# Stoichiometric and Energetic Considerations and the Role of Entropy

The statements of Jennings as published in (Jennings et al., 2005; 2006; 2007) suggest the possibility that the second law of thermodynamics might stand in contradiction to the primary photochemistry of plants. Such an opinion seems to be rather challenging, if not to say wrong as we will seek to show in the following.

We will try to briefly discuss the basic underlying thermodynamic concepts and their relation to photosynthesis. In fact, it is hard to prove that photosynthesis does not proceed with negative entropy formation, but it is equally difficult to prove the opposite. As for all thermodynamic processes that might violate the second law of thermodynamics, no plants are observable that grow with a universe overall negative entropy formation. It turns out that plants grow in local nonequilibrium situations in the stationary overall equilibrium of the energy and entropy flux from the sun that interacts with the earth and leads to a high entropy production on the earth surface. We strongly believe and make suggestions for a proof that no process violates the second law of thermodynamics. The second law of thermodynamics mainly reveals that unlikely conditions relax to the most probable distribution. The appearance of plants is just a kind of nonequilibrium dynamics in such process that leads from a local strong nonequilibrium to the equilibrated situation. Plants slow down the pathway of a system into its most probable state. This aspect of physics and living matter is shortly treated in the following subsection.

The photosynthetic formation of sugar from carbon dioxide and water as denoted by chemical equation 1 is highly endergonic and therefore needs an input as driving force, the Gibbs free energy of the absorbed photons. The endothermic character is shown by the difference of the enthalpy (H) which is AH = +2808 kJ/mol. The reduction of entropy (S) can be calculated to AS = -259.1 J/(mol*K) (see Muller, 2005). Therefore at room temperature (T) (295 K) one gets *TAS* = -76.4 kJ/mol.

Generally a process occurs spontaneously if the change of Gibbs free energy (G) is negative (i.e. if the Gibbs free energy of the product compounds is smaller than the Gibbs free energy of the reaction products, AG < 0):

Without the photon contribution the resulting Gibbs free energy of the photosynthetic reaction described in chemical equation 1 calculates to A*G* = +2884.4 kJ/mol according to equation 64 indicating the endergonic character of the reaction. The equilibrium of the chemical equation 1 is far at the left side of the reaction, i.e. at the side of the chemical educts, carbon dioxide and water.

The analysis of the photosynthetic processes leading to the production of one mol glucose shows that at least 60 photons are absorbed per single molecule glucose that is generated (Hader, 1999; Campbell and Reece, 2009).

The absolute minimal value for the energy uptake should at least correspond to the absorption of 4 photons for the oxidation of one water molecule (two turnover cycles of PS I and PS II, each) and therefore to 48 photons per mol glucose. The splitting of 2H_{2}O to 4H+ and O_{2}

is releasing 4 electrons in the photosystem II (PS II) only. These electrons have to be pumped from the PS II via PS I to the place where NAD+ is reduced (see Figure 51 and Renger and Renger, 2008; Renger 2007, 2008, 2008b; Kern and Renger, 2007). PS I and PS II are working hand in hand which doubles the absorbed number of photon quanta per mol glucose. In fact, 8 photons are absorbed to split 2 water molecules, i.e. 48 photons are absorbed per mol glucose (see chemical equation 2).

This leads to a photonic contribution of at least *AH ^{phot}* = 48-

*N*per mol that is delivered by the photon energy (Jennings et al., 2005). The longest absorbed wavelength in the PS I is about 700 nm and therefore an additional

_{A}hv*AH*= 48-

^{phot}*N*8200 kJ/mol is involved into chemical equation 1.

_{A}hv&For detailed thermodynamic considerations the eq. 64 has to be evaluated for each single chemical step inside a plant. Such accurate thermodynamic analysis of photosynthesis is a very complicated task. It might be the reason for the recent discussions whether or not photosynthesis might violate the second law of thermodynamics (Jennings et al., 2005; 2006; 2007; Lavergne, 2006).

