Change in heat of photosynthesis

Consider a plant that uses the energy it receives from the sun to do work (to undergo photosynthesis) to produce glucose. The minimum amount of work this plant can do to produce 1 mole of glucose is equal to the change in Gibbs free energy that goes into producing one mole of glucose. I calculated the change in Gibbs free energy that goes into producing one mole of glucose to be $$\Delta G =2878.94\ \mathrm{kJ}$$. Therefore the plant does $$2878.94\ \mathrm{kJ}$$ of work. Also if we calculate the change in enthalpy for the same process we get, $$\Delta H = 2803.04\ \mathrm{kJ}$$. Therefore the heat from photosynthesis under these conditions is, $$Q=\Delta H - \Delta W = -69.9\ \mathrm{kJ}$$.

This confuses me because it implies that photosynthesis under these conditions gives off heat. How can this be?

• The reaction is still endothermic. The sun does the work on the reaction, which is endothermic and gives off heat. It only sounds confusing if you are more used to discussing processes that run in the absence of work. The heat release is important to comply with the 2nd law. – Karsten Theis May 1 at 18:20

Generally speaking (without need to refer to your particular problem), the sign of the free energy change and thus the magnitude of maximum possible work does not depend on $$\Delta H$$ or $$\Delta S$$ alone, it depends on the combination:
$$w_{max} = \Delta G = \Delta H - T \Delta S$$
Therefore one need not perceive an inconsistency only based on the thermicity of a reaction. The thermicity describes the change in entropy of the surroundings, since $$\Delta H = -T\Delta S_{surr}$$. This allows you to rewrite the above in terms of entropies of the system and the surroundings (at constant T and P):
$$w_{max} = -T(\Delta S_{surr} + \Delta S_{sys})$$
In fact, as the above makes clear, exothermicity ($$\Delta H<0$$) generally makes a process more likely, not less, since it may counter a reduction in the entropy of the system.