Lecture Slides on Collision Theory - Introductory Physical Chemistry | CH 331, Study notes of Physical Chemistry

Download Lecture Slides on Collision Theory - Introductory Physical Chemistry | CH 331 and more Study notes Physical Chemistry in PDF only on Docsity! 1 Lecture 25 Collision Theory NC State University Chemistry 331 Kinetics Gas Phase Kinetics Elementary Reactions Collision Theory Experimental Rate Law Application to Formation of NO2 Elementary Reactions Reactions usually occur in a series of steps involving one or two molecules. The molecularity of an elementary reaction is equal to the number of molecules that come together in an elementary step. The reaction order is an empirical quantity that does not have to equal the molecularity. Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E 2 Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH E 5 Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier. Br BrH The reaction: H + Br2 HBr + Br occurs because the kinetic energy is larger than Ea. Ea E Relationship to Arrhenius theory For the preceeding reaction the molecularity is two. It is a bimolecular reaction. Thus, rate of collision ∝ [H][Br2] But, as we saw the collisions must occur with an energy of Ea or greater for a reaction to occur. The kinetic theory of gases tells us that there is a connection between the kinetic energy and the temperature. The rate law and the probability The probability that a given reaction will occur at given temperature is f= e-Ea/RT Thus, the rate of reaction ∝ [H][Br2]e-Ea/RT If we compare to the second order rate law rate of reaction = k[H][Br2] If follows that k ∝ e-E a/RT. The pre-exponential factor The value of A can be calculated from the kinetic theory of gases µ = m1m2/(m1 + m2) is the reduced mass σ is the collision cross section There is also an orientational factor P. If not all orientations lead to products then P < 1. A = σ 8k BTπµ 1/2 NA 2 The formulation of rate laws The empirical rate law for the reaction 2 NO(g) + O2(g) 2 NO 2(g) is: Rate of formation of NO2 = k[NO]2[O2] Does this rate law mean that NO 2 is formed in a termolecular process? Such collisions have very low probability. A combination of bimolecular processes is much more likely. The following reaction mechanism has been proposed. Elementary steps in the formation of NO2 Step 1. Two NO molecular collide to form a dimer ka NO + NO N 2O2 ka’ Step 2. The O2 molecule collides with the dimer to form NO 2 kb O2 + N2O2 NO2 + NO 2 For step 2 the rate of consumption of N 2O2 is kb[N2O2][O2] and the rate of formation of NO 2 is 2kb[N2O2][O2]. 6 Applying the steady state approximation to intermediate N2O2 The rate of formation of NO2 = 2kb[N2O2][O2] is not acceptable as a rate law since it contains the concentration of the intermediate N2O2. d[N2O2] = ka[NO] 2 - ka’[N2O2] - kb[N2O2][O2] dt Now we use the steady state approximation: 0 = ka[NO]2 - ka’[N2O2] - kb[N2O2][O2] How to deal with the intermediate N2O2 Solving the steady state equation we find: so that the rate of formation of NO2 can be written: Note that this agrees with the empirical rate law provided ka’ >> kb[O2]. Thus, Rate of formation of [NO2] = (2kakb/ka’)[NO]2[O2]. [N 2O2] = k a[NO] 2 k a ′ + k b[O2] v = 2k b[N 2O2][O2] = 2k ak b[NO ] 2[O2] k a ′ + k b[O2] Rate limiting step If the concentration of O2 is increased to a very high level then the second step is very rapid compared to the first. The first step becomes the rate limiting step. Under these conditions the rate law becomes: Such conditions are used experimentally to determine the order of NO in the reaction. k = 2k a[NO] 2