REMARKS OK THE PHOTOCHEMISTRY OF POLYATONIC 3IOLECULESl JAMES FRANCK
Department o f Physics, T h e Johns Hopkins Cnlversity, Baltzmore, illaryland ASD
KARL F. HERZFELD
Department ai"Physics, T h e Catholic Cniversity, V a s h i n g t o n , D . C . Recezved October 16, 1996 I.
ISTRODUCTIOS
In a recent discussion (18) of photosynthesis in plants K. Wohl arrives a t the following conclusion: The assumption made u p to now that the photochemical part of photosynthesis is described by Chlorophyll
+ 4hv + CO2 + 3H20
-+
HKO
+ 2H202 + chlorophyll (1)
cannot be reconciled with the facts, since the energy supplied by the absorption of 4 quanta of red light is too small. He uses as wave-length 680 mp, the long wave-length limit within the absorption of chlorophyll which still gives photosynthesis, and finds the energy of four such quanta to be about 20 kg-cal. per mole smaller than the energy difference between initial and end products. The difference in free energy is also found to be slightly greater than the energy supplied by the photons. Assuming, as has always been done hitherto, that the energy differences alone are the determining quantities in a photochemical reaction and that one can neglect the contribution from thermal energy at room temperature, he has to reject the possibility of the chemical reaction 1. The problem whether the assumption generally made, that the thermal energy plays no r81e in photochemical reactions at room temperature, is also valid for polyatonic molecules able to give fluorescence, ha5 not yet been inrestigated. This and its possible application for the problem of photosynthesis make it worthn-hile to undertake a systematic diqcussion. We treat first the general case, then briefly its application to photosynthesis, and leave the mathematical proofs to the appendix. 1 Presented a t t h e Syniposium on AIolecular Structure, held at Princeton Lniversit?, Princeton, Sen. Jersey, December 31, 1936 t o January 2,1937, under the auspices of the Division of Physical and Inorganic Chemistry of the -4merican Chemical Society .
97 T H E JOCRXAL OF P R Y S I C A L CHEMISTRT. V O L .
41, SO. 1
98
JBMES F R A S C K AXD KARL F. HERZFELD 11. GEKERAL DISCUSSIOX
We start out with photochemical reactions in the gaseous state and limit ourselves mainly to room temperature. Of course it has long been known that the primary photochemical reaction might be followed by all kinds of secondary ones. We discuss here, however, the special case (not mentioned before) where the secondary reaction is endothermic and follow the primary one in a very short time. The usual view, mentioned in the introduction, has its origin in the case of photochemical reactions of diatomic molecules in the gaseous state, where it is entirely justified. If a light quantum is absorbed having insufficient energy to dissociate the molecule, the excited state will last to sec. The molecule has thermal energy in its degrees of freedom of translation and rotation, but the first are completely, the second almost (Le., except for the centrifugal expansion) useless for dissociation. I n the single degree of oscillation, there might be, a t room temperature, a t the utmost 1 or 2 quanta (see appendix, a), which however are so small that they affect the dissociation energy very little, so that the real limit of the photochemical process might be lifted by 1/20 of a volt or less for 200OC. One can of course gain more energy if one makes use of the few molecules having more quanta. They are (at least partly) responsible for the absorption at the long wavelength side of the main absorption band, which absorption is however very weak, because of the small number of such molecules. The lifetime of the optically excited molecule, however, is too small to acquire more vibrational energy by collision with other molecules z(see appendix, f ) , and therefore, if the light quantum is too small, it is lost (by reemission or collisions of the second kind) before a reaction can occur. The situation is quite different for very large molecules in the gaseous state or in solutions in which fluorescence is possible. It is still true that, as far as the primary transition into the excited state is concerned, the vibrational energy of not more than one eigen-vibration, i.e., about 1 or 2 vibrational quanta, can be used in general, so that if the light energy is smaller than the heat of dissociation, the system gets into an excited state below the dissociation. But while it is in that state, there is now a good chance that it may get enough thermal energy into the bond that has to break. The energy necessary t o supply the difference between bond strength and optically excited states gets into the important bond through fluctuations between different vibrations (Wigner and Polhnyi (17)). The time required increases strongly with the amount of the necessary energy and 2
At 1 atmosphere pressure the time t o acquire by collision 4 kg-cal. per mole is
10-7 s e c . , a n d correspondingly longer a t lovxr pressures.
