= rate =

rate constant

derived

definition:

Rate law can be written from molecularity

Lowers the Ea by making a new ______________

Why? Experimental

ly derived rate laws show that all reactions aren’t elementary

k = zpe-Ea/RT

k =ln k=

Relating [ ] vs. time

1st order: 2nd order:

0 order:

Ch. 12 A.P. Chem.

Chemical Kinetics

Reaction Rates Mechanisms

-[ ]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

[ ]0/2kt½

series of steps by which a reaction occurs

Elementary steps

Inter-mediates RDS catalyst

s

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-Ea/RT

(-Ea/R)1/T + lnA

Arrhenius Equation(s)

order

Ch. 12 Review Problem

The transfer of an O atom from NO2 to CO has been studied at 480 K,

CO(g) + NO2(g) ----> CO2(g) + NO(g)

and the following data was collected,

---initial conc. (mol/L)--- initial rate (mol L-1 hr-1)

[CO] [NO2]

#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 NO2 + CO --> CO2 + NO

#2

(slow) NO2 + NO2-->NO3 + NO (fast) NO3 + CO-->NO2 + CO2

#3

(slow) NO2 --> NO +O (fast) CO + O --> CO2

Rate = k [CO]m [NO2]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= =0.34m

m= 0

Same ↓ Doubles ↓ Quadruples ↓

Rate = k [NO2]2

3.4 x 10-8 mol L-1hr-1 = k [0.35 x 10-4 mol/L]2 k = 28 L/mol·hr t½= 1/k[NO2]0 =1.0x103 hr

1st order in NO2 and CO

2nd order in NO2 and zero order in CO

OK 1st order in NO2 and zero order in CO

Therefore n=2

-[C4H10]time

-[ O2 ]time

2C4H10 + 13O2 → 8CO2 + 10 H2O

Are these equal?

-6.5[C4H10]time

-[ O2 ]time

2C4H10 + 13O2 → 8CO2 + 10 H2O

Are these equal?=