VendingMenace

cience of mind is what brings that which cannot be named, because structure hinders the studier from comprehending the mind in it's true state, as opposed to a bag of satsumas.5 kilos and assorted nuts and bolts

well put YT! there is definately a difference between creating a force that will act opisite the force of gravity (like picking something up with our hands) and creating something that actually nullifies the cause of gravity (antigravity).

Perhaps i am just a bit undereducated about this whole christianity thing, but i think when the area contained in the nation of Isreal hears about this, they are going to be pissed! I mean gee wiz man, you have a guy that starts a religion and he is; a) Decended from your coutry (Nazereth) a1) Decended from a king of your coutry (David) b) Born in your country (betheham i belive) c) Dedicated to his paternal faith in your coutry (Jerusalem) d) Baptized in your coutry (Jordan river, was it?) e) Preaching in your coutry (everywhere, but alot around galile) f) Died in you coutry (Jerusalem again) g) Has his followers start their preaching in your country (Jerusalem again, musta been a happening place) h) Have the center of his faith located in your country for the first 70-100 some odd years (Jerusalem again, i belive) i) Oh yeah, and there is absolutely no evidence whatsoever, that he was ever in Iraq and then people just go and ascribe the birthplace of one of the major religions to another country?! And it is IRAQ??!! I mean, shoot if i was Isreal i would like WTF!? OK that is it, now we are going to freakin' bomb iraq, stealing our religious birthplace bragging rights. Shoot, man! Anyways, like i said, i am open to the possiblility that i am wrong, but i kinda think Christianity was born in Isreal. Is it just me?

I think you are missing my point. My point is this; you can have a reaction in which there no heat of reaction. That is, the net delta H for the total reaction is zero. That is what I am saying. You, however, seem to be addressing the process of the reaction. You are saying (mostly correctly) that stripping away an electron from a neutral atom or bonding orbital of a molecule requires energy, while adding an electron into a bonding orbital or a positive ion will give off energy. Since we move electrons about during the course of a reaction it is fair to assume that energy will be given off or required or both, during the course of the reaction. Like I said this is mostly correct. However, this does not effect my assertion at all, nor does it affect the example I proposed. My assertion, that delta H for a reaction can be zero deals with the state of the reaction at the beginning and the end. It is not concerned with how you got there. I am not claiming that energy is not moved about during the process, I am merely claiming that when all is said and done the energy that you started with in the reactants is equal to the energy that you end with in the products (chemically speaking -- that is ignoring vibrational, rotational, and translational energy). Perhaps you are unfamiliar with the idea of a state variable. I will try to give a quick run down of what they are (I apologize if you are familiar with them). Basically, a state variable is a value whose value is path-independent. Elevation is an example of a state variable. That is, if I start at point A and move to point B, then there is a definite difference in elevation between my starting and ending positions. It does not matter how I arrived at point B, the difference in elevation between point A and point B is the same. It also happens that enthalpy is a state variable. That means that it does not matter what path we take between reactants and products, the change in enthalpy for a certain chemical reaction will be the same. Since change in enthalpy is path-independent, then we can ignore all processes that occur during the reaction. The way in which the reaction happens cannot change the value of enthalpy change. Thus, in my reaction that I propose has a delta H of zero, it does not matter what we do during the reaction, if we have a zero change in enthalpy, then we will always have a change in enthalpy of zero. The path of the reaction cannot change this. So I am not ignoring the energy change in the middle of the reaction. Rather, I am saying that the sum of all the energy changes that occur during the course of the reaction must add up to give us the total change in energy for the reaction as a whole. Thus, when I say that delta H is zero, we can ignore the course of the reaction (and any energy transfers that occur in the middle) as far as thermodynamics are concerned. This is because once we know what delta H is for the reaction, we already know what the sum of all energy changes during the reaction must add up to. I hope that this helps somewhat in explaining why it is valid to ignore the middle of a reaction when we are talking about things such as change in enthalpy, which are state variables. Thus, I am not claiming that ‘just because you don't look at the energy transfer it isn’t there.’ Rather, I am saying that there is no need to concern yourself with these energy transfers and we are justified in ignoring them. Now of course this only goes for thermodynamic considerations. If you want to talk about the kenetics of the reaction then that is a whole ‘nother ballgame. Then it is the energy “in the middle” of the reaction that is intimately important to the problem rather than the total change in energy. Hope that helps

