Lectures 29-30

# Lectures 29-30 - The Boltzmann distribution: the...

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Unformatted text preview: The Boltzmann distribution: the probability to have an energy ε i is given by: pi ∝ Ωi exp(−ε i / kBT ) Examples of this 1. A piece of chalk on the table has the probability to be above the table ~exp( ­mgh/kBT). The surrounding molecules can bump into the chalk lifting it off the table. A “typical” elevation of chalk above the table is found from the condition mgh / kBT ~ 1 For m = 1g this gives h ~10 ­23m, many orders of magnitude smaller than a typical molecular size. So we don’t have to worry about chalk jumping off the table. However if we substitute a typical molecular mass we will find h ~ 103m. This explains why molecules of air are not all lying on the surface of the Earth. Furthermore, you can recognize this distribution in the barometric formula: P ∝ exp[−Mgh / RT ] = exp[−mgh / kBT ] where m is the mass of one molecule. 2. You can also recognize this distribution in the solution of the Clapeyron ­Clausius equation (molecules are less likely to be found in the vapor phase because they have a higher energy in the gas). The lower energy of the molecule in the liquid phase as compared to the gas phase, the less likely it is to find the molecule in the vapor phase. That’s why Al has a very low vapor pressure while, e.g., the vapor pressure of Hg is much higher. 3. we can also calculate the typical velocity v at which a molecule moves from the condition mv 2 ~ 1 2 k BT v ~ kBT / m ~ 300m / s (for something like O2). Irreversible second order reactions Those are typically bimolecular reactions of the type: A + B → C + D 2A → C + D The concentration of the reactant A evolves according to: d[A]/dt =  ­k[A][B] for the reaction (1) and d[A]/dt =  ­k[A]2 for the reaction (2). Note that, according to our convention about the definition of the rate, it would make more sense to divide the lhs of the last equation by two. However the above form is more commonly used. Here k is the rate constant. Why such a rate equation? In order to react, A and B must collide. The probability of such a collision is proportional to the concentrations of A and B, therefore it is proportional to [A] [B] and so the rate is proportional to [A][B]. Sometimes a reaction that is described by Eq. 1 or 2 is not 2nd order. This is because when writing Eqs. 1 or 2 we might be ignorant what’s really going on and so these equations may not represent the actual reaction mechanism. Here is an example: H2+I2 →2HI we might think it is a 2nd order reaction. It is not. The actual mechanism is: I2 → 2I H2+I → HI + H H + I2 → HI + I … (2) (1) We can solve the kinetic equations for the reaction (1) analytically. Introducing the extent of reaction η, we can write [A](t) = [A](0)  ­ η, [B](t) = [B](0)  ­ η, so that  ­dη/dt =  ­k([A](0)  ­ η)([B](0)  ­ η) k dt = dη/(([A](0)  ­ η)([B](0)  ­ η)) or kt = C + ∫ dη/(([A](0)  ­ η)([B](0)  ­ η)) To perform the integral, note that 1/(([A](0)  ­ η)([B](0)  ­ η)) = (([A](0)  ­ η) ­1  ­ ([B](0)  ­ η) ­1)/([B](0) – [A](0)), from which we find: kt = C + Using the initial conditions, C =  ­ so that kt = 1 [ B](0) − η 1 [ B](t ) = ln ln [ B](0) − [ A](0) [ A](0) − η [ B](0) − [ A](0) [ A](t ) (3) 1 [ B](0) ln [ B](0) − [ A](0) [ A](0) 1 ([ B](0) − η )[ A](0) ln [ B](0) − [ A](0) ([ A](0) − η )[ B](0) Eq. 3 implies that if I plot ln([B]/[A]) as a function of t I should get a linear plot. This fact can be used in order to establish whether or not a given reaction is second order. There is an interesting limiting case of the above equations. If we assume that [B](0) >> [A](0) then ([B](0) ­η) /[Β](0) ≈ 1 and also [B](0) – [A](0) ≈ [B](0) so we find: kt = (1/[B](0)) ln([A](0)/[A](t)) i.e. [A](t) = [A](0) exp( ­k[B](0) t) In the limit where the concentration of B is much larger than that of A, [A] follows first order kinetics with an effective rate constant equal to [B](0) k. At the same time, the concentration of [B] stays approximately constant and equal to [B](0). Although the change of [B] is precisely equal to that of [A], the relative change in [B] is negligible. This kind of situation is called pseudo ­first order kinetics. (The kinetics are not really first order but effectively they are). We could obtain the same result from the rate equation d[A]/dt =  ­k[A][B] of we assume that [B] = [B](0). Now consider the reaction (2). We have d[A]/dt =  ­k[A]2, d[A]/[A]2 =  ­k dt, or, integrating,  ­1/[A] =  ­kt + C The constant is equal to C =  ­1/[A](0) so we find kt = 1/[A](t) – 1/[A](0) Finding order of a chemical reaction from kinetic data Suppose for simplicity that you are dealing with an irreversible chemical reaction and that you have measured the dependence of the reactants’ concentrations [A](t), [B](t), … as functions