Lecture 33 Chapter 19 Kinetics: Transition State Theory Announce: • Outline: Transition State Theory Review Reversible Reactions ln([A] ! [A]eq )= !(k1 + k!1 )t + ln([A]0 ! [A]eq ) so plotting ln(A-Aeq) vs. t gives a line with the slope that is the sum of the rates. But, we also know that at equilibrium k1[A]eq = k -1[Z]eq k1 [Z]eq so = = Keq k!1 [A]eq (Detailed Balance!) Therefore, we can find both the sum and ratio of the rates and know both of them. Temperature dependence of rates E " a k = Ae RT (Arrhenius 1889) transition state ‡ ! E E Reactants a !U Rxn Products Rxn coordinate 1 Can we use theory to figure out what A and k should be for some reaction? Well, kinetics is still a pretty young science and is therefore heavily based in experiment. Theories of reaction dynamics are a pretty big challenge and have been a focus of the Nobel committee fairly recently: Recent Chemistry Nobel Prizes in Kinetics/Dynamics/Catalysis 2001 – Catalysis 1999 – Reaction dynamics 1995 – Ozone kinetics Thus, kinetics is recognized as a difficult but important problem. Next time we’ll dive into a simple kinetics theory that is reasonably successful, called Transition State Theory. Transition state theory There are a number of theories for reaction kinetics. One of the more successful simple theories is transition state theory. In it, we assume that the transition state lives long enough to be in equilibrium with the reactants. (For the reaction: A + B Products) Define the equilibrium between ‡ and reactants (A and B) [‡] 0 0 ‡ c [‡]c K = = [A] [B] [A][B] or c0 c 0 K ‡[A][B] [‡] = c 0 Also define a frequency with which transition states cross over barrier to products (this is just a first order rxn of transition state to products. f is rate constant) d(products) fK ‡[A][B] v(t) = = f [‡] = dt c 0 assuming we have a simple rate law, v = k[A][B] and then fK ‡ k = c 0 ! 2 k T From the partition function stuff we did in chapt 13… f = B h The frequency factor (yes h is Planck’s constant). This is not a trivial result. It depends on three important assumptions: 1) Molecules in transition state obey Maxwell-Botzmann statistics with the reactants (and among themselves). 2) Molecules that have crossed barrier cannot turn around and recross. (This can be loosened by introducing the transmission coefficient, k‡ = κf. κ approx 1 for gas phase… κ < 1 for condensed phase.) 3) Motion along the reaction coordinate may be separated from other motions and treated classically as translation. Note that the Primary Isotope Effect comes about because a C-H bond (for instance) has a very different ground state vibrational energy than a C- D bond. We assume that they both have roughly the same vibrational energy at the transition state, so the ΔG‡ and therefore K‡ (see below) is very different for the two isotopes. Finally, the free energy difference can be related to the equilibrium constant just like we did a few chapters ago… ‡ ‡ "G = !RT ln K ‡ "!G so K ‡ = e RT Plugging back in !G‡ !S ‡ !H ‡ k BT " RT k BT R " RT k = e = e e hc 0 hc 0 This is the Eyring equation (also simultaneously Evans and Polanyi) Another detail from statistical mechanics relates the activation energy to enthalpy: ‡ is that Ea = Δ H + 2RT ‡ or Δ H = Ea − 2RT (The factor of 2 is related to the change in # moles.) Depending on the exact derivation, this number sometimes is 1, or even 0 (Ea >> 2RT is often valid). (Δn through ΔH = ΔU + Δ(PV) = ΔU + ΔnRT.) ‡ We’ll do as is often done and just assume Δ H ≈ Ea, (i.e. that 2RT is 0 negligible compared to Ea) and that c = 1: 3 ‡ E k T "S ! a k = B e R e RT h comparing to our empirical relation of Arrhenius gives: k T !S‡ A = B e R h So, Arrhenius factor depends on entropy difference between reactants and transition state. Note that there is a temperature dependence in the Arrhenius prefactor, not just in the activation energy term (but linear factor is mild compared to exponential factor) Elementary reactions Reactions are not usually single step: Reactants −−> Intermediates −−> Products Each simple step is called an elementary reaction and we classify them as unimolecular, bimolecular, or termolcular depending on the number of reactants. N2O4 ⇒ 2 NO2 unimolecular NO2 + F2 ⇒ NO2F + F bimolecular O + O2 + NO ⇒ O3 + NO termolecular (note