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Pub #brownchem: @alpha137
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LizK: Hello, I have a question about the homework
alpha137: hello
LizK: I am very confused
alpha137: OK shoot
LizK: Ok, THis is what happened, first I tried to balance the methane equation half reactions and this is what I got
LizK: 2H_20 + CH_4 --> CO_2 +8e- + 8H+ and 8H+ +2O_2+8e- --> 4H_2O
LizK: and those balanced out to form CH_4 + 2O_2 --> CO_2 + 2H_2)
LizK: H_2O
LizK: and then I tried to find how that related to volts and so I found
on my little chart the voltage for the O_2 half reaction (I don't know why it was
different from the equation you gave me yesterday but it was) and that was fine
I probably made a mistake. No, actually I did not. You choose an acid solution
and I chose a basic one. That is the difference between yours and mine. It just a
matter of choice.
LizK: Then I had to find the total voltage and my problems began
alpha137: OK. From the O_2 half reaction that you have to use there are 8 electrons. Right?
LizK: I had found that the change in free energy was -818 kJ/mol last week so I used that for the G value, and I assumed that the number of moles of electrons was probably 8 (because there were 8 on each side of the equation, but it could be 16) and found that the vlots was-.001065 and that must be wrong, it is way too small.
LizK: And that is even with out subtracting the O_2 part
alpha137: No, it is eight electrons.
LizK: Thats what I used
LizK: I mean, couldn't it be 8 moles of electrons just as eaisily as 8 electrons?
LizK: Or am I supposed to divide 8 by avogadro's number to find the number of moles of elecrons?
alpha137: Well, if you use one molecule then you have eight electrons, but if you use one mole of methane then you have eight moles of e's
alpha137: Lets work in moles because that is a practical unit.
alpha137: Everything is cool here.
alpha137: You have to keep track of powers of ten-that is kJ and J, etc.
LizK: OK, but where did I mess up
alpha137: The voltage should be 1.0596 V.
alpha137: So, you are off by 1000.
alpha137: Where do you think you messed up?
LizK: I don't know thats what I was wondering, I was thinking about where I put in the number to the G = (number of moles etc)(96000)*V
LizK: equation
alpha137: If you use Joules for G on the left you will come out right.
LizK: like I was supposed to divide 96000 by 1000 because I was working in kJ not J
LizK: Ojh, that makes even more sense,
alpha137: Right, you should use 96.5
LizK: Ok, so I was on the right track at least
alpha137: The voltage of our methane fuel cell will be 1.06V per cell.
alpha137: We can stack them up to get a higher voltage.
LizK: How does that compare with the voltage of gas or other alternate fuel
alpha137: Just like a car battery.
LizK: I see
alpha137: I have question. If we have one mole of methane or ten moles of methane will the voltage be the same, larger, or what?
LizK: Well, wouldn't it be the same for each mole, but more because you have more moles?
alpha137: No!
LizK: OK, so what would it be?
LizK: THe same?
alpha137: What is the word for V, what is it called besided voltage?
alpha137: besides
LizK: Volts, ?
LizK: Volume?
LizK: What are you looking for?
alpha137: another term-electrical term
alpha137: Potential.
LizK: Like Amps?
alpha137: No, like potential.
LizK: Well, I wrote amps before I saw what you had written
alpha137: Similar to gravational potential, or potential energy.
LizK: OK, so potential,
alpha137: Potential does not depend on the amount.
LizK: So its the same
alpha137: Right.
alpha137: The Free Energy depends on the amount of stuff.
LizK: BUt you said you could stack fuel cells and get even more power
alpha137: No, I can stack them and get more volts.
alpha137: Power is energy per unit time.
And that depends on the amount of chemicals.
LizK: HOw can you get more volts if volts are potential and the same for one or ten?
alpha137: Because we put these in series like any battery and the voltages add up.
LizK: Ok, i get it
alpha137: The power will depend on the numbers of moles we react in a given time.
alpha137: The more moles the more free energy.
LizK: THen why are the volts the same?
alpha137: Remember the equation Change in G (kJ/mole) = (number of moles)(96.5 kCoul/mole)*Volts.
LizK: Ok, I remember
alpha137: The "moles" cancel out on both sides of this relationship.
LizK: Yes,
alpha137: Anyway, as in a gravational field, the potential is measured relative to something. Here we are measuring, defining, the potential relative to hydrogen ion.
LizK: Oh, I see
alpha137: Potential, has potential so to speak, while energy, and free energy depends on the amount of substance.
LizK: I see, so numebr of moles effects (or affects?) the free energy but not the potential
alpha137: Right!
alpha137: The power to drive our car will depend on the rate we can draw free energy from the reaction.
alpha137: Moles per second.
LizK: So potential isn't all that important
LizK: ?
alpha137: No, potential is not all that important.
