1. Mr. Bartelt presents How fast??? Chapter 12 Chemical Kinetics

2. Chemical kinetics Thus far we have looked exclusively at reactions from a before and after stand point. Now we look at the reactions as they happen. This should help complete the picture.

3. Will it happen? Look at the equation below. 2NO 2(g)  2NO (g) + O 2(g) We can use ΔH and ΔS to determine if the reaction will occur at room temperature. Let’s do that: H f of NO 2(g) = 34, S = 240 H f of NO (g) = 90, S = 211

4. SHIVA or  2NO 2(g)  2NO (g) + O 2(g) H f of NO 2(g) = 34, S = 240 H f of NO (g) = 90, S = 211 S of O 2 = 205 ΔH = (2*90)-(2*34) ΔH = 112 kJ/mol  ΔS = (2*211+205)-(2*240) ΔS = (627)-(480)=147 J/molK ΔS = (627)-(480)=147 J/molK ΔS = 147 J/molK = 0.147 kJ/molK

5. Temperature controls it 2NO 2(g)  2NO (g) + O 2(g) ΔH = 112 kJ/mol  ΔS = 0.147 kJ/molK ΔS = 0.147 kJ/molK ΔG = ΔH - T ΔS ΔG = 112 – (20+273)(0.147) ΔG = 112 – (43) = 69 kJ/mol  This will need to be heated or forced by other means to occur.

6. Stoichiometry 2NO 2(g)  2NO (g) + O 2(g) Thermodynamics tells us if it will happen. Stoichiometry tells us how much of each product/reactant is needed/produced. For instance, we know that 4 moles of NO 2 will produce 4 moles of NO and 2 moles of O 2.

7. What we don’t know… We have no idea how long it will take those 4 moles of NO 2 to react completely. It doesn’t happen instantly, it takes time. Look at the sample data on the following page.

8. 2NO 2(g)  2NO (g) + O 2(g) From now on [bracket] will be used to express concentration (M). Note: “S” tells us that [NO 2 ] + [NO] = [NO 2 ] initial [O 2 ] = [NO]/2 Graphs to come Time (sec)[NO 2 ][NO][O 2 ] 00.01000.0000 500.00790.00210.0011 1000.00650.00350.0018 1500.00550.00450.0023 2000.00480.00520.0026 2500.00430.00570.0029 3000.00380.00620.0031 3500.00340.00660.0033 4000.00310.00690.0035

9. The graph

10. Back to math The rate of reaction is defined as the change in concentration vs. the change in time. Rate of reaction is the slope of the line on the graph. Note: Slope will always be expressed as a positive number because a negative rate of reaction doesn’t have a lot of physical signficance

11. Rate calculation Rate of oxygen production: From 0  50 sec Time (sec)[NO 2 ][NO][O 2 ] 00.01000.0000 500.00790.00210.0011 1000.00650.00350.0018 1500.00550.00450.0023 2000.00480.00520.0026 2500.00430.00570.0029 3000.00380.00620.0031 3500.00340.00660.0033 4000.00310.00690.0035 This is the average rate of oxygen production from time 0  50 sec. This is also the “instantanious rate” for time t=25 sec

12. Rate vs. Time Rate vs. time lies at the heart of chemical kinetics. We need to know how the rate of the reaction is affected by time. Note: the rate of reaction slows as time moves forward. Graph on next slide Time (sec)[O 2 ]Time Inst. Rate 00.0000252.10E-05 500.0011751.40E-05 1000.00181251.00E-05 1500.00231757.00E-06 2000.00262255.00E-06 2500.00292755.00E-06 3000.00313254.00E-06 3500.00333753.00E-06 4000.0035

13. Why the slow down Recall that the thermo data predicts that this reaction will be spontaneous in the opposite direction. Hence we force the reaction forward by chemical means. But once NO and O 2 are produced the will react to form NO 2 in the reverse reaction. 2NO 2(g)  2NO (g) + O 2(g) The rate will ultimately flatten and the concentrations will remain constant. This is called equilibrium ΔG=0 Le Chatelier's principal etc. More on that later

14. Rate equation The rate equation relates the rate to the concentration of reactants. For the above example it would be expressed: Rate = k[NO 2 ] n The value of “k” is a constant unique to every experiement. The value of “n” is 0, a whole #, or a fraction. BOTH “n” and “k” MUST BE DETERMINED BY MANIPULATION OF EXPERIMENTAL DATA!!! I’ll show you how.

