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Sorry if this is too elementary, but can someone explain why the following reaction can occur without a net influx of energy from some other process?

$C_6H_{12}O_6 + 6O_2 \to 6 CO_2+6H_2O$

And why does this one need an influx of energy?

$6CO_2+6H_2O\to C_6H_{12}O_6+6O_2$

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  • $\begingroup$ Wait is the second one hydrolysis? $\endgroup$ – suomynonA Aug 24 '16 at 23:26
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    $\begingroup$ Changed "influx" to "requirement" in your title, as we don't really use that word in metabolism in the sense you describe. The energy is generally provided by the use of ATP in reactions, which is hardly coming from the outside, as influx implies. Left it in the question body, but change makes a better title for indexing etc. Hope that's OK with you. $\endgroup$ – David Aug 25 '16 at 8:39
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I’m not quite sure which of two questions you are asking here, but as this is basic to understanding metabolism, I think an answer to both questions will be of most general utility, even if you personally already understand the answer to the first.

[1] The Gibbs Free Energy Change in a chemical reaction determines whether it will proceed spontaneously

If a reaction involves a negative change in Gibbs Free Energy (ΔG), it will proceed spontaneously; if not, an input of energy is needed to drive it.

The breakdown of glucose to carbon dioxide and water has a large –ve ΔG and therefore does not require input of energy — the reverse reaction has an equivalent +ve ΔG and therefore requires an input of energy to drive it.

This basic aspect of chemical reactions emerges from the laws of thermodynamics, and is explained in detail in section 1.3.3 of Berg et al., available on-line.

[2] Reactions that break bonds tend to involve negative changes in free energy

Why does the oxidation of glucose have a –ve ΔG, and its synthesis a positive –ve ΔG? This is because glucose oxidation involves breaking covalent bonds, and its synthesis making covalent bonds.

To understand the thermodynamic contribution of bond energies, the account in section 2.4 of Lodish et al. (available on-line) is recommended. In brief, ΔG = ΔH – TΔS, where H is enthalpy and S is entropy. The bond energies contribute to ΔH, and there is also generally in increase in entropy when two moieties become one, as mentioned in the answer by @MangoPrincess.

Footnotes

  1. Not all chemical or biochemical reactions involve net formation or breakage of bonds — for example molecular rearrangements. In these cases more sophisticated chemical considerations are needed to understand the experimentally determined changes in free energy.

  2. The importance in discussing biochemical reactions in terms of Gibbs Free Energy changes is that an energetically favourable reaction can be coupled to an energetically unfavourable one to drive the latter, and it is the quantitation of the ΔG that allow one to predict or understand this. It is also important as ΔG can be quantitatively linked to oxidation–reduction potentials.

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  • $\begingroup$ Is my second equation an example of hydrolysis? and the first one a dehydration reaction? Studying them more closely, that is what it seems like. $\endgroup$ – suomynonA Aug 26 '16 at 4:50
  • $\begingroup$ @Anonymous 1. Each equation is just a representation of the overall balance of constituents of glucose. You cannot consider them as a reaction as they are the balance of many reactions. 2. The overall reactions are oxidation of glucose and reduction of carbon dioxide — not hydrolysis and dehydration. 3. The equations omit the oxidising and reducing agents NAD+/NADH and FAD/FADH2 — not to mention ATP and ADP — so are incomplete. 4. I‘ll look at the individual reactions in another comment as I’m running out of characters. $\endgroup$ – David Aug 26 '16 at 20:57
  • $\begingroup$ @Anonymous 4 cont. You can see tables which classify the inidividual reactions of glycolysis and the TCA cycle (for the oxidation of glucose) in Berg et al. Table 16.3 and Table 17.2. The reactions that break C–C bonds are not hydrolyses but an aldol cleavage and decarboxylation/oxidations. Hydrolysis breaks a bond by adding water, for example the peptide bond ~CO–NH~ becomes ~COOH + ~NH2. Other examples are hydrolysis of phosphodiester and glycosidic bonds. $\endgroup$ – David Aug 26 '16 at 21:13
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Oxygen has a high electronegativity, in other terms it really likes electrons.

In this scale neon has the highest electronegativity of all elements, followed by fluorine, helium, and oxygen.

These are redox reactions. In the first reaction carbon was oxidized 0 -> +4 and oxygen was reduced 0 -> -2. In the second reaction the reverse happens. Nomen est omen oxygen is an oxidizing agent, so it does not like the reactions where it will be oxidized, since by oxidation it loses electrons. And it likes electrons. Really.

Forcing somebody to do something they don't want, always requires energy, and oxygen is not an exception. ;-)

Just to mention there is always an activation energy regardless of the exothermicity of the reaction. Remember; sugar doesn't burn spontaneously at room temperature... These are multi step reactions and enzymes lower the activation energy of each step, that's why you don't need to invest much by the first reaction. Ofc. a similar reaction can happen by burning sugar, but in that reaction the activation energy is much higher, and it is coming from the heat of the gas flame.

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  • $\begingroup$ It's a bit misleading to say "these are redox reactions". The equations representing the overall process are, but they represent a whole series of reactions, only some of which are redox reactions. And the principle of whether a reaction requires energy is applicable regardless of whether the reaction is redox. The point about activation energy is important, but one needs to stress that in metabolism the reactions proceed at room temperature etc. because of the enzyme lowering the activation energy (or raising the ground state). $\endgroup$ – David Aug 25 '16 at 8:44
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    $\begingroup$ @David I did not want to be perfectly accurate with this answer. Ofc. I could have described it in more detail, but I don't think that would be necessary. If you feel the need you can always edit it, but please try to keep the same length and simplicity. $\endgroup$ – inf3rno Aug 25 '16 at 13:44
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The first reaction is catabolism. It does require some energy input in the beginning ("energy investment" of glycolysis) but ultimately leads to net energy release through "energy payoff" of glycolysis, citric acid cycle, and finally electron transport chain and oxidative phosphorylation.

The final result is oxidation of glucose, and reduction of oxygen.. At various points, electrons transfer to NAD+ and FAD. Then, the reduced forms of these 2 electron carriers (NADH & FADH2) go to the electron transport chain, where electrons transfer to oxygen at the end (reduction of oxygen). Meanwhile, protons were pumped into the intermembrane space to create the proton gradient that drives ATP synthesis.

The second reaction is anabolism. The Calvin Cycle requires energy from ATP and NADPH to create the necessary bonds to build up sugars.

In terms of bioenergetics, degradation of a molecule is exergonic. This is favorable if you think about the second law of thermodynamics: there is more entropy when there are multiple, separate sub-units, while there is more order when one molecule made up of those sub-units. Synthesis of a molecule is endergonic (requiring energy). It IS favorable in the sense that we get sugar to eat and burn in the future.

In summary:

  • Breaking down a molecule = catabolism = release energy from chemical bonds
  • Building up a molecule = anabolism = absorb energy to FORM chemical bonds (think of the resources needed to construct a new building)

Note my answer leaves out specific details about the processes, but I hope this is basic enough to answer your question!

Source: Lehninger Principles of Biology, David Nelson & Michael Cox

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    $\begingroup$ Welcome to Biology.SE. Please add some references that give both evidence for your claims and input for further reading. With these basic processes, Wikipedia articles should suffice. $\endgroup$ – AlexDeLarge Aug 24 '16 at 10:48
  • $\begingroup$ But if I didn't recognize those processes as glycolysis and the citric acid cycle, how would I know? I'm searching for a more generalized way of figuring out my question. $\endgroup$ – suomynonA Aug 26 '16 at 4:48

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