PHOTOCHEMISTRY O F POLYATOMIC MOLECULES
99
depends in addition somewhat on the particular form of this complex (see appendix, c). Energies betn-een 2.4 and 9.2 kg-cal. per mole might occur within 0.5 x lop8sec. in the bond through interference. The range 2 to 9 kg-cal. is PO large because some particular configuration of the molecule might be necessary to make dissociation possible (steric hindrance). I n general, 7 kg-cal. is a n average value. As mentioned before, the lifetime of the excited state is too short for any considerable amount of energy to be supplied by collision in the gas, even at 1 atmosphere. Therefore, the molecule must be large enough to contain sufficient thermal energy in its normal state; i.e., it must have, for a thermal energy of 7 kg-cal., at least 7: 0.002 X 300 = 12 useful degrees of freedom . 3 There is, howel-er, an important difference between this case and the usual case of a purely thermal reaction. I n the latter, the reaction rate falls off at low gas pressures, because the energy supply through collisions is then not sufficient to keep up the number of molecules with higher thermal energy. The higher the number of degrees of freedom, the later will this falling off come, according to theories of Politnyi, 0. K. Rice, and Kassel (10, 11). Here however, the possession of the critical thermal energy, say 7 kgcal., is not sufficient for decay; the molecule must also be excited optically. Kom the combined reaction rate, photochemical excitation plus thermal reaction, is, with the usual light intensities, enormously slower than would be an entirely thermal dissociation with a heat of dissociation of 7 kg-cal. (which latter ~ o u l dproceed at room temperature with a reaction constant lo8 see.?). Therefore the energy supply by collision has all the time of between optical excitations and will be amply sufficient even at very low pressures. If, for example, every molecule receiTes lo3 quanta per second, which is J-ery high, a pressure of 0.06 mm. n-ould still be sufficient to keep up the supply. I n solutions, however, the latter problem does not come up at all. It is, on the other hand, difficult to decide which number of degrees of freedom a niolecule in liquids has to have so that the fluctuation can occur un-
-
3 The equilibrium probability t h a t an energy > AC resides in a certain bond is the same for the diatomic and for the large molecule. But the distribution is different. Disregard first the possibility of dissociation. Then (in gas of 1 atmosphere AU . pressure) the diatomic molecule nil1 gain L U about 1O1O exp. - - times a second,
-
RT
keep i t
sec. and lose it again. A C
.
In the polyatomic molecule, the bond will get the
same energy 1013esp. - - times a second and lose it after lO-13sec. RT If we now periiiit dissociation, the first time the molecule gains sufficient energy will permit dissociation; it is not a question of the whole time the molecule has the high energy, b u t of ho\T long it has to wait for it.
100
JAMES FRBXCK AND KARL F. HERZFELD
hindered. We know very little about the forces betneen molecules in the liquid. If we could treat the whole liquid as one molecule, as far as vibrations are concerned, there would be no necessity of storing all the energy in the particular chemical niolecule to be dissociated. That would mean that even a diatomic molecule, if dissolved, would under proper conditions (Le., the right position and form of the absorption curve and the possibility of keeping excitation long enough, which shows in the presence of fluorescence) only need an optical energy which is by 7 kg-cal. less than the dissociation energy. On the other hand, it seems probable that the forces betxeen molecules are several times less than the forces within molecules, so that the boundaries of niolecules act as a kind of barrier preventing quite as effective interference of the ela?ticwaves, i.e., for the same amount of energy, the accumulation in the bondwill be less frequent. That means that we might expect even diatomic niolecules in solution to react with a n insufficient amount of photon energy, but only if the difference is less than 7 kg-cal. Up to now, there was no mention of the difference in free energy between initial and end products. What r61e does it play? It determines the rate of the back reaction. Gaffron and Wohl (7) have recognized this, but have not given a quantitative discussion. I n chemical equilibrium