A diamond is forever.

# A diamond is forever.

## A diamond is forever.

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##### Presentation Transcript

1. De Beers A diamond is forever. Is It So?

2. Consider: Gibb’s Free Energy ? C diamond C graphite ΔG = ∑ ΔGproducts - ∑ ΔGreactants ΔG = ΔGgraphite - ΔGdiamond ΔG = (0) - (3 kJ/mol) ΔG = -3 kJ/mol

3. Look quick, before it turns into graphite.

4. While it’s true her diamond is spontaneously turning into graphite before her eyes, it’s happening very slowly. Don’t hold your breath waiting to see any change. It takes billions of years.

5. While thermodynamics answers the question as to whether or not a reaction or event is spontaneous, it DOES NOT tell how fast a reaction goes. This is what kinetics does....describes the rate of the reaction.

6. Chemical Kinetics Chemical Kinetics are about the only tool a chemist has to probe the actual mechanism of a chemical reaction. It is the study of reaction rates.

7. Reaction Rates NO2 NO + O2 Reaction Rate: The slope of a Concentration vs Time graph.

8. Reaction Rates Factors affecting reaction rates: • The nature of the reaction mechanism • The concentration of the reactants • The temperature at which the reaction occurs • The presence of a catalyst • The surface area of solid or liquid reactants or catalysts

9. Red  Blue

10. Rate Laws describe reaction rates mathematically Rate = k[reactant 1]m [reactant 2]n…

11. Rate Laws To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations. There are two forms of the rate law… Differential rate laws: Shows how the rate depends on concentration. Sometimes called just the “rate law”. Integrated rates laws: Shows how the concentration depends on time.

12. The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated. Differential Integrated

13. The choice of which rate law to use depends on the type of data that can be collected conveniently and accurately. Once you know one type, the other can be calculated. Differential Integrated This of course requires the use of my calculus.

14. Other Stuff to Know: Reaction Order The differential rate law for most reactions has the general form…. Rate = k[reactant 1]m [reactant 2]n… The exponents m and n are called reaction orders. Their sum (m + n) is called the overall reaction order.

15. Other Stuff to Know: Reaction Order Usually the values of the reaction order must be determined experimentally, they cannot be predicted by looking at the chemical reaction. In most rate laws, reaction orders are 0, 1, or 2. However, occasionally they are a fraction or even negative. The rate of a reaction depends on concentration; however the rate constant does not.

16. Finding The Differential Rate Law The most common method for experimentally determining the differential rate law is the method of initial rates. In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t = 0 as possible)

17. Using Initial Rates to Determine the Form of the Rate Law Example: A + B  C • Exp # [A] [B] Initial Rate (M/s) • .100M .100M 4x10-5 • .100M .200M 4x10-5 • .200M .100M 16x10-5 From this data, find the form of the rate law.. Rate = k[A]m[B]n

18. Exp # [A] [B] Initial Rate (M/s) • .100M .100M 4x10-5 • .100M .200M 4x10-5 • .200M .100M 16x10-5 Rate = k[A]m[B]n n = 0 [B]0 = 1

19. Exp # [A] [B] Initial Rate (M/s) • .100M .100M 4x10-5 • .100M .200M 4x10-5 • .200M .100M 16x10-5 m = 2 Rate = k[A]2[B]0 = k [A]2

20. Exp # [A] [B] Initial Rate (M/s) • .100M .100M 4x10-5 • .100M .200M 4x10-5 • .200M .100M 16x10-5 Now, solve for k… Rate = 4x10-3 [A]2

21. First Order Reactions Rate = Integrating, etc. leads to… [A] = [A]o e-kt Ln [A]t – ln [A]0 = -kt

22. Derive the Integrated First Order Rate Law: Given a first order differential rate law… That is, the rate of change of the concentration of reagent A is proportional to how much A there is. Let x = [A] (the concentration of A), Then… The minus sign is because the value of [A], or x, is decreasing.

23. Transforming to calculus notation… Separate the variables… Integrate both sides…

24. Integrating… Solving for x…

25. Now use initial conditions to determine C. At t = 0, x = [A]o [A]o = initial concentration of A Recall that..

26. First Order Reactions Ln [A]t – ln [A]0 = -kt • These equations can be used to determine … • The concentration of a reactant remaining at any time after the reaction has started. • The time required for a given fraction of a sample to react. • The time required for a reactant concentration to reach a certain level.

27. Half Life (First Order Process)

28. Half Life of a Reaction The half life of a reaction, t½ , is the time required for the concentration of a reactant to drop to one half of its initial value. ln (½) = -kt½ T½ = 0.693/k Half life for a first order rate law is independent of the initial concentration of reactant. Thus, the half life is the same at any time during the reaction.

