Can in any case the faster step of the reaction be rate determining? - Chemistry Stack Exchange most recent 30 from chemistry.stackexchange.com 2019-09-22T16:47:42Z https://chemistry.stackexchange.com/feeds/question/15974 https://creativecommons.org/licenses/by-sa/4.0/rdf https://chemistry.stackexchange.com/q/15974 20 Can in any case the faster step of the reaction be rate determining? DSinghvi https://chemistry.stackexchange.com/users/5424 2014-09-06T10:21:16Z 2014-09-21T17:23:08Z <p>I found a sentence in book which states the reverse step of this reaction ( I forgot what was it!) has the faster step as rate determining .</p> <p>Even <a href="http://en.wikipedia.org/wiki/Rate-determining_step">Rate determining step-Wikipedia</a> states: </p> <blockquote> <p>In chemical kinetics, the rate (or velocity) of a reaction mechanism with several steps is "OFTEN" determined by the slowest step .</p> </blockquote> <p>which I think 'OFTEN' means not 'everytime' so,</p> <p>if faster step of the reaction is rate determining. Then why it is? </p> https://chemistry.stackexchange.com/questions/15974/-/16529#16529 8 Answer by Nicolau Saker Neto for Can in any case the faster step of the reaction be rate determining? Nicolau Saker Neto https://chemistry.stackexchange.com/users/1499 2014-09-20T18:13:43Z 2014-09-20T18:13:43Z <p>In the past I was also explicitly told that the slowest reaction is "often" the rate determining step, and like you I figured some day I would find a reaction where the fastest step determines the rate. Thinking over it a bit more now, however, I believe we may have misinterpreted what was meant. Rather than looking for an opposite situation, it is likely we were indirectly told that in some cases the rate-determining-step picture simply isn't applicable in the first place. </p> <p>In general, reactions can't be modelled though such simple kinetic theory; in truth, when a bunch of reactants are brought together, <em>all</em> steps in <em>every</em> possible reaction route matter. It just happens that for several simple but relevant chemical systems there are few possible steps and few accessible reaction routes, and they have such wildly different rate constants that we can approximate by looking only at the slowest step of the fastest route. A more complex reaction with several steps, several side-reactions and similar rate constants will likely be poorly described by a RDS framework, so one can say that the reaction rate is not determined by the slowest step because <em>the reaction rate won't be well determined by</em> <strong>any</strong> <em>single step</em>.</p> https://chemistry.stackexchange.com/questions/15974/-/16541#16541 10 Answer by RBW for Can in any case the faster step of the reaction be rate determining? RBW https://chemistry.stackexchange.com/users/7159 2014-09-21T11:33:45Z 2014-09-21T17:23:08Z <p>I can provide you an example. The oxidation of formate ion by peroxydisulfate in water solution: \$\ce{HCOO^{-} + S_2O_8^{2-} -&gt; CO_2 + 2SO_4^{2-} + H^+}\$</p> <p>has the following mechanism: <img src="https://i.stack.imgur.com/OdWpn.gif" alt="enter image description here"></p> <p>As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: \$r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]\$. The first reaction is very slow, so most of peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.</p> <p>(This isn't the true mechanism but it is good for showing the point. A more complex mechanism includes the formation of OH radicals and several chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)</p>