Timeline for Proof that Equilibrium Constant is a Ratio of Rate Constants
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29 events
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| Oct 13, 2014 at 17:36 | comment | added | Yandle | @jonsca My apologies I will do this in the future. The only question I had that was not related to my original question was the last one regarding the relationship between between β and global rate constants. Should I post this question? | |
| Oct 13, 2014 at 9:41 | comment | added | jonsca | @Yandle Please ask new questions as these issues come up. You can link your subsequent questions to this one if they are follow-ups. It doesn't help anyone in the future if you discuss them beyond asking for a clarification on what was posted. | |
| Oct 13, 2014 at 5:41 | comment | added | t.c | 1.Yes. 2. Yes, if you look at the other example in the link I provided, Br has the power of 3/2. | |
| Oct 13, 2014 at 5:32 | comment | added | Yandle | Okay, so although β value will imply whether reactants/products are favoured, it does not imply anything about the global rate constant itself? In the link you provided, the theoretical derivation showed [N2O5] was raised to the power of 1 in the global rate equation. I know empirical curve fit powers won't be integers, but is it possible for theoretical derivations result in fractional values? | |
| Oct 13, 2014 at 5:16 | comment | added | t.c | I'm not sure about that. If the units of $k$ are unknown - you may be taking the ratio of two dimensionally unequal constants(e.g. one may have the units $s^{-1}$, while the other $mol^{2/3}s^{-1/3}$ (just an example)). Qualitatively, I'm also not sure whether there is a definite relationship between $\beta$ and $kfg/krg$. It may be tempting to believe that if $kfg$ is higher than $krg$ then the reaction may proceed to the right and $\beta$ would be higher, but mathematically I'm not sure how to prove that, without knowing $kfg$ or$ krg$ | |
| Oct 13, 2014 at 4:46 | comment | added | Yandle | I think I understand what you are saying. Basically $k_{fg}$ and $k_{rg}$ don't exist because they are dependent on other variables (elementary rate constants)? If I still want to use the two as simplifying representations, can we still qualitatively say, the β we have is very large, hence $k_{fg}$ is likely bigger $k_{rg}$ (not attaching numbers to anything) and forward reaction is favoured? | |
| Oct 13, 2014 at 4:36 | comment | added | t.c | The Kfg in your book will probably comprises of many elementary rate constants such as those posted in the link - they merely simplify it to Kfg so that it looks neater - but it doesn't mean anything since it's a pseudo rate constant. | |
| Oct 13, 2014 at 4:00 | comment | added | Yandle | Sorry in my previous question I was using global rate constants $k_{fg}$ and $k_{rg}$ in the context of my combustion book where Foward Rate = $k_fg[A]^m[B]^n$; in the link posted $k_{fg} = k = k_fk_2/(k_b+2k_2)$. | |
| Oct 13, 2014 at 3:44 | comment | added | t.c | No, there is no such thing as $k_{fg}$ or $k_{rg}$, take a look at science.uwaterloo.ca/~cchieh/cact/c123/steadyst.html Example 1 | |
| Oct 12, 2014 at 19:40 | comment | added | Yandle | So a large $β$ doesn't necessarily mean that the global forward rate constant ($k_{fg}$) is higher than the global reverse ($k_{rg}$), or did I mis-interpret as this implies the equilibrium favours a direction opposite to that implied by the global rate constants? Physically, why does ($k_{fg}/k_{rg}$) deviate from $β$ where for elementary reactions K = rate constant ratio? In other words, what physical effects cause the global rate constant ratio to not simply be a product of all the elementary reaction rate constant ratio. | |
| Oct 12, 2014 at 6:28 | comment | added | t.c | Interesting question. I'm not sure whether there is any direct proportionality relationship between $\beta{}$ and $k_g$, (since $k_g$ is different for different situations) but I would be careful before assuming that there is. And yes, $\beta{} = K_1K_2..$ so you are right in saying it is the product of kf/kr of each elementary reaction, although the kr of the first reaction does not necessarily equal to the kf of the second reaction. | |
| Oct 12, 2014 at 6:25 | comment | added | Yandle | What I meant to ask is, although $β≠k_{f,overall}/k_{r,overall}$ mathematically, can I still interpret $β$ to qualitatively reflect the ratio of the global rate constants (i.e. large $β$ means $k_{f,overall}>k_{r,overall}$), but not using any quantitative definitions between $β$ and global rate constants. In addition, is $β$ also basically the product of kf/kr of each elementary reaction? | |
| Oct 12, 2014 at 3:49 | comment | added | t.c | And yes, your m and n will probably be fractions - since that particular rate equation is not an elementary equation. | |
