SpectraCat said:
Anything you can back up with some facts is fine by me. If think you can "spin" my analysis another way, then am at least a little interested to see your idea.
OK, but you aren't going to like this...
You have presented the following equation in support of your position:
AgBr ----> Ag + ½ Br2 ΔH = +99 kJ/mol
You point out that the reaction is endothermic and therefore cannot procede without the input of energy from a photon.
It’s a good argument, and it should be a good argument because it’s my argument. I was the one who suggested, in the face of some initial ridicule, that we could settle the question by looking at the thermodynamics of the process. It turns out that I was right to look at the thermodynamics but wrong to think that we would settle the question in this way.
The Gibbs Free Energy is, of course, the parameter which normally tells us if a reaction procedes to the right or to the left. You have used the enthalpy instead in presenting your numbers; but no matter. The correction for the entropy is in any event rather small; only a little more than 3 kJ/mol in this case. Nothing decisive.
The important factor you have neglected is concentration. The Gibbs Free Energy equation gives us the change in free energy only when the reactants and their products are present in stoichiometric ratios. In the photographic process, a typical crystal may have, after being exposed to light, only a literal handful of silver atoms out of trillions. It is apparent that the silver and silver halide species are very far from their stoichiometric proportions, and therefore a more careful analysis is required.
The themodynamically correct method must be to treat the crystal as a solid solution of silver and silver bromide. In its initial state, the crystal is 100% AgBr. Is the formation of a single silver atom thermodynamically spontaneous, or does it require a net input of free energy?
For convenience, we will double the reaction to clear fractions:
2AgBr -------------> 2Ag + Br2 ΔG = +192 kJ
We can calculate the familiar equilibrium constant from first-year chemistry with the formula
K = exp(-ΔG/RT)
With RT = 2.2 kJ (approx) we get
k=exp(-87) = 10^{-38}
With this information we can write the chemical equilibrium equation:
K = 10^{-38} = \frac{[Ag]^2[Br_2]} {[AgBr]^2}
This equation easily solves for the equilibrium concentration of Ag being approximately one part in ten trillion (10^-13). This is within an order of magnitude or so of typical concentrations in an actual exposed crystal. It is apparent that for an initially pure crystal of AgBr, the spontaneous creation of trace amounts of silver is thermodynamically favored. The stimulation of the incident light wave (I’m going to stop calling it a photon) merely speeds up the process, but is not strictly necessary energetically.
We can do the same calculation a different way. Use the Gibbs Free Energy to calculate the conversion of one part per trillion (10^-12) of silver. It comes to 192 nanojoules. This is the input of free energy required to drive the process. But this assumes that the two species are unmixed. In fact, we ought to treat the silver as being in solution. Then we can show that, within the accuracy of this calculation, the needed energy is available from the entropy of mixing.
The entropy of mixing is given by the formula
ΔS = nRm*ln(m)
where n is the number of moles and m is concentration of the mixed species. Multiplication by T gives you the free energy of mixing:
ΔG = T ΔS = nRTm*ln(m)
For two moles at a concentration of one part per trillion, the free energy comes to 121 nanojoules; and make no mistake, it is in the right direction to drive the reaction forward. It is true that with the numbers I have chosen we are just a little short of the 192 nanojoules we said we needed, but remember we haven’t yet accounted for the contribution from the mixing of the bromine. In any case, you only need to drop the concentration another factor of ten to tilt the reaction decisively to the right.
The case becomes even more convincing (convincing to me, you understand: I know you're still not convinced) when we recall that the crystal in its pure form is considered to be a poor photodetector. In practise the material must be doctored by the addition of impurities, dislocations, and what are called “electron traps” to become really effective. It’s not hard to imagine (OK, for
me to imagine) that the energy needed at the trace concentrations we are dealing with comes at least partly if not in large measure from the “doping” of the crystal. In other words, there is plenty of chemical energy available to drive the transition from silver bromide to silver without needing the energy of a photon. This is in line with my original description of the process as proceding from a metastable state to one of lower energy.
