Chem Unit 10- Kinetics
if a catalyst is added, increasing temp has
Ea is lowered bc a lower energy transition state is formed - less energy required for molecules to react when collide, so K increases no effect on Ea- rather changes amount of energy particles have - increasing temp just means more particles with energy greater than Ea so more successful and increase rate
unimolecular bimolecular termolecular
elementary step: A—-> products, rate law: Rate = k[A] elementary step: A + A —-> Products or A + B —-> Products Rate law: Rate = K [A]2 or K[A][B] elementary step: A + A + B —-> Products A + A + A ——> Products A + B + C —-> Products Rate law: Rate= K[A]2[B], K[A]3, K[A][B][C]
Mathematically the rate constant k is explained as there is connection between basically where A= we can make rate constant by making temp in uncstalyzed pathway any time reduce Ea Ea is wherever you ___ to the A =
(Arrhenius equation) k= Ae ^ -Ea/RT rate constant of a reaction and temp- for any reaction rate is proportional to K[A]m[B]n, but if K increases, rate increases- increase rate constant K, rate goes up- temperature is in denominator of a negative exponent- in terms of exponent, activation energy is ALWAYS positive- R, gas constant in this case 8.31 J/mol•l (always positive), a temp in K always positive- given all of that- that exponent always negative- negative and three positive values- always negative- temp in denominator of negative exponent, having temp go up makes negative exponent smaller, which makes K value larger- so increase in temp always corresponds to increase in K value which increase rate. make temp larger, denominator is bigger, exponent gets smaller, rate constant increase larger, or rate constant bigger by making Ea smaller- that's what you do when add catalyst start to peak of leakiest peak normal Ea- but in catalyzed pathway, changes mechanism- when catalyze go thru diff pathway- no difference in delta H of catalyzed vs non catalyze reaction- still start and end in same spot- but now have curvy humps) whats different is that has lower Ea energy - reducing Ea which makes k value larger which makes rate faster- in go thru catalysis, diff pathway that requires less Ea and increase K value correspond to increase in K value and increase in rate frequency factor- product of collision frequency z and an orientation probability factor p
If the rate quadruples when [NO2-] doubles, y is proportional to ____- if that was the case, that would tell me In RDS for above, theres ____- what if doubled NO2 while holding ammonium constant and rate doesn't change ex exponent tells you RDS that involves one molecule to react meaningfully have to faster reaction when increase____, because
(ammonium held constant), if rate quadrupled, y is proportional to x^2, if that was case, that would tell me in RDS there were 2 NO2s involved. (cuz if have 2 NO2s and double that would make four aka quadruple)- [NO2-]^2 so exponentwould be 2- that would be second order- in RDS there's two of them (NO2-)(doubling concrenteation aka having two of them quadruples rate would be second order) (two first order reactants) in RDS there's two of them (NO2-)(doubling concrenteation aka having two of them quadruples rate would be second order) (two first order reactants) effecting concentration of that reactant didnnot affect rate, so that reactant is not in RDS - 0th order- any concentration if put to 0th power is one, so factors out of rate law rate= k[C4H9Cl] [H20]0 - in slow step, one particle of cl stuff, there's zero next to water which means in RDS, there are zero waters so take it out (only one particle in RDS, there is no collision- so RDS with only one particle- that's decomposition) order unimolecular collide and collide w proper orientation and energy temp bc increase speed particles and frequency and energy of collisions so can pass activation energy and react more quickly
For a first order reaction, in the RDS, there is a rate constant always unit for rate
collision between 1 particle and 1 particles k- rate constant true for a given reaction at a given T- vary on circumstance mol/L•S or M/S (bc M is mol/l)
thius for the ex above, rate= concentration of reactant will concentration of product will how quickly reactant is decreasing is related to if coefficients not all one and had C4H9Cl + H20—> 2C4H9OH + Hcl and imagine u made 26 of those
-delta [C4H9Cl] / delta t = delta [C4H9OH]/ delta t (relating rates) decrease over course of reaction bc it's a reactant- increase over course of reaction bc it's a product to how quickly product increasing- so if 1:1 ratio, lose one reactant gain one product if made 26 of those, that would involve losing 13 of C4H9Cl- every time i lose one of C4H9Cl, i make 2 of C4H9OH.
Rate of reaction is equal to so then for B rate of reaction for reactant is Product is ___ and reactant is ____ in 1 to 1, if lose 20 reactants At time t1, concentration A1, at time t2, concentration A2 SO OVERALL equation
-delta[A]/ delta t (units of mol/L•s) rate = delta[B]/ delta t if following product negative of change in concentration of A over change in time increasing and reactant decreasing energy gain 20 products A2<A1 -delta [reactant]/ delta T = delta [product]/delta t
second order reaction
-rate is proportional to either the concentrations of 2 reactants or to the square of the concentrations of a single reactant (second order in only one reactant, first order in two reactants) (so can be overall second order if two first orders, or second order when doubling concentration of reactant quadrupled rate, indicating 2 of those) *Rate = k[A]1[B]1 or rate = k [A]2 or rate = k [B]2
Given the following equation N2O5 —> N2 + 5/2 O2 and the following plots ln[N2O5] vs time straight line down 1/ [N2O5] vs time exponential up [N2O5] vs time curved line down Determine order of reaction wrt [N2O5] Reaction half life- all a half life is is Half life- half life expression depends on Because [A] vs t1/2 is It is necessary to be able to derive only half life expression on equation sheet is whatever initial concentration was, at half life, First order plot-
1st order, bc it's linear when plot ln[N2O5] vs time how long it takes to reach half the initial reactant concentration- cut in half. time required to reach half the initial reactant concentration overall order of reaction one half of the original [A], [A]t1/2= .5 [A]o (half of initial) the half life expressions from the integrated rate laws 1st order -.963/k (aka ln2/k) that concentration now half decreasing l shape- notice the general change of the concentration bs time. the half life is constant.. t1/2= ln2/k (concentration vs time for first order isn't linear- ln concentration vs time would be straight line bc this process is first order)
consider this reaction: 2Hi——>?H2 + I2 __:__:___, so rate of reaction = In general, for a reaction, (equation with little and big letters) rate equation: take coefficient and if ratio 1:3, every time lose 5 reactants, reactants are ___, products are ___ whatever coefficient is acts as Express rate in terms of changes in concentrations with time for each substance- 4NO + O2 —-> 2N2O3 how fast is [O2] decreasing when [NO] is decreasing at 1.60 x 10-4 mol/L•S?
