reaction rates rate laws reaction mechanisms catalysts AP Unit 8: Kinetics reaction rates rate laws reaction mechanisms catalysts

introduction in thermodynamics, we looked at a reaction such as: 2 C8H18 + 25 O2  16 CO2 + 18 H2O we calculated how much energy would be released IF the reaction happened but we never asked how fast (or how completely) the reaction would happen. with the above reaction, for instance, does a puddle of gasoline usually react with atmospheric oxygen quickly? NO!!

the key event in reactions Collisions!!! Reacting molecules must collide Molecules must be properly aligned Molecules must meet with enough energy to break the existing bonds Reaction rate depends upon: how frequently reactant molecules collide what fraction of the collisions are effective (i.e. have proper energy and alignment)

frequency of effective collisions is increased by increasing temperature larger fraction of molecules have sufficient KE (to provide activation E) at higher T increasing solution concentration molecules collide more often at higher C increasing gas pressure higher P (or smaller V) is gas equivalent of increased solution concentration increasing surface area of solid (A:V ratio) dividing solid into smaller pieces increases fraction of surface molecules, allowing them to be struck more often adding a catalyst

reaction rates Is “rate” the same thing as “time”? Does “6 hours” tell you how fast a car moves? No!! RATE involves “something”  time for a car: miles/hour, meters/sec, etc for a reaction: grams/second moles/hour etc....

reaction rates which reaction is faster? 5000 molecules 3000 molecules 50 seconds 20 seconds 5000 molecules in 50 seconds 3000 molecules in 20 seconds can’t be answered! can’t be answered! = 100 molec/s = 150 molec/s faster→

NH4+ + NO2–  N2 (g) + 2 H2O (l) rate = k [NH4+]x [NO2–]y find x of N2 production 7 0.500 0.250 ??? rate = k [NH4+]x [NO2–]y find x find initial rates of H2O production (1-7) find y find k (with units) find initial rates of NH4+ (and NO2–) consumption find rate7

3 A + 2 B  C + 4 D –15.0×10-4 +8.0×10-7

2 NO (g) + 2 H2 (g)  N2 (g) + 2 H2O (g) of N2 production 4 0.33 0.50 ??? rate = k [NO]x [H2]y find x find rates of H2O production (1-4) find y find k (with units) find rates of NO (and H2) consumption (1-4) find rate4

change in concentration (and reaction rate) over time reaction rate often is a function of the concentration of reactant(s) rate therefore decreases as reactants are consumed

change in concentration (and reaction rate) over time just how fast the rate decreases (if it does) is not always obvious by simple inspection of graph or data

change in concentration (and reaction rate) over time we will need a reliable strategy to identify the degree of rate change over time

integrated rate laws zero order: first order: second order: additional written work was done in class to make this complete integrated rate laws zero order: first order: second order:

Arrhenius equation the experimental rate constant, k, has three underlying contributors A Ea T additional written work was done in class to make this complete

activation energy * previously we examined H now we will examine the energy “hump” energy must be added to break reactants’ bonds this added energy is called the activation energy, EA EA provided by the kinetic energy of colliding molecules * represents the activated complex * reactants products time potential energy EA H

* the activated complex also called the transition state is a weird intermediate “molecule” not a normal reactant or product molecule very unstable has high energy (reactant PE + collision KE) very short-lived probably has an “illegal” Lewis structure exists (very briefly) after reactant molecules collide and before they separate into product molecules

catalysts catalysts decrease the activation energy with lower activation energy, more collisions succeed collisions succeed more often even without T increase reaction rate thus increases catalysts cause NO change in a reaction’s ΔH! without catalyst reactants products time potential energy with catalyst

catalyst calculations without catalyst catalyst decreases the activation energy: from by to reactants products time potential energy 100 kJ 300– 260– with catalyst 200– 40 kJ 75– 60 kJ ΔH remains unchanged at –125 kJ

reaction mechanism is a model of what happens to atoms and electrons (bonds) step-by-step as reactant molecules collide an activated complex forms product molecules are released from the activated complex example: CH4 + 2 O2  CO2 + 2 H2O

reaction mechanism

reaction mechanism

reaction mechanism

reaction mechanism

reaction mechanism (as reactions rather than animations) CH4 + O2  “CH4O2” step 1: “CH4O2” + O2  CO2 + 2 H2O step 2: net: CH4 + O2 + “CH4O2” + O2  “CH4O2” + CO2 + 2 H2O CH4 + 2 O2  CO2 + 2 H2O

relationship between reaction mechanism and rate law the stoichiometric coefficients in the rate-limiting elementary step of the reaction mechanism exactly match the respective exponents in the rate law

relationship between reaction mechanism and rate law CH4 + O2  “CH4O2” step 1: slow fast  “CH4O2” + O2  CO2 + 2 H2O step 2: fast slow option 1: step 1 is slow option 2: step 2 is slow rate = k1[CH4]1[O2]1 rate = k2[CH4O2]1[O2]1 But CH4O2 is not a valid reactant molecule. Since step 1 is fast, equilibrium is established: option 3: some other mechanism rate1,for = rate1,rev k1[CH4][O2] = k -1[CH4O2] O2+O2... step 1: k1[CH4][O2] [CH4O2] = k -1 rate = k [O2]2 k1[CH4][O2] rate=k2 [O2] =kx[CH4][O2]2 k -1

catalysts catalysts cancel in overall reaction stoichiometry catalyst ultimately is not consumed in a reaction it enters as a reactant it is a product in a later step it is thus recycled in the next reaction cycle one catalyst atom/molecule therefore can catalyze GAZILLIONS of cycles of the reaction since the catalyst is regenerated during each reaction cycle, ready to be used again

catalyst example ozone is destroyed in the upper atmosphere MUCH faster because Cl atoms (from CFCs) catalyze the reaction Cl + O3  ClO + O2 (step 1) ClO + O  Cl + O2 (step 2) Cl+O3+ClO+O  ClO+O2+Cl+O2 O3 + O  2 O2

catalysts vs. intermediates both catalysts and intermediates cancel in net reaction intermediate (ClO): generated in an earlier step, consumed in a later step catalyst (Cl): consumed in an earlier step, released (regenerated) in a later step

homework example forward reverse H EA EA = 25 kJ H=–80 kJ EA,rev = reactants products time potential energy (kJ) H=–80kJ EA=25 kJ reactants products time potential energy (kJ) H EA 125– 125– 100– 100– 20– 20– forward EA = 25 kJ H=–80 kJ reverse EA,rev = H= 105 kJ +80 kJ

homework example forward reverse H= EA= EA = H= EA,rev = Hrev= 195– reactants products time potential energy (kJ) H= EA= 195– 150– 30– forward EA = H= reverse EA,rev = Hrev=

homework example forward reverse H EA EA = H= EA,rev = Hrev= 250– reactants products time potential energy (kJ) H EA 250– 220– 70– forward EA = H= reverse EA,rev = Hrev=

reverse reactions: easy or difficult? easier more difficult reactants products time potential energy H EA reactants products time potential energy H EA forward exothermic EA relatively small reverse endothermic EA quite large but vice versa for endothermic forward reaction: forward reaction more difficult; reverse reaction easier

reverse reactions theoretically, any reaction can be reversed, i.e. the products can be turned back into the reactants practically speaking, however, this is very difficult, especially for exothermic reactions, because the reverse reaction is endothermic (which is not spontaneous) the activation energy is very large for the reverse reaction