Simple Rate Equations Consider a chemical reaction: There are two common trends for the time rate of change of the concentration, C A. first order reaction kinetics second order reaction kinetics order refers to the exponent on C A and k is the rate constant. If C A o is the initial concentration then both of these equations are readily integrated to yield: first order reaction kinetics second order reaction kinetics Radioactive decay is the prototypical first order reaction, but there are many molecular first order reactions as well. Isomerization of cyclopropane to proprene is a good example of a system exhibiting second order kinetics: This latter kind of reaction is interesting because it begs the question of where the energy to drive such a reaction comes from. Surely an isolated molecule will not spontaneously absorb such energy it will require an interaction with other molecules, i.e., through collisions. Collisions are associated with second order kinetics where the rate is proportional to the product of (a) the probability of finding a molecule at some position times (b) the probability of find a second molecule at the same position. Secondly, it is also noted that for many reactions, at smaller concentrations of the molecule, the reaction rates switch from first order (at higher concentrations) to second order (at lower concentrations). We need a unified picture. Processes close to equilibrium Consider the chemical reaction with rate constant k + for the forward reaction and k - for the back reaction. The system consists of a mixture of components A and B. The chemical potentials of each of the components can be written as: The difference in chemical potentials is (A) At equilibrium the chemical potentials of each species In the mix are equal; Therefore: This steady state equilibrium also mandates This dynamic equilibrium says that the forward and reverse reaction rates are equal. From chemical equilibrium we recall that the equilibrium constant, K, is defined by, relating the equilibrium constant to the kinetic quantities. This then allows us to write the rate equation as or using (A) This last expression expresses that the reaction rate depends on the difference in chemical potentials between products and reactants. There are two limiting cases for this expression: In chemical reactions usually the chemical potential of B is much smaller than that of A corresponding to A large driving force for the reaction: In this case In materials processes such as diffusion, In this case, A B Reaction coordinate The Lindemann Hypothesis This hypothesis evolved the first model for connecting 1 st and 2 nd order reaction kinetics. The central idea is that a reaction A B involves the formation of an energized or activated state A* resulting from collisions between molecules of A. In the Lindemann model the reaction is broken into parts: First two reactants collide forming an activated complex The activated complex may be hit by another A molecule de-energizing it or, the activated complex may disintegrate into the product, There are three associated rate laws for each component of the mixture: (a) (b) (c) Typically, we assume that the concentration of the activated complex, A*, is in steady state, With this assumption we solve for C A* using equation (b), Substitution for C A* into equation (a) yield the primary result, Now we consider several limiting cases: Decay is rate limiting: Then: Activation is rate limiting: Then: We can see that: For large values of C A we can expect 1 st order kinetics. For small values of C A we can expect 2 nd order kinetics. Rates of processes in condensed phases A B A* Diffusion within the solid state can be considered as corresponding to the Lindemann mechanism. The atoms gain energy through collisions with lattice vibrations (solid) or from nearby molecules (liquid). The atom may then momentarily reside in an activated state from which it may fall back to its original position, or finish a diffusive hop. If this is correct, then whats the rate limiting step, activation or decay? Typically a diffusion event in a solid has an activation energy of several eV. Lets say E* = 4 eV. Temperature provides an energy of k B T/2 per molecule per degree of freedom (DOF). The equipartition theorem says that for N atoms each having 3 DOF the total system energy is (3/2)Nk B T. At room temperature, k B T ~ (1/40) eV. We can see that a system only need contain ~ 100 atoms to provide 4eV of energy. [(3/2)N(1/40) = 4] Most systems even in todays nano-world contain many more atoms so activation is never rate limiting. For this reason, a discussion of effective rate constants for processes such as diffusion in liquids or solids is not well described by 2 nd order reaction kinetics. Rather for a transition between A and B through a transition state A*, we should use the 1 st order scheme. This is particularly reasonable if there is a steady state equilibrium between A and A*, i.e., such that k*> f* In this case the rate is limited by the formation of A*. Furthermore, the concentrations of reactants in the transition state is just some small fraction of reactants, i.e., Case 2: k* >k* Physical interpretation of the net rate constant The net rate constant is a product of a rate constant k* and an equilibrium constant K*. K* may be viewed as the probability of finding the reactant in the activated state and k* corresponds to an attempt frequency, i.e., rate of formation of B (#/sec.) = attempt freq.(#/sec.) x prob. of success k net = k* x K* We can write this in the more conventional form, The general form Arrhenius equation is referred to as the Arrhenius equation. H* is called the activation enthalpy (or energy). If one measures the rate of a process as a function of temperature and plots rate versus reciprocal temperature, both the pre-factor and the activation energy may be obtained. Ln (rate) 1/T slope = - H*/R