the double arrow designating elementary rxn) In an elementary reaction molecularity = order. So, bimolecular reaction is always second order. Empirical reaction orders (complex reactions) may contain several elementary reactions, so the converse is not necessarily true. Again, because an elementary reaction forms no intermediates, the reactants must collide simultaneously and go straight to products. Therefore, we can write a rate law directly from the stoichiometry of the elementary reaction. From the second reaction above?: v = k[NO2][F2] Tetramolecular elementary reactions are very rare. Why? The need for simultaneous collision means that anything beyond termolecular is highly unlikely. 4 Steady-State Approximation Recall our two step, sequential reaction A !!k1 " I !!k2 " P that I really briefly mentioned last time. We’ll use this to introduce a common approximation. Note that our rates of change of each component must be: d[A] = !k [A] dt 1 d[I] = k1[A] ! k2[I] dt d[P] = k [I] dt 2 Now, this is a set of coupled, linear differential equations. You can either go to your DiffEQ book, or take Chem395 next semester to figure out how to solve this exactly. But, in many cases it turns out that the concentration of intermediate is fairly constant. Or, in other words? dI/dt = 0. That will make things very easy. Let’s see how… !k1t Solving the d[A]/dt eqtn gives? [A] = [A]0 e So, then our steady-state approximation allows us to say d[I]/dt = 0, or k1 k1 !k1t [I]SS = [A] = [A]0 e k2 k2 Plugging this into d[P]/dt equation and integrating gives [P] t ' k1 !k1t $ d[P] = k2 % [A]0 e "dt (0 (0 % " & k2 # t [P] = ! [A ]e!k1t SS 0 0 !k1t [P]SS = [A]0 (1! e ) 5 As a sanity check, let’s see what d[I]ss/dt is to see if it really is roughly 0 k1 !k1t [I]SS = [A]0 e k2 taking this derivative gives us: 2 d[I]SS k1 !k1t = ! [A]0 e dt k2 So, our assumption that d[I]/dt = 0 is reasonable when? 2 if k2 >> k1 [A]0, this is basically what we thought coming in. (Show plot comparing exact solution to steady-state approx.) Note that in this case, the S-S approx didn’t do much for us, but in more complicated cases, it can simplify the math substantially. **Note also, you must have enough information to make the approximation valid. Several Mechanisms Often Fit One Rate Law We’ve talked about the importance of kinetics in helping us to determine reaction mechanisms. But one must be careful – just because a mechanism is consistent, does not mean it is correct. For example, let’s look at the reaction in the text kobs 2NO + O2 !!!" 2NO2 if this is an elementary reaction, what is the rate law? d[NO2 ] 2 = 2k [NO] [O ] dt obs 2 It turns out that this is consistent with experiment. So, this must be right, right? Not so fast, let’s also consider two other mechanisms. k1 First: NO + O2 #!! !!" NO3 rapid k$1 equilibrium k2 NO + NO3 !!" 2NO2 rate determining 6 So we are assuming k2 << k1 and k −1 Because the first step is at equilibrium, we can relate to equil const? k1 [NO3 ] Keq,1 = = k!1 [NO][O2 ] and the second step then is? d[NO2 ] = 2k [NO ][NO] dt 2 3 We want to get rid of intermediates, so solve first equation for NO3 [NO3 ] = Keq,1[NO][O2 ] and substitute d[NO2 ] 2 = 2k K [NO] [O ] dt 2 eq,1 2 Note that this mechanism is also consistent with our experimentally observed rate law. Let’s check a second mechanism: k1 Second: NO + NO #!! !!" N 2O2 k$1 k2 N 2O2 + O2 !!" 2NO2 In this case, we will make use of the steady-state approximation. This can happen two ways, either k2 or k-1 must be >> k1 therefore d[N2O2]/dt approx. 0. (I won’t do this in detail, but it turns out that we get something similar.) 7 d[NO] = !2k [NO]2 + k [N O ] dt 1 !1 2 2 d[N 2O2 ] 2 = k1[NO] ! k!1[N 2O2 ] ! k2[N 2O2 ][O2 ] dt d[NO ] 2 = 2k [N O ][O ] dt 2 2 2 2 Steady-state approximation allows us to solve second eq. for [N2O2]. 2 k1[NO] [N 2O2 ] = k!1 + k2[O2 ] Now, we have to make a leap here. Steady-state approx requires that k-1 >> k1, but it can also be the case that k-1 >> k2. If this is true, than k1 2 [N 2O2 ] = [NO] k!1 Plugging this back in gives: d[NO2 ] k1 2 = 2k2 [NO] [O2 ] dt k!1 So, we again get the same rate law. In the last case, we had to make some extra assumptions, but the point is that we have three very different mechanisms that all give the same rate law.