29. Half Life for a First Order Reaction (Garden Variety Half Life – Radioactivity)

30. First order reaction example The first order rate constant for the decomposition of a certain insecticide in water at 12oC is 1.45 g/mLyr. A quantity of this insecticide is washed into a lake on June 1, leading to a concentration of 5.0x10-7 g/ml water. Assume that the effective temperature of the lake is 12oC. a. What is the concentration of the insecticide on June 1 of the following year? b. How long will it take for the concentration to drop to 3.0x10-7 g/ml?

31. a. What is the concentration of the insecticide on June 1 of the following year? Ln [A]t – ln [A]0 = -kt Ln [A]t = -kt + ln [A0]t=0 ln [insecticide] t=1 year = - (1.45g/mLyr)(1 yr) + ln (5.0x10-7 g/ml) ln [insecticide] t=1 year = - (1.45g/mL) + (-14.51g/mL) ln [insecticide] t=1 year = -15.96 g/mL [insecticide] t=1 year = e-15.96 [insecticide] t=1 year = 1.2x10-7 g/mL

32. b. How long will it take for the concentration of the insecticide to drop to 3.0x10-7 g/ml? ln [A]t = -kt + ln [A0]t=0 ln (3.0x10-7 g/ml) = - (1.45 yr-1)(t) + ln (5.0x10-7 g/ml) t = -[ln (3.0x10-7g/ml) – ln (5.0x10-7g/ml)]/1.45yr-1 t = - [-15.02 + 14.51]/1.45yr-1 t = 0.35 yr

33. Second Order Reactions A second order reaction is one whose rate depends on the reactant concentration raised to the second power, or on the concentrations of two different reactants, each raised to the first power. For a reaction that is second order in just one reactant, A, the rate law is given by… Rate = k[A]2

34. Second Order Reactions Rate = k[A]2 Integrating and stuff leads to the Integrated Rate Law…

35. Second Order Half Life’s Setting [A] = ½[A]o and solving for time gives… Unlike the half life of first order reactions, the half life of a second order reaction is dependent on the initial concentration of the reactant.

36. Half Life for a Second Order Reaction

37. Zero Order Reaction A zero order reaction is one whose rate depends on the reactant concentration raised to the zero power, or in other words, it does not depend on the concentration of the reactant as long as some reactant is present. For the reaction that is zero order in just one reactant, A, the rate law is given by… Rate = k[A]0 Integrating… [A]t = -kt + [A]0 Solving for t½ T½ = [A]0/2k

38. Half Life for a Zero Order Reaction: (Rate doesn’t depend on concentration)

39. Rate Laws: A Summary To simplify the rate laws, we assume conditions where only the forward reaction is important. This produces rate laws that only contain reactant concentrations. There are two types of rate laws… Differential rate laws: Shows how the rate depends on concentration. Integrated rate laws: Shows how concentration depends on time.

40. Rate Laws: A Summary The most common method for experimentally determining the differential rate law is the method of initial rates. In this method several experiments are run at different initial concentrations and the instantaneous rates are determined for each at the same value of time (as near t=0 as possible). The point is to evaluate the rate before the concentrations change significantly from the initial values. From a comparison of the initial rates and the initial concentrations the dependence of the rate on the concentrations of various reactants can be obtained – that is the order of each reactant can be determined.

41. Rate Laws: Summary To experimentally determine the integrated rate law for a given reaction, concentrations are measured at various values of time as the reaction proceeds. Then the job is to see which integrated rate law correctly fits the data. Typically this is done visually, by ascertaining which type of plot gives a straight line. Once the correct straight line plot is found, the correct integrated rate law can be chosen and the value of the rate constant, k, can be obtained from the slope of the plot. Also, the differential rate law for the reaction can be calculated.

42. All the Good Stuff

43. We can use this information to experimentally find the rate law for a reaction.

44. The gas phase decomposition of NO2 is studied at 383oC giving the following data. NO2 NO + O2

45. Graph the ln (concentration) vs Time. It the plot is a straight line, the reaction is first order. Original Data Graph

46. Graph the 1/(concentration) vs Time. It the plot is a straight line, the reaction is second order. Reaction constant k = 10.16 Original Data Straight line  Second Order Reaction

47. Remember that the rate equation between two substances A and B looks like this… Rate = k[A]m[B]n The equation shows the effect of changing the concentrations of the reactants on the rate of the reaction. What about all the other things like temperature and catalysts, for example which also change rates of reaction? Where do these fit into this equation? These are all included in the so-called rate constant – which is only actually constant if all you are changing is the concentration or the reactants. If you change the temperature or the catalyst the rate constant changes. These changes of the rate constant are shown mathematically in the Arrhenius Equation.

48. Svante August Arrhenius Svante August Arrhenius was a Swedish physical chemist best known for his theory that electrolytes, certain substances that dissolve in water to yield a solution that conducts electricity, are separated, or dissociated, into electrically charged particles, or ions, even when there is no current flowing through the solution. In 1903 he was awarded the Nobel Prize for Chemistry. 1859-1927

49. The Arrhenius Equation K = rate constant A = frequency factor Ea = activation energy R = the gas constant (8.314 J/mol·K) T = temperature in Kelvin