[Rutherford]

What was the 7th problem  ? TOF&TON?

[Big-Daddy]

No it was simple equilibrium, we had a discussion about the second half of it with Borek and curiouscat if I remember right.

Done a bit of Problem 2 and I already disagree with your answers in Problem 2.1.

ΔH_r°=(Sum of ΔH_comb°[Reactant]·v[Reactant])-(Sum of ΔH_comb°[Product]·v[Product]) is a well known result of Hess' law, I skipped writing out the reactions fully but it's the same result

ΔH_r°=0+0-(0+(1/6)·ΔH_comb°[C₆H₁₂O₆])
ΔH_r°=-(1/6)·ΔH_comb°[C₆H₁₂O₆]=-(1/6)·(-2805)=+467.5 kJ·mol^-1

Seems like the answer to me. I think you have forgotten a minus somewhere. Or maybe it is me 

Obviously this affects the ΔG_r° answer (mine agrees with yours otherwise), coming out to +480.53 kJ·mol^-1 from my value.

How do we work out minimum number of photons needed? My intuition would say it's all about how energetic the photons are (i.e. their frequency) but clearly there is some minimum number required ...

[Rutherford]

Yeah, I made a mistake. I remember that problem. For the number of photons use: ΔG/N_a=N·h·ν (N_a-Avogadro's number, N-number of photons, h-Plank's constant, ν-frequency). I think that the answer needs to be an integer.

[Big-Daddy]

Yeah, I made a mistake. I remember that problem. For the number of photons use: ΔG/N_a=N·h·ν (N_a-Avogadro's number, N-number of photons, h-Plank's constant, ν-frequency). I think that the answer needs to be an integer.

1) What's the logic behind this formula? Many things I don't understand about it ... Comes from E=h·frequency clearly but how have you derived it?

2) However I try to use your formula to work out how many photons are necessary, the answer doesn't come out reasonable. Show the use please.

[Rutherford]

1) It is derived from E=h·ν. As I remember they ask how many photons are needed to produce 1 molecule of CH₂O. ΔG is the energy needed so the reaction can proceed, as it is expressed through kJ/mol, you need to divide it by avogadro's number to obtain the energy required to form one molecule (also convert kJ to J!).

Now you have the energy needed for one molecule. A photon will have the energy E=h·ν and you need N photons, thus ΔG/N_a=E·N=h·ν·N. Understood now?


2) I think the wavelength was 700nm, right? Using it I got that N=2.8=3 photons.

[Big-Daddy]

1) It is derived from E=h·ν. As I remember they ask how many photons are needed to produce 1 molecule of CH₂O. ΔG is the energy needed so the reaction can proceed, as it is expressed through kJ/mol, you need to divide it by avogadro's number to obtain the energy required to form one molecule (also convert kJ to J!).

Now you have the energy needed for one molecule. A photon will have the energy E=h·ν and you need N photons, thus ΔG/N_a=E·N=h·ν·N. Understood now?


2) I think the wavelength was 700nm, right? Using it I got that N=2.8=3 photons.

Got it, thanks. λ=680 nm so N[photons]=2.73 molecule^-1. But that's if all energy is absorbed - according to the problem, only 10% is absorbed so we'll have to multiply our answer by 1/0.1=10, 27.3 photons. Not the integer we were looking for but actually your suggested working seems decently robust. What I didn't recognize is that ΔG_r° is the energy required for the reaction to proceed. (Also I did not see that λ is provided)

Edit: Actually maybe all energy is absorbed. (in which case 2.73 is the answer again) The statement "green plants utilize 10% of the available solar energy" is unclear ... does it mean that, of all the photons that fall, green plants utilize 10% of them, or does it mean that of the photons which the green plants take in, 10% of their energy is absorbed? I'm guessing the former but seemed ambiguous to me.

[Big-Daddy]

I get the impression you cannot see the problem so here is the text:

Photosynthesis is believed to be an efficient way of light energy conversion. Let’s check this
statement from various points of view. Consider the overall chemical equation of photosynthesis
performed by green plants in the form:
H2O + CO2  CH2O + O2
where CH2O denotes the formed carbohydrates. Though glucose is not the main organic product
of photosynthesis, it is quite common to consider CH2O as 1/6(glucose). Using the information
presented below, answer the following questions.
1. Calculate the standard enthalpy and standard Gibbs energy of the above reaction at 298
K. Assuming that the reaction is driven by light energy only, determine the minimum number of
photons necessary to produce one molecule of oxygen.
2. Standard Gibbs energy corresponds to standard partial pressures of all gases (1 bar). In
atmosphere, the average partial pressure of oxygen is 0.21 bar and that of carbon dioxide – 310–4
bar. Calculate the Gibbs energy of the above reaction under these conditions (temperature 298
K).
3. Actually, liberation of one oxygen molecule by green plants requires not less than 10
photons. What percent of the absorbed solar energy is stored in the form of Gibbs energy? This
value can be considered as the efficiency of the solar energy conversion.
4. How many photons will be absorbed and how much biomass (in kg) and oxygen (in m3 at
25 oC and 1 atm) will be formed:
a) in Moscow during 10 days of IChO;
b) in the MSU campus during the theoretical examination (5 hours)?
5. What percent of the solar energy absorbed by the total area will be converted to chemical
energy:
a) in Moscow;
b) in MSU?
This is another measure of photosynthesis efficiency.
Necessary information:
Average (over 24 h) solar energy absorbed by Moscow region in summer time – 150 Wm–2;
Moscow area – 1070 km2, percentage of green plants area – 18%;
MSU campus area – 1.7 km2, percentage of green plants area – 54%;
green plants utilize ~10% of the available solar energy (average wavelength is 680 nm)

Then some enthalpy of combustion/entropy stuff which we just used already.

