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 [ ] = concentrationrate 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 of reactants = -∆ [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 ∆td ∆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 the rate change is the same as the conc. change, it is first order (2 1, 3 1, 4 1 ) ex: conc. doubles, rate doubles *if the rate change is the square of the conc. change, it is second order (2 2, 3 2, 4 2 ) ex: conc. doubles, rate quadruples *if the rate change does not change as the conc. changes, it is zero order -do the same holding each reactant constant

Example: look at page 563 Table 14.2 rate = k[A] m [B] n *find order of each reactant 5.4x10 -7 M/s = k [NH 4 + ] m [NO 2 - ] n *m and n are both first order based on data = k(0.0100M)(0.200M) k = 2.7x10 -4 /M∙s

*can use rate law and k to calculate rate for any set of conc. ex: rate = 2.7x10 -4 /M∙s(0.100M)(0.100M) = 2.7x10 -6 M/s

-a large value of k (≥ 10 9 ) means a fast reaction -a small value of k (10 or lower) means a slow reaction -units of k change depending on overall order of reaction zero order = Ms -1 1 st order = s -1 2 nd order = M -1 s -1 -overall order of reaction found by adding together orders of each reactant

First Order Reactions -rate is directly proportional to a single reactant rate = -∆ [A] = k[A] ∆t Integrated Rate Law -relationship between initial 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] t is 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 rate = -∆ [A] = k[A] 2 ∆t -more sensitive to conc. of the reactants than first order Integrated Rate Law 1/[A] t = kt + 1/[A] 0 [A] t = 1/ (kt + 1/[A] 0 ) *if when t vs. 1/[A] t is graphed and a straight line is produced then it is second order

Zero-Order Reactions -rate of reaction is independent of the [A] rate = -∆ [A] = k ∆t -rate is the same at any concentration Integrated Rate Law [A] t = -kt + [A] 0 *if when t vs. [A] t is 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 t 1/2 = [A] 0 /2k First-Order t 1/2 = 0.693/k Second-Order t 1/2 = 1/ k[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 (E a ) *as E a inc. reaction rate dec.

activated complex/transition state- a high energy intermediate step between reactants and products -page 577 figure for reverse reaction E a = ∆E + E a *must change sign on ∆E because going from right -E a will dec. with inc. temp -page 578 figure the fraction of molecules that have an energy ≥ E a 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 E a or greater b) # of collisions per second 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 page 579 Rank slowest to fastest 2 < 3 < 1 Ea=Ea= #2= 25kJ/mol #3= 20kJ/mol #1= 15kJ/mol Practice Exercise Reverse from slowest to fastest 2 < 1 < 3 E a = ∆E + E a *change sign on E a 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 page 582 a)molecularity = uni and bi b) equation for overall reaction 2O 3 (g)  3O 2 (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 for elementary reactions *table 14.3 page 584 Sample Problem page 584

Multistep Mechanisms -each step in a mechanism has its own rate constant and E a -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 *pg 585 -since step 1 is slow it is rate-determining step -rate of overall reaction depends on rate of step 1 -rate law equals rate law of step 1 (k 1 ) -step 1 is bimolecular so using page 584 it is second order rate = k 1 [NO 2 ] 2

Mechanisms with fast initial step *pg 586 and 587 -step 1 has forward (k 1 ) and reverse (k -1 ) reaction -step 2 is slow and determines overall rate rate = k 2 [NOBr 2 ][NO] -NOBr 2 is an intermediate (unstable, low conc.) -rate law depends on unknown conc. -not desirable -want to express rate law in terms of reactants and products