1. Chemical Kinetics H 2 O 2 decomposition in an insect H 2 O 2 decomposition catalyzed by MnO 2 Day 2: Rate equations

2. Collision Theory Collisions Energy Collisions Energy Orientation NO YES

3. REACTION RATES RR =  [P ] = -  [R ]  t  t P = products R = reactants

4.  Three “types” of rates  initial rate  average rate  instantaneous rate Reaction Rates Section 15.1

6. Relative Rates Reactant 4A  2B + 3C -  [A ] =  [ B ] =  [ C ] 4  t 2  t 3  t

7. Rate Expressions

8. To determine a reaction mechanism, we study the effect of concentration on the rate of the reaction To determine a reaction mechanism, we study the effect of concentration on the rate of the reaction

9. Rate of reaction is proportional to [reactant] We express this as a RATE LAW Rate of reaction = k [ X ] where k = rate constant k is dependent only on T Rate of reaction is proportional to [reactant] We express this as a RATE LAW Rate of reaction = k [ X ] where k = rate constant k is dependent only on T

10. Rate contant: Arrhenius equation Rate constant Temp (K) 8.31 x 10 -3 kJ/Kmol Activation energy Frequency factor Frequency factor = frequency of collisions with correct geometry. Rate constant is dependent on only the activation energy and temperature

11. REACTION ORDER In general, reaction: a A + b B --> x X with a catalyst “C” Rate = k [A] m [B] n [C] p The exponents m, n, and p are the reaction order can be 0, 1, 2 or fractions must be determined by experiment! must be determined by experiment!

12. Rate = k [x] m If m = 0, Reaction is zero order. Rate = k [X] 0 = k *1 If [X] doubles; rate = K

13. if Rate = k [X] m If m = 1, Reaction is 1st order Rate = k [X] 1 If [3X] 1, then rate goes up by 3 1 If [2X] 1, then rate goes up by 2 1

14. 14 FIRST ORDER REACTIONS The rate law is

15. Rate = k [X] m If m = 2, Reaction is 2nd order. Rate = k [X] 2 [2X] 2 increases rate by 2 2 = 4 [3X] 2 increases rate by 3 2 = 9

16. 16 SECOND ORDER REACTIONS

17. Deriving Rate Laws Expt. [CH 3 CHO] Disappear of CH 3 CHO (mol/L) (mol/Lsec) RATE 1 0.100.020 2 0.200.081 3 0.300.182 4 0.400.318 Derive rate law and k for CH 3 CHO(g) --> CH 4 (g) + CO(g) from experimental DATA for rate of disappearance of CH 3 CHO

18. Deriving Rate Laws Rate = k [CH 3 CHO] 2 Here the rate goes up by 4 when initial conc. doubles. Therefore, we say this reaction is SECOND order. The value of k. Use expt. #3 data— 0.182 mol/Ls = k (0.30 mol/L) 2 k = 2.0 (L / mols) k = 2.0 (L / mols) Using k you can calc. rate at other values of [CH 3 CHO] at same T.

20. SOLVED PROBLEM: pg141

21. 21 practice: pg151

22. 22 FIRST ORDER REACTIONS The rate law is