I was thinking that perhaps some people would not be satisfied by this mathematical and physical explination given above? So I decided to work through an example real quick, for those continueing doubters (I am not trying to be mean or antagonistic here. I think that it is good to insist on explinations until you feel that you have a good understanding of something). Let’s see then, what question should we use. Oh yes, lets answer the following; Lets assume that you go ahead and do it incorrectly, and convert from C to F, do your calc and then concert back to C again. Well we have already seen that the answer you will get then is, -8.9 C. Now let’s see if this answer holds up to the math, shall we? Again we have the equation; v=[(3)(kT/m)]1/2 as proved above, we must use K in this equation, so lets convert -8.9C = 264.1 K (I just used 273 as the conversion, for simplicity) pluging in the numbers (boltzmann’s constant, k=1.38 x 10^-23 J/K). The identity of the material doesn’t matter, right, so lets assume that it has a mass of 1x10^23 kg. v=[3(1.38x10-23 J/K)(264.1 K)/(1x10-23 kg)]1/2 v=33.1 m/s Ok, so we have the velocity of the sample at -8.9 C. WE using the same exact process, we can calculate the sample’s temperature at 0 C, that is, its starting temperature. 0 C =273K. SO we find that the velocity of the molecules at 0 C would be… v=33.6 m/s now the original question was aking for half the temperature right? That means that nothing but the temperature changes. So we look at the equation; v=[(3)(kT/m)]1/2 and ask ourselves what will happen to the speed of the molecules when we drop the temperature by half? I think that it is fairly easily seen that the speed will decrease by a factor of (1/2)1/2. So the average velocity after cooling should be (1/2)1/2 the original velocity at 0 C. So, we see that we should have; (1/2)(1/2)(velocity at 0 C) = (velocity at -8.9 ) plugging in our values we obtained above, we have; (1/2)(1/2)(33.6 m/s) = (33.1 m/s) 23.75 m/s = 33.1 m/s This is obviously not true. And so we see that we do not arrive at the correct answer by converting to F then carring out calc and converting back to C. Whew! That was a lot of typing, and I think I am done for now. I hope that answers your questions. The take home message is that you must convert to K before carrying out the calculations. You will run into similare problems everytme you do not do this. This is partially why the Kelvin scale was invented. Basically, the Kelvin scale takes into account the baseline energy for you, so that you do not have to think about it. Fail to do this, and you run into serious problems. Well that is it for now!

