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# Kinetics/Estimating Rule Cem 152 (P. Hunt)

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This 3 page Class Notes was uploaded by JessicaJ on Wednesday February 24, 2016. The Class Notes belongs to Cem 152 (P. Hunt) at Michigan State University taught by P. Hunt in Spring 2016. Since its upload, it has received 38 views. For similar materials see Principles of Chemistry in Chemistry at Michigan State University.

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Date Created: 02/24/16

Rxn = reation 2/22/16 Kinetics (cont.) Average Rate of rxn = (change in concentration of reactant or product)/(unit time) o A B Concentration of B = [B] –T[B] /0(t-0) = ( ∆[B]¿/∆T Instantaneous rate obtained in limit ∆T 0 [B]0= initial Conservation of Mass o [A] = -[B] o Special Case [A] = -[B] = 0 equilibrium (steady state) o Generally Take aA + bB cC + dD Rate = (-1/a)*( ∆ [A]/∆t¿ = (-1/b)*( ∆[B /∆t¿ = (-1/c) *( ∆ [C /∆t¿ = (-1/d)*( ∆ [D /∆t ¿ Always in the limit ∆ t 0 Comments: Concentrations of rxns; partial pressure for gas reactions Rate is positive if net product is formed as rxn written Empirically, rates are determined in the form o Rate = k[A] [B] n K = rate coefficient m,n are typically small while numbers [u-0, 1, or 2] Reaction is ‘after m’ in A and ‘after n’ in B Reaction has overall order of m & n A may not equal m!!!! B may not equal n!!! Sometimes m or n can be negative Size of k gives limit as to reaction speed o > 10^9 (whatever)/seconds fast o <10^1 (whatever)/seconds slow First order rxn AB o –A[A]/dt = K[A] = instant rate [A]t= [A] 0xp(-kt) o Half Life Want to find t =t 1/2uch that [A] t/2 ½[A] 0 [A]t/2 ½[A] e0p(-kt ) 1/2 = exp(-kt 1/2 Ln(1/2) = -kt1/2 T1/2= 0.693/k Second order in A o Rate = k[A]^2 o 1/[A] t kt + 1/[A] 0 Zero Order o Rate = k[A]^b = k o [A] = -kt + [A] 0 nd Half life for 2 order o 1/[A]t= kt + 1/[A] 0 o [A] = kt + 1/[A] t/2 1/2 0 2/24/16 Arrhenius Equation K = Aexp[-E /Ra] A = frequency of collisions in an elementary reaction. Weakly dependent on T E =aactivation energy Observe rate and hence the rate of the coefficient at the values of T Calculate E a Lnk – Lnk = -E /k[1/T – 1/T ] T1 T2 a 1 2 Rln(k 1k2)/(1/T1– 1/T 2 = E a Estimating Rule Raising T by 10K (or10C) roughly doubles the speed of the reaction. Let k =22k 1 T =1298k T =2308k -1 -1 R = 1.987 calK mol = 4.184 kJ/mol Ea= 12.9 kJ/mol = 52.9 kJ/mol Estimating reaction occurs in a single collision. A reaction mechanism may include a sequence of elementary stops, and a sequence of collisions For a simple collision, we use molecularity Unimolecules Bimolecules Trimolecules Rate equations for elementary steps have the exponents of the concentrations/pressure equal to the molecularity A B elementary step Rate = k[A] or kP A A B not elementary Not necessarily, must be [empirical] A + B C + D elementary Rate = k[A][B] or kP P A B A + B C + D not elementary not necessarily What is the slow step? Case A = slow step first 2NO 2(g) NO 3(g)+ NO (g) Slow NO 3(g)+ CO (g) NO (2)(g)CO 2(g) Fast NO 2(g)+ CO (g) NO (g)+ CO 2(g) Rate of the composite reaction hacks the rate of the slow step 2 Rate = k[NO ]2

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