Textbook Notes (362,820)
Canada (158,064)
Chemistry (261)
CHEM 1AA3 (18)
Pippa Lock (13)
Chapter 14

Textbook Chapter 14 - Chem 1AA3

19 Pages
Unlock Document

McMaster University
Pippa Lock

Chem 1AA3 Chapter 14: Chemical Kinetics Preface  Chemical Kinetics – concerns how rates of chemical reactions are measured, how they can be predicted and how reaction-rate data are used to deduce probable reaction mechanisms  The 3 Big Questions in Any Chemical Reaction 1) What are the products? 2) What is the equilibrium position? 3) How fast is the reaction?  Why we measure rates 1) Predict/Control Reactions: o Industrial syntheses o Environmental reactions (eg/ smog formation, ozone layer breakdown) 2) Monitor Biological or Chemical Systems o Clinical diagnostics (eg/ liver damage) o Polymerization (eg/ length and composition of polymers) 3) Understand reaction mechanisms o Reaction order shows if a reaction in N 1 orNS 2 o Build general structure/function relationships, such as rate vs. leaving group abilitN in S 2 reactions 4) Tell the temperature by measuring the rate of crickets chirping: Chem 1AA3 14.1 The Rate of a Chemical Reaction  Rate = change in concentration of reactants and products over time  Eg/ o A  G o 1 minute after starting reaction, [G] = 1M o What is the rate of the reaction over 1 min? o Average Rate = Δ[G]/Δt = 1M/1min o For A  G average rate = Δ[G] / Δt  Rates are always positive: add a negative sign if using decrease of starting material to describe rate: o Average rate = - Δ[A] / Δt  Stoichiometry is important: if A  2G o Average rate = - Δ[A] = ½ Δ[G] Δt Δt  In general: o aA + bB  gG + hH o Average rate = -1 Δ[A] = -1 Δ[B] = 1 Δ[G] = 1 Δ[H] = a Δt b Δt g Δt h Δt Key Concepts  Rate is change in concentration over time  Rates are always positive  Stoichiometry matters 14.2 Measuring Reaction Rates  Rates can change during a reaction, as seen in the food coloring exp  Many reactions may give an average rate of 1M/min for 1 min o average rate = 1M/min over first min for all 4 curves o average rates are of limited use – prefer instantaneous rates  Instantaneous rate (v): the rate at a specific time t  v = the slope of the tangent to the curve of [G] vs. time Chem 1AA3  Measuring Δ[A] or Δ[G] over a finite time interval, Δt, gives the average rate for that interval (eg/ 1M/min)  We would prefer the instantaneous rate, v, at time = t o This is equivalent to the average rate over an infinitely short time interval about t (eg/ 0.69 M/min)  Taking an infinitely short time intervals is the same as differentiating an equation in calculus  Average rate = 1 Δ[G] g Δt  v = lim 1 Δ[G] = 1 d[G] Δt0 g Δt g dt Average vs. Instantaneous and Initial Rates  Experimentally measured rates are always average rates  To best approximate instantaneous rates, we measure Δ[A] or Δ[G] over the smallest possible Δt  We ideally get instantaneous rates at t=0 o To get such initial rates, we have to measure Δ[G] very close to t=0  At t=0 reactions are uni-directional and do not suffer from product inhibition – best place to study their mechanism  The resulting initial rate,0v , can reveal much about the reaction mechanism  Food Coloring Expt: Key Concepts  Want to know initial rates  Experimental rates are average rates  To approach initial rates experimentally, use short measurements times Δt & measure rates near t=0 14.3 Effect of Concentration on Reaction Rates: The Rate Law  v m0y be independent of [A]: v ∝ 0A] 0 Chem 1AA3 1  v m0y increase linearly with [A]: v0∝ [A] o Food coloring demo 2  v m0y increase with the square of [A]: v0∝ [A]  For any given reaction, aA + bB  gG + hH, the dependence of v on 0A] and [B] reflects its mechanism  The rate law, or rate