Physically this is incorrect. I will try my best to give a detailed explination of why, it will out of nessesity use a fair amount of math. However, it will only be basic algebra, so while it might be somewhat tedious, please follow along if you really want to understand. First off there are some things that we must realize; First, Temperature is a measure of average kenetic energy of the molecules in teh substance you are measuring the temperature of. The eqation that relate kenetic energy to temperature is as follows; KE=(3/2)kT Where: KE=kenetic energy k=boltzmann’s constant T= temperature (in Kelvin) You may wonder why the temperature has to be in kelvins for this equation. Wait just a bit and you will understand. (the answer to this is why you cannot just convert to F from C, perform the cal then convert back). Second, The kentic energy of an object is related to the speed at wich it is traveling. The eqation that relates energy to speed is as follows... KE=(1/2)(mv2) Where: KE=kenetic energy m = mass of particle v = velocity (but since the velocity is squared, then the only important part is the speed and not the direction.) looking at this equation, we find that if we have two particles of a given mass and one is traveling twice as fast as the other, then the faster travelling particel will have 4 times as much energy. Correct? Ok, now that we have got the preliminaries out of the way, we can get down to business. First off, we can combine these two equations for kenetic energy together. If we realize that we must arrive at the same value for the energy whether we calculate the kenetic energy of sample by mearuring its temperature or figuring out the averate speed of the molecules in it, then we can set the two equations equal to eachother. That is; (3/2)kT= KE =(1/2)(mv2) (3/2)kT=(1/2)(mv2) using this new equation, we can solve for how the speed of molecules vary if we were to change the temperature. We find that (3/2)kT= (1/2)(mv2) (2)(3/2)kT=(mv2) (3)(kT/m)=(v2) [(3)(kT/m)]1/2=v so we find that the speed of particles in a substance is related to the temperature by the following way… v=[(3)(kT/m)]1/2 cool. HERE IS THE ANSWER Now we see why we must use the Kelvin scale to calculate energies. If we used either C or F we could have a negative temperature on the right side of this equation. The boltzman constant is a positive number as well as the mass. Thus, if the temperature were allowed to be negative, we would be trying to take the square root of a negative number! This would lead us to a an answer than had an imaginary number. But speed is not an imaginary thing, so we cannot ever have an imaginary value. This is, in essence, why you must use Kelvin when calculating “how cold is half” and indeed, if you do not wish to make a mistake, you should always use Kelvin when dealing with questions that ask you to manipulate or use temperature. Well, I really hope that clears up the questions you all have. It certainly was a lively debate. And feel free to ask more questions, questions are good!

yeah, that was exactly what i was reffering to earlier, but i couldn't find it! Thanks muffin! Sweet!

interesting question... i think first you have to decide whether anything will happen or not. So as far as that goes, here is my thoughts... in a proccess in wich the right side of the equation has more mole of gas than the left side, increasing the pressure in reaction conditions will favor the left side of the equation. IN this case there is certainly more moles of gas in the products, so as the pressure increases the products will become more and more favored. In the begginning of the experiment we would be at standard temp and pressure (just for the sake of argument). As the reaction starts, it would proceed until a certain pressure was reached, and then it would stop. Durring this time the temperature would also be rising. So, at the point when the reaction stops i feel that all prducts would remain gases. As the system starts to cool, you would observe condesation of the gases perhaps, due to the intesnely high pressures. As the gas starts to condense, the pressure would drop and the reaction could again proceed in the forward direction, provided that there was enough heat left in the system for the reaction to occur. This would continue to occur until there was not enough energy left in the system for the reaction to occur or until all the reactants were used up. (assuming that the reagents could "find" eachother in solution". WEll, that is my thoughts for now. They are not ment to be 100 accurate or a definite timeline. Most likely, all of the prcesses would be occuring at different times, not just in the sequence that i have provided. Also, i am implicitely assumed that the liquid products take up less volume than the solid reagents. Cool. Feel free to add to or corrrect errors or whatever

LOL! i get it now. Ok, that was funny. Thanks for putting up with me.

I am not claiming that it does, i am merely trying to show that to get an answer other than -136C you must ignore the physical reality of the system. Once you do that, the only thing left to let you solve the problem is mathmatics. Thus, and answer gartered this way will be a purely mathmatical answer (ie. no physical significance). No, it is a physical priciple that happens to use mathmatics. Energy is not a mathmatical priciple any more than mass is. The things we measure are physical properties wich we express using mathmatics. I am not claiming that a mathematitian would ignore the physicallity of the problem. Again, i am merely trying to show how to logically arive at an answer other than -135 C and the only way is to ignore the physics of the problem. Of course, as sayonara pointed out, this doesn't really answer the question, but more on this later....ok, now. perhaps i was unclear, i am sorry. This is the point i was trying to make. They only way you could every come up with an answer other than -136C is to ignore the physical significance of the question. Then you may get another answer. Of course, this answer has no physical meaning, that is why i said the answer would be 0. Not 0 degrees C. If you try to answer the question with just pure math, then you loose the physcis of it and left with an answer that does not really answer the question :/ Thanks for making that really clear though SWEET!

