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Lecture 18 (Ch 18)HW: Ch 18: 1, 3, 15, 41

Kinetics pt 2: Temperature Dependence of Rate ConstantsTemperature and RateThe rates of most chemical reactions increase with temperature.

How is this temperature dependence reflected in the rate expression?

Rates increase with temperature because rate constants increase with temperature. An example is the 1st order reaction: CH3NC ---> CH3CNVariation in k with temperature.

Collision ModelThe collision model makes sense of this. This model is based on kinetic theory

Weve seen that the thermal energy of a molecule is converted to kinetic energy in order to facilitate motion, and weve seen that the velocity of a molecule increases with T.

The central idea of the collision model is that molecules must collide to react. The more collisions per second, the faster the reaction goes.

This model also rationalizes the concentration dependence. The more molecules present, the more collisions you have.

Activation EnergyOf course, there is more to a chemical reaction than just collisions of molecules.

Molecules must have some minimum energy in order to overcome the energy barrier of reaction

Upon colliding, the kinetic energy of molecules is used to stretch, bend, and break bonds in order to cause a reaction.

But if the molecules dont have enough kinetic energy, they simply bounce off one another.

This minimum energy that molecules must have is called the activation energy,(Ea)

E too lowActivation Energy Needed to Reach Transition State

Lets go back to the reaction CH3NC ---> CH3CN

The molecule passes through a transition state, a high-energy intermediate in which the CN bond gets rotated 90o.

This is an unstable configuration and requires an activation energy to be reached. Once this transition state is reached, the process is energetically downhill. The transition state is always the highest energy point in the reaction pathway. As drawn, this reaction is exothermic because the energy of the products is less than the energy of the reactants.Transition statenitrogencarbonHEnthalpyArrhenius EquationThe fraction of molecules that have an energy equal to or greater than the activation energy is given by the expression:

Arrhenius noted that reaction-rate data depended on three aspects(1) the fraction of molecules possessing an energy Ea or greater(2) the number of collisions per second(3) the fraction of molecules oriented in the right way for a reaction to occur

These factors are incorporated into the Arrhenius Equation

This equation relates k to temperature. A is the frequency factor, and is related to aspects (2) and (3) listed above.Rearranging the Arrhenius EquationTaking the natural log of both sides, the Arrhenius equation becomes:As you see, we once again have an equation in y=mx+b form. Plotting ln k vs 1/T yields a linear plot with slope Ea/R and a y-intercept of ln AShowing Changes in k GraphicallyFrom the data below, determine the activation energy of the 1st order reactionTemperature, oCK (s-1)189.72.52 x 10-5198.95.25 x 10-5230.36.30 x 10-4251.23.16 x 10-3Plot ln k vs 1/T to get the slope. We know R, so we can calculate Ea. T must be in Kelvin.1/T (oK-1)ln K 0.002161-10.58870.002119-9.85470.001987-7.369790.001908-5.75718Calculating Changes in k MathematicallyThe Arrhenius equation can be rearranged once again to compare k values at different temperatures

Please note the following:

ExampleThe reaction above has an activation energy of 43.5 kJ/mol. The reaction occurs at 298 K with a rate constant of 110 s-1. What will the rate constant be if we increase the temperature to 308 K?Reaction MechanismsIn a previous lecture, I alluded to the following reaction:

In looking at this equation, and applying your understanding of collision theory, you would assume that this reaction proceeds when NO2 and CO collideThe activation energy is exceededThe atoms rearrange to form NO and CO2Thus, we would expect the rate of the reaction to depend on both [NO2] and [CO]

Reaction MechanismsHowever, according to experimental data, the reaction is 2nd order with respect to NO2 and zero order with respect to COThis means that the rate of the reaction does not depend on [CO] at all. How can this be? What does this tell us about the reaction mechanism?This particular reaction must proceed through multiple steps, and the rate-determining step must only depend on NO2Reaction MechanismsFor this particular reaction, there are two steps:This step-by-step description of the molecular pathway is the reaction mechanism. From reaction (1), we can physically understand why the reaction is 2nd order with respect to [NO2].

The 1st step is substantially slower than the 2nd. Therefore, step (1) acts as the reaction bottleneck, and the overall reaction speed can only be as fast as its slowest step.

Thus, step (1) is the rate determining step.13Each step in a reaction mechanism is called an elementary reaction.

An elementary reaction is any reaction that proceeds in a single step. Mechanisms consist of sums of elementary reactions.

Going back to rate laws, each elementary reaction must have a corresponding rate constant

*** For elementary reactions only, you can assume that the rate law depends directly on the number of species present. Thus, the rate law of an elementary reaction follows the stoichiometry.k1k2rate of overall reactionintermediate (transition step)Reaction Mechanisms Involving a Fast Equilibrium StepLets consider reaction mechanisms with a fast equilibrium step.

To date, we have discussed reactions occurring in only one direction. However, in many cases, the forward and back reactions are both important. For example:k1k-1When a reaction is in equilibrium, like the one shown above, the forward and back rates are equal, so there is no change in the amount of reactant or product once equilibrium is established.For many reaction mechanisms, a rapid equilibrium step is involved. Lets look at the following reaction:

This reaction goes through two elementary steps:

Overall reaction rate would be written as:

[N2O2] is an intermediate. Intermediates are never included in the overall rate law because they are too short-lived, so their concentrations cant easily be measured.k1k-1k2Reaction Mechanisms Involving a Fast Equilibrium StepMathematical Substitution For [Intermediate]k1k-1k2Since we have an equilibrium in step (1), the forward and back rates are equal

Solving for [N2O2], we get:

Substitute this term into our rate law expression:may be written in the book as simply KExamplesWrite the rate law for the following reaction given the elementary reaction steps of the mechanism:k1k2Write the rate law for the following reaction given the elementary reaction steps of the mechanism. Substitute any intermediates.k1k-1k218