equation, describes this relationship between rate and concentration for each reaction: o v =0k[A] [B] n o m and n  The reaction orders with regard to A and B  Not necessarily equal to a and b, the stoichiometric coefficients  Determined experimentally  Usually small, positive integers (0, 1 or 2) o Overall order of a reaction is (m+n), while  m is the order of the reaction with regard to A  n is the order of the reaction with regard to B  We determine m and n through the method of initial rates: measure v at d0fferent [A] and 0B] , and0 extract m and n  Reactions may be: o Zero-order m, n = 0 no effect of [A] or [B] o First-order m = 1 v0∝ [A] o Second-order (m + n) = 2 v ∝ [A] or [A][B] 0  Once m and n are known, you can solve the rate law v for0the rate constant, k  k is a fundamental properly of each reaction o Depends on temperature, catalyst, solvent, but not on [A] and [B] o Larger k  faster reaction Method of initial Rates  For a reaction: 2A + B  C + D o v =0k[A] [B]  what are m and n? Chem 1AA3  Set up three reactions:  Extract m and n from rations of v 0(exp2)v0(exp1) o V (0)/V (10 = k(2M) (1M) = 2 = 4  m=2 o V (0)/V (10 = k(2M) (1M) = 2 = 1  n=0  The order of the reaction tells us how many species are present in the rate-limiting step of the reaction  This information helps us determine reaction mechanisms Key Concepts:  Order ≠ stoichiometry  Order is determined experimentally  Order = number of species involved at transition state of the rate-limiting step 14.4 Zero-Order Reactions  v i0 independent of [A]  v = k [A] = k = constant 0  Units for v are always concentration/time (M/s) Examples  Zero-Order Processes: o Evaporation/sublimation with constant surface area  Pseudo Zero-Order reactions: o Reactions where catalysts or enzyme is saturated with reactants (eg/ drug/alcohol metabolism) Key Concepts:  Rate = rate constant (k)  True zero-order reactions are rare Integrated Rate Law  v =0k or -Δ[A]/Δt = k  The integrated version of this rate law is: o [A] = [A] -kt t 0 Chem 1AA3 Reaction Orders In Nucleophilic Substitutions  Nucleophilic Substitution – an electronegative atom or group bonded to a carbon atom is replaced by another o Can be first order or second order overall, depending on the chemical structure 14.5 First Order Reactions  Experiments have shown that the above reaction of F with tBu-Cl is first order in tBu-Cl o i.e. that v ∝ [tBu-Cl] 0 - o Measurement of v at di0ferent [tBu-Cl], [F ] (method of initial rates), and checking the dependence of v on 0A], [B] st Reaction order w.r.t tBu-Cl 1 order Reaction order w.r.t F - 0 order Overall Reaction Order 1 order Rate Equation v = k[tBu-Cl] = k[tBu-Cl] [F ] 0 o Why first order?  Step 1: Slow C-Cl bond cleavage to form carbocation  Step 2 Fast attack of F on carbocation   Overall Rate Controlled by Step 1: C-Cl bond cleavage o F does not enter until after the rate-limiting step, hence no effect from [F ] - o One molecule in the rate-limiting step, hence 1 order st Chem 1AA3  v0= k[A] o Reaction is called S 1N  S = substitution  N = Nucleophilic  1 = one species in the rate limiting step  Concept Check o What are the units of k for a first-order reaction?  