How to relate a reaction barrier to the time the reaction needs to proceed? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-18T02:52:58Z https://chemistry.stackexchange.com/feeds/question/61602 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/61602 21 How to relate a reaction barrier to the time the reaction needs to proceed? Martin - マーチン https://chemistry.stackexchange.com/users/4945 2016-10-26T07:15:50Z 2017-04-25T11:50:43Z <p>As I am writing this I am at a conference and one of the participants just asked a question where he linked reaction barriers to durations for the reaction to complete. To paraphrase:</p> <blockquote> <p>From our experience a reaction with an activation barrier of 15 kcal/mol should occur instantaneous at room temperature. An activation barrier of 20 kcal/mol takes about one to two minutes and an activation barrier of 25 kcal/mol needs about 10 hours.</p> </blockquote> <p>I would like to rationalise this statement as it seems quite hand-weaving to me. How can I judge from (a possibly also computed) activation barrier how long a reaction will need to complete? For the the sake of the argument, let's only consider reactions that proceed in one step; generalisations may also be implied, but may be too complex.</p> <p>In the comments (and the already existing answer) the Eyring equation is mentioned. If the connection between the energy values and the duration can be made with that, an illustrative example would be nice.</p> https://chemistry.stackexchange.com/questions/61602/-/62570#62570 3 Answer by DSVA for How to relate a reaction barrier to the time the reaction needs to proceed? DSVA https://chemistry.stackexchange.com/users/37313 2016-11-13T00:16:23Z 2016-11-13T00:16:23Z <p>The activation barrier you got, is it $\Delta G^‡$ or $E_a$ (the Arrhenius activation energy)? Depending on this you might either use the Eyring equation or the Arrhenius equation.</p> <p>It is actually quite common to use the Eyring equation to calculate $\Delta H^‡$ and $\Delta S^‡$ and with that $\Delta G^‡$ from experimental rate constants, so doing it the other way round and using the equation for prediction is completly fine. But I would not say something like </p> <blockquote> <p>An activation barrier of 20 kcal/mol takes about one to two minutes </p> </blockquote> <p>but rather use half-lifes or 95% conversion or something similar. Just be careful if the reaction is not 1<sup>st</sup> order, since only there the half-life is independend on the concentration.</p> <p>As an example see <a href="http://pubs.acs.org/doi/abs/10.1021/ja201844c" rel="nofollow noreferrer">this paper from Joe Fox</a>. At page two, bottom of the left column they predict one reaction to be 29-times faster than another one, purely based on calculated $\Delta G^‡$ values, using the eyring equation to calculate those rate constants. On page 21 of the supporting information you can see that they used the eyring equation to calculate $\Delta G^‡$ from experimental data.</p> <blockquote> <p>As an extension to this, a similar approach should be possible to also judge the temperature of a reaction and linking it to the reaction duration and reaction barrier.</p> </blockquote> <p>Yes it is, if you assume that your $\Delta G^‡$ is constant over that that temperature range.</p> https://chemistry.stackexchange.com/questions/61602/-/65688#65688 7 Answer by Zhe for How to relate a reaction barrier to the time the reaction needs to proceed? Zhe https://chemistry.stackexchange.com/users/35555 2017-01-04T15:50:11Z 2017-01-04T15:50:11Z <p>Here are some back of the envelope numbers from using the Eyring equation:</p> <p>$$k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$$</p> <p>Let's just assume we're at $298\ \mathrm{K}$ for the reaction, and the reaction is relatively simple:</p> <p>$$\ce{A-&gt;B}$$</p> <p>I constructed the following table by plugging in values. $t_{1/2} = \frac{\ln 2}{k}$</p> <p>\begin{array}{|c|c|} \hline \Delta G^{\ddagger}\ (\mathrm{kcal\,mol}^{-1})&amp; k\ (\mathrm{s}^{-1}) &amp; t_{1/2} \\ \hline\hline 15 &amp; 63.4 &amp; 10.9\ \mathrm{ms} \\ 20 &amp; 0.0138 &amp; 50.2\ \mathrm{s} \\ 25 &amp; 2.98\cdot 10^{-6} &amp; 64.6\ \mathrm{h} \\\hline \end{array}</p> <p>The values for 15 and 20 $\mathrm{kcal\,mol}^{-1}$ seem pretty consistent with your rule. The top value is a bit off, but we're working with a very small numbers at this point, and there may be other sources of error that we're not accounting for in the model.</p> https://chemistry.stackexchange.com/questions/61602/-/65727#65727 5 Answer by snurden for How to relate a reaction barrier to the time the reaction needs to proceed? snurden https://chemistry.stackexchange.com/users/17102 2017-01-05T08:13:07Z 2017-04-25T11:50:43Z <p>If I understand your last statement correctly, what you would like to have is the reaction time $t$ as a function of the reaction barrier $\Delta G$. However, $t$ also depends on the conversion $c$ (for (pseudo)first order reactions, as is the assumption in the Eyring equation, conversion can never be 100%) and temperature $T$.</p> <p>Although it has been mentioned, just for the sake of completion, here is the Eyring equation giving us the rate constant $k$:</p> <p>$$k = \frac{k_B T}{h}e^{-\frac{\Delta G^\ddagger}{RT}}$$</p> <p>We know that the half life $\lambda$ is:</p> <p>$$\lambda = \frac{\ln(2)}{k}$$</p> <p>The conversion $c(t)$ is related to this:</p> <p>$$c = 1 - \frac{1}{2^{\frac{t}{\lambda}}}$$</p> <p>If we solve this for $t$, we get:</p> <p>$$t = \frac{\ln(\frac{1}{1-c})}{\ln(2)}\lambda = \frac{\ln(\frac{1}{1-c})}{k}$$</p> <p>Where we can insert the Eyring equation for $k$, to get this final result:</p> <p>$$t(\Delta G, c, T) = \frac{h \cdot \ln(\frac{1}{1-c})}{k_BT} \cdot e^{-\frac{\Delta G^{\ddagger}}{RT}}$$</p> <p>Here is a plot of the reaction time for some typical conversion rates at room temperature: <a href="https://i.stack.imgur.com/gRYfc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gRYfc.png" alt="tvG"></a> And another logarithmic plot, which makes it easier to get the involved time scales, from 10$^{-9}$ h (3.6 $\mu$s) to 100 h: <a href="https://i.stack.imgur.com/zMnKC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zMnKC.png" alt="enter image description here"></a> As you can see, reactions around 20 kcal/mol lie around the "typical" regime, from seconds to several hours, while reactions with $\Delta G^{\ddagger}$ &lt; 15 kcal/mol proceed within milliseconds and reactions with $\Delta G^{\ddagger}$ > 25 kcal/mol may take days or weeks to complete.</p>