15. Concentration vs. Rate If you use excel this is really easy. The computer does all the work for you. I’ll show you how right now. Doing this by hand is not hard, but it’s very tedious. Learn to use excel!

16. Zero or first order? Rate = k[NO 2 ] n A zero order equation has an “n” value of zero. Hence, rate = k This would give a linear concentration vs. rate graph A 1 st order reaction has an “n” value of 1. Hence the concentration vs. rate graph is linear.

17. Here’s another example Is this first order. Time (sec)[N 2 O 5 ] 01 2000.88 4000.78 6000.69 8000.61 10000.54 12000.48 14000.43 16000.38 18000.34 20000.3

18. Use Excel or your calculator Sadly, you won’t have access to excel on the AP exam. Excel makes it easy to calculate slope (rate) Time[N2O5] 01 2000.88 4000.78 6000.69 8000.61 10000.54 12000.48 14000.43 16000.38 18000.34 20000.3 [N2O5]Inst. Rate 0.94-6.00E-04 0.83-5.00E-04 0.735-4.50E-04 0.65-4.00E-04 0.575-3.50E-04 0.51-3.00E-04 0.455-2.50E-04 0.405-2.50E-04 0.36-2.00E-04 0.32-2.00E-04 Note that on the concentration vs. rate table the relationship is directly proportional. 0.83  -5.00E-4 and 0.405  -2.50E-4 This show that when the concentration is halved so to is the rate.

19. If you don’t have excel You need to calculate the slope using your calculator. This is tedious, but you need to do it. Rate = k[N 2 O 5 ] n The directly proportional relationship we just established means that “n” = 1. “k” can be established as well, it’s the slope of the concentration vs. rate graph. You NEED to get good at calculating slope. There is another way to determine “n”…

20. The Method of initial rate Reactions can be set up so that the reverse reaction will be initially negligible. NH 4 + (aq) + NO 2 - (aq)  N 2(g) + 2H 2 O (l) Note, the N 2(g) will escape from the reaction vessel into the atmosphere. In accordance with Le Chatelier's Principal the reaction will continue to be driven to the right until all the reactants are spent. Thus the reaction rate should be unaffected by the reverse reaction early in the reaction.

21. What can you control??? You can control the initial concentrations of all reactants at the start of the reaction, and determine the rate of the creation. Consider the data below: Experiment [NH 4 + ] 0 [NO 2 - ] 0 Initial Rate 10.1000.0051.35E-07 20.1000.0102.70E-07 30.2000.0105.40E-07 Note: A [concentration] 0 means initial concentration

22. How do I interpret this? Compare experiments where only the concentration of ONLY ONE reactant is changed. You can only compare Exp1 to Exp 2 and Exp 2 to Exp 3. Exp1 CANNOT be compared to Exp 3 Experiment [NH 4 + ] 0 [NO 2 - ] 0 Initial Rate 10.1000.0051.35E-07 20.1000.0102.70E-07 30.2000.0105.40E-07

23. How do I interpret this? The rate equation will look like this: Rate = k[NH 4 + ] n [NO 2 - ] m Note: When Exp1 is compared to Exp2 you find that when [NO 2 - ] is doubled so to is rate. Hence [NO 2 - ] and rate are directly proportional and m=1. Because both [NH 4 + ] and [NO 2 - ] change from Exp1 to Exp3 the two can’t be compared. Use Exp2 and Exp3 to determine n Experiment [NH 4 + ] 0 [NO 2 - ] 0 Initial Rate 10.1000.0051.35E-07 20.1000.0102.70E-07 30.2000.0105.40E-07

24. How do I interpret this? 1.What happens to [NH 4 + ]? 2.What happens to rate? 3.What does that tell us about n? 4.What is the final form of the rate equation? Experiment [NH 4 + ] 0 [NO 2 - ] 0 Initial Rate 10.1000.0051.35E-07 20.1000.0102.70E-07 30.2000.0105.40E-07

25. The total order of the equation Finding the total order of the rate equation is easy. It’s simply the sum of all the exponents. Since n=1, and m=1, the total order of the equation is 2. It’s a second order equation.

