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H 80 kj mol 1 d h 80 kj mol 1 e h

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Unformatted text preview: e mechanism for the following reaction. H2(g) + 2 ICl(g) I2(g) + 2 HCl(g) Experimental investigation shows that HI is a reaction intermediate and the rate law is: rate = kobs[H2][ICl] CHM1311 Chemical Kinetics 62 Example: Postulating a mechanism H2(g) + 2 ICl(g) I2(g) + 2 HCl(g) SLOW --> IT IS THE RDS rate = kobs[H2][ICl] H2(g) + ICl(g) HI(g) + ICl(g) HI(g) + HCl(g) Rate = k1 [H2][ICL] FAST I2(g) + HCl(g) Rate = k[ICL][HI] CAN'T USE OVERALL REACTION TO GET RATE LAWS, MUST USE ELEMENTARY STEPS because it can't be simplified any further H2(g) + 2 ICl(g) I2(g) + 2 HCl(g) CHM1311 Chemical Kinetics 63 Reaction profile for this mechanism: A Slow Step Followed by a Fast Step CHM1311 Chemical Kinetics 64 Example: Postulating a mechanism Postulate a plausible mechanism for the following reaction. 2NO(g) + O2(g) 2 NO2(g) Experimental investigation shows that N2O2 is a reaction intermediate that forms in a rapid equilibrium and the rate law is: rate = kobs[NO]2[O2] CHM1311 Chemical Kinetics 65 Example: Postulating a mechanism 2NO(g) + O2(g) 2 NO2(g) rate = kobs[NO]2[O2] Step 1: rapid equilibrium 2NO(g) K-1 k2 K1 N2O2(g) Rate (forward) = k[NO]^2 Rate (reverse)= k-1 (N2O2) Keq= [N2O2]/[NO]^2 = K1/k-1 N2O2= [K1/K-1] [NO2] 2 Step 2: N2O2(g) + O2(g) OVERALL REACTION Slow - RDS 2NO2(g) Rate= K2 [N2O2][O2] = K2 [(K1/K-1)][NO2]^2[O2] Rate= Kobs [NO2]^2[O2] 2NO(g) + O2(g) 2 NO2(g) EXP DATA: equil is fast, therefore K2<< k1 & k-1 CHM1311 Chemical Kinetics 66 Your Turn... What is the correct expected rate law for the overall reaction, based on the following proposed mechanism and experimental data? A + B X X C C + A D k1 <<< k2 < k3 A. rate = k[A][C] B. rate = k[A]2[C] C. rate = k[A][B] D. rate = k[X] CHM1311 Chemical Kinetics 67 The Steady State Approximation In complex multistep processes more than one step may be important for the reaction rate! Reconsider the previous reaction: k1 2NO(g) N2O2(g) k-1 2NO(g) N2O2(g) k3 N2O2(g) + O2(g) 2NO2(g) k1 2NO(g) N2O2(g) rate=k1[NO]^2 k2 N2O2(g) 2NO(g) rate=k2[n2o2] k3 N2O2(g) + O2(g) 2NO2(g) rate= k3[n2o2][o2] rates that form n202 - rates consume n202 d[NO2] = k3[N2O2][O2] =0 dt CHM1311 Chemical Kinetics 68 The Steady State Approximation used to eliminate intermediates from rate laws k1 k2 A X C rate of formation of X = rate of disappearance of X or: -D[X]/Dt=0 rates form x- rates consume x =0 k1[A] - k2[x]= 0 (now you can rearrange for x) CHM1311 Chemical Kinetics 69 The Steady State Approximation k1 2NO(g) N2O2(g) k2 N2O2(g) 2NO(g) k3 N2O2(g) + O2(g) 2NO2(g) -d[N2O2] dt CHM1311 = k1[NO]2 k2[N2O2] k3[N2O2][O2] = 0 Chemical Kinetics 70 The Steady State Approximation d[N2O2] = k1[NO]2 k2[N2O2] k3[N2O2][O2] = 0 dt Rearrange: k1[NO]2 = [N2O2](k2 + k3[O2]) [N2O2] = k1[NO]2 (k2 + k3[O2]) Now replace the [N2O2] in the rate expression: -d[NO2] dt CHM1311 = k3[N2O2][O2] = k1k3[NO]2[O2] (k2 + k3[O2]) 71 Chemical Kinetics Kinetic Consequences of Assumptions k 1 d[NO2] dt = k1k3[NO]2[O2] (k2 + k3[O2]) 2NO(g) N2O2(g) k2 N2O2(g) 2NO(g) How can we interpret this rate?? If k2 << k3 d[NO2] dt d[NO2] dt = k3 N2O2(g) + O2(g) 2NO2(g) k1k3[NO]2[O2] ( k3[O2]) k1k3[NO]2[O2] ( k2) = k1[NO]2 If k2 >> k3 = = k1k3 k2 [NO]2[O2] CHM1311 Chemical Kinetics...
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