2:1:1; so rate of reaction = -1/2 delta[HI]/ delta t = delta[I2]/ delta t = delta [H2]/ delta t ;; 1/2 because it's decreasing twice as fast as products r increasing- every time make 5 of H2, it involves losing 10 of HI aA + bB —-> cC + dD (little letters r coefficient. and big letters r reactants and products) -1/a delta [A]/ delta t = -1/b delta [B]/ delta t =-1/c delta [C]/ delta t = 1/d delta [D]/ delta t drop into denominator (coefficient 3 would be 1/3) make 15 products (if it's coefficient is 3) -, products + denominators rate= -1/4 delta [NO] / delta t = -delta [O2]/ delta t = 1/2 delta [N2O3]/ delta t 4.00 x 10^-5 mol/l•s
Consider the mechanism for the breakdown of ozone in the stratosphere. 2O3 —-> 3 O2 to be a reaction mechanism, important part is
A two step mechanism has been proposed for this reaction- The first step is a unimolecular reaction O3 —-> O2 + O second step is a bimolecular reaction O3 + O —> 2O2 looking at those two steps, rxn intermediate is O- bc it is created in early step and consumed in later step (as product then as reactant) also O is a diradical- 2 individual electrons unpaired- unstable - doesn't show up in overall net rxn- add up rest 2O3—> 3O2 the two elementary steps add tg to give overall process- important requirement in the construction of a reaction mechanism -must add up to overall net
integrated rate law 0th order integrated rate law 1st order integrated rate law second order integrated rate lawsare in format y vs x: in 0th order, in 1rst order, in 2nd order, we can measure how slope is
Rate law :rate= k, integrated rate law = [A]t = -kt + [A]o, linear plot [A] vs t, slope of plot -k, half life (t1/2) —— [A]o/2k Rate law: rate= k[A], integrated rate law = ln[A]t= - kt + ln[A]o, linear plot ln[A] vs t, slope of plot -k, half life (t1/2) ——— ln2/k Rate law: rate= k[A]2, integrated rate law 1/[A]t= kt + 1/[A]o, linear plot 1/[A]t vs t, slope of plot +k, half life (t1/2) ——— 1/k[A]o y=mx + b if follow along how decreasing reactants in reaction, can plot 3 things and see which is linear- whichever is linear is one that is correct in 0th order, concentration of a reaction vs time - if you plot this and see straight line, tells you this reaction is 0th order if plot natural log concentration vs time, and that gives straight line- tells u first order bc that matches up w first order integrated rate law being a linear plot if plot 1/ [A]t vs time, and give straight line, that recess is second order wrt [A] things decrease over time if we can evaluate reactant and track along, plot it's change in concentration over time, can plot data as concentration vs time- if straight line already, that process is 0th order wrt that component, but if not straight, next step take all concentration value and plug in ln of those values, give second data set, so if take natural log of concentration vs time and plot that, if give you straight line, correspond to process first order wrt component- if still not straight line, take same set of original data and rather than plot in concentrations, plot 1 over each concentration and compare w time- if that gives straight line, process second order (take raw data and do concentration vs time, ln concentration vs time, 1/concentration vs time, whichever linear tells u order wrt that component rate constant (determine rate constant of from slope)
Consider the following reaction: 2A + B —-> E + F Reaction intermediates so overall this mechanism is slowest will determine The molecularity of a process tells how many
a reaction mechanism for this reaction might involve these three simple steps A + B —-> C C+ A —-> D D —-> E+F when add tg and look at overall total, see if any thing on both sides of arrow that cancel- get rid of C bc don't show up in overall net, can get rid of D, so C and D are reaction intermediates- high energy components- generated in early step, used in later step- can tell C is reaction intermediate bc cancelled out AND bc didn't have any C at v beginning- had to MAKE it in an early step and CONSUME it in a later step- there are circumstances where have catalysts present as well- imagine had G + A + B—-> C C + A—-> D D —-> E + F + G - g present both right and left so cancel, but had at beginning AND reproduced at end, so functioning as role of a catalyst- just bc cancel things out doesn't mean reaction intermediates- have to be created in early step and used in later step- if cancel things out but had at beginning and reproduce later, that's role of catalyst C and D, exists only for short period of time, highly energetic, do not appear in balanced equation (created early and used up later and get cancelled out) (ex's of intermediates are carbocations and free radicals bc highly energetic) 2A + B —> E + F 3 step process - one of those three is gonna be slower - whichever is slower determines rate of reaction rate law molecules are involved in the process
Graph of concentration vs time (look at it) can tell we're monitoring concentration of reactant when the graph is ____ the curved slope downward is straight line that goes from top to bottom of graph ALL WE CARE ABOUT IN KINETICS IS smaller line near decrease curve smaller the time window, so straightish line at bottom (tangent line) that aligns w curve is _____- back in time changes _____, forward in time changes ____, but right at this moments its the ____. most thing we're concerned about is
always plotting [x] over time (in mol/L) decreasing how concentration of reactant changes wrt time avg rate- over first 60 seconds, rate of change of concentration reactant is rep by line- meaningless- slope of line if connect two pts- Instantaneous rates- specifically instantaneous rates that occur st beginning of reaction avg rate over smaller time - avg rate first 10 secs more match up w curve- instantaneous rate of change- how is it changing right at that exact moment- back in time changing faster forward in time changing slower- but right at moment it's the instantaneous change instantaneous ratebof change right at beginning of reaction- line that slopes v downward- initial RATE-
only particles able to dissolve are ones that are To go into solution, solute must take alla seltzer that's ground up vs tablet surface area makes huge difference in ____, and you cant dissolve until _____, so grinding it up _____ temperature also _____. Hot vs cold water, ____ smal changes in temp
are touching water- contacting solvent be in contact with solvent ground up will react faster- bc surface area is way greater so more touching solvent how quickly something is exposed to solution, and can't dissolve until exposed to solution- grinding up expose more surface area which allows more points of contact for collisions, and if more points of contact have faster rate effects rate of reaction - if hot vs cold water, in hot water will happen more quickly bc hot water has higher temp and on average particles have higher KE and move at faster speeds so have more frequently and collisions of higher energy- so chances of effective collisions are greater and rate increases can dramatically change rate of process
Arrhenius As the temp increases, the fraction of collisions with Activation energy is increase temp more particles that have an energy greater than Ea = increasing temp doesn't always
colliding particles must have an energy greater than a threshold energy called the activation energy. energy > activation energy increases certain barrier we have to overcome, only molecules above that particular barrier can be successful- to the right of line have enough energy to react shift curve and flattens it out, so more particles sit to right of barrier so higher proportion of particles over barrier means more frequent collisions and more importantly collisions that will supersede Ea will occur more often, increasing rate increase in rate increase rate- situation would be if had certain process that had Ea barrier way further right than curve- rate = 0 bc no particles past energy barrier, still when increase temp and move to right, rate still 0 bc still hasn't reached Ea- eventually will- but you can have a rate of 0 and still after increase temp have rate of 0 if process has high enough Ea barrier - but as long as some particles on right of barrier, increase T will increase rate
