entropy

DG = (6 mol)(-394 kJmol^-1) + (6 mol)(-229 kJmol^-1) - (-910 kJmol^-1) - (6 mol)(0 kJmol^-1)

= -2830 kJ (H₂O(g)) Spontaneous

[answers when water is a liquid]
DH      = (6 mol)(-394 kJmol-1)  +  (6 mol)(-285 kJmol-1)  -  (-1274 kJmol-1)  - (6 mol)(0 kJmol-1) 
                        =  -2806 kJ (H2O(l))   Exothermic

DS      = (6 mol)(214 Jmol-1)  +  (6 mol)(70 Jmol-1)  -  (212 Jmol-1)  - (6 mol)(205 Jmol-1) 
                        =  262 J (H2O(l))        Driven by increasing entropy

DG      = (6 mol)(-394 kJmol-1)  +  (6 mol)(-237 kJmol-1)  -  (-910 kJmol-1)  - (6 mol)(0 kJmol-1) 
                        =  -2880 kJ (H2O(l))  Spontaneous

C. So what is keeping you alive? What is the original source of energy that you "consume" as glucose?

Where does glucose come from? How is it made?

Glucose keeps us alive, which we get from plants grown by photosynthesis.
Photosynthesis—the reverse reaction of glucose metabolism:

Write it down.

6CO₂(g) + 6H₂O(g) or (l) + Energy => C6H12O6(s) + 6O₂(g)

What is the value of DGrxno for photosynthesis? Is it spontaneous?

DG_{Photosynthesis} = - DG_{GlucoseMetabolism} = 2880 kJ Not Spontaneous

But you know it happens? How can that be?

Light from the sun provides energy to make the reaction go.

We're lucky: we get to eat sugar for energy! J

Some organisms "eat" sulfur compounds for their needs. Purple and green sulfur bacteria, which can be found in anaerobic mud in ponds and lakes, use H2S instead of water to make glucose:

3 H2S + 6 CO2à 3 HSO4^- + C6H12O6

- Write the balanced reaction for glucose formation using aqueous hydrogen sulfide, H2S.

6 H2O + 3 H2S + 6 CO2à 3 HSO4^- + C6H12O6 + 3H+

- Calculate DH, DS and DG for this reaction. Is this reaction spontaneous? Make a guess as to how these bacteria do it?!

Glucose HSO4 H+ H2S H2O CO2

DH = (-1274 kJmol^-1) + (3)(-887 kJmol^-1) + (3)(0 kJmol^-1) - (3 )(-39.7 kJmol^-1) - (6 )(-285 kJmol^-1) - (6)(-394 kJmol^-1)

= + 258 kJ (H₂O(l)) Endothermic

Glucose HSO4 H+ CO2 H2S H2O

DS = (212 JK^-1mol^-1) +(3)(131.8 JK^-1mol^-1) +(3 mol)(0 JK^-1mol^-1) - (6 )(214 JK^-1mol^-1) - (3)(121 JK^-1mol^-11) - (6)(69.9 JK^-1mol^-11) -

= -1459 JK^-1mol^-1 (H₂O(l))

Glucose HSO4 H+ H2S CO2 H2O

DG = (-910 kJmol^-1) - (3)(-755.9 kJmol^-1) +(3)(0 kJmol^-1) – (3)(- 27.8 kJmol^-1 ) - (6 )(-394 kJmol^-1) - (6)(-237 kJmol^-1)

= + 692 kJ (H₂O(g)) NON - Spontaneous

maybe the bacteria also use photosynthesis to get the non-spontaneous reaction to go? Yes they do ! They are photosynthetic bacteria, probably the first organism to do photosynthesis and produce our oxygen atmosphere 1000’s of years ago.

- How do the values of DH, DS and DG compare with those for photosynthesis?

Overall, the reaction is not as bad, not as positive a free energy value. Does not take as much photo-energy.