LizK: Ok,
LizK: So I have a question,
alpha137: We just need to design the electrical motor to work on a practical potential.
alpha137: What is it?
LizK: Nevermind, I figured it out
alpha137: OK. We will let the electrical engineer design the electric motor. Back to chem.
LizK: OK
LizK: Fine whithme
alpha137: By the way, amps would be connected to the change in numbers of moles with time.
alpha137: Amps would be related to power.
LizK: I see, I knew it wasn't the right answer, but that was the only electrical term i could think of
alpha137: As I remember, volts times amps = power.
LizK: Probably
alpha137: Power is Joules/sec = watts
LizK: oH, SO VOLTS TIMES amps = watts
LizK: I remember that
alpha137: Don't you think that our methane reaction looks like a low pollution deal?
LizK: Yes, but, isn't CO_2 bad too?
alpha137: Well, we are not going to help global warming with this excepting if it is more efficient than burning methane in an internal combustion engine.
alpha137: What is the point of making a fuel cell driven car anyway?
LizK: Ok, so its a trade off, pollution now, or global warmin glater
LizK: Less pollution
alpha137: Guess so. We need to consider the efficiency question. Another basic chemistry question and an important environmental question as well.
LizK: THats true,
alpha137: We need to figure out the maximum work we can extract from these processes.
LizK: How do we do that?
alpha137: The first process is simply burning methane (or hydrocarbon) and transfering heat.
alpha137: The second process is using the free energy.
We are using the free energy directly by virtue of taping
into the electrical work. The real problem is in designing the cells to make this
happen.
LizK: Isn't that what I figured out last week? and tried to last night?
LizK: Like the free energy and then the volts?
alpha137: Yes, but we have not figured the efficiency yet and compared the two processes.
alpha137: I have question: If there are two bodies, one hotter than the other, and they are placed in thermal contact what happens?
LizK: The heat will transfer from the hotter one to the colder one
alpha137: Does it ever go the other way?
LizK: NOt unless its forced, like in a refrigerator
alpha137: How does the refigerator do that?
LizK: IT was in thge book, let me find it, there was those coils
LizK: were
alpha137: There was something more important than coils.
LizK: THe compressor
alpha137: Right.
alpha137: The electric company supplied electricity to drive that compressor.
LizK: Yes
alpha137: And that was what?
LizK: WOuld the car have the same type of thing
LizK: What was what? THe electricity? The compressor?
alpha137: I think that all these transformations of energy and heat are the same.
LizK: Makes sense
alpha137: Well, there was work done on the compressor supplied by the electric company.
LizK: In the form of electricity
alpha137: You know it from your answer before. In order to get heat to flow from a colder body to a hotter body work must be done.
LizK: Yes, that follows logically
alpha137: Now, we want to apply the First Law of Thermo to the compressor.
LizK: Ok, so thats enthalpy
alpha137: Change in E (internal energy of comprssor) = heat added less the work done on the surroundings
LizK: Ok,
alpha137: Not so fast, enthalpy in a minute.
LizK: What do we know?
alpha137: If the compressor is perfect then there should be not internal energy change.
LizK: But there will be
alpha137: What is comming into the compressor and what is going out from the compressor?
LizK: Like low pressure gas vs. high pressure gas? or electricty works on the low press. gas to make it have more pressure
alpha137: Remember, we are transporting heat from inside the fridg to outside.
alpha137: Forget the details of how the compressor works. That is another environmental problem.
LizK: So there is a heat transfer from the inside to the outside, and that takes work because its backwards (from cold to hot)
alpha137: Right, and the so called First Law keeps track of all this.
LizK: I see
alpha137: If we have a perfect comprssor then 0 = heat change plus work done ON the comprssor.
alpha137: That is, internal energy = E = 0
LizK: What work is done on the compressor?
alpha137: You plug the fridg into the electrical outlet, right?
LizK: Yes
alpha137: That runs the motor that runs the comprssor, right?
LizK: Yes
LizK: And that is the work done on the compressor
alpha137: That is the work done on the compressor, right.
LizK: Ok
alpha137: You pay for it in the electrical bill at the end of the month.
LizK: Oh, so that
LizK: works
alpha137: We need to add up the heat that is transfered. Let the heat in the fridg be Q1 and the heat transfered to the room be Q2. Give me the balance.
LizK: In a perfect frige it would be Q1-Q2 = 0, but that doesn
LizK: t really happem
alpha137: You forgot the work needed to perform this heat transfer.
alpha137: Anyway, this is going to be the very best one could do!
LizK: But in reality, the Q1 would be bigger right?
alpha137: You know, a perfect, if nonexistant, heat engine.
LizK: Yes, i got that much
alpha137: We are not saying anything about the size of Q1 and Q2.