the free energies of the initial and end products are equal. So are the forward and back reactions. If one diminishes the concentration of the end products, one does two things: (1) one diminishes the free energy of the end products, Le., arranges the process so that there is a decrease in free energy when a reaction takes place from initial into end products, and (2) one diminishes the back reaction. In other words, the free energy difference in a reaction occurring spontaneously measures t8he ratio between forward and backward reaction (see appendix, d). For an actual calculation of the back reaction, one can either use the difference in free energy calculated for the concentrations as they actually occur, or one can use the free energy difference for standard conditions as they are found in the tables, and introduce the concentrations explicitly. The difference is, of course, purely formal; the latter method seems simpler. If one uses the first method, the fraction of the products which is lost by back reaction is, for an uninterrupted process, according to equation 9 of the appendix AF
exp - RT if the forward reaction is purely thermal. If the speed of the forxard reaction is determined by light absorption at the rate Jzl’cl instead of t,he thermal reaction
PHOTOCHEMISTRY O F POLTAIOMIC MOLECULES
101
AU A’ exp - RT the ratio lost by back reaction is
A‘
AU - A F
x.1
RT
7exp
or
A’ A4 exp - kl ET
(4’)
Therefore, in a photochemical reaction AF = 0 has no immediate significance. The necessary difference in free energies depends on the strength of the light. From now on we use the second method. Consider a dissociation 1+2+3
(2)
Assume that the product 2 is not lost and will be present in amounts comparable to those of 1. The only way to slow up the back reaction in comparison to the forward reaction is to diminish the amount of 3. If we want the product of a photochemical dissociation to be ayailable for subsequent reactions, photochemical or thermal, it is therefore necessary to withdraw the molecule 3 as fast as possible. Up to now we have assumed that the thermal reaction following the primary photoreaction is a dissociation. If it is only an internal change, Le., of such a nature that the back reaction is also monomolecular, the difference between the change in free energy and the change in total energy is a small one (9, 5 ) , provided only one degree of freedom is involved. T h a t is,if this difference is 1.5 cal. in the right sense,if the forward reaction is endothermic and the back reaction has no heat of activation, the speed of the back reaction would be 1 O l o to 10l2set.-'. Only if a very large number of degrees of freedom were involved, if, for example, a very large number of vibrations would simultaneously go over into rotations, could we expect a considerable increase in entropy, and the back reaction would be correspondingly slow. That means kinetically that many degrees of freedom would have to get into rather special positions, which is a slow process. 111. APPLICATION TO PHOTOSYNTHESIS
The application of these considerations to photosynthesis gives us a greater freedom in the hypotheses which we can make for possible steps of the photochemical reactions. One is not any more limited strictly to steps in which hv 2 Q, where Q is the heat of reaction. Thus it appears
102
JAMES FRBNCK AND KARL F. HERZFELD
that the energy relations do not exclude the possibility that hydrogen peroxide and formaldehyde are formed as intermediate products by the absorption of 4 light quanta of 6600 A.U. On the other hand, special assumptions are necessary if one accepts this hypothesis. Khile other hypotheses may be preferable, one might just as n-ell treat this one as an example for the application of our results. Wohl calculates the energy necessary for thiq reaction to be 192 kg-cal. To find the energy of the light quanta, Kohl uses aq wave length 6800 A.U., which gives 167.5 kg-cal. But the main absorption (12) lies a t 6600, and the very weak absorption at 6800 might be due to molecules of high thermal energy, as explained a t the beginning of qection 11. T a r b u r g (16) also uses 6600 A.U. for his estimates. For this latter wavelength, the 4 quanta give 172 kg-cal. The missing 20 cal. might easily be supplied by thermal energy; this would necessitate about 5 kg-cal. per step. Wc can, however, expect about 8 to be available. First, it might be posqible to utilize about 1 kg-cal. from vibrational energy in the ground state. To wait for an additional ’7 kg-cal. in the upper state the molecule needs a lifetime of somewhat more than see., if we take A to be Now Prins (12) has measured the total absorption of chlorophyll in the red and found a n “electron number” of about 1/8, which would correspond to a transition probability from the excited state back to the normal state under fluorescence of