——Mechanism with fast initial step—— THIS IS V IMPORTANT!!!!! If the slow step is not first in the mechanism, it acts as a product of fast initial step eventually, __ How to handle a fast initial step Consider following reaction: Net reaction: 2NO + O2 —-> 2No2 Experimental determined rate law is k[NO]2[O2] (could this reaction be one step?) Is the following proposed mechanism valid for the overall reaction? 1) NO + O2 <==> NO3 (fast, reversible) 2) NO3 + NO —-> 2NO2 (slow)
bottle neck— backup- builds up and begins to revert back into reactants (so have both forward and reverse reactions for step one - that means this process becomes an equilibrium process- useful bc can use equilibrium to cancel stuff out) equilibrium is reached, with the equilibrium becoming part of the overall rate law -write rate laws for both directions of the fast and slow step (slow step only worry about one direction) -show that the slow steps rate law is equal to overall rate law by determining intermediated (eliminating them bc not allowed in rate law) -substitute into the slow step rate law yes bc coefficients in original equation match rate law- propose this cuz could be detecting presence of NO3 and detect intermediate somewhere so must be multi bc intermediate- rate law matches molecularity of overall net so could be one step, but since it says it's not we know that probably detected something to imply its multi step 1) involves 3 or fewer molecules for each step so physically reasonable 2) adds to overall net 3) slow step rate law match experimental- rn doesn't but we have to consider that NO3 is intermediate so cant remain in rate law- get rid of- so don't know yet and have to use substitution- bc fast step reversible and have backup, have to say first and second step gonna match eachother relative to rate- rate of forward reaction and rate of reverse reaction are equal, so write rate law for forward and reverse of first step and for second step. rate1= k[NO][O2] rate -1 (reverse)= Rate = K[NO3] rate 2: rate = k[NO3][NO] recognize that we will use rate 1 and reverse rate 1 to élimante rate 2 bc don't know whether RDS matches overall rate law bc intermediate in it and it's not allowed to stay if at equilibrium, rate of step 1 = rate step -1 (fwd and reverse =) so can take rate 1 and set equal to -1. gonna have diff rate constants k1, k-1, k2 k1[NO][O2] = K-1[NO3] now try get NO3 by self so divide by K and get 1 so [NO3] = k1/k-1 [NO][O2] now, take it and substitue for NO3 in rate 2!!! so rate 2 = k2 k1/ k-1 [NO]2[O2] so just call overall k, so rate= k[NO]2[O2] which matches overall rate law and slow step so yay so VALID (possible) (could have few intermediated so be prepared)
Effect of temperature on rate constant in general as temp increases if double rate constant if increase temp endo Collision Theory/Transitional state theory- Both work in conjunction to They both utilize the idea of
change temp of reactant, change rate constant- exponentially related (up curve)- increase of 10 degrees celsius, height increase by double ... which roughly doubles rate so does the reaction rate- if have a reaction that has such high Ea barrier, rate of reaction can actually be 0- might increase temp but still not have any particles cross EA barrier- eA barrier is measure of how easy it is for particles to get over jump to go from reactants to products- if too high barrier, none can go to products and cross- if can't cross barrier reaction can't happen- even if increase temp, just high energy particle smacking into barrier- double rate- so reaction at 30 vs 40 degrees celsius, bc increase temp by 10 degrees celsius, roughly doubling rate constant, roughly doubling rate of reaction- happen twice as fast- so do process at 50 degrees warmer, doubled 5 times faster (32 x faster) increase products explain observed kinetic phenomenons an activation energy (Ea) to explain data, but approach it from a slightly different perspective
—-factors affecting reaction rates—- Concentration- particles must ______. For collision to be successful, you have to have ___ and ____ The speed of reaction is proportional to why need sufficient energy? if you don't have enough ___ or enough ___,
collide with eachother- molecule A must run into B to react have to have sufficient energy and have to occur in proper orientation- have to occur in certain manner collision frequency which is proportional to concentration we need to make sure we can exceed activation energy barrier and make sure we can get our reactants over the transition state of a reaction collisions or energy, don't collide
Consider the reaction below O2 + 2NO —> 2NO2, rate law is k[O2]m[NO]n 1: Init. [O2] 1.10x10^-2 Init. [NO] 1.30 x10^-2 Initial rate of No2 production: 3.21 x10^-3 2: Init. [O2] 2.20 x10^-2 Init. [NO] 1.30 x10^-2 Initial rate of No2 production: 6.40 x10^-3 3: Init. [O2] 1.10 x10^-2 Init. [NO] 2.60 x10^-2 Initial rate of No2 production: 12.8 x10^-3 If look at 1 and 2, if look at 1 and 3, So rate order is with math... on test explain
concentration oxygen doubles, [No] held same, rate doubles, so tells you first order wrt [O2] [O2] same, doubling [NO] causes rate to quadruple, rate is proportional to [NO]^2, so indication 2nd order wrt [NO] k[O2][NO]2 -process could be elementary (all one step- reactants to products) (if triple causes triple still first order, if triple causes nontupling (3^2) so second order, if triple causes no change, 0th) do faster over shower- rate 2/rate 1= k[O2]m[NO]n / k [O2]m[NO]n (in trial one and two, NO held constant so can get rid of) and k get rid of bc reaction occur at same temp so rate 2/ rate 1= [O2]m/ [O2]m so [6.4 x10^-3 M/s/ 3.21 x 10^-3 M/S = (2.20 x 10^-2 M/ 1.10 x 10^-2 M) ^m 2= 2 ^m so only 1 rate 3/rate 1= K [O2]m[No]n/K[O2]m/[NO]n, get rid of k and o2 bc o2 held constant so rate 3/ rate 1 = [No]n/ [NO]n 12.8 x 10^-3 M/S/ 3.21 x 10^-3 M/S = (2.6 x 10^-2 M/ 1.30 x 10^-2 M ) ^n or 4= 2^m = second order comparing trials one and two, when double concentration of O2 while holding concentration of NO constant, rate of reaction doubled- because doubling of councentratiok of O2 cause rate to double, i know it's first order relative to O2- comparing experiments 3 and 1, held O2 constant, doubles NO, rate quadruples, bc doubling of one component cause reaction to quadruple, know process is second order relative to NO
Factors affecting rates
concentration, physical state, surface area of reactant, temperature
high concentration goes ___ over time graph of t on x axis and reactant on y graph of t on x and product on y as T increase, you always ____ because particles are
down will decrease till level off and reach equilibrium ( L decay shape) exponentially increase till level off at equilibrium- always increase rate of reaction whether making more products or not- particles moving faster , more frequent collisions w higher energy, chance of efficiency of reaction increases
Ea is if look at energy diagram and starts high and ends low, if increase barrier, Activation energy is Collision theory bimolecular vs termolecular What it's saying is that any time you have a ___ or ____ termolecular réactions rare bc postulates that majority of collisions only those in which have ___ will have successful collisions if EITHER or BOTH of the two pre requisites above