LizK: But when you relate them- balance them- there must be some idea of quantity
alpha137: I need an equation that takes everything into account from an energy/heat/work point of view.
LizK: Sounds like thermo stuff to me- but i don't think thats right
alpha137: When we do a calculation we can worry about numbers. We can also worry about the relative sizes of Q1 and Q2 because that will determine wether we have to do work or get work out.
LizK: Ok
alpha137: What are the three quantities we have here for this fridg?
alpha137: Q1, Q2 and ?
LizK: the work done on the firge
alpha137: Remember, E = 0
alpha137: Right, work done. So put these together to balance everything.
LizK: So Q1 + Q2 + Work done = 0?
alpha137: Right!
LizK: Oh, well that was easy
alpha137: This is for a perfect frig.
LizK: But lifes not perfect,. what about a real one?
alpha137: This is the very best that would be possible in a perfect world.
LizK: Q1 +Q2 + work done = internal energy?
alpha137: That is an engineering question.
LizK: OK, NEVEr mine
alpha137: Right, to your last equation.
alpha137: We may write Q2 - Q1 = w
LizK: ok
LizK: thats more consciece
LizK: shorter
alpha137: In other words, it takes w work to transfer Q2 - Q1 of heat.
LizK: I Yes, but what happened to the internal energy?
alpha137: Earlier, you agreed that we could not have heat flow from a colder body to a hotter without doing some work. This is it.
LizK: Oh, I se
LizK: e
alpha137: The internal energy is zero because this compressor works without friction and works reversibly. This is an ideal.
LizK: BUt in a real one, there would be IE somewhere
alpha137: We are just keeping track of "energy" here for this perfect process.
LizK: Ok
alpha137: Yes, in a real heat engine (fridge, air conditioner, etc) there are losses.
LizK: Ok
alpha137: We do not know how to compute these at the moment. This is the best we
alpha137: can do.
alpha137: The first law of thermo is a conservation law.
LizK: Ok
LizK: yes about the 1st law of theremo
LizK: thermo
alpha137: A defining of the relationship between heat, work and internal energy.
alpha137: But we also know that nature likes to become more disordered.
LizK: yes, and its all balanced with the free energy
alpha137: We are not at free energy yet. What is the concept, term, describing this order/disorder thing?
LizK: Entropy
alpha137: Right. We need to take this into account if we are to figure out how much work is required to transfer heat using our compressor.
alpha137: Rather, the efficieny.
LizK: Oh, so thats how you figure out efficiencyu
alpha137: Lets just state the Second Law of Thermo.
LizK: THings in nature tend to go to states of more disorder
alpha137: Entropy = S = Q(reversible)/T defines entropy. Where Q(rev) is the heat added in a reversible way.
alpha137: Right about your stmmt.
LizK: ok
alpha137: Reversible means that the process is carried out at equilibrium every step of the way-from start to finish.
LizK: Ok
alpha137: The Second Law says that the sum of all the entropy changes is either zero or positive-it is never negative overall.
LizK: Yes
alpha137: It could be that in some part of a process there could be a negative entropy change, but when you add them all up there is either no change or a positive change.
LizK: Ok
alpha137: The Second Law says that for an equilibrium process the total entropy change is zero.
LizK: YEs
alpha137: That is S1 + S2 = 0.
alpha137: But, S = Q/T for reversible processes.
LizK: ok
alpha137: Here is a homework problem. You have just obtained Q2 - Q1 = w. Now, find the entropy change for our fridge assuming Q1 is at temp T1 and Q2 at temp T2.
LizK: Do I know T1 and T2?
alpha137: Assuming everything is perfect-reversible that is.
alpha137: Plug into S1 + S2 = 0 and combine with the first law heat/work balance.
LizK: ok.
alpha137: Give me an equation that tells me w/Q1 related to only temperature differences and is a dimensionless ratio. This will be an efficiency like thing.
alpha137: In other words, w work will be required to deliver Q1 heat.
alpha137: This will be the best we can do with a perfect heat engine.
LizK: Ok
alpha137: This will be the best we can do by burning gas in a car too.
LizK: SO then onece we have both we can compare efficicency
alpha137: The idea is that the temperature in the auto cylinder is at one temp and the exhaust gas out the tail pipe is at another temp.
alpha137: Right we can get the efficiency.
LizK: And the difference is important in finding work
alpha137: Then we need to get the efficiency from the free energy-extract work from the free energy thing. Compare the fuel cell with the internal combustion engine.
LizK: oh, i understand'
alpha137: Right, we can find the amount of useful work that can be extracted from these two ways of getting eneryg out of methane.
alpha137: Which one will win?
alpha137: Talk to you tomarrow? Same time?
LizK: That depends on free energy right?
LizK: Yes same time tommorow
alpha137: Right, the second depends on G
alpha137: bye
LizK: bye
LizK: thank you