That is, if fluorescence were the only possible way to get back to the normal state, the lifetime of the excited state mould be T’ = 2 X lO-’sec. Even if we assume that only 5 per cent of the transitions give fluorescence (Le., if the fluorescent light is 5 per cent of the absorbed light), the lifetime would still be l o p 8 sec., the total transition probability being then twenty tinies as great as that under fluorescence alone. M’e would therefore still have a n activation energy of about 3 kg-cal. per step available. Kext, we have to ask whether the back reaction is too fast. Gaffron and ITohl (7) hare already given some consideration to that matter, and believe that i t presents great difficulties for equation 1. Actually, we do get into difficulties if we assume that product 3 in the forward reaction, which n-e identify with hydrogen peroxide, is destroyed by a monomolecular reaction. If we assume that in the forward reaction we have a gain of 5 kg-cal. in the entropy per step under standard conditions (this 5 kg-cal. being one-fourth of the 20 kg-cal. which is the difference between total energy change and free energy change, according to
PHOTOCHEMISTRY O F POLYATOMIC MOLECULES
103
Wohl (18)), we find for the lifetime of product 2, say formaldehyde, equation 14 of the appendix. Let us now consider a case where the production of oxygen is about 1 per cent of the saturation value. According to Arnold and Emerson (1, 2, 3 ) we have then 4 quanta absorbed by a chlorophyll molecule every 2000 sec. (for half-saturation, they calculate 20 sec.). There is about 0.01 mole of chlorophyll per liter of cells (moles per liter are the units used for the standard free energy calculations), so that 2 x 0.01 : 2000 = 10-6 mole hydrogen peroxide are produced per second per liter under these conditions. If we take the reaction destroying hydrogen peroxide to be the Blackman reaction, which has a n average duration of 0.02 see., and assume it to be monomolecular, we have 7i: = 10-5, ks = 50 in equation 15 and therefore c3 = 2 x 10-7 and the lifetime of formaldehyde comes out to be 3 X lop3sec., n hich is probably too short. On the other hand it is not quite impossible that for moderately high concentrations of hydrogen peroxide, its destruction by catalase might be of zero order. Then we can calculate the rate of destruction of hydrogen peroxide from the time of this destruction, 0.02 sec. (Blackman reactions) and the amount of oxygen developed during a light flash, which amount is twice that of hydrogen peroxide decomposed. This amount of oxygen, Arnold and Emerson (1, 2, 3 ) find, js about 5 X mm.3 of oxygen per flash per 1 mm.3 cell, or 5 X mole of hydrogen peroxide decomposed per liter cell per flash, which decomposition lasts 0.02 sec. Therefore, the speed of decomposition is 2.5 X moles per liter per second, which is sufficient up t o a n intensity a t which it takes 40 sec. to have 4 quanta absorbed per chlorophyll molecule. At light intensities comparable with this one (which is very close to the value a t half-saturation mentioned by Gaffron and Kohl), the influence of back reactions will become important. TTe will discuss this problem e l s e ~ h e r ein connection n-ith other possibilities for the theory of photosynthesis. The main result of this paper is the fact that in photochemical reactions of polyatomic nioleculeq able to emit fluorescent light the thermal energy can contribute to a much greater amount than it was formerly assumed. One gains hereby a greater freedom for assumptions on possible intermediate steps in photosynthesis. We wish to thank Professor E. Teller of George 1J7ashington University for helpful discussions. REFERENCES
(1) ARNOLD, W.,. ~ S D EMERSOS, R.: J. Gen. Physiol. 16,391 (1932). (2) ARSOLD,IT.,ASD EMERSON, R.:J. Gen. Physiol. 16, 191 (1932). (3) ARSOLD,IT.,BSD K O H N :J. Gen. Physiol. 18, 109 (1934). (4) EYRING, H . : J. Chem. Physics 3, 107 (1935); Chem. Rev. 17, 65 (1935). ( 5 ) ETRISG, H.: J. Chem. Physics 3, 107 (1935).