| Oct 12, 2014 at 3:42 | comment | added | t.c | No, there is no such thing as global rate constant, so you are right in saying that it is incorrect to interpret it the same way as elementary rate constants. $\beta{}≠k_{f,overall}/k_{r,overall}$ | |
| Oct 11, 2014 at 17:21 | comment | added | Yandle | 2. So m and n in my original equation would also be something complex for the same reason? How should I interpret the relationship (hand-wavy wise) between global equilibrium coefficient and global rate constant? I used to interpret in the same way as an elementary reaction (kf/kr) which is incorrect. | |
| Oct 11, 2014 at 17:13 | comment | added | t.c | 1. Yes. Remember kf/kr must be of an elementary reaction. 2. The global rate constants is most likely made up of elementary rate constants. usually to avoid writing something complicated like k1k2'/k2-k1, we assign it to $k_g$ | |
| Oct 11, 2014 at 15:53 | comment | added | Yandle | 1. So only the K=kf/kr derivation assumes elementary reactions? 2. In my combustion text, they do define a global rate constant $k_g$. For F+aOx->bPr it is $d[F]/dt=-k_g[F]^m[Ox]^n$ where the equation is (m+n)th order and m and n are from curve fitting experimental data.. | |
| Oct 11, 2014 at 7:43 | comment | added | t.c | 1. Yes you are right, -RTlnβ = -RT(lnK2-lnK1) = ΔG2-ΔG1 = $ΔG_{overall}$. 2. Nope, there is no relationship between β and kf/kr simply because there an overall rate constant doesn not exist. | |
| Oct 11, 2014 at 6:54 | vote | accept | Yandle | ||
| Oct 11, 2014 at 6:54 | comment | added | Yandle | If we can simply go β = K1K2 etc., then in the derivation for K from Gibbs free energy, does it even matter whether the reaction is elementary or global, since the intermediate terms would cancel? How come global K (or β) isn't defined as ratio of overall rate constants and is there another relationship between β and kf/kr (global)? I always thought equilibrium constant is found by rearranging kf[A]^w[B]^x=kr[C]^y[D]^z where w,x,y,z may or may not be the stoichiometric coefficients. | |
| Oct 11, 2014 at 6:33 | comment | added | t.c | 1. Yes your assumption that it is an elementary reaction is correct. 2. Yes for global reaction (A+B->C+D), you would not be able to use K = kf/kr. 3. While you can multiply the eqm constants to obtain the global eqm constant, you cannot multiply the rate constants to obtain the global rate constant (in fact there is no such thing since the overall rate is dependant on many rate constants, for example overall rate = k1/(k2-k1) [A]^3/2 [B]^1/2)). Therefore, you cannot use $\beta{}=kf_{overall}/kr_{overall}$ - it would not be mathematically equivalent. K = kf/kr must be of a reaction step. | |
| Oct 11, 2014 at 6:18 | comment | added | Yandle | So when we derive K from Gibbs free energy (like in the Wikipedia link), the assumption is that the reaction is an elementary reaction? My confusion: For global reaction (A+B->C+D), Rate = kf[A]^x[B]^y where x and y are not the stoichiometric coefficients and hence K=kf/kr would be ratios of the concentrations raised to empirically determined values x,y etc.. However, if I go B=K1K2... I still end up with concentrations raised to the stoichiometric coefficients of the global reaction as if the global reaction was elementary, since intermediate terms cancel. | |
| Oct 11, 2014 at 5:50 | history | edited | t.c | CC BY-SA 3.0 |
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| Oct 11, 2014 at 5:44 | comment | added | t.c | @Yandle, I've edited my answer to explain further. | |
| Oct 11, 2014 at 5:43 | history | edited | t.c | CC BY-SA 3.0 |
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| Oct 11, 2014 at 5:07 | comment | added | Yandle | I am confused now. I looked at the derivation for free energy here and it doesn't assume the reaction as elementary or global. This answer and an example in my combustion book for CO2 dissociation (which I do not believe is an elementary reaction) also uses stoichiometric coefficients for Kp. But in another chapter the book states that Rate = k[A]^x[B]^y where x and y are empirically determined. | |
| Oct 11, 2014 at 1:35 | comment | added | Yandle | Is a reaction step an elementary reaction? For an elementary reaction aA+bB->cC+dD, Kc = kf/kr = ([C]^c[D]^d)/([A]^a[B]^b), but for a global reaction qQ+pP->tT, Kc = [T]^x/([Q]^y[P]^z) where x,y,z are empirically determined and no longer equal to the stoichiometric coefficients (for H2O formation this would be Kc = [H2O]^x/([H2]^y[O2]^z))? So the fact that Kc = kf/kr is always true, but Kc is not necessarily a ratio of the concentrations raised to their stoichiometric coefficients? | |
| Oct 10, 2014 at 17:47 | history | edited | t.c | CC BY-SA 3.0 |
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| Oct 10, 2014 at 17:42 | history | answered | t.c | CC BY-SA 3.0 |