Steady state approximation

Transcription

1 Steady state approximation Supplementary notes for the course Chemistry for Physicists Course coordinator: Prof. Dr. Mathias Nest Teaching assistant: Dr. Raghunathan Ramakrishnan contact: Department Chemie Technische Universität München Summer Term 013 1

2 1 When is steady state approximation useful? We apply steady state approximation for reactions where the of the reaction intermediate can be considered to be constant. Steady state means an chemical equilibrium, i.e. the rate of formation equals the rate of consumption. Why do we need steady state approximation? We need steady state approximation to simplify the derivation of rate laws of many step reactions. 3 general reaction to show the applicability of the steady state approximation Consider the following reaction Let us say that the reaction occurs in two steps (1) The differential rate laws are given by k 1 () (3) dc t () dc t () dc t () = k 1 c t () (4) = k 1 c t () c t () (5) = c t () (6) The integrated rate laws which are obtained by solving the coupled differential equations are given by c t () = c 0 ()e k 1t (7) k c0 () 1 ( c t () = k 1 e k! t e t ) ; k 1 c 0 ()k 1 te k1t (8) ; k 1 = c t () = c 0 () c t () c t () (9) The s of the reactant, intermediate and the product for the cases i) k 1, ii) k 1 =, and iii) k 1 are shown below k 1 >> k 1 = k 1 <<

3 Clearly the steady state approximation is applicable for the case k 1 where the of the intermediate is small and it varies slowly and can be considered to be constant most of the during the reaction (i.e. to a good approximation, dct() = 0). The situation k 1 also means that the intermediate is very reactive and this step is very fast. lso note that the first step is relatively very slow hence the first step determines the rate of the entire reaction. Now let us apply the steady state approximation for this case to derive the rate law. n most of the cases, we look for an expression for c t () in terms of the rate constants and the of reactants. Let us apply dct() = 0 which is the steady state approximation to eq. 5 dc t () = 0 c t () = k 1 c t () (10) Using the above equation we can rewrite the rate laws (differential and integral) as follows and dc t () dc t () dc t () = k 1 c t () (11) = 0 (1) = k 1 c t () (13) c t () = c 0 ()e k 1t (14) c t () = constant (15) c t () = c 0 () c t () k 1 c t () k = c 0 () = c 0 () where in the last equation we have used k 1 1 c 0 ()e k 1t k 1 c 0 ()e k 1t 1 c 0 ()e k 1t } (16) = 0 because is large. 4 n this course we use steady state approximation only to derive the differential rate laws of many step reactions Let us try an example (p. 60 of Chemie by Morr, Müller 9. uflage). PLESE RED THS SECTON N THE OOK (p.60) EFORE REDNG THE FOLLOWNG. The complete reaction is given by H 3 COH + H + + r H 3 Cr + H O (17) For this reaction the experimentally derived rate law is given by The proposed mechanism has three steps v = k c (H 3 COH) c ( H +) c ( r ) (18) step 1 : H 3 COH + H + k 1 H3 COH + fast (19) } 3

4 step : H 3 COH + H3 COH + H + fast (0) step 3 : r + H 3 COH + k 3 H3 Cr + H O slow (1) Since the third step is the rate determining step, we can write the rate of the reaction as (remember, the slowest step determines the rate of the entire reaction) v = v 3 = k 3 c ( H 3 COH + ) ( c r ) () This expression involves the of the intermediate. Recall that we usually write the rate law in terms of the of the reactant only. Let us use the stationary state approximation and see if we can eliminate c ( H 3 COH + ) in eq.. Let us equate the net rate of formation of the intermediate to zero dc ( H 3 COH + ) = k 1 c (H 3 COH) c ( H +) c ( H 3 COH + ) k3 c ( H 3 COH + ) ( c r ) = 0 (3) k 1 c (H 3 COH) c ( H +) c ( H 3 COH + ) [ k + k 3 c ( r )] = 0 (4) c ( H 3 COH + ) k 1 c (H 3 COH) c (H + ) = + k 3 c (r (5) ) Note that the third step is the rate determining step, hence k 3 or + k 3 c (r ) c ( H 3 COH + ) k 1 c (H 3 COH) c (H + ) (6) y substituting the above equation in eq., we arrive at the experimental rate law v = v 3 = k 1k 3 c (H 3 COH) c ( H +) c ( r ) (7) = k c (H 3 COH) c ( H +) c ( r ) (8) Now let us do a bit of analysis. For the two-step reaction given in page of this document we saw that steady state approximation is valid if the intermediate is very reactive, right?. ut in this three-step reaction the third step is a slow step, i.e., in the third step the intermediate reacts slowly. Then why did we use steady state approximation here? The reason is that we assume that is very large, i.e., the intermediate undergoes a very rapid reverse reaction as in step. n this example if k 1 k 3 the of the intermediate increases very rapidly at the beginning of the reaction, for a very short and then for most of the reaction it varies very slowly, i.e., dc() 0. This is illustrated in the following figure. Note that in the bottom figure (short ) the of the intermediate increases very rapidly, but it stays essentially constant through out the reaction. 4

5 CH 3 r = H O >> K 1 >> k 3 CH 3 OH = H + (red) + r - (green) CH 3 OH