not at all. the reaction was ment to be an example of a reaction in which the reactants and the products sides were identical. This applies if there is only one product and only one reactant as well. For example... HYDRONIUM + HYDRONIUM --> HYDRONIUM + HYDRONIUM in this case, the pairs of hydronuims can swap protons. Now you only have one species on the right and the same species on the left -- yet a reaction could still occur. I think it is fairly clear that delta G for this reaction should be zero. Perhaps i am wrong? I apologize for any confusion my attempt at gereralization may have caused. But i still stand by my original acertion and that is reactions can in which the reagents and products are indistiuguishable. That is all

i would agree with jordan here. The reason why we cannot have perpetual motion is that any machine must loose energy to the environment outside of it. Thus, every machine eventually grinds to a halt due to its loosing energy. But to where would the universe loose energy? It is is loosing energy, it must be loosing it to something, and that something would by definition would be part of the universe. It seems (at least to me) that the universe should always have the same amount of energy (energy cannot be created nor destroyed). Thus, if is a closed system, it would seem that by definition it would be a perpetual motion machine. Thatr is just how it seems to me -- perhaps my reasoning is wrong :/

LMFAO!!!!!!!!!!!!!!! holy nut! i swear i was crying when i read this. Sooo funny. Yeah, it is kinda expensive to maintain a large army -- or something. Anyways, it is quite ironic (though not classic irony i suppose, but funny nonetheless).

I am going to take this back now. If we assume that the electron exists in an infintely high energy state within the atom/molecule, then it will take zero energy to strip it away from the molecule. Likewise, if an electron falls into an infinitely high state within the atom/molecule, then it will release no energy in this process. Of course, the probablility for this happening is somewhat low (understatement of the year) but at least it is theoretically possible. cool, i think that is all for now

of course not. But that speaks only to the process of the reaction. While i was talking about the beginning vs ending products. the change in heat or free energy if you will for the reaction. if that is all we are concerned about, then we can ignore the energy that was required or put out durring the reaction, as both delta H and delta G are state variables. that is, the value of delta H and delta G are not dependent on the path you take to get there

Nope, i only counted 3. Oh well, i guess i am not a genius after all

Not true. Consider the following... H3O+ + H2O --> H2O + H3O+ This is ment to be proton transer in a acidic solution. here the reactants and the products are the same, however, we know that proton transfer occurs (or rather, we have strong evidence for this). Thus, a reaction is occuring. The problem is observing the reaction, that is there is no way to garuntee that a reaction has occured, even though it is quite likely that it has. So, though we suspect that there are reaction where the reactants and the products are identical, we have no way of testing this :/ would you care to explain this position more. I do not understand how this statement proves that energies of different molecules must be different. Cool

WOW, what a debate. I will try to explain this problem some. Ok, i will play your game so you claim that there are two answers, the physical one and the mathmatical one. For the physics one, i think we can all agree that you must convert to kelvin first. (at least i hope this should be clear). This is becuase temperature is a measure of average kenetic energy of the molecules in the system. Half the temp means half the energy. Period. The easiest way to approach this is to convert temp to kelvin and divide by two. This is how the problem must be solved physically speaking. Ok then, lets look at a mathmatical answer. In a purely mathmatical system, physical ideas hold no significance, we are working purely in the realm of numbers. As such, celcius, farenhieht and kelvin hold no meaning -- we must ignore them. Becuase of this, we cannot convert between any two temp scales (if we tried, we would be admiting that the numbers have physical significance, wich they don't in a purely mathmatical appraoch). Thus, we are ask what is half of 0. The only answer can be 0/2 or 0. Thus, -8.9C is incorrect, as that is an answer that has physical significance. The only purely mathmatical answer is 0. The only phsically correct answer is -136.575C. I hope that makes sense.

yeah, i only saw three, and i used the mouse cursor like YT! lol. I think what it means is that normally we tend to skip over the word "of" when we read. As we become familiar with reading, our minds get used to patterns and we learn to process these patterns rather than the letters themselves. "OF" is so short and so prevelent in english that you mind recognizes extrememly fast -- so fast that it usualy does not even register. I think this is a similar phenomenon to that one research article that talked about how all the human mind really needs to recongnze a word is the first and last letters. All the other letters of a word can be scrambled around and you will still understand what the word is, pretty easily too. So i think that is the point. People who have alot of expreience reading recongnize word patterns so well that the letters of the words loose their meaning. Cool.

