Timeline for Kinetics of interstellar chemistry: Applying steady state to formation of H₂
Current License: CC BY-SA 3.0
19 events
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| Mar 25, 2018 at 9:34 | history | edited | orthocresol | CC BY-SA 3.0 |
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| Mar 25, 2018 at 9:14 | history | edited | orthocresol | CC BY-SA 3.0 |
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| Jul 9, 2014 at 0:24 | history | bounty ended | 1110101001 | ||
| Jul 7, 2014 at 6:10 | history | edited | Martin - マーチン♦ | CC BY-SA 3.0 |
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| Jul 7, 2014 at 0:47 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 6, 2014 at 23:38 | comment | added | ron | Yes, everything you said is correct, but...that was my point above. When working with large numbers like Avogadro's number, $6 \cdot 10^{23}$ (let's call it $N$), the difference between $N^2$ and $N(N-1)$ is sooo small, let's make life easy and just call it $N^2$. But for very low concentrations like on the dust particles, then there is a big difference between $2\cdot 2$ (which is $m^2$) and $2\cdot 1$ (which is $m(m-1)$). So in this low-concentration case we need to use the tedious, but accurate, $m(m-1)$ expression. | |
| Jul 6, 2014 at 23:31 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 6, 2014 at 23:28 | comment | added | 1110101001 |
Thanks for taking the time to answer :D. With regards to the "When dealing with a mole of molecules and one reacts in a bimolecular reaction with itself ($\ce{H + H -> H2}$), no need to write (6*10^23)(6*10^23 - 1)", doesn't the step still occur one atom at a time? As is rather than one mol of H atoms reacting with another mol of H atoms all at once, it happens one atom at a time? So then why do we only take the leading order term ($[H]^2$) instead of writing the more precise $n(n-1)$?
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| Jul 6, 2014 at 18:26 | comment | added | ron | Taking the approach that "m" represents concentration is much easier to understand than the statistical approach. The statistical approach is correct but uses (IMO) much more cumbersome language than the concentration approach. When I get a few minutes I will rework the answer removing the probability approach and replacing it with the concentration analogy. Thanks for continuing to ask questions until we arrived at a comfortable place. | |
| Jul 6, 2014 at 18:26 | comment | added | ron | Good questions. There are 2 ways to form $\ce{H_{2}}$. Label the atoms A and B, then we can go via the AB or the BA route. m(m-1) is exactly analogous to [H][H], it's just not a square because there is a significant difference between 2*2 and 2*1. When dealing with a mole of molecules (N = Avogadro's number, 6*10^23) and one reacts in a bimolecular reaction with itself, no need to write (6*10^23)((6*10^23)-1), in our case the difference matters. BTW, the more I think about it, the more I like the approach in point 5) above. | |
| Jul 6, 2014 at 17:20 | comment | added | 1110101001 | But it is $_{m}P_{2}$ that gives a net result of $m(m-1)$. If you have two H atoms, $m(m-1)$ gives that there are two ways to form $H_2$, but there is clearly only one way. Shouldn't you use $_{m}C_{2}$ instead, which would give $1$?. Also, if you were normally writing the rate expression for $\ce{H + H -> H2}$ you would have something like $k[H]^2$. Why do you not have the squared term here and instead have the permutation? | |
| Jul 6, 2014 at 16:28 | comment | added | ron | 1) I do use combination instead of permutation, permutation is used when order matters, combination is used when order doesn't matter, all of the H atoms are identical so order does not matter 2) the H atoms are being picked one at a time, not two at a time 3) there is no order of choosing, you pick one atom out of two and then one out of one 4) yes, order doesn't matter, the H atoms are identical, you get the same hydrogen molecule 5) "m" is representing the "concentration" of H atoms in the reaction and desorption cases, the "concentration" is either 0, 1 or 2. | |
| Jul 6, 2014 at 16:14 | vote | accept | 1110101001 | ||
| Jul 6, 2014 at 16:11 | comment | added | 1110101001 | Why wouldn't you use combination instead of permutation? If you want to choose 2 H atoms to react and you have two H atoms on a dust particle shouldn't it just be 2C2 = 1? Why do you take into account the order of choosing which would give 2*1 = 2? Don't you get the same molecule $\ce{H2}$ regardless of the order the atoms are chosen? Also, I've never seen permutations used in rate expressions. Why are there usually no permutations (just the rate constant times concentration to a power), but in this case you have one? | |
| Jul 6, 2014 at 14:25 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 6, 2014 at 2:57 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 5, 2014 at 13:55 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 5, 2014 at 13:47 | history | edited | ron | CC BY-SA 3.0 |
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| Jul 5, 2014 at 2:16 | history | answered | ron | CC BY-SA 3.0 |