26. You practice Use the table below to determine n,m,p and the overall order of the reaction below: BrO 3 - (aq) + 5Br - (aq) + 6H + (aq)  3Br 2(l) +3H 2 O (l) Experiment[BrO 3 - ] 0 [Br - ] 0 [H + ] 0 Initial Rate 10.10 8.00E-04 20.200.10 1.60E-03 30.20 0.103.20E-03 40.10 0.203.20E-03 Rate= k[BrO 3 - ] n [Br - ] m [H + ] p

27. Integrated rate equation We can calculate rate, and hence we can produce a rate vs. time graph. When it comes to rate vs. time graphs, it’s important to find a relationship between rate and time. This can be accomplished in several ways. The easiest way is to use excel. I’ll show you.

28. Harder You should also memorize how different graphs can be “straightened”. I’ll show some examples on slides to come. These relationships must be memorized, check the hall…

29. Linear (zero order) The linear graph is easy to spot, the lines are straight. This is a zero order relationship

30. Exponential (1 st order) This exponential graph is harder to spot. It looks a lot like the inverse graph on the next slide.

31. Inverse (2 nd order) As you can see, this look similar to the previous graph. When confronted with a rate vs. time graph like this, you need to “straighten” it. We’ll do this in the library tomorrow.

32. Straighten your line This takes time, but you need know how to do it. It’s easy with excel, but it’s a pain to do it with your calculator. Just be glad you don’t need to use a slide ruler. Not that I know what that is, I’m only 27.

33. Consider the data from before You can calculate the rate and make a rate vs. time graph. If it’s linear, then you have a first order equation. You can also calculate ln[N 2 O 5 ] and graph that. Time (sec)[N 2 O 5 ] 01 2000.88 4000.78 6000.69 8000.61 10000.54 12000.48 14000.43 16000.38 18000.34 20000.3

34. Manipulated concentration vs. time If a graph of ln[con.] vs. time is linear then the reaction has 1 st order kinetics. Why?Calculus For first order equations ln[con.]=-kt+ln[con.] 0 ln is the natural log -k is a constant t is time Time[N2O5]Ln[N2O5]1/[N2O5] 0101 2000.88-0.127831.136364 4000.78-0.248461.282051 6000.69-0.371061.449275 8000.61-0.49431.639344 10000.54-0.616191.851852 12000.48-0.733972.083333 14000.43-0.843972.325581 16000.38-0.967582.631579 18000.34-1.078812.941176 20000.3-1.203973.333333

35. When it’s graphed Note that this graph is a straight. You can prove this by calculating the slope of the first two and final two points. If they’re the same you’re good. If that doesn’t work plot time vs. 1/concentration.* *If that doesn’t work you need to check your work because I’ll only give you 0, 1 st, or 2 nd order reactions.

36. Look at the 1 st order equation ln[con.]=-kt+ln[con.] 0 Memorize this!!! Note that this is of the y=mx+b format -k is the slope and ln[con.] 0 is b This is useful because once you establish the order of the equation you can determine the concentration at any time.

37. And now… second order Memorize this!!! Note that this is of the y=mx+b format k is the slope and 1/[con.] 0 is b This is useful because once you establish the order of the equation you can determine the concentration at any time.

38. First or second order? Is this a first or second order reaction? 1.Create a ln[C 4 H 6 ] and a 1/[C 4 H 6 ] 2.Graph both vs. time. 3.Choose the linear one. Time (sec)[C4H6] 00.01000 10000.00625 18000.00476 28000.00370 36000.00313 44000.00270 52000.00241 62000.00208

39. What is the rate constant? Now that you’ve established that it’s a second order equation: 1.Plug in 0.01000 for [C 4 H 6 ] 0 2.Plug in couple of values from the table for t and [C 4 H 6 ] 3.Solve for k using math, it’s not that hard 4.Or find slope of the line, that’s k as well

40. Half life Half life is defined as the amount of time it takes for half of a sample to decompose/decay. There are equations on page 578 that will allow you to calculate half life if you have k and [con] 0