1. 2 NO(g) + O 2 (g)  2 NO 2 (g) Determine the rate expression and the value of the rate constant from the data below. [NO] (mol L -1 ) [O 2 ](mol L -1 ) initial rate (mol L -1 s -1 ) (1) 1.0 x 10 -4 1.0 x 10 -4 2.8 x 10 -6 (2) 1.0 x 10 -4 3.0 x 10 -4 8.4 x 10 -6 (3) 2.0 x 10 -4 3.0 x 10 -4 3.4 x 10 -5 Rate = k [O 2 ] m [NO] n To determine the rate law from the data, first determine the dependence of the rate on each reactant separately. rate 2 /rate 1 = k [O 2 ] 2 m [NO] 2 n / k [O 2 ] 1 m [NO] 1 n 8.4 x 10 -6 / 2.8 x 10 -6 = (3.0 x 10 -4 ) m / (1.0 x 10 -4 ) m 3= 3 m => m = 1; 1st order in O 2 Determining Rate Laws

2. rate 3 /rate 2 = k [O 2 ] 3 m [NO] 3 n / k [O 2 ] 2 m [NO] 2 n 3.4 x 10 -5 / 8.4 x 10 -6 = (2.0 x 10 -4 ) n / (1.0 x 10 -4 ) n 4= 2 n => n = 2; 2nd order in NO Rate = k [O 2 ][NO] 2 Order of reaction = 3 2.8 x 10 -6 mol L -1 s -1 = k [1.0 x 10 -4 mol L -1 ] [1.0 x 10 -4 mol L -1 ] 2 k = 2.8 x 10 6 L 2 mol -2 s -1

3. Concentration and Time - Integrated Rate Laws Integrated rate laws: variation of concentration of reactants or products at any time Derived from the experimental rate laws Zero order reactions A  P - d[A]/ dt = k d[A] = - k dt [A] t = [A] o - k t integrated rate law [A] o : concentration of A at t = 0 Slope = -k

4. First Order Reactions Rate = - d[A]/ dt = k [A]first order reaction Units of k for a 1st order reaction is time -1 d[A]/ dt = - k [A] d[A]/[A] = - k dt Solution is: ln [A] t [A] o = - kt [A] o where is the initial concentration of A at time t = 0 [A] t = [A] o e -kt

5. ln[A] t = ln [A] o - k t [A] t = [A] o e -kt N 2 O 5 (g)  N 2 O 4 (g) + 1/2 O 2 (g)

7. Half life of a 1st order reaction Half life : time it takes for the concentration of the reactant A to fall to half its initial value t 1/2 when [A] t = [A] o /2 ln[A] t = ln [A] o - kt ln [A] o /2 = ln [A] o - k t 1/2 ln(1/2) = - k t 1/2 ln(2) = k t 1/2 t 1/2 = ln(2) / k t 1/2 = 0.693 k

9. Mercury (II) is eliminated from the body by a first-order process that has a half life of 6 days. If a person accidentally ingests mercury(II) by eating contaminated grain, what percentage of mercury (II) would remain in the body after 30 days if therapeutic measures were not taken? k = ln 2 / t 1/2 = (ln 2) / (6 days) Fraction remaining after 30 days [A] t / [A] 0 = e -kt = 0.03 Answer: 3% remaining in the body after 30 days.

10. Radioactive decay is a first order process N t = N o e -kt where N t is the number of radioactive nuclei at time t N o is the initial number of radioactive nuclei k is called the decay constant t 1/2 = ln 2/ k

11. Carbon-14 dating uses the decay of 14 C 14 C is produced in the atmosphere at an almost constant rate As a result proportion of 14 C to 12 C is ~ constant 14 C enters living systems as 14 CO 2 ; all living systems have a fixed ratio of 14 C to 12 C; about 1 14 C to 10 12 12 C When the organism dies, there is no longer exchange of C with surroundings, but 14 C already in the organism continues to decay with a constant half life, so ratio of 14 C to 12 C decreases.

12. Second order reactions Rate = k[A] 2 - d[A]/ dt = k [A] 2 2nd order reaction for which the rate depends on one reactant rate = k [A] [B] or rate = k [A] 2 d[A]/[A] 2 = - k dt

13. The half-life of a 2nd order reaction when [A] = [A] o /2

14. ln[A] t = ln [A] o - kt 1st order 2nd order

15. slope = k 2C 2 F 4  C 4 F 8 ln[C 2 F 4 ] vs time - nonlinear rate = k [C 2 F 4 ] 2

16. Reactions often proceed in a series of steps - elementary reactions For example: O 2 + light  O 2 * O 2 *  O. + O. 2(O. + O 2 + M  O 3 + M) Overall reaction: 3O 2 + light  2O 3 Reaction Mechanisms O. is an intermediate species; involved in a step in the mechanism, but does not appear in the overall reaction Determines rate laws; use experimental rate law to determine mechanism

17. The rate of an elementary reaction is directly proportional to the product of the concentrations of the reactants, each raised to a power equal to its coefficient in the balanced equation for that step The number of reacting species in an elementary reaction - molecularity Overall reaction 6 Fe 2+ (aq) + Cr 2 O 7 2- (aq) + 14H + (aq)  6 Fe 3+ (aq) + 2 Cr 3+ (aq) + 7 H 2 O(l)

18. Unimolecular reaction; molecularity = 1 O 2 *  O + O Rate = k[O 2 *] Bimolecular reaction; molecularity = 2 NO(g) + O 3 (g)  NO 2 (g) + O 2 (g) Rate = k [NO] [O 3 ] Termolecular reaction; molecularity = 3 O + O 2 + M  O 3 + M rate = k [O] [O 2 ] [M] Termolecular reactions are low probability reactions; require three species to come together simultaneously Types of elementary reactions

19. From rate law to reaction mechanism Products of a reaction can never be produced faster than the rate of the slowest elementary reaction - rate determining step Experimental data for the reaction between NO 2 and F 2 indicate a second-order rate Overall reaction: 2 NO 2 (g) + F 2 (g)  2FNO 2 (g) Rate = k [NO 2 ] [F 2 ] How can a mechanism be deduced from the rate law? Rate law indicates that the reaction cannot take place in one step