Unformatted text preview: The temperature dependence of the rate constant For most chemical reactions, the temperature dependence of the rate constant is given by the Arrhenius law: k(T) = a Tn exp(
Ea/RT) where the activation energy Ea is expressed in J/mol. To understand the physical origin of this, consider dissociation of a diatomic molecule, e.g., H2 = 2H. We can model a diatomic molecule as two spheres connected by a spring. Consider a container filled with a gas that consists of such contraptions and also with inert gas atoms modeled as spheres. If a diatomic molecule suffers a collision with an inert atom, the relative velocity of its two spheres connected by a spring may increase. If the energy of their relative motion along the spring is large enough (say larger than the bonding energy E) then the bond will be broken. The higher the temperature, the faster the molecules move. We have previously learned that the probability of having energy E is proportional to exp(
E/kBT). If E is expressed in J/mol then it will be exp(
E/RT). The number of molecules that have enough energy to dissociate is then proportional to exp(
E/RT) and so the rate constant should be proportional to exp(
E/RT). Transition state theory Transition state theory provides a quantitative framework for the above ideas. First, let us introduce the idea of a potential energy surface. From physics, you know that the potential energy of a spring connecting two spheres, A and B, is equal to U(x) = γx2/2 where γ is the spring constant and x is the extension of the spring. If we plot this as a function of x we get a parabola. However this plot does not reflect the fact that if we stretch the spring too much the bond will be broken. A more realistic plot will look like this: The depth of the well, ED is called the dissociation energy. The probability to dissociate is proportional to have the energy around ED, exp(
ED/kBT), so we expect the dissociation rate constant to be of the form k = ν e− ED / kBT How can we estimate the prefactor ν? One idea is to consider the limit where ED / kBT → 0. It’s possible to convince yourself that in this limit one gets something like ν = ω / 2π where ω is the vibration frequency for the diatomic molecule. ω can be measured spectroscopically because the molecules will absorb electromagnetic waves of that frequency. Our estimate for the rate constant is however not quite right because of (1) quantum mechanical effects and (2) the fact that the molecule can only dissociate if it collides with something else or possibly absorbs a photon – therefore ν should generally depend on other factors. Now consider a more complex reaction: k1 AB + C A + BC k
1 We will assume that this can only happen when A,B, and C are arranged along the same line (“collinear reaction”). Suppose that A and C are far apart and I am dragging B relative to A and C. In the course of this reaction I need to break the spring AB and to form the spring BC. The resulting potential energy profile will look like this: Plots like this are common for chemical reactions. Generally, we plot the potential energy as a function of the reaction coordinate, which represents the progress when going from the reactants to the products. The choice of the reaction coordinate depends on the reaction we consider. One problem with the above picture is that we have fixed the distance between A and C. In general, the potential energy of the system will depend on both xAB and xBC. We can represent potential energy U ( xAB , xBC ) as a surface, whose contour plot is shown below. AB+C RBC (ABC) A+BC RAB
The reaction involves starting in the “reactants valley” (AB+C), surmounting a potential barrier and getting into the “products valley” A +BC. The lowest energy path goes over the saddle point indicated by a cross in the Figure. (Imagine surmounting a mountain ridge: You’re better off heading for a mountain pass, which, mathematically, is a saddle point). We can think of the minimum energy path that goes over the saddle point as a reaction coordinate. Along this coordinate, the potential energy looks like this: and the rate constants are: k1 = ν1 exp(−( E ∗ − Er ) / kBT ) and k−1 = ν −1 exp(−( E ∗ − E p ) / kBT ) so the activation energy is E ∗ − Er ( p ) . Rigorous estimates for ν 1 and ν −1 are beyond the scope of our class but we mention that these quantities generally depend on the “activation entropy” that is missing in the simple one
dimensional picture above. Think of gas particles escaping from a container through a small hole: There’s no activation barrier in this case; however there is an “entropic barrier” related to the fact that a molecule has to pass through a narrow constriction. ...
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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.
 Spring '11
 Makarov

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