One mol of absorbed red photons with 700 nm wavelength contains an energy of 171 kJ. It is not fully consistently answered in the literature how much entropy a single photon contains or even if it is possible to define the entropy of a single photon. The entropy of the photon ensemble can be described by the entropy of the Planck spectrum and therefore thermal single photons exhibit a probability distribution that carries the corresponding amount of entropy. The absorbed light energy of 8200 kJ/(mol glucose) seems to carry a huge amount of “excess” energy in comparison to the energy consumption AG = +2884,4 kJ/mol in chemical equation 2.

Some of the relevant literature suggests that this “excess energy” shows that photosynthesis could be much more efficient. We believe that this point of view is too simple: how the directed transfer of photon energy to the chemical Gibbs free energy of glucose is possible is an important question. During the reaction sequence highly energetic compounds are formed as ATP that is urgently necessary to drive the functional work of the cell. It is not possible to drive the photosynthetic processes without energy dissipation.

The transformation of solar radiation to free energy is possible due to the low entropy per enthalpy (H) ratio *S/H* of the sunlight. If the volume and pressure do not change during a thermodynamic process, as we assume for the absorption of a photon, then *dH* = *dU* and *1/T* = S/H.

The inverse temperature of the solar radiation and therefore (S/H)^{solar }is much lower than the (S/H)^{earth} ratio of thermal photons that are released after absorption on the earth surface. The wavelength maximum (A_{max}) of the emitted black body radiation determines the so called colour temperature (Tc) of radiation which is associated with the *S/H* ratio according to Wien's law:

Equation 65 shows, that the wavelength maximum (Amax) of electromagnetic radiation is proportional to the entropy per enthalpy ratio. Therefore *S/H* is much lower for solar light than for thermal radiation of a 280 K black body.

In that sense the plant uses a pool of “negative entropy” from sunlight. There exist different estimations of the effective colour temperature Tc determined by the light spectrum that is available for photosynthetic organisms taking into account scattered light or light absorbed by the earth atmosphere (Muller, 2005), but for all estimations the available light spectrum corresponds to Tc » 1000 K in marked contrast to the typical entropy content of thermal baths in the sea or the earth atmosphere at *T* = 280 K. The formation of highly organized structures is a dissipative process in accordance to the second law of thermodynamics because the process occurs during the equilibration of sun radiation and earth temperature. The plants grow driven by a fractally local nonequilibrium.

As we have seen the photosynthesis is a quantum process which is driven by absorption of single photons. Entropy is a statistical quantity which is not well defined for single particles.

Generally the equation

holds where the equality denotes reversible processes when the uptake of heat goes along an isothermal path while the inequality is characteristic for irreversible processes, i.e. when the system is dissipative and the dynamics is not restricted to a thermodynamic path.

At this point it has to be clearly stated that the photosynthetic process does not occur in the equilibrium. The equations of thermodynamics are only valid assuming local equilibrium conditions. This might not be possible for light absorption and the interaction of excited states with the surrounding environment. The reader should therefore keep in mind that our considerations suggest extensions of formulas that are valid in equilibria conditions in spite of strong nonequilibrium conditions.

Evaluating equation 65 we can make a rough suggestion for the average entropy of a single photon by identifying it's energy with enthalpy and heat: *Q H = h* .

The equality of heat and enthalpy is fulfilled for an isobar thermodynamic process if the pressure remains constant which can be assumed to be valid for reactions of the photon gas.

From that simple consideration we would suggest

as contribution of the probability distribution of the single photon to the entropy. Interestingly *S** _{phot}* as given by eq. 67 is a fixed value independent from the photon wavelength and qualitatively comparable to the value

*S*1.38-10

_{phot}*k_{B}=^{-23}J/K as presented in (Kirwan, 2004).

Kirwan Jr. (2004) suggests a comparable value for the photon entropy as shown here, while Gudkov (1998) takes the viewpoint that “light is a form of high grade energy which carries no thermodynamic entropy”. Already Planck calculated the black body radiation spectrum considering the thermodynamic entropy of the radiation field. The thermal black body radiation which is in thermal equilibrium with the environment carries no Gibbs free energy: **A**G = 0 which corresponds to **A**H = *AQ = TAS* according to eq. 64.