energy needed to proceed from reactants to products process is exo- delta H is diff btwn final minus initial- negative —/ Ea is diff btwn where we start and peak of the hill barrier towards reaction occur- it's gotta be cleared to finish over there less particles can go from reactant to products, decrease reaction rate based on kinetic theory and assumes a collision between reactants before a reaction can take place- assumption of a collision is for any bimolecualr or termolecular process- involving 2 or 3 particles- if have unimolecular step, it only involves one molecule so no collision needed- bimolecular or termolecular step, where 2 or 3 particles collide, will involve a collision- if you go beyond 3 particles, have steps thy are physically unreasonable- bc 4 things colliding at same time w proper orientation and enough energy too complicated majority of collisions do not lead to a reaction- bc if two particles collide but don't have enough energy, not gonna lead to productive collision- if two particles collide and don't have proper orientation, will be lack of successful collision- a KE greater than a certain minimum, called activation energy (Ea) and have to have proper orientation wrt eachother (correct reaction geometry and sufficient Ke) are not met, then then collision does not lead to the formation of products and the reactants simply bounce off each-other
even tho not look at linear graph, can tell first order bc this is very big first order integrated rate law / how to do half life problems [A]t is rlly second order plot DO QUESTION any time have radioactive decay at some point
for first order process, the half life is a consistent increment of time- if took hour to cut in half, to cut in half again takes another hour.:.. only true for first order plot- so looking at raw data u can tell clue to telling order!!! (unique to first order if half life consistent) ln[A]t= -kt + ln[A]o , since we talking bout half life, ln[A]t is half of what it was originally, so ln[A]t can be represented by 1/2[A]o, so want to get time by self, so first take kt= ln[A]o -ln[A]t and divide by k....so t1/2 = ln[A]o - ln[A]t /k ... can make it ln [A]o/[A]t /k so we know ln ([A]o/ 1/2[A]o) / k ...cancel out... ln 1/1/2 = ln2/k (don't need that) 1/2 [A]o can tell second order bc if took hour to cut in half, to cut in half again takes two hours... cut it again takes four hours... eight hours.. length of half life doubles each time —Notice the steep change of the concentration vs time at the beginning. the plot flattens at the end. the half life lengthens as reaction proceeds - t1/2= 1/k[A]o- identity secod order decay bc doubling half life length each time as continue equation for half life first order t1/2= ln2/k....inverse hours .. hr^-1 = .0529 hr^-1 38.5% complete says we've made 38.5 % product... but we concerned w react not product .. A —->. B I 1000. 0 C -385. +385 E. 616. 385 so [A]t = .615 [A]o (bc we have .385 of products so have .615 of reactants left so .615 times initial) .. so k = 1.01 x 10 ^-3 s^-1 c.) 684 s for half life... so for 95 percent complete.;. initially had 1000 and none... to be 95 percent complete, have 950 bs and 50 As... have .05 of what had originally... bc 95 % of complete means .05 of whatbhad orignal ... so t= ln ([A]o / [A]t) /k = ln ([A]o/ .05 [A]o) /k = ln 1/.05 /k = 2970 seconds always first order bc just emitting particles doesn't require collision if reaction proportional to concentration of reactants, if keep reduce concentration of reactants keep reducing rate
Collision theory transition state theory Transition state theory describes an it envisions the the activated complex forms only if have to achieve the activation energy is proposes that ever reaction
for particles to react have to collide w eneough energy that super cèdes energy barrier- if collide w enough energy and proper orientation then can have successful collision- but most collisions are unsuccessful- interact w wrong orientation or have collision below Ea whenever have reaction, there is in inbetween state between reactants and products, called either activated complex, or transition state activated complex, or transition state of a reaction kinetic energy of the particles changing to potential energy during a collision the energy > Ea, and the orientation of the collision is correct - (only forms of energy of a collision is greater than Ea) if gonna clear peak of hill that's where transition state is achieve transition state to go from reactants to products is used to stretch and deform bonds in order to reach the transition state- used to make awkward inverween reactants and products where have bonds that are partially broken and partially formed - transition state has new bonds forming in half formed stage and old bonds breaking in half broken stage- longer connections btwn atoms- in between what it would like to be reactants or products- if reach transition state don't know if always gonna proceed forward- goes through it's own transition state, from which the reaction can go in either direction (all reactions are reversible) once reach transition state, can proceed in either direction- go blue to purple or blue back to blue ex
Kinetics is the study of it is the study of changes in One way to investigate how quickly a reaction takes place: or, you can monitor Rates can go in a wide range of rates:
reaction rates -(speeds of reaction) in concentrations of reactants or products as a function of time monitor how quickly reactants go down monitor in reaction how quickly the products increase explosions (fast and energetic) rusting (slow) aging (v slow)
REACTION INTERMEDIATES ARE ... -formed as a ____.. - don't -aren't allowed to -usually For a reaction mechanism to work, the following must be true: if all above are met,
formed as a product in one step and become a reactant in a following step -appear in the balanced equation (not allowed) (cancelled out) -to remain in a rate law (must be eliminated) (situation where would be in rate law must eliminate them by substitution) -unstable vs reactants and products thus fleeting (radicals, carbocations) (overall exo hill ends lower) if look at energy diagram (see pic) what's present in middle are intermediates- valley of graph- higher than both reactants and products energy- not gonna sit around for length amount of time- bc can proceed fwd or backwards to make themselves more stable, peaks are transition states- if two humps, two steps — highest eA step is RDS!! cuz have to proceed they that barrier to proceed from reactants to products!! elementary steps must add to the overall balanced equation the elementary steps must be physically reasonable (no more than termolecular) the slow step rate law must match the experimental rate law proposed mechanism is valid (could work) for an overall process (if mechanism is valid means POSSIBLE)
——Look at potential energy curve- transition state theory——- For every reaction there exists a___ and ___ with the difference between the two activation energies is the The size of the Ea will usually give information on reactants co and No2 products co2 and No both forward and reverse reactions have
forward (Ea,f) and reverse activation energy (Ea,r) with the values rarely ever equal heat of the reaction (delta H rxn) which direction the reaction will favor (forward or reverse) transition state has connection between two components that were separated at beginning- bond partially formed and one in process of being broken- LINE drawn longer indicate bonds in process of being broken- carbon oxygen bond longer too bc longer bond- once get to that stage, can either roll to right and finish or roll left and go back to reactants- activation energies- the Ea of the forward reaction is wherever you start to peak of leakiest peak- Ea of reverse reaction is wherever it's starting to go backwards to peak do leakiest peak (so like if exo, Ea for reverse reaction is at bottom of hill to top) - if wanna know delta H, just relate Ea energies to eachother- take fwd Ea and subtract reverse Ea, give you delta H of rxn. forward and reverse Ea are almost never the same-(if are same, delta H 0- ones where have reactants and products that are identical)
increase temp doesn't always but, it always causes Representing rate of reaction equation (simple) Bc A is a reactant, over course of reaction one way to follow rate of reaction is by following the _____. It goes _____, so the delta A is gonna be so ___/___ allows us to relate the rate of ___ SO WHAT UR MEASURING IS Rate of reaction = [A] = why measure how quickly reactants decrease vs how fast products increase?