[Big-Daddy]

2.2. Answer is 496.76 kJ·mol^-1, taken from ΔG_r=ΔG_r°+R·T·log_e(Q)=ΔG_r°+R·T·log_e(P[O₂]/P[CO₂]), ΔG_r° as before (480.53 kJ·mol^-1), R modified for kJ instead of J. The only constituents of Q are the gases, since they are the only ones which contribute to the partial pressures (question gives that Gibbs' energy depends here on partial pressures of gases, not on how much of the other liquids or solids are available).

[Rutherford]

Yes, I cannot see the problems, I did it only how I memorized.

4. and 5. are very ambiguous for me. Couldn't solve them when I tried. I have the same dilemma you have. We could use some help here .

For the integer, it seemed more logic to me to have e.g. 3 photons than 3.127 photons (what is that 0.127?).

[Big-Daddy]

Yes, I cannot see the problems, I did it only how I memorized.

4. and 5. are very ambiguous for me. Couldn't solve them when I tried. I have the same dilemma you have. We could use some help here .

For the integer, it seemed more logic to me to have e.g. 3 photons than 3.127 photons (what is that 0.127?).

An average, maybe?

I'll take a look at 2.3 onwards later, need to study a bit for my boards in 10 days now ...

[Rutherford]

That seems illogical. Like if you said, an atom has 3.5 electrons. How can a number of particles not be a whole number?

[Big-Daddy]

That seems illogical. Like if you said, an atom has 3.5 electrons. How can a number of particles not be a whole number?

This is a pretty big discussion but light isn't a particle. In this case it maybe better to think of it as a stream of energy. So maybe it can be averaged, ΔG_r° after all is a statistical average over 6.0221·10²³ reactions and the amount of energy needed for each.

I'm not sure, perhaps someone else can clarify why a) we keep getting non-integer answers (I am not satisfied with saying 2.73=3, IChO does not leave things that far open) b) if it really has to be an integer in all these cases.

An example for you: what's the oxidation state of sulphur in S₄O₆^2-? Must be ((6*2)-2)/4=+2.5, right? Not that one of the sulphur atoms really gets 0.5 electrons ... it's an average! (Hope this example isn't too bad - I realize it doesn't relate directly to our case but the way I see it, seems possible that the number of photons wouldn't be discrete)

Or possibly it IS discrete and we just have to round up the number of photons, even if the minimum number of photons is calculated as 2.2 we'll still need 3 to provide enough energy, etc.

[Corribus]

Edit: The statement "green plants utilize 10% of the available solar energy" is unclear ... does it mean that, of all the photons that fall, green plants utilize 10% of them, or does it mean that of the photons which the green plants take in, 10% of their energy is absorbed? I'm guessing the former but seemed ambiguous to me.

Both, actually.  The sun puts out photons of all different energies and photons of different wavelengths are absorbed with different efficiency by the various pigments involved (otherwise, leaves wouldn't be green).  And not all of that energy is utilized efficiently.  There are always losses because transfer efficiency is never 100%.  For any photophysical process or photochemical process, we call this the quantum yield.  I would say most of the losses come from the fact that many photons are simply not absorbed.  But even if you neglect the energy that isn't absorbed, plants can't use all the absorbed energy with 100% efficiency.  I'm not sure what the actual percentage is off the top of my head, but you can probably look it up.

And photons are both particles AND waves.

[Big-Daddy]

Edit: The statement "green plants utilize 10% of the available solar energy" is unclear ... does it mean that, of all the photons that fall, green plants utilize 10% of them, or does it mean that of the photons which the green plants take in, 10% of their energy is absorbed? I'm guessing the former but seemed ambiguous to me.

Both, actually.  The sun puts out photons of all different energies and photons of different wavelengths are absorbed with different efficiency by the various pigments involved (otherwise, leaves wouldn't be green).  And not all of that energy is utilized efficiently.  There are always losses because transfer efficiency is never 100%.  For any photophysical process or photochemical process, we call this the quantum yield.  I would say most of the losses come from the fact that many photons are simply not absorbed.  But even if you neglect the energy that isn't absorbed, plants can't use all the absorbed energy with 100% efficiency.  I'm not sure what the actual percentage is off the top of my head, but you can probably look it up.

And photons are both particles AND waves.

Thank God one of you guys still read this thread!

Can you help with this part?

2.1. Determine the minimum number of photons necessary to produce one molecule of oxygen.

The reaction is H₂O + CO₂  CH₂O + O₂. The ΔG_r^°, equivalent to the amount of energy required to make the reaction proceed (as Raderford said), is +480.53 kJ·mol^-1. λ for green light (the only type plants can absorb, assumed) is 680 nm.

Our method was basically that ΔG_r^°=E[photon]·N[photons]=(hc/λ)·N[photons], solve for N[photons] and that's how many we need to make 1 mole of the reaction proceed. To produce 1 molecule of O2 we divide this by N_A. How does this sound? And is 2.73 an acceptable and correct answer? (Given the potential hang-up that only integer numbers of photons can be correct.)

[Corribus]

I haven't read the whole thing, just skimmed the last few posts.

So with that in mind, your logic seems reasonable, although since you can't have a fraction of a photon, I'd say the right answer is 3.

Also this answer is based only on the energy difference between reactants and products and neglects the fact that additional energy is required to surmound the activation barrier.  Also it assumes 100% absorption efficiency.  So in reality it would take more than 3 photons.