1. Kinetics  The study of reaction rates.  Spontaneous reactions are reactions that will happen - but we can’t tell how fast.  Graphite will spontaneously turn to diamond– eventually.  Reaction mechanism- the steps by which a reaction takes place.

2. Review- Collision Theory  Particles have to collide to react.  Have to hit hard enough  Things that increase this increase rate  High temp – faster reaction  High concentration – faster reaction  Small particles = greater surface area means faster reaction  Catalyst

3. Reaction Rate  Rate = Conc. of A at t 2 -Conc. of A at t 1 t 2 - t 1  Rate =  [A] Dt  Change in concentration per unit time  For this reaction  N 2 + 3H 2 2NH 3

4.  As the reaction progresses the concentration H 2 goes down ConcentrationConcentration Time [H 2 ] N 2 + 3H 2 → 2NH 3

5.  As the reaction progresses the concentration N 2 goes down 1/3 as fast ConcentrationConcentration Time [H 2 ] [N 2 ] N 2 + 3H 2 → 2NH 3

6.  As the reaction progresses the concentration NH 3 goes up 2/3 times ConcentrationConcentration Time [H 2 ] [N 2 ] [NH 3 ] N 2 + 3H 2 → 2NH 3

7. Calculating Rates  Average rates are taken over long intervals  Instantaneous rates are determined by finding the slope of a line tangent to the curve at any given point because the rate can change over time  Derivative.

8.  Average slope method ConcentrationConcentration Time D[H 2 ] DtDtDtDt

9.  Instantaneous slope method. ConcentrationConcentration Time  [H 2 ] D t D t d[H 2 ] dt dt

10. Defining Rate  We can define rate in terms of the disappearance of the reactant or in terms of the rate of appearance of the product.  In our example N 2 + 3H 2 2NH 3   [H 2 ] = 3  [N 2 ]  t  t   [NH 3 ] = -2  [N 2 ]  t  t

11. Rate Laws  Reactions are reversible.  As products accumulate they can begin to turn back into reactants.  Early on the rate will depend on only the amount of reactants present.  We want to measure the reactants as soon as they are mixed.  This is called the Initial rate method.

12.  Two key points  The concentration of the products do not appear in the rate law because this is an initial rate.  The order (exponent) must be determined experimentally,  can’t be obtained from the equation Rate Laws

13.  You will find that the rate will only depend on the concentration of the reactants. (Initially)  Rate = k[NO 2 ] n  This is called a rate law expression.  k is called the rate constant.  n is the order of the reactant -usually a positive integer. 2 NO 2 2 NO + O 2

14.  The rate of appearance of O 2 can be said to be.  Rate' = D[O 2 ] = k'[NO 2 ] Dt  Because there are 2 NO 2 for each O 2  Rate = 2 x Rate'  So k[NO 2 ] n = 2 x k'[NO 2 ] n  So k = 2 x k' 2 NO 2 2 NO + O 2

15. What are the units for k? It depends upon the order of the reaction. Rate = mol/L sec Zero Order: the value of x = 0 Rate = k[A] 0 = k therefore k=mol/L sec

16. What are the units for k? First Order: the value of x = 1 Rate = k[A] 1 = mol/L sec = k (mol/L) therefore k=1/sec

17. What are the units for k? Second Order: the value of x = 2 Rate = k[A] 2 = mol/L sec = k (mol/L) 2 therefore k=L/mol sec

18. What are the units for k? Third Order: the value of x = 3 Rate = k[A] 3 = mol/L sec = k (mol/L) 3 therefore k=L 2 /mol 2 sec

19. Types of Rate Laws  Differential Rate law - describes how rate depends on concentration.  Integrated Rate Law - Describes how concentration depends on time.  For each type of differential rate law there is an integrated rate law and vice versa.  Rate laws can help us better understand reaction mechanisms.

20. Determining Rate Laws  The first step is to determine the form of the rate law (especially its order).  Must be determined from experimental data.  For this reaction 2 N 2 O 5 (aq) 4NO 2 (aq) + O 2 (g)  The reverse reaction won’t play a role because the gas leaves

21. [N 2 O 5 ] (mol/L) Time (s) 1.000 0.88200 0.78400 0.69600 0.61800 0.541000 0.481200 0.431400 0.381600 0.341800 0.302000 Now graph the data

22.  To find rate we have to find the slope at two points  We will use the tangent method.

23. At.80 M the rate is (.88 -.68) = 0.20 =- 5.0x 10 -4 (200-600) -400

24. At.40 M the rate is (.52 -.32) = 0.20 =- 2.5 x 10 -4 (1000-1800) -800

25.  Since the rate at twice as fast when the concentration is twice as big the rate law must be..  First power  Rate = -D[N 2 O 5 ] = k[N 2 O 5 ] 1 = k[N 2 O 5 ] Dt  We say this reaction is first order in N 2 O 5  The only way to determine order is to run the experiment.