Thermodynamics of photochemical reactions
$h\nu$ is not a thermal energy so thermodynamic considerations do not apply. For example, you cannot predict the probability of being in an electronic excited state from equilibrium thermodynamics.  The main reason is that the excitation is not a result of (classical) thermal energy transfer, but rather the adsorption of a photon which changes the quantum state.

With a constant light source 

$$\ce{CO2} + h\nu \ce{-&gt; CO + C}$$ 

will reach a steady state with constant concentrations that will be different from the thermal equilibrium concentrations.  However, you can not extract a standard free energy difference by
$$ \Delta G^\circ=-RT\ln \left( \frac{p_{\ce{CO}}p_{\ce{O}}}{p_{\ce{CO2}}} \right) $$
because the products and reactant are not in thermal equilibrium.