give you more products(exo) will have less ALWAYS causes reaction to happen faster- get to equilibrium more quickly (give u more or less deending on heat of solution) A (reactant) —-> B (product) amount of A will decrease and B will increase decrease of concentration of reactant- it goes down, so delta A is gonna be a negative change in concentration- so negative concentration of A over change in time allows us to relate rate of decrease of a reactant to rate of increase of product how does concentration of A change over a certain amount of time, and if take negative of that u get rate of reaction- always negative [ ] for reactant - delta concen. A/ delta T concentration reactant in mol/L ur gonna look at what's easiest to observe- something easy to watch occur - watch A or B, whatever is easier
two types of catalysts and what they are _____- exists in ____ with the _____-speeds up a issue w solid catalyst: ex of heterogenous ANOTHER EX acetone in copper catalyst
homogeneous- exists in solution with the reaction mixture - catalyst same phase as reactants (ex gas reactants and catalyst has) heterogeneous- speeds up a reaction that occurs in a separate phase—- diff phase between catalyst and reactants- usually means solid catalyst) ; most often a solid interacting with gaseous or liquid reactants, needs large surface area to be effective—- bc solid interact with liquids or gases, SA critical- fact that are high SA components important- high SA component where on surface things can take place- connecting pt issue w solid catalyst- solid catalyst can get clogged up- solid catalysts don't regenerate normally and have lifespan- become less effective over time- but still not reactant or product Metal-catalyzed hydrogenation or hydrocarbons (aka catalytic hydrogenation) our way of conversing alkenes and alkynes into alkanes - if had alkene and hydrogen gas and palladium or nickel or platinum- those metals often serve in catalytic roles- convert this into the alkane by having them interact w surface of catalyst- without catalyst v slow- orientation of appropriate collisions and likelihood of happening small- 2 molecules diff mass in same container have one v fast moving and one v smal moving- (if at same temp) so to have our effective collisions, work better to get H to sit still- so when H adsorbs to catalyst- H atoms bind to catalyst and not moving and can have meaningful connection in terms of how supposed to interact w eachother catalytic converter in car on bottom- burn fuel in car but many gases toxic so we have to convert into less harmful gases- catalytic metals that convert those gases into other things - spread out metal catalyst convert to ketene and methane (copper quick to heat up and quick to low down bc conduct heat and electricity v well but also low speficif heat)
slow step is what determines how quickly rate law can be determined from rate law of above an example ; Step 1: O3 + NO —-> O2 + NO2 Step 2: NO2—->NO + O Step 3: O + O3 —-> 2 O2
how quickly product produced molecularity of RDS k[No2]2, bc first step slow step and also it's second order reaction NO2 and O are reaction intermediates NO is not bc it is present at beginning and used later and is a catalyst (catalyst used (present) at beginning and reproduced later)
——Rate limiting steps of a reaction mechanism—— elementary reactions difffer in usually, one step is much slow step is the Rate law for the ___ represents the rate law for the ____ Rate limiting step is synonym for figure out which step is slow and look at its molecularity, Consider the following: No2 + CO —-> NO + CO2 If the overall reaction was an elementary reaction (mechanism consists of ___) we would immediately however, if experiment shows actual rate law is rate = k[NO2]2 , we know how experiment? so what does rate= K[No2]2 tell us? only thing can increase to make reaction faster is
in rates h slower and limits the rate of reaction RDS rate determine step represents the rate law for the overall reaction RDS tells you rate law overall process (mechanism consists of only one step) , immediately write rate law- rate = K[NO2][CO] (would tell us occurred in one step so know rate law) immediately that the overall reaction at the beginning can not be elementary do experiment w number of diff trials where take certain number of reactants and hold constant while doubling other reactants, see if rate doesn't change doubles or quadruples - indicate order wrt reactants when doubling NO2 while holding CO constant causes rate to quadruple, indicating second order wrt to NO2. CO isnt in rate law, so doubling CO while hold NO2 constant caused no change to rate and therefore CO not in RDS. Or can plot concentration vs time points w three diff graphs ([] vs t, ln [] vs time, 1/[] vs time) and whichever linear gives you order - if plot CO vs time 0th order reactant in RDS
increase temp, Frequency factor gives idea of A= P is P= The fraction of particules with an E > Ea is given by function
increase K increase rate how often collisions are occurring (z) and orientation probability factor- how likely are collisions to be productive-(p) pz (as orientstion prob factor decrease, does so bc molecules are more complex-(if had a lock and key like collision- gonna bounce of eachother unproductively- more complex molecules are, smaller percent of effective collisions is, smaller K value and lowers rate) (bc smaller percent of successful collisions, smaller frequency factor, smaller K) percent of effective collisions number effective collisions / total number collisions f= e^-Ea/RT
Increasing temperature of particles So graph of collision energy/ number collisions Increasing temp is just when molecules collide in correct orientation with sufficient energy, only way to get from beginning to end, is by going thru structure that has a to completely react from beginning to end, have to,
increases fraction of molecules that collide w energy greater than Ea when increase temp aka energy, whole graph gonna shift right and flatten out- increase number of particles w enough energy to react- push more molecules to right side of barrier and have more successful collisions pushing more molecules to fraction with enough energy to react (more particles can pas energy barrier and react) and collisions will be more effective when they happen transition state is formed- reaction proceeds- bonds are partially formed and bonds are partially broken has a bond that's partially formed and a bond that's partially broken as we show what in between structure would look like to get there have to go through in between phase to get to end- if don't put enough energy in, can't make transition- just bc achieve transition state doesn't mean a process proceeds forwards from reactants to products - if made to top of Ea, still could have not enough energy to transition
VERY IMPORTANT ON GRAPH- It is simply the rate at beginning is the ____ and most ____- if have collisions start to take place, A plot of __ vs ___ provides info about reaction rates Average rate- occurs over a ___. Found by Instantaneous rate-any initial rate = initial rate is the The initial rate gives the
initial rate!!! initial rate is simply the instantaneous rate of change at time 0 - at very beginning of reaction, this is rate taking place fastest and most trustworthy- at v beginning b4 anything is reacting, you have max amount of reactants and 0 products- so when two particles collide they're gonna be reactant particles-produce some product- so now when two things collide now it could be reactant with reactant or could be reactant with product, so gonna be more things getting in the way of effective collisions- so most reliable data is initial rates concentration v time a time period- the slope of a line joining any two points (ur time interval) along the curve. the slope of line b is avg rate over the first 60 secs of reaction any tangent line- rate at a particular instant during reaction- found by the slope of a line tangent to the curve at a particular point . Line d shows instantaneous rate at 35 secs into reaction tangent line of curve at time = 0 instantaneous rate at time = 0 (slope of tangent line is initial rate) tangent to the curve at t= 0. the initial rate gives the best kinetic data. At this point there is no product in the way to affect the data. all kinetic rates are found through initial rate.
colder system has warm system- increase temp, ____,____,____ surface area Temperature- particles must increase temp, _____, which means endothermic reaction to dissolve more both of those processes will reach So no matter what, increasing temp
less frequent collisions and less energy, slower rate of reaction more collisions, more energy collisions, increased rate more surface area, more contact btwn reactants (only relevant for solid reactants), more contact faster rate of reaction bc collide more often collide w enough energy to react more collisions per second (more frequent collisions), which means faster reactions that are more energetic (more likely to be successful) increase temp bc heat is reactant and so too many reactants and shift to products so increase solubility as increase heat- but heating exo makes it exo- adding heat means too many products and turn into reactants- equilibrium faster if heat them up increases rate of reaction (both endo or exo will reach equilibrium faster if increase temp)
Can tell if multistep by if not in rate law, Proposed mechanism: 1) NO2 + NO2 —-> NO3 + NO (slow) 2) NO3 + CO —-> NO2 + CO2 (fast) Slow step rate law (which must match___) is K[NO2]2 Since the proposed mechanism NO3 in reaction is To be valid mechanism rate lawof NO2 + NO2 —-> NO3 + NO
looking at coefficients- if coefficients coincide w rate law, single step elementary process. if not, multistep consumed in fast step experimental rate law- adds to the overall equation and the slow step rate law matches the experimental rate law, its a valid mechanism reaction intermediate (formed and used up in overall mechanism) 1. experimental determined rate law (trials) has to match rate of slow step (molecularity of slow step must match up w experimental rate law) 2. all steps must add to net reaction (also can have substance that cancels but isn't RI or catalyst just a reactant) 3.) All steps must be physically reasonable - has to be three or few particles- in collision all particles must collide w enough energy and proper orientation, if more than 3 particles not physically reasonable (termolecular or less for each step) k[NO2]2
