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Photosynthesis is obviously an endothermic reaction, I mean, what would be the point otherwise, right? This is probably just sheer and utter stupidity on my behalf, but why does Wikipedia say the heat of formation of glucose is negative? As you can probably see I'm on a quest of understanding enthalpy as a concept and I'm still quite new, but some sort of ultranoob clarification would be very much welcome.

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Consider that when $1$ molecule of glucose forms, this happens at the cost of $6$ molecules of $\ce{CO2}$ and $6$ molecules of $\ce{H2O}$ :

$ \ce{6CO2(g) + 6H2O(l) -> C6H12O6(s) + 6O2(g)} $

Considering that at $25\ \mathrm{^\circ C}$ and $1\ \text{atm}$

$ \Delta H_{\text{f},\ce{CO2}}^\circ=-393.5\ \text{kJ/mol} $

$ \Delta H_{\text{f},\ce{H2O}}^\circ=-285.8\ \text{kJ/mol} $

$ \Delta H_{\text{f},\ce{C6H12O6}}^\circ=-1273.3\ \text{kJ/mol} $

You get, by summing the enthalpies of formation for each of the involved species and considering that for $\ce{O2}$ that enthalpy is zero:

$ \Delta H_\text{r}^\circ=6\times393.5\ \text{kJ/mol}+6\times285.8\ \text{kJ/mol}-1273.3\ \text{kJ/mol}=2802.5\ \text{kJ/mol} $

Note that, for $\ce{CO2}$ and $\ce{H2O}$, the change of sign is required, since you "lose" them, while you form glucose.

So, finally $\Delta H_\text{r}^\circ$ (heat to be absorbed to form one mole of glucose) is positive, indicating that photosynthesis is indeed endothermic.

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  • $\begingroup$ If you're still there, a tiny by-the-side question: those 2802.5 kJ, they're stored in the bonds of the 1 mol of glucose, right? Is that number (2802.5kJ) synonymous to the amount of photon-energy that hit the plant? $\endgroup$ – Uffe Apr 4 '14 at 14:01
  • $\begingroup$ @Uffe that's a good approximation, considering the complexity of such a process. $\endgroup$ – mannaia Apr 4 '14 at 14:15
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    $\begingroup$ This is a great answer (+1). I think people would be less confused by this topic in general if people included photons in their reaction. The photon reactants would have an indicated wavelength (and thus energy), and would either be converted to nothing or to photons of lower wavelenths (thermal infrared). If written this way with actually observed photon stoichiometries, it would also indicate in one reaction how inefficient photosynthesis is from an energy standpoint. $\endgroup$ – Curt F. Apr 17 '15 at 18:37
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The heat of formation is the difference in energy between the current state (in this case, glucose) and the state of the elements at 25 °C (carbon, oxygen, hydrogen, etc).

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Endothermic, exothermic. The difference is that an endothermic molecule is unstable. From the moment you remove the extra energy, it reverts back to its original structure. From original author, Park S. Nobel, 1974, it all depends on the manner you get the four protons in order to get the monomer. (five or six monomers to build a sugar). You get either four protons with two or four water molecules. Up to now, it is utilized two water molecules. So you say that the oxygen of photosynthesis is endothermic. And that can't be. (Park S. Nobel is the father of the expression Electron Chain Transport). He also proposed the possibility that the origin of the four protons is due to four molecules of water but for that one he did not propose a model, so the winning model is Two Water Molecules, only because the other model was not described.

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