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CHE 112 - MODULE 3

CHE 112 - MODULE 3. CHAPTER 15 LECTURE NOTES. 0. 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

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CHE 112 - MODULE 3

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  1. CHE 112 - MODULE 3 CHAPTER 15 LECTURE NOTES

  2. 0 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. 0 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. 0 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. 0 Decomposition of N2O5 2 N2O5 4 NO2 + O2 • We look at the disappearance of N2O5 • Rate of Reaction = change in [N2O5] change in time • Rate of Reaction =  [N2O5]  t

  6. 0 Stoichiometry 2 N2O5 4 NO2 + O2 2moles : 4moles : 1mole Therefore, for every 2 moles of N2O5decomposed, you have 1 mole of O2 formed. The rate of formation of O2 is equal to ½ the rate of decomposition of N2O5 [O2]/t = -½ [N2O5]/ t

  7. 0 Decomposition of N2O5 • Therefore we can determine the rate of decomposition by the rate of formation of either NO2 or the O2 with consideration to stoichiometry. 2 N2O5 4 NO2 + O2 Rf =  [NO2] = -2Rd Rf =  [O2] = -½ Rd  t t

  8. 0 Calculations 2 N2O5 4 NO2 + O2 • Where ti = 600s and tf = 1200s and the concentration of N2O5 decomposed from 1.24 x 10-2 M to 0.93 x 10-2 M • Rate of decomposition = [N2O5] /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 O2 = -½ [-5.2 x 10-5 M/s] = 2.6 x 10-5 M/s

  9. 0 Plot of Concentration vs. Time • Figure 15.2 shows the graph of the disappearance of N2O5. • 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. 0 Rate Law • Rate Law – equation that relates the rate of reaction to the concentration of reactants raised to various powers Rate = k [N2O5]x • where k is the rate constant and x is the order of the reaction

  11. 0 Determining Rate Law We can perform an experiment to decompose N2O5. 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/l•sec Exp.2 2x10-2 mol/l 9.6x10-6 mol/l•sec Observation: The rate of reaction doubled.

  12. 0 Determining Order • When a reaction has this observable result, it is said to be first order. • As we double the concentration, the rate is 2x, where x is the classification of the order of that particular reaction. • Or R2/R1 = 2x, where x=1, then we can see that it is clearly first order.

  13. 0 Overall Rate Law for N2O5 We previously stated that: Rate = k [N2O5]x Where x = 1 determined experimentally, therefore this reaction is a first order reaction and follows the rules of first order kinetics.

  14. 0 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 (2x=4, x=2) second order Double rate = 2 (2x=2, x=1) first order No change = 1 (2x=1, x=0) zero order Half rate = 1/2 (2x=1/2, x=-1) -1 order 1.4 the rate = 1.4 (2x=1.4, x=1/2) half order

  15. 0 Review of Logarithms • Logs are the inverse function of an exponential function (y = 3x) • Algebraically it is written as x = 3y or y = log3x where y is the exponent on 3 that results in a value of x • ex. 52 = 25 or log5 25 = 2 • Written as a function logs are expressed as f(x) = logbx • If b is not designated, b=10 is assumed. • If b = e it is called the natural log (ln).

  16. 0 First Order Kinetics 2 N2O5 4 NO2 + O2 • Rate of decomposition = - [N2O5]  t • Rate = kt • Therefore - [N2O5] = -kt (by substitution)  t • ln [N2O5]t = -kt + ln [N2O5]0 (derive by integration)

  17. 0 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. 0 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. 0 Half-life of Reaction • Half-life (t½) = the time it takes for the initial concentration of a reaction to disintegrate to half its original concentration • First order reaction: t½ = 0.693/k

  20. 0 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. 0 Activation Energy • Ea = 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 - Ea/R (1/T) ln (k 2/k1)= - Ea/R [1/T2 - 1/T1]

  22. 0 Conditions Increasing Rates • Temperature - increases the KE and enough energy to overcome Ea • Presence of a catalyst - provides different pathways with lower Ea • Surface area - increases the probability of a collision with proper orientation

  23. 0 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. 0 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 634

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  26. 0 Rate-determining Step Step 1: A + B  X + M; k1 is slow Step 2: M + A  Y; k2 is fast Overall Rxn. 2A + B  X + Y Because step 2 is fast, it does not contribute to the overall rate, therefore: R = k1[A][B]

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