If molecules move too slow, energy must be supplied to stretch thus, the barrier represents rate constant affected by
merely bounce off one another without changing. bond and allow a group in transition phase to rotate - rotating allows for bonds to begin to form , energy of molecule drops. the energy necessary to force molecule through the relatively unstable intermediate state to the final product temp, catalysts- not affected by concentration of reactants ONLY constant when only factor you change is concentrations (if increase temp, increase K)
collision theory ea used to Central idea of collision model two things moleculesmmust have for successful collision This energy comes from the upon collision, the KE of molecules can be used to ____ is used to change PE of a molecule
molecules must collide in order to react, must have correct orientation, must have enough enough (e > ea) (it have enough energy will vibrate strongly to break bonds) deform bonds and stretch them in order to reach transition state molecules must collide to react proper orientation- orientations molecules during their collisions determine whether or not suitably positioned to form new bonds energy- must collide w enough energy to react kinetic energies of the colliding molecules can be used to stretch, bend, and ultimately break bonds- leading to chemical reactions KE
doubling concentration of a component will cause the rate either to ___,__,or___ if doubling doesn't change rate, that's ___, because if doubling causes rate to quadruple, rate would not be ____- it would be Rate constant is figure out k value Units of K value tell you 2nd order overall means Do Predetermine rate law ex's
not change, double, quadruple 0th order- bc if have exponent of 0 next to reactant, anything to 0th power is 1 and eliminate from rate law rate would not be proportional to that reactant- would be proportional to that reactant squared- so 2nd order- unique to each reaction at given temp- change temp, change rate of reaction, change rate constant conduct trials at same temp and same reactants, that means k value for each of these experiments is same- once you know orders wrt components, you pick trial and plug in numbers and solve for k- so pick trial one, say 5.4 x10^-7 M/S= K (.0100 M) (.200M) so that equals 2.7 x 10^-4 1/M•S overall order of reaction- so it's 2nd order overall- anytime 2nd order process, units are 1/M•s In the RDS, there are 2 molecules colliding- one molecule of each one bc both have first order component- rate law tells us exactly what's happening in RDS- inslow step, there's an ammonium and there's a nitrate colliding- one of each makes it first order relative to each, 2 total makes it second order overall [NO2] is first order, [F2] first order, second order overall (overal 2nd order is bimecular- saying there are two particles colliding in rds) 1st order wrt CH3 thing and 0th order wrt H20, so overall first order (unimolecular -in RDS only one molecule) if exponents don't match coefficients we know not elementary- can just rule it out (if did match would tell u POSSIBLY)
however, as temp increases, not only does The combination of these two effects need enough energy to eveb tho overall enthalpy for exo is favorable, still have if not enough energy, if Ea at 10.5 and turn up heat to 10.4,
number of collisions per second increase (and hence the number of collisions in the correct orientation per second) (more frequent collisions.. increase temp, particles have more energy and travel faster and collide more often and with more energy so number of collisions and number of successful collisions increase, so rate increase- but also the proportion of collisions that have KE equal to or greater than Ea, the activation energy (proportion of successful collisions increase, rate increase) and particularly the second, is used to explain the marked effect of temperature on reaction rate in the collision theory. overcome chemical bonds and exceed EA and turn into products some reactions still unsuccessful bc don't clear overall energy barrier won't be able to overcome minimum- more energy, but still collide unsuccessfully- if go to 10.5, then can collide successfully- if keep going, still successful but just happens quicker
imagine toll booth when slow step occurs first, slow step 1: 2NO2 —-> NO3 + NO fasts step 2: NO3 + CO ——> NO2 + CO2 overall : NO2 + CO —> NO + CO2 molecularity of an elementary step is determined by the
old lady is the first toll booth and plaza B is speedy person- cars slowed at plaza A and backed up- that's slow step- determining how quickly things can get thru- no backup occuring makes determination for rate law v simple- if steps are reverse and have fast step first while slow step second, quickly going through first step and then getting backed up in second step (slow step) —- so have to do substitution in math to find rate law- in first one if slow step happen first, just look at what molecularity is of slow step and that tells you rate law- if slow step second, think about it diff (ex funnels- if have small funnel above big funnel, can only have stuff proceed through as fast first step can manage them- what limits how quickly water gets to bottom is only affected by first step- only look at first step and look at its molecularity and determine rate law bc unaffected by fast step if its second- if switched funnels, fast step first- slow step is second- start to have backup- think about equilibrium situation- so much back-up, possibility for first step to be reversible first step happens first and its bimolecular and rate law for it = K[NO2]2, so bimolecular process, and that's how build rate law- take coefficients of reactants in slow step and that's rate law No3 reaction intermediate- but NO2 doesn't completely cancel bc 2 -1 = 1 NO2 number of molecules free atoms or radicals that collide in the step - reaction order of elementary step is same as its molecularity (if elementary step, coefficients dictate its molecularity)
if individual step involves one particle converting into products, overall unimolecular reaction means RDS has each individual part of process could have The overall order of reaction tells you termolecular process, exponents add to
one into products means only molecule was present- unimolecular one molecule own molecularity- from previous example, first step bimolecular, second step is bimolecular third step is unimolecular, overall molecularity of this process depends on which one of the steps is RDS!!!!!!!! (slowest) (so if say this reaction bimolecular, then you know RDS was step one) molecularity for the process 3- most sensible is K[A]3 bc already have 3 diff particles colliding, 3 diff particles colliding with enough energy and proper orientation has pretty low probability to go from beginning to end- if had to have 3 diff particles colliding w enough energy and proper orientation, likelihood in physical feasibility much lower- so when have termolecular process, common for it to be same component three times bc otherwise have to have lot of different things falling into place to work well
for first order integrated rate law: time is only slope of plot will give us for second order integrated rate law: rate constant must be positive DO INTEGRATED RATE LAW EX if you know rate constant, an elementary process only
positive- so -k, rate constant is positive as well, so indicate slope of line here is going down - Y variable is ln[A] and on x axis time - line going down- intercept of that line is ln[A]o(initial concentration of a), slope is -k. rate constant now plotting 1/[A] vs t (on x), y intercept will be diff- line must go up because slope is not -k- and have y intercept that corresponds to 1/[A]o you know order. units tell you. so 2.4x10^-21 L/mol•s aka 1/M•s, corresponds to second order rate constant , so use second order integrated rate law = 4.63 x10^21 seconds takes one step- if only take one step- balanced equation itself would represent order wrt to each component
Overall order of reaction units of k: 0th 1st 2nd 3rd Units of k equation if you are given K, it's units Integrated rate laws are Found by
rate = K (rate must be morality per seconds) SO mol/L•s Rate = K[A] SO 1/s Rate= K[A]2 L/mol•s L^2/mol^2•s or 1/m^2•s units of k = (L/mol) ^ovetall order- 1 / time tell you overall order mathematical equations for the reactant concentration as a function of time integrating concentration over its boundary conditions t=0 and t=t
Reaction Mechanism for a reaction mechanism, it could be a___, it could be a ___, Reactions may occur all at ___ or each of these processes is known as kinetics for a reaction are determined by every time you have an individual step
the sequence of events that describes the step by due process by which reactants become products one step process, two step process or three step- can be serveral steps once or through several discrete steps- if occur al at once, once step process (elementary) an elementary reaction or an elementary process RDS- always slowest and longest- in multi step process will be one step slower than others, that's what determines the kinetics- also possibility for one step process- reaction mechanism is what dictates that- how do I get form reactants to products- sometimes it's one step sometimes multi step where making intermediates and convert those intermediated into other things an individual step is an elementary reaction or elementary process- if you have process that is one step overall that entire process is an elementary process- if you have multi step process, there are several elementary steps that are a part of that
One can gain information about the rate of a reaction by seeing how the a fast step doesn't effect the way you investigate is by taking ____ of ____ and look and graph table and make inferences ex: NH4+ + NO2- —-> N2 + 2H20 INTERPRET THIS DATA- TRIAL 1: Initial [NH4+]: .0100, Initial [NO2-]: .200 Obserrved initial rate : 5.4x10^-7 TRIAL 2: Initial [NH4+]: .0200, Initial [NO2-]: .200 Obserrved initial rate : 10.8 x 10^-7 NOW COMPARE- TRIAL 4: Initial [NH4+]: .200, Initial [NO2-]: .0202 Obserrved initial rate : 10.8 x 10^-7 TRIAL 5: Initial [NH4+]: .200, Initial [NO2-]: .0404 Obserrved initial rate : 21.6 x 10^-7 so here's what we know: so orders In summary, exponents tell us
