1. CHE 112 - MODULE 3 CHAPTER 15 LECTURE NOTES

2. Chemical Kinetics  Chemical kinetics - study of the rates of chemical reactions and is dependent on the characteristics of the reactants  Reaction mechanisms - detailed pathway that atoms and molecules take as a chemical reaction proceeds  CD-ROM Screen 6.2 & 15.2

3. Effects on Chemical Kinetics  Concentration of reactants – usually the rate of a reaction increases with increased concentration of reactants  Concentration of a catalyst - catalyst speeds up the reaction  Temperature – rate will increase with an increase in temperature (increased KE)  Surface area – as the surface area increases, the reaction will proceed at a faster rate

4. Rates of Chemical Reactions  Rates - the change in concentration of a chemical per unit of time ex. (M/sec) –much like the speed of a car ex. (miles/hr) –much like interest rates ex. (5.4%/year) –much like sales taxes ex. (7cents/$)  Rates are just a ratio or fraction of one thing changing with respect to another

5. Decomposition of N 2 O 5 2 N 2 O 5  4 NO 2 + O 2  We look at the disappearance of N 2 O 5  Rate of Reaction = change in [N 2 O 5 ] change in time  Rate of Reaction =  [N 2 O 5 ]  t

6. Stoichiometry 2 N 2 O 5  4 NO 2 + O 2 2moles : 4moles : 1mole Therefore, for every 2 moles of N 2 O 5 decomposed, you have 1 mole of O 2 formed. The rate of formation of O 2 is equal to ½ the rate of decomposition of N 2 O 5  [O 2 ]/  t = -½  [N 2 O 5 ]/  t

7. Decomposition of N 2 O 5  Therefore we can determine the rate of decomposition by the rate of formation of either NO 2 or the O 2 with consideration to stoichiometry. 2 N 2 O 5  4 NO 2 + O 2 R f =  [NO 2 ] = -2R d R f =  [O 2 ] = -½ R d  t  t

8. Calculations 2 N 2 O 5  4 NO 2 + O 2  Where t i = 600s and t f = 1200s and the concentration of N 2 O 5 decomposed from 1.24 x 10 -2 M to 0.93 x 10 -2 M  Rate of decomposition =  [N 2 O 5 ] /  t = (0.93 x 10 -2 M -1.24 x 10 -2 M) (1200s – 600s) = -5.2 x 10 -5 M/s  Rate of formation O 2 = -½ [-5.2 x 10 -5 M/s] = 2.6 x 10 -5 M/s

9. Plot of Concentration vs. Time  Figure 15.2 on pg 702 shows the graph of the disappearance of N 2 O 5.  Average rate = change in concentration over an interval in time (  c/  t)  Instantaneous rate = concentration at an instant in time; tangent to the curve at a particular point in time (dc/dt)

10. Rate Law  Rate Law – equation that relates the rate of reaction to the concentration of reactants raised to various powers Rate = k [N 2 O 5 ] x –where k is the rate constant and x is the order of the reaction

11. Determining Rate Law We can perform an experiment to decompose N 2 O 5. If we doubled the concentration of the reactant we can observe the change in rate as follows: Initial Conc. Rate of decomposition Exp.1 1x10 -2 mol/l 4.8x10 -6 mol/lsec Exp.2 2x10 -2 mol/l 9.6x10 -6 mol/lsec Observation: The rate of reaction doubled.

12. Determining Order  When a reaction has this observable result, it is said to be first order.  As we double the concentration, the rate is 2 x, where x is the classification of the order of that particular reaction.  Or R 2 /R 1 = 2 x, where x=1, then we can see that it is clearly first order.

13. Overall Rate Law for N 2 O 5 We previously stated that: Rate = k [N 2 O 5 ] x Where x = 1 determined experimentally, therefore this reaction is a first order reaction and follows the rules of first order kinetics.

14. Order of Reactions Considering other reactions where we double the initial reactant concentration, we can observe the rate and order as follows: Quadruple rate = 4 (2 x =4, x=2) second order Double rate = 2 (2 x =2, x=1) first order No change = 1 (2 x =1, x=0) zero order Half rate = 1/2 (2 x =1/2, x=-1) -1 order 1.4 the rate = 1.4 (2 x =1.4, x=1/2) half order

15. Review of Logarithms  Logs are the inverse function of an exponential function (y = 3 x )  Algebraically it is written as x = 3 y or y = log 3 x where y is the exponent on 3 that results in a value of x –ex. 5 2 = 25 or log 5 25 = 2  Written as a function logs are expressed as f(x) = log b x  If b is not designated, b=10 is assumed.  If b = e it is called the natural log (ln).

16. First Order Kinetics 2 N 2 O 5  4 NO 2 + O 2  Rate of decomposition = -  [N 2 O 5 ]  t  Rate = kt  Therefore -  [N 2 O 5 ] = -kt (by substitution)  t  ln [N 2 O 5 ] t = -kt + ln [N 2 O 5 ] 0 (derive by integration)

17. Integrated Rate Equations  For a zero order reaction: [A] t = -kt + [A] 0  For a first order reaction: ln [A] t = -kt + ln [A] 0  For a second order reaction: 1/[A] t = kt + 1/[A] 0

18. Iodine Clock Reaction Prep Each solution is prepared in a 100ml volumetric flask and qs with distilled water. Accurately weigh the following: 2.998g of NaI = 0.2M 1.169g of NaCl = 0.2M 0.2482g of sodium thiosulfate = 0.01M 3.485g potassium sulfate = 0.2M 5.404g potassium persulfate = 0.2M

19. Half-life of Reaction  Half-life (t ½ ) = the time it takes for the initial concentration of a reaction to disintegrate to half its original concentration  First order reaction: t ½ = 0.693/k

20. Collision Theory  Three conditions must be met for a reaction to take place: –The reacting molecules must collide with one another –Molecules must collide with sufficient energy –Molecules must collide with the proper orientation

21. Activation Energy  E a = the minimum amount of energy that must be absorbed by a system to cause it to react where k = Ae -Ea/RT  Activation energy can be determined from the Arrhenius Equation: ln k = ln A - E a /R (1/T) ln (k 2 /k 1 ) = - E a /R [1/T 2 - 1/T 1 ]

22. Conditions Increasing Rates  Temperature - increases the KE and enough energy to overcome E a  Presence of a catalyst - provides different pathways with lower E a  Surface area - increases the probability of a collision with proper orientation

23. Reaction Mechanisms  Reaction mechanisms = sequence of steps that show the intermediates formed ( bonds broken and bonds formed) between the reactant side and product side of any reaction  See CD-ROM Screens 15.12 – 15.13

24. Rate Equations for Elementary Steps  Elementary Step - a singular molecular event classified by the # of reactant molecules involved (molecularity) –unimolecular: A  P where R = k[A] –bimolecular: A + B  P where R = k[A][B] or A + A  P where R = k[A] 2 –termolecular: 2A +B  P where R = k[A] 2 [B]  See Table at the bottom of page 734

25. Rate-determining Step Step 1: A + B  X + M; k 1 is slow Step 2: M + A  Y; k 2 is fast Overall Rxn. 2A + B  X + Y Because step 2 is fast, it does not contribute to the overall rate, therefore: R = k 1 [A][B]