of time. How can you deduce the order of the reaction from such kinetic data? Finding the order of a reaction may be useful as this may suggest the reaction mechanism. Basically, the way to find the order of the reaction is to try to fit the dependences [A](t) by each of the formulas that we have derived for the 0th order, 1st order, 2nd order reaction etc. and see which one works best. If none, then perhaps the reaction mechanism is a lot more complicated than you thought. A useful trick is to plot some function of [A](t) vs t in such a way that the resulting plot is a straight line. For instance, if we are dealing with a 2nd order reaction of the second type then we just saw that 1/[A](t) plotted vs t will be represented by a straight line. Thus if you suspect that your reaction has a 2nd order mechanism, the first thing to do is to plot 1/[A](t) as a function of t and see if this can be fitted by a straight line. In practice, because of experimental errors the fit may not be perfect so you will need to decide whether it is good enough. You may want to try see if other rate laws would do a better job describing the same data. The table below summarizes what we know about irreversible reactions: order 0 1 2 2 More complex reaction mechanisms & the concept of rate limiting step. Successive reactions: Consider the kinetic scheme A → B → C. The rate constant of the 1st reaction is k1, the 2nd, k2. The rate equations are d[A]/dt =  ­k1[A] d[B]/dt = k1[A] – k2[B] d[C]/dt = k2[B] kinetic equation d[A]/dt =  ­k d[A]/dt= ­k[A] d[A]/dt = ­k[A]2 d[A]/dt = ­k[A][B] linear plot [A] vs t ln[A] vs t 1/[A] vs t ln([B]/[A]) vs t If [B](0) = [C](0)=0 then one can find: [A](t) = [A](0) exp( ­k1t) [B](t) = [A](0) k1 -k t (e-k 2 t -e 1 ) k1 -k 2 k1e-k 2 t -k 2e-k1t [C](t) = [A](0)(1) k1 -k 2 These three functions are plotted below for [A](0)=1, k1=1, k2=2. We see that the concentration of C goes first up and then down and the concentrations of A and B, respectively, decrease and increase. 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 I would like to use the above solution to demonstrate a couple of useful limits and approximations. First of all, I will demonstrate the idea of a rate limiting step. This is a very important concept, particularly in biochemistry, because the kinetic equations a biochemist has to deal with are usually very complicated and their analysis is simplified greatly if a rate limiting step can be identified. Suppose that in our experiment we can only monitor the concentrations of A and C. Suppose first that k2 << k1. This means that A is converted to B very quickly. This means that the population of A will be depleted very quickly and the formation of the product C will take place with a rate constant k2. Indeed, we see that the exact solution for C in this limit can be represented as [C](t) = [A](0)(1 – exp( ­k2t)). We say that the rate limiting step in the formation of C is the slow reaction B → C whose rate constant is k1. 1 0.8 0.6 0.4 0.2 A plot of [A](t), [B](t), and [C](t) in this limit (k2=1, k1 = 50) is given below 0.5 1 1.5 2 Now consider the opposite limit: k1<<k2. In this case B is quickly converted to C so that there is no significant accumulation of the intermediate C. For example, a plot of [A](t), [B](t), and [C](t) for k2=15, k1 = 1 is as follows: 1 0.8 0.6 0.4 0.2 0.5 1 1.5 2 The formation of the product C is then limited by the transition from A to B and so we expect [C](t) = [A](0)(1 ­exp( ­k1t)). The rate limiting step here is the conversion of A to B. For the case of a short ­lived intermediate such as B in the example above one often uses the steady state approximation. In this approximation one assumes that since B is depleted as soon as it is populated, then [B](t) should stay constant. Thus one writes d[B]/dt = 0 which gives k1[A] – k2[B] = 0 or [B] = k1[A]/k2. Now for the concentration pf C we get d[C]/dt = k2[B] = k2 k1[A]/k2 = k1[A], which tells us that we can think of this process as conversion of A directly to C with a rate constant k1. The steady state approximation is often used to describe reactions that involve catalysts. Consider the reaction R → P The same reaction performed in the presence of a catalyst K is often written as k1 k2 R + K RK → P + K k ­1 Assuming the steady state approximation for the intermediate RK, we find d[RK]/dt =  ­[RK] (k2+k ­1) + k1[R][K] = 0 so that [RK] = k1[R][K]/ (k2+k ­1) and d[P]/dt = k2[RK] = k2k1[R][K]/ (k2+k ­1) The above reaction mechanism is often used to describe how enzymes work (Michaelis & Menten mechanism). ...
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## This note was uploaded on 04/24/2011 for the course CH 52635 taught by Professor Makarov during the Spring '11 term at University of Texas at Austin.

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