rate changes with changes in concentration rate (if run to car, then drive 40 minutes running doesn't do anything) taking concentrations of reactants and measuring initial rate of reaction and then varying concentrations of reactants while holding others constant same initial concentration of NO2 (.200 M) they have (in experiment 2) double concentration of ammonium- (from .100 M to .200 M) then look st what happened to initial rate- originally 5.4 x 10^-7, now 10.8 x 10^-7, means initial rate DOUBLED. [N2], when says initial rate, measuring slope of tangent line at time 0 (if product and up graph, tangent line goes up) So while holding [NO2] constant, doubling [NH4+] doubled rate- if doubling causes doubling, that means directly related (linear) and tells me NH4+ is involved in RDS bc affecting its concentration changed rate- affected in linear fashion- doubling caused doubling- that tells me on slow step, one particle of Nh4 involved in collision . hold Nh4+ constant, take [No2-] and double it, causes rate to double. doubling [Nh4+] doubles rate , doubling [NO2-] doubles rate — that tells me there is one particle of ammonium involved in a collision in the RDS and one particle NO2- colliding in RDS, so that means rate law= K[NH4+][NO2-] (doubling caused doubling means one particle in there) 1st order wrt [NH4+], 1st order wrt [NO2-], so overall this reaction is 2nd order- (this processes is bimocilar and means RDS (slow) involves two molecules colliding- 1 NH4 and 1 NO2- in the RDS there is an Nh4 and NO2 they are colliding - 2 particles total when NH4+ doubles, initial rate doubles (1st order wrt to [NH4+] so rate is proportional to [NH4+]1, also when NO2 doubles, initial rate doubles so 1st order wrt [NO2-] and rate is proportional to [NO2-]1 so- rate is proportional [NH4+][NO2-] so rate = k[NH4+][NO2-] if double both, that means rate would go up by a factor of four, so overall process is second order (so if double both first order reactants) (both 1st order add and get 2) order wrt each reactant and sum of exponents indicates the overall order and molecularity (number of molecules that collide during RDS)
unimolecular process- rate law is if have two reactants, and at end of process only one shows up in rate law, you know if you have more water, reaction rate if one step reaction, that tells us Affecting RDS affects some rates occur in _____, some occur in ____ if ratio of reactants is 1:1, rate of ____ of a ___ is equal to rate of ____ of the _____. so rate =
rate determining step involves one molecule- equation that describes the rate of reaction relative to the concentration of reactants in process water is not a part of RDS- RDS is process that is slowest and longest- highest Activation energy- (you know that the one that doesnt show up wasnt involved in RDS) won't change bc only depends on one involving RDS RDS involved one molecule rate of reaction, and vise versa one step- rate determine step- some occur in multi step- longest is RDS lose one gain one, so rate of disappearance of a reactant is equal to rate of appearance of the corresponding product (think as C4H9Cl goes to C4H9OH) -delta [A]/ delta t = delta [B]/ delta t
Mechanisms with slow step first Mechanism example: 1) NO2 + F2 —-> NO2F + F (slow) 2) NO2 + F —-> NO2F (fast) determine overall equation write rate law identify reaction intermediate draw pot energy diagram
rate law comes from first step, no bottleneck occurs, no special treatment needed overall equation: 2NO2 + F2 —-> 2NO2F (F reaction intermediate) rate = k[NO2][F2] (1st order wrt No2, 1st order wrt F2, second order overall) (bimolecular bc i volve two particles colliding in RDS) F (free radical v unstable) 2 transition phase humps and start higher end lower (negative delta H) (y is energy and x is progress) (2 steps, so each step corresponds to one hump- RDS is step w higher hump so first step higher hump) in between hump have valley that's higher than reactant and product energy (Ea is from where start to peak, delta H from product energy to reactant energy (diff)
ex C4H9Cl + H20 —-> C4H9OH + HCl (rate law?) ex grandma To go faster, have to effect if reaction is multi step, you have to effect the so if change amount of water in ex, water isnt____, so reaction rate having more concentrated reactant involved in RDS when look at reaction that's 1:1, if you wanna compare rates,
rate law is gonna be K[C4H9Cl]— only one reactant molecule present so that tells me this reaction is multi step- both reactants involved, but if only one involved in rate law, tells me water is not a part of the RDS 1gour 15 mins and 13 seconds- 13 seconds doesn't matter - totally ignore fast step of process step that takes longest (rds) longest step to effect it changing water isn't part of RDS so reaction rate won't change- if include more c4H9cl, since it's involved in RDS, the reaction happens more quickly (its multi step process)(higher concentration more collisions faster rate bc its involved in RDS) increase rate of reaction if wanna compare rates-easy- every time a reactant shows up the product shows up- coefficients guide us as to how products increase and reactants decrease
physical state affects why does physical state affect it? Particles must___- you have to have _____, and state relates to ____ which corresponds to _____. so if have solid, they _____ and you have particles that are In solids, you dont have liquids and _____ allow for gases - concentrations are ___- not___, but have particles with Surface area of reactant-
reaction rate particles must mix to collide- you have to have mixing- and state relates to motion which corresponds to energy rarely react with other solids (no mixing) and you have particles v close to eachother and slightly vibrating- if you have particles that are very slightly vibrating, if you have collisions, chances are those collisions won't have enough energy - don't have forceful enough collisions forceful enough collisions and aqueous solutions allow for mixing with frequent contact- particles move around more and faster and have lots of contact and higher energy collisions and react well low - not frequent contact- but have particles w very high energy, so more effective collisions when you do have them bc energy is way higher more surface area, more contact, faster reaction-
Rate laws- a rate law shows the the exponents tell us the ____ of the ____ (ex) rate= k[NH4+][NO2-] For a general reaction: K= it is ___ for a given It doesn't ___ as ____, but it does ____. m and n are essentially "___" I'd rate doubles when [A] doubles, If rate quadruples when [B] doubles, Components of the rate law must be found by _____ a and b of the balanced equation (coefficients), are not necessarily coefficients same as exponents if there are no ___ in rate law because
relationship between the reaction rate and the concentration of reactants order of the reaction with respect to each reactant -sum of exponents gives overall order first order in NH4, 1st order in NO2-, second order overall aA + bB—> cC + dD, rate law has a form Rate= k[A]m[B]n rate constant specific for a given reaction at a given temperature not change as a reaction proceeds (as long as temp doesn't change), but it does change with temperature reaction orders- how rate is affected by concentration of each reactant "how many of each particle ultimately collide in the slowest step of the reaction" (thats m and n) rate depends on [A]1 , m =1 rate depends on (is proportional to) [B]2 , n=2 experimentation where we compare diff trials related to reaction orders m and n- not necessarily any connection between coefficients of a and b and exponents of m and n- but could be- coefficients same as exponents if a reaction occurs in only one step- one step process means you go directly from reactants to products- only one collision, has to involve all the reactant particles and the numbers involved there- elementary process- one step reactants turn into products- coeff do match exponents- but no way by looking at reaction to know whether one step or two step based on overall net reaction (only would be if process was an elementary process) if a reaction occurs in only one step- one step process means you go directly from reactants to products- only one collision, has to involve all the reactant particles and the numbers involved there- elementary process- one step reactants turn into products- coeff do match exponents- but no way by looking at reaction to know whether one step or two step based on overall net reaction (only would be if process was an elementary process) products, not part of what's happening in collision in RDS
Catalysts are what they do is the change the ____ up both how? often accomplishes this by they provide a they cause a reaction to they do this by where we start and end lower Ea, if have energy diagram w two steps, Ea is determined by
substances that increase the rate of reaction without being consumed - (not reactants or products)— decrease Ea, by decrease Ea, increase K and increase rate- arrhenius equation- K= Ae^-Ea/RT — k value is rate constant- equal to frequency factor which has to do with how often collisions are occurring and how probable it is that those collisions will be successful- if wanna make K bigger, either increase temp bc make temp bigger (in denominator of negative exponent make negative exponent smaller) or make Ea lower - so catalyst makes K value larger therefore increasing rate through pathway of making Ea lower. reaction mechanism by which the reaction occurs —changes pathway- different mechanism with and without catalyst- with catalyst maybe 2 and without maybe 3 step for ex- go through diff route to get there speeds up both forward and reverse reaction of a process - not only faster to get left to right thru catalyzed mechanism, but faster to go from right to left- speed up fwd and reverse- so when a catalyst is present will get TO EQUILIBRIUM FASTER! speeds it up by making Ea lower and rate constant higher making collisions more effective - think enzymes- grabs molecules and hold still in right orientation to have successful collision- different pathway for the reaction to occur (lowering the energy pathway) faster- catalysts not actually a part of over all net reaction- not gonna show up as reactant or product- cancelled out thruecourse of reaction interacting w Ea- provide alternate pathway for reaction same- just smaller hill w lower Ea particles can travel thru that path more easier and corresponds to faster ratev highest peak bc that's RDS!!
