Chemical Kinetics: Factors Affecting Reaction Rates and Rate Calculations

Chapter 14 Chemical Kinetics *concerned with speed or rates of chemical reactions reaction rate- the speed at which a chemical reaction occurs reaction mechanism- step-by-step pathway from reactants to products

Factors That Affect Reaction Rates 1. Physical state of the reactants -if reactants are in different phases, the reaction is limited by the area of contact *inc surface area of solid to inc rate 2. Reactant concentrations *inc conc = inc collisions = inc rate 3. Reaction temp -inc temp = inc KE/collisions = inc rate 4. Presence of a catalyst *inc rate without being used up

rate of appearance of product = [product] time [ ] = concentration rate units = M/s *conc of reactants dec with time while the conc of product inc *rate of disappearance of reactants = rate of appearance of product rate of disappearance = - [reactant] time

Sample Exercise 14.1 page 560 Using data in figure 14.3 pg 559, calculate the rate at which A disappears over the time interval from 20s to 40s. rate of disappearance = - [reactant] time rate = -(0.30M-0.54M) (40s-20s) = 0.012M/s

Practice Exercise rate of product = (0.70M - 0M) (40s - 0s) = 0.018M/s

-it is typical for rates to dec over time WHY? -the conc of reactants dec page 561 Table 14.1 and Fig 14.4

instantaneous rate- rate at a particular instant during a reaction -determined from the slope of the curve at a particular point in time -tangent lines are drawn to find the rate initial rate- instantaneous rate at t=0 Page 561 figure 14.4 *Problems page 562

Rate = - [reactant] = [product] t t aA + bB cC + dD -a,b,c,d are the coefficients from the balanced equation rate = -1 [A] = -1 [B] = +1 [C] = +1 [D] a t b t c t d t

Rate Law -relationship between the rate of the reaction and the conc of the reactants -consider this reaction: aA + bB cC + dD -the rate law would be: rate = k[A]m[B]n k = rate constant m and n = reaction order (small whole numbers) *the value of m and n determines how the rate depends on the conc of the reactant- can only be found experimentally

-to find reaction order, look at two values of a reactant that do not change -see what happens to the rate when the other reactant is changed *if rate change is same as conc change, it is first order (21, 31, 41) ex: conc doubles, rate doubles *if rate change is the square of conc change, it is second order (22, 32, 42) ex: conc doubles, rate quadruples *if rate does not change as the conc changes, it is zero order -do this for each reactant

Example: look at page 563 Table 14.2 rate = k[NH4+]m[NO2-]n *find order of each reactant *m and n are both first order based on data rate = k[NH4+]1[NO2-]1 -solve for k -using experiment #1 data 5.4x10-7M/s = k(0.0100M)1(0.200M)1 k = 2.7x10-4/M s

*can use rate law and k to calculate rate for any set on conc Ex: -find rate of same reaction if each conc is 0.100M rate = 2.7x10-4/M s(0.100M)1(0.100M)1 = 2.7x10-6M/s

-a larger value of k means a fast reaction -a smaller value of k means a slow reaction -units of k change depending on overall order of reaction zero order = Ms-1 1storder = s-1 2ndorder = M-1s-1 -overall order of reaction found by adding together orders of each reactant

First Order Reactions -rate is directly proportional to a single reactant Integrated Rate Law -relationship between intial conc of reactant and conc at any other time ln[A]t = -kt + ln[A]0 y = mx + b *has form of general equation for a straight line *if when t vs. ln[A]tis graphed and a straight line is produced then it is first order

Second-Order Reactions -rate is proportional to the square of the [A] or to reactants each raised to first power -more sensitive to conc of reactants than first order Integrated Rate Law 1 [A]t= kt + [A]0 [A]t= (kt +1/[A]0) *if when t vs. produced then it is second order 1 1 1 [A]tis graphed and a straight line is

Zero-Order Reactions -rate of reaction is independent of the [A] -rate is the same at any concentration Integrated Rate Law [A]t= -kt + [A]0 *if when t vs. [A]tis graphed and a straight line is produced then it is zero order

Half-life -time required for the conc of a reactant to fall to one half of its initial value Zero-Order t1/2 = [A]0 2? First-Order t1/2= 0.693 ? Second-Order 1 t1/2= ?[A]0

Temperature and Rate -as temp inc rate inc Ex- Glow sticks collision model- molecules must collide in order to react -the greater the # of collisions/sec the greater the reaction rate -inc temp and conc of reactants inc the collisions which inc the rate

-for a reaction to occur more is required than just collisions -molecules must be oriented in a certain way to make sure they are positioned to form new bonds -page 576 figure 14.15

-molecules must also possess a certain amount of energy to react -this energy comes from collisions -KE is used to stretch, bend and break bonds activation energy/energy barrier- minimum amount of energy required to initiate a chemical reaction (Ea) -as Eainc reaction rate dec

activated complex/transition state- a high energy intermediate step between reactants and products -page 577 figure 14.17 -for reverse reaction Ea= E + Ea *must change sign on E because going from right -Eawill dec with inc temp -page 578 figure 14.18 -the fraction of molecules that have an energy Ea is given by: f = e-Ea /RT

Arrhenius found that most reaction data obeyed an equation based on: a) the fraction of molecules possessing energy Eaor greater b) # of collisions/sec c) the fraction of collisions that have the appropriate orientation Arrhenius Equation: k = Ae-Ea /RT A= frequency factor related to frequency of collisions and the probability that the collisions are favorably oriented

Sample Exercise 14.10 page 579 Rank slowest to fastest 2 < 3 < 1 Ea= #2= 25kJ/mol #3= 20kJ/mol #1= 15kJ/mol Practice Exercise Reverse from slowest to fastest 2 < 1 < 3 Ea= E + Ea*change sign on E b/c reversing #2= 40kJ/mol #1= 25kJ/mol #3= 15kJ/mol

Reaction Mechanisms -the steps by which a reaction occurs Elementary Reactions -when a reaction occurs in a single event or step molecularity- defined by the # of reactants in an elementary reaction unimolecular- single reactant is involved bimolecular- collision of two reactants termolecular- collision between three molecules (not as probable as uni or bi)

Multistep Mechanisms -sequence of elementary equations ex- page 582 -the elementary reactions in a multistep mechanism must always add to give equation of overall process intermediate- not a reactant or a product, formed in one elementary reaction and consumed in the next

Sample Problem 14.12 page 582 a) Molecularity = uni and bi b) Write equation for overall reaction 2O3(g) 3O2(g) c) Identify intermediates O(g)

Practice Exercise a) yes it is consistent b/c both equations add to give overall reaction b) Molecularity = uni and bi c) Intermediates = Mo(CO)5

Rate Laws for Elementary Reactions -rate law is based on molecularity Table 14.3 page 584 Sample Problem 14.13 page 584

Multistep Mechanisms -each step in a mechanism has its own rate constant and Ea -often one step is much slower than the others -the overall rate cannot exceed the rate of the slowest elementary step rate determining step/rate limiting step- limits overall reaction rate -determines the rate law

Mechanisms with slow initial step Page 585 -since step 1 is slow it is rate-determining step -rate law equals rate law of step 1 rate = k1[NO2]2

Mechanisms with fast initial step Page 587 -step 1 has forward and reverse (k-1) reaction -step 2 is slow and determines overall rate rate = k2[NOBr2][NO] -NOBr2is an intermediate (unstable, low conc.) -rate law depends on unknown conc. -not desirable -want to express rate law in terms of reactants and products