Unites for k are 1/time (eg/ s )1 o Recall the characteristics of k: 1. Constant regardless of concentration 2. Depends on the identity of the reactants, temperature, catalyst, solvent Integrated Rate Law for First-Order Reactions o Rate equation gives v at every instant o Integrating v vs. time gives [A] consumed (or [G] formed) o Integrate d[A]/dt with respect to time: o ln[A] =t-kt or ln[A] t -kt + ln[A] 0 [A]0 o ln[A] is unitless, so kt is unitless  Rearrange equation o [A] =t[A] 0 -kt Half Lives, t1/2  Half Life – the time required for one-half of a reactant to be consumed  First order reactions have constant half-lives, or t 1/2 o This means [A]t= ½  [A] 0 at t1/2 o Substitute into [A]t= [A]0e -kt o To obtain ½  [A] = [A] e -kt 0 0 o Apply ln and rearrange to get ln(½ [A] 0 = -kt 1/2 and ln( ½ ) = -kt1/2 [A]0 t1/2 -(ln1/2) = ln2 = 0.69 k k k Reactions involving gases  Same as solution reactions, but use partial pressures (P ) inAplace of [A] o Eg/ ln (P )A t-kt (P A 0  Examples: 1. Radioactive Decay 2. S N  Key Concepts o v 0 [A] o t 1/2= constant o Rate and integrated rate equations Chem 1AA3 14.6 Second-Order Reactions  A second order reaction: o S 2NNucleophilic substitution o S 2N  S = substitution  N = Nucleophilic  2 = two species in the rate-limiting step st Reaction order w.r.t Me-Cl 1 order Reaction order w.r.t F- 1 order Overall Reaction Order 2 order Rate Equation v = k[Me-Cl] = k[Me-Cl] [F ] 1 o Why second order?  Single converted step with both reactants present at rate-limiting step  Rate depends on the presence of both reactants -  v0= k[F ][Me-Cl]  Units for v are concentration/time (eg/ M/s)  Units of [F ] and [Me-Cl] are concentration (eg/ M)  Concept Check: o What are the units of k for a second-order reaction?  Units for k are 1/(concentration  time) (eg/ M s )1  Example from the textbook: o 2A  products o Where”  v0= k[A] 2  S 2Nexample: o A + B  products o Where  v0= k[A][B]  Preview: Ester Hydrolysis o Esters are synthesized from carboxylic acids and alcohols Chem 1AA3 o Esters are broken down by hydrolysis o Reversible reaction; equilibrium position is determined by Le Chatelier principle  Stoichiometric ester hydrolysis is a 2 order reaction which becomes very slow as the reactants are depleted o Eg/  Eg/ reaction of 0.1 M ester and 0.1 M H O 2n acetone:  Turn 2 Order Reactions into Pseudo 1 Order Reactions o Run ester hydrolysis reaction in dilute aqueous acid o Now the change in [H O] 2s negligible, and the reaction behaves like a 1 order irreversible reaction:  v = k[ester]  Key Concepts o Rate depends on both reactants (or [A] ) 2 o Make pseudo 1 order with one excess reagent 14.7 Reaction Kinetics: A Summary  Things you can calculate: 1. v when the rate law is known: 0 m n  v0= k[A] [B] … 2. Instantaneous rate (v) from:  Tangent to line in graph of [A] vs t  -Δ[A]/Δt (average rate) for short Δt  Rate law 3. Order of reaction from Chem 1AA3  v0vs. concentration  Graph that gives a straight line  Constant t 1/2(1 order)  Integrated rate law which gives a constant k 4. K from:  Slope of line in an appropriate graph  The appropriate integrated rate law st  t1/2(1 order) 5. [A] tnd [G] ftom k and [A] , 0sing the integrated rate law 14.8 Theoretical Models of Chemical Kinetics  Collision Theory o While molecules can collide at high frequency (~10 /s) only very few (~10 M s )have -1 -1 enough energy to react  Reactions are slow because they require transfer of kinetic energy into the reacting bonds  Molecules must be moving fast; activation energy must be reached  Molecules must be correctly oriented to be able to react  Transition State Theory o The transition state – the energy divide between the reactant and the product  i.e. the highest point along the reaction energy profile o The transition state species or activated complex – the species present at the transition state
More Less

Related notes for CHEM 1AA3

Log In


Don't have an account?

Join OneClass

Access over 10 million pages of study
documents for 1.3 million courses.

Sign up

Join to view


By registering, I agree to the Terms and Privacy Policies
Already have an account?
Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.

Add your courses

Get notes from the top students in your class.