So how does Arrhenius equation show an increase of k with increasing T? therefore, if k is determined experimentally at several temperatures, Ea can be calculated from Use Arrhenius equation to figure out How determine Ea from A from Arrhenius equation lnk=
taking natural log of both sides, equation becomes ln k= - Ea/ R (1/T) + ln A slope of a plot of ln k vs 1/T figure out Ea for reaction- how tall hump is- analyze equation- ln of entire equation makes k ln of k, on other side, makes ln of A and -Ea/R (1/T) do it this way bc y=mx+b - y intercept correspond to natural log of frequency factor- so if want to figure out Ea, determine slope of line (-Ea/R) (r value constant 8.31 j/mol k) (knowing slope can figure out Ea) So make a plot with points of a few k values at a couple of diff temps (y and x values)- do experiment and monitor concentration vs time - when monitor this, for reactant will trend as [A] vs t, ln[A] vs t, and 1/[A] vs time- only one will give u linear line- linear is correct- once figure out linear, when plot it, slope of those will give me k value- so do experiment, do three plots, figure out which is straight, and look at line and figure out slope (k value) - so now instead of plot of concentration vs time, do plot of ln K vs 1/T- so do same experiment at diff temps, so get diff k values, take ln of those k values plot them against 1/T gives you linear relationship- from slope of this line can find Ea bc u always know R (gas constant) PLOT lnK vs 1/T —- Ln A is y intercept, slope of line is -Ea/R To construct plot of lnk vs 1/T, first must collect [ ] vs time data for your experiment. then determine order of the reaction via integrated law plots (which is linear), slope of linear integrated law plot produces k, change T, do steps 1-3 again, manipulate data to obtain lnk, -/T (when plot w multiple temps, find multiple K values and then plot the ln of those multiple k values vs 1/T) , plot lnk vs 1/T data, slope gives -Ea/R, y intercept gives lnA ln A -Ea/RT
Solid catalysts work by using____ Effect on a catalyst on the PE curve and catalzyed vs uncatalyzed- for uncatalyzed path, ____ look at graph would slope of a lnk vs 1/T plot for a catalyzed reaction be more or less negative than the slope of the lnk vs 1/T plot for an uncatalyzed reaction? lnk vs 1/T
using SA to react w molecules catalyzed vs un catalyzed- for uncatalyzed path- Ea to get from where started to over top of hill is wherever we started to peak of peakiedt peak- for reverse reaction, same idea, but Ea fwd is less than Ea reverse- means when do Ea fwd- Ea reverse, get delta H, negative, correspond to exo i'm catalyzed pathway- ea forward is energy of where u started to peak of peakiest peak- so tallest hump (RDS) so Ea fwd and Ea reverse will revolve around the RDS catalyst lowers Ea- neg exp. smaller, so numerator of slope getting smaller, numerator smaller means line getting flatter and less negative (look at ex of graphs) take natura log of arrhenius (K= Ae ^ -Ea/RT), get ln k = -Ea/R (1/T) + lnA plotting that gives u Ea
use first or second integrated rate law deending on half life expression incorporate use of integrated rate law plots over course of experiment First Order integrated rate law: If a reaction is first order, a plot of Second order integrated rate law If a reaction is second order, a plot of
whether process is first or second order wrt that component- specifically whatever reactant "A" happens to be (1st order wrt A use first integrated rate law) ln2/k to find half-life of first order process, radioactive decay- first order- one thing thing skip apart- look at rate constant for a process to determine to determine order can monitor concentration as changes relative to time- measure pressure change, loss of mass, how color changes...if plot concentration vs time and linear, tells me process is 0th order... if plot ln concentration vs time... first order if straight line.... 1/cincrenteation vs time ... if works straight line, 2nd order ln[A]t = -kt + ln[A]o [ln] vs t will yield a straight line, and the slope of the line will be -k 1/[A]t = kt + 1/[A]o 1/[A] vs t will yield a straight line, and the slope of that line is k
——Multi Step Mechanisms—— In a multi step process, one of the steps The overall reaction can not occur Thus, the overall rate law Rate law for overall process is if slow step was first one in previous O3 reaction if second step was slow step, complicated... why?
will be slower than all others (RDS) faster than the slowest RDS ALWAYS comes from the slow step and it's rate law!! (from rds rate law) rate law of slow step we would say rate law was rate= k[O3]- process would be unimolecular bc that's the molecularity of RDS and would say overall reaction first order in second step, have O3 collide w O- problem is O is the reaction intermediate, which isn't allowed to show up in rate law- rate law is always K times concentration of REACTants to powers- O isn't reactant it's an intermediate - if slow step is later on its more complicated
When you have more particles, you have ____, ___, and a ____ reaction double concentration, particles will _____- more particles means having more reactant particles (_____), causes a increasing concentration of reactants relates to ______- usually. (the reactant you are describing must be____) when you have a one step process, if you have multi step process,
you have more collisions, more frequent collisions, and faster reaction collide twice as often- more particles means more collisions and chances that you have collisions of sufficient energy and proper orientation happen more often and reaction rate will go up (higher concentrations) causes a faster rate increase of rate- usually- reactant ur describing must be involved in rate determining step one step is rate determine step only one of those steps is meaningful in how long the reaction takes- bc one step always way longer than others- rate determining step is longest