104 (6) (7) (8) (9)
(IO) (11) (12) (13) (14) (15) (16) (17) (18)
JAMES FRAKCK b K D KARL F. HERZFELD
EYRIKG, H . , GERSHIKOWITZ, H., ASD SUS, C. E. : J. Chem. Physics 3,786 (1935). GAFFROSAXD WOHL,IC: Naturwissenschaften 24, 81, 103 (1936). GERSHINOWITZ, H . , AND E Y R I N GH.: , J. Am. Chem. SOC.67,985 (1935). HERZFELD, K . F.: Physik. Z. 23, 95 (1922); Kinetische Theorie der Jl'arme, p. 193 ff. Braunschweig (1924). KASSEL,L.: Kinetics of Honiogeneous Gas Reactions, S e w York (1932). P O L ~ NM.: Y I , Z. Physik 1,337 (1920). PRINS, J. A.: Nature 134,467 (1934). RICE, 0. K . , A N D GERSHISOTVITZ, H.: J. Chem. Physics 2, 273, 853 (1934); 3, 473 (1935). SHERMAN, A,, SUN,C. E., ASD EYRING, H . : J. C hem. Physics 3 , 4 9 (1935). STEARNS, A. E., AKD EYRING, H.: J. Chem. Physics 3,778 (1935). WARBURG, 0. : u b e r die katalytische Wirkung der lebenden Substanz. Berlin (1928). WIGNER,E., AND P O L ~ YM. I , : Z. physik. Chem. 139,439 (1929). WOHL,K . : Z. physik. Chem. 31B, I52 (1935). APPENDIX
(a) The probability that the molecule has, at the moment of light absorption, the necessary thermal energy (either as a whole, if i t is diatomic, or in the particular bond, if it is polyatomic) is
exp
- AU RT
(2)
i.e., for T = 300" Abs. a n d AU = 6.5 kg-cal., 2 X 10-j. ( b ) The number of collisions which one n~oleculemakes in a second with a kinetic energy larger than E is
E RT
2 = exp - -
(3)
where 2, the total number of collisions, is of the order 2
-
1010 p-'
p being the total gas pressure in atmospheres. Therefore, if E = 6.5 kg-cal. per mole, equation 3 has the value 2
x
105 p-1
(c) The fraction of molecules dissociating thermally per second is
AU A exp - RT
(4)
where AC is the minimum amount of energy needed, plus a possible heat of activation. A depends slightly on the temperature, is on the average 1013sec.-l but might vary between 1O1O to depending on the particular reaction.
105
PHOTOCHEMISTRY O F POLYATOMIC MOLECULES
Eyring and 0. K. Rice and Gershinowitz have given detailed theories of A (4, 6, 8, 13, 14, 15). If we want a molecule to dissociate in the average during the time aX sec., the following values of AU are permitted for 300" Abs.
'1
U p p e r l a m t of C an kg-cal
A
10'0
1013
10"
+ 1.4 lug
CY
9.6
The general formula is
T
AU -
=
~y
)(
10-8 = A-1 elCT
( d ) Consider the dissociation
(5)
1+2+3 Measure the concentrations in units of a standard state. t o denote quantitie3 in that standard state, so that
AF
=
P,
Use circumflexes
- ( P , + P,)
(6)
is the charge in free energy in the standard state, as taken from the tables. Then, if the velocities of the forward and backward reactions have the form
(7)
k1c1; kzczcs
v e have, in well-known expressions AP
I n -kl = I n k = RT kz
If the actual concentrations are
c1,
c2,
c 3 , we have
- RTlncz - RTlncB)
1 RT
= -AF
where LF is the change in free energy in the actual reaction with the given concent rationo. (e) Coiiqider a ('aye in which the hack reaction has no heat of kctivation, so that k , = A exp -
Write
AU RT
(4')
106
J A M E S FRBSCK -4ND KARL F. HERZFELD
Le., with L-, as the light energy A F = - C , - Ab7
+ A@
(10)
in the notation we have used. Therefore A@ (positive) is the amount of energy by which the free energy balance is more favorable to the reaction (in the standard state) than the total energy balance. Therefore
Now the life T z of a product molecule, Le., the time before it recombines, is given by
or
If, for example, A@ is 5 kg-cal. and A
Tz
=
we have
=
lo-''
(14)
~3-l
If product 3 is produced at the (photochemical) rate k : and destroyed by a third process at3the rate k3c3, we have
I n this formula a n assumption is implied, namely, that if the whole fomard reaction went thermally instead of part photochemically, part thermally, the velocity constant would be
k:'
= kl exp
-
CV
-=
RT
A exp
-
+ AC
c-lJ
RT
If a heakof activation E" is present in the back reaction, this increases the lifetime of product 2, making it A*-
1 Tz = - e
A
c
"
7
C,'
(13')
But at, t,he same time it increases the energy necessary in the forward t,hermal reaction by the same aniount.
PHOTOCHEMISTRY O F POLYATOMIC MOLECULES
107
(f) If the destruction of product 3, however, goes in a reaction of zero order (catalytic on the walls or on some catalyst present only in small amounts), we have to change the calculation leading to equation 15. Assume it were known that C' moles of 3 are destroyed within t see. Then the rate of destruction of unit concentration is I
3
C' t
= -
Then, the stationary concentration of product 3 is zero, if kS1 > k1'
while for k1'
> k,'
there is no stationary state, but the amount of product 3 present increases continuously (that is, if we neglect the back reaction) as production exceeds destruction.