That was kinda my point. The chemical species must be different in order for us to say that a reaction has taken place, becuase if the starting materials and products are the same, then we would not know if anything had happened. But this says nothing (and i am not sure if anything does) about the relative energies of the starting material and products. Specifically, i am not sure that they are nessesarily of different energies. The reason why i bring this us is becuase of this quote... that is all, just a response to that quote.

cool, i would like to two coppers too Lets assume that you take some starting material run a reaction and the look at the products. The starting material must be different from the products otherwise we would be unable to tell wether or not there had been a reaction right? (for convienence, i will call the starting material "A" and the products "B" from now on). Now we know that A and B must be different chemical species correct? But does this nessesarily mean that they are at different energies? NOt really. It could be (though extremely unlikely) that both A and B have exaclty the same energy. Thus, converting from A to B would seem as it this reaction should not liberate heat at all. So can anyone prove that two chemicals must have different energies? Just a question. I will have to think about it some too.

DO NOT CHEAT FOLLOW THE DIRECTIONS, It's Funny!!! Quick Eye Exam... This will blow your mind...! Just do it -- don't cheat!!!!!!!!!!!! Try this its actually quite good. But don't cheat! Count the number of F's in the following text: FINISHED FILES ARE THE RESULT OF YEARS OF SCIENTIFIC STUDY COMBINED WITH THE EXPERIENCE OF YEARS Managed it? Scroll down only after you have counted them! OK? How many? Three? Wrong, there are six -- no joke! Read again! FINISHED FILES ARE THE RESULT OF YEARS OF SCIENTIFIC STUDY COMBINED WITH THE EXPERIENCE OF YEARS The reasoning is further down... The brain cannot process the word "OF". Incredible or what? Anyone who counts all six F's on the first go is a genius Three is normal.

well, it appears that my posts are all appearing all over the place, perhaps this thread is just not for me

No worries. But if you think it is wrong, then perhaps you could point out my mistake so that i could learn from it? Cool. Well, there is a correct answer and it is fafalone's (-136.575). The easiest way to arrive at this answer is to convert to Kelvin and then divide by two. Why? Well, if you look closer at what is happening if you instead convert to Farienhieht what you find is that you have changed the temperature by 16 degrees. However, the farienhieht scale does not end at 0 degrees. THere are many many more degrees that are negative. Thus saying that the temperature is 32 and dividing by two to get 16 only takes into account the positive temperature. HOwever, there are a whole lotta negative ones that need to be considered as well. So, you need to figure out how many degrees are between 0 degrees F and absolute zereo and figure these in as well. Of course, if you just use the kelvin scale and divide by two then all this work is done for you and you immediately arrive at the correct answer. But great questoin YT! It really helps illuminate why the kelvin scale was invented at all. It really simplyfies calculations and helps to eliminate conceptual mistakes like the one you and wolfson made. But i hope you don't feel bad, i think anyone that has worked in science before and been caught off gaurd by this very same thing, so you are in good company

yeah, lennigrad was a major hit for germany, hitler wasted alot of resources over a pile of ruble for what amounted to a issue of pride for him -- bad move. I think classically, the end germany is considered to be the opening of the eastern front. Before hitler attacked russia, he had an ally in stalin (or at least a non-agressor) however, when he attacked russia, he not only lost an ally but spread his resources quite thin -- another bad move. there were many other bad moves though. If you ask me (which i guess you kinda did) i would say that it was when Hitler stopped listening to his war advisors and started only listening to himelf. Having only one opinion (esp if it is your own) is never a good idea. And it is this arogance of Hitler's that (in my opinion) cost him the war.