Question

Given... `6CO_2(g) + 6H_2O(g)rarrC_6H_12O_6(aq) + 6O_2(g)` `DeltaH_"rxn"^@=+2803*kJ*mol^-1` What energy is associated with the formation of glucose from ONE mole of carbon dioxide?

Given...
`6CO_2(g) + 6H_2O(g)rarrC_6H_12O_6(aq) + 6O_2(g)`
`DeltaH_"rxn"^@=+2803*kJ*mol^-1`

What energy is associated with the formation of glucose from ONE mole of carbon dioxide?

Answer 1

Given your equation `(2803*kJ*mol^-1)/6=467.2*kJ`

Explanation:

Enthalpy change is always written per mole of reaction as written.

For the photosynthesis reaction.....

`6CO_2(g) + 6H_2O(g)rarrC_6H_12O_6(aq) + 6O_2(g)`

`DeltaH_"rxn"^@=+2803*kJ*mol^-1`

`6CO_2(g) + 6H_2O(g)+2803*kJrarrC_6H_12O_6(aq) + 6O_2(g)`

On the other hand for the combustion reaction.....

`C_6H_12O_6(s) + 6O_2(g) rarr 6CO_2(g)+6H_2O(aq)`

`DeltaH_"combustion"^@=-2803*kJ*mol^-1`

Or alternatively I could write......

`C_6H_12O_6(s) + 6O_2(g) rarr 6CO_2(g)+6H_2O(aq)+2803*kJ`

...for the COMBUSTION.

The sign of the enthalpy term determines whether it is RELEASED (`"minus sign"`) or ABSORBED (`"positive sign"`) by the reaction. In other words the sign designates whether the reaction is `"exothermic"` or `"endothermic"`.

And thus for the conversion of `1*mol` of carbon dioxide into glucose, an energy input of approx. `470*kJ` is required. Are you happy with this? From where does this energy come; the local power station?