# When thermochemical data are quoted, often the units are DeltaH*mol^-1. In the photosynthesis reaction, why is DeltaH_"rxn"^@=2800*kJ*mol^-1?

Sep 8, 2016

This is a good question. Thermodynamic data are quoted per mole of REACTION as WRITTEN.

#### Explanation:

$6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right) \rightarrow {C}_{6} {H}_{12} {O}_{6} \left(a q\right) + 6 {O}_{2} \left(g\right)$

$\Delta H = 2800 \cdot k J \cdot m o {l}^{-} 1$

You have grasped why this reaction is endothermic. It is a bond breaking reaction, that requires the cleavage of strong $C = O$ and $H - O$ bonds. Thermodynamic data are always quoted per mole of reaction as written. That is the reaction of 6 mol quantities of carbon dioxide and water gives molar quantities of sugar and 6 mol quantities of dioxygen gas, with the consumption of 2800 kJ of energy.

Note that it is not only carbon dioxide cleavage that requires this energy; the reaction requires the breaking of strong $O - H$ bonds, and this adds to the energy budget of the reaction.

When we do the reverse reaction, i.e. the combustion reaction, conservation of energy requires that for,

${C}_{6} {H}_{12} {O}_{6} \left(a q\right) + 6 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$

$\Delta H = - 2800 \cdot k J \cdot m o {l}^{-} 1$,

i.e. a molar quantity of glucose gives this energy on combustion (by formation of 6 molar quantities of carbon dioxide and water).

Of course, I am just writing this fact down here. But I can assure you that this combustion reaction has been experimentally measured ad infinitum, and $2800 \cdot k J \cdot m o {l}^{-} 1$ (of glucose) results from the combustion.