Thermodynamics of photochemical reactions

Suppose I have a photochemical reaction in gas phase, such as $$\ce{CO2 + {h\nu} -> CO + O}.$$ I would like to work through the thermodynamics of such a reaction and understand the meaning of every term.

For a non-photochemical reaction, such as $\ce{CO + O -> CO2}$, I have a good understanding of how the change in Gibbs energy breaks down into enthalpy, internal entropy and a log-concentration term corresponding to the entropy of mixing. If I ignore the photon in the first reaction above, I can regard it as simply the reverse of this reaction, which due to its photochemical driving force is able to reduce the Gibbs energy by the same amount, $\Delta G$. All of this is unproblematic.

However, $h\nu$ itself is an energy change, which I can calculate, assuming I know the frequency $\nu$ of the absorbed light. I would like to know how to think about the thermodynamics of the whole system, including the coupling of the chemistry to the radiation field. In particular, my questions are:

1. How does the energy $h\nu$ relate to the other terms with energy units, such as $\Delta H$ and $\Delta G^0$?

2. I know that while a single photon has no internal entropy, a beam of light of a particular frequency can be thought of as having a temperature and an entropy. How should I think about the role of radiation entropy in the thermodynamics of photochemistry, and in particular, how can I do second-law calculations for photochemical reactions?

• Not sure what you mean by second-law calculations. Can you clarify? – Jan Jensen May 27 '15 at 7:13
• @JanJensen sure - I mean either of the following two closely related things: (i) how can I calculate the total rate of increase of entropy in the universe, including the entropy of the incoming and outgoing radiation fields, per mole reacted; or (ii) for a reversible photochemical reaction, what are the conditions for the chemical system to be in thermodynamic equilibrium with the radiation field? – Nathaniel May 27 '15 at 8:27
• Photochemistry doesn't work by thermally exciting the reacting molecules. You can't simply convert $h\nu$ to energy or entropy and treat it as a thermodynamic variable – Jan Jensen May 27 '15 at 8:57
• Your first statement is true but the second is a non sequitur. $h\nu$ is an energy. $E=h\nu$ is the equation for the energy of a photon, that's why we write it that way. It's completely uncontroversial that the heat given off by a photochemical reaction is $h\nu-\Delta H$. But that's only the first-law energy balance. I'm asking how to do the corresponding second-law calculations, that's all. – Nathaniel May 27 '15 at 12:00
• I have only just seen this post so by now you have probably worked it out. If not look at G. Porter Journal of Chemical Society, Faraday Transactions 2, 1983, vol 79 , p 473-482 which discusses the thermodynamics of photochemical reactions. You can also find the article in the book 'Chemistry in Microtime' by the same author. – porphyrin Jul 29 '16 at 13:26

$h\nu$ 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.
$$\ce{CO2} + h\nu \ce{-> CO + C}$$
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.