1. = rate = rate constant derived definition: Rate law can be written from molecularity Lowers the E a by making a new ______________ Why? Experimentally derived rate laws show that all reactions aren’t elementary k = zpe -E a /RT k = ln k= Relating [ ] vs. time 1st order: 2nd order: 0 order: Ch. 12 A.P. Chem. Chemical Kinetics Reaction RatesMechanisms -  [ ]  time Avg. Rates Instantaneous Rates Slope of the line tangent at that moment Rate Law k [ ] x experimentally Method of initial rates Constant rates [ ]’s for certain substances high Zero order in one substance Compare runs with all but one [ ] constant Integrated Rate Laws ln[ ] = -kt + ln[ ] 0 1/[ ] = kt + 1/[ ] 0 [ ] = -kt +[ ] 0 ln([ ]/[ ] 0 ) = -kt 0.693/k 1/k[ ] 0 [ ] 0 /2k t½t½ series of steps by which a reaction occurs Elementary steps Inter- mediates RDS catalysts mechanism: Used in one step, produced in another Produced in one step, used in another with a new and faster (dictates) Speeds up a reaction without being consumed collision theory Effective collisions enough energy Break and make new bonds transitional particle activated complex...seen on P.E. diagram Proper orientation Ae -E a /RT (-E a /R)1/T + lnA Arrhenius Equation(s) order

2. Ch. 12 Review Problem The transfer of an O atom from NO 2 to CO has been studied at 480 K, CO (g) + NO 2(g) ----> CO 2(g) + NO (g) and the following data was collected, ---initial conc. (mol/L)--- initial rate (mol L -1 hr -1 ) [CO] [NO 2 ] #1) 5.1 x10 -4 0.35 x 10 -4 3.4 x 10 -8 #2) 5.1 x10 -4 0.70 x 10 -4 1.36 x 10 -7 #3) 5.1 x10 -4 0.18 x 10 -4 8.5 x 10 -8 #4) 1.5 x10 -3 0.35 x 10 -4 3.4 x 10 -8 A. Write the generic rate law for this reaction. B. Find the order of the reaction with respect to each reactant using the data provided. C. Calculate the rate constant from the data provided. D. Find the half-life for the reaction for run #1 above. F. Which mechanism below best agrees with the experimentally derived rate law: #1 NO 2 + CO --> CO 2 + NO #2 (slow) NO 2 + NO 2 -->NO 3 + NO (fast) NO 3 + CO-->NO 2 + CO 2 #3 (slow) NO 2 --> NO +O (fast) CO + O --> CO 2 Rate = k [CO] m [NO 2 ] n Rate1= 3.4x10 -8 = k[5.1x10 -4 ] m [0.35x10 -4 ] n Rate4= 3.4x10 -8 = k[1.5x10 -3 ] m [0.35x10 -4 ] n 1= =0.34 m m= 0 Same ↓ Doubles ↓ Quadruples ↓ Rate = k [NO 2 ] 2 3.4 x 10 -8 mol L -1 hr -1 = k [0.35 x 10 -4 mol/L] 2 k = 28 L/mol·hr t ½ = 1/k[NO 2 ] 0 =1.0x10 3 hr 1 st order in NO 2 and CO 2 nd order in NO 2 and zero order in CO OK 1 st order in NO 2 and zero order in CO Therefore n=2

3. -  [C 4 H 10 ]  time -  [ O 2 ]  time 2C 4 H 10 + 13O 2 → 8CO 2 + 10 H 2 O Are these equal?

4. -6.5  [C 4 H 10 ]  time -  [ O 2 ]  time 2C 4 H 10 + 13O 2 → 8CO 2 + 10 H 2 O Are these equal? =