The conversion of the energy of 1 mol photons in the solar radiation field at Tc = 5000 K into thermal radiation generates about 17 mol pho-

1

tons at *T _{C}* = 280 K according to Wien's law — and the

*T _{C} 2897,8ftmK*

*hc*

fact that *E _{photon}=* — . As about half of the radiation is converted to excited states in the plant while the rest of the incoming spectrum dissipates the generation of 1 mol glucose it at least correlated with the emission of 8 • 48 » 400 mol thermal photons. These photons might carry away an entropy of

*AS*

_{therm}*d*

*w*400-

*N*

_{A}*-k*

_{B}*in*3.3 kJ/(mol-K) which is easily compensating

**A**S

_{glucose}« -259.1 J/(mol-K) necessary to be reduced for the production of one mol glucose as shown above.

**A**S

_{t}, , exceeds

**A**S ,

*thermal glu* cos *e*

by a factor of ten. The rather rough and incomplete estimation as presented here might finally lead to the conclusion that a plant can not grow without dissipating a large fraction of its absorbed energy as this is necessary to preserve the second law of thermodynamics.

The necessary rise of the overall entropy when a plant grows is correlated with dissipation of energy into the environment. If a system absorbs photons and then relaxes to a final state emitting more photons than previously absorbed (in particular the production of phonons or any bosons fullfills the necessary increase of entropy) then the final state can be of lower entropy than the initial state in full accordance to the second law of thermodynamics because the environment takes up entropy.

At the moment there is no reason to assume that photosynthesis violates the second law of thermodynamics as done by Jennings et al. (Jennings et al., 2005) and answered by detailed description of the processes which are in line with the second law of thermodynamics by Lavergne (Lavergne, 2006) which let to further comments of Jennings (Jennings et al., 2006, 2007).

Generally entropy is a quantity of an ensemble describing the probability of an ensemble's state. The most general formulation of this property is given by the Shannon entropy

calculating the entropy from the single probabilities *p.* of each state *i *that occurs in the thermodynamic equilibrium (*k _{B}* is the Boltzmann constant). A more general formulation where is time dependent

and the system is not necessarily in an equilibrium is suggested by Haken (1990).

In eq. 69 *pjt^j* might be calculated from rate equations. We will use equation 68 and equation 69 to estimate the probability of a forward in comparison to backward steps in the rate equations as given by eq. 8 (see chap. 1.3). Equation 69 is a generalization of the problem of calculating entropy in the nonequilibrium case which helps us to understand the correlation of dynamics of probabilities and entropy when a photosynthetic complex relaxes after light absorption.

In the thermodynamic equilibrium equation 68 and equation 65 connect the macroscopic observables of thermodynamics (here: temperature *T* as given in eq. 65) with the microscopic probabilities of certain distributions of space and momentum in an ensemble of states (equation 68). Therefore the combination of equation 68 and 64 is the most important step for a statistical motivation of thermodynamics.

Eq. 8 and 9 (see chap. 1.3) can be derived from eq. 68 if we use the second law of thermodynamics which postulates that the Shannon entropy function of eq. 68 is at maximum in the equilibrium case of a closed system.

Then

where *Л* is an arbitrary Lagrange parameter with *SX = 0* that can be added because *^p** _{j}* = 1 and

*f}*is a Lagrange Parameter that can be

*i*

added because _{j}E_{j}.

Accomplishing the variation one gets:

Taking eq. 66 (the equality for the reversible processes) and eq. 68 one gets:

For the thermally equilibrated
system without work *(E^ = Q* one finds: - .

*k _{B}T*

And therefore

where *Z* denotes the standard canonical partition function.

From eq. 70 the proposed Boltzmann distribution for state populations according to eq. 9 and for rate constants according to eq. 8 follows directly.