1. Chem 1310: Introduction to physical chemistry Part 2b: rates

2. Kinetics How fast does a reaction go? (and why?) Solution of crystal violet poured into solution of NaOH.

3. What is crystal violet? Large organic molecule, deep purple. Also known as "Gram's stain", used to distinguish types of bacteria ("Gram-negative" and "Gram- positive"). Disinfectant and toxic! Do not get on skin!

4. What is crystal violet? + CV +

5. How does it react with OH - ? + + - CV + OH -

6. How does it react with OH - ? + + - CV + OH - CVOH deep purplecolorless

7. Following the progress of the reaction Visually: color changes: purple  pink  colorless Quantitative: colorimetry measure transmitted light

8. Colorimetry Law of Lambert-Beer more convenient: use Absorbance A: I0I0 ItIt

9. Following the progress of the reaction t[CV + ] smol/L 0.05.00E-05 10.03.68E-05 20.02.71E-05 30.01.99E-05 40.01.46E-05 50.01.08E-05 60.07.93E-06 80.04.29E-06 100.02.32E-06

10. What is a rate ? The rate of a chemical reaction is the speed at which it transforms reagents into products. It is a measure of how fast things change with time. Compare with the speed of a car: measures how fast its position changes with time. In chemistry we look at concentrations and changes in them. A heterogeneous reaction

11. Calculating rates  t large: average rate over interval  t  t very small: instantaneous rate note the "-" sign! average rate over t = 10.. 20 s

12. Decrease of concentration follows a smooth curve. At each point, the rate is the (negative of the) slope of the curve. Average and instantaneous rates

13. We cannot measure rates directly. We can only measure concentrations. Rates can be estimated by drawing a smooth curve and estimating the slopes (tangents) by using average rates over real (but small) time intervals

14. Average and instantaneous rates The instantaneous rate is given by the slope (tangent)

15. Average and instantaneous rates The average rate over a larger interval differs significantly from the instantaneous rates.

16. Following the progress of the reaction t[CV + ]av rate smol/Lmol L -1 s -1 0.05.00E-05 1.32E-06 10.03.68E-05 9.70E-07 20.02.71E-05 7.20E-07 30.01.99E-05 5.30E-07 40.01.46E-05 3.80E-07 50.01.08E-05 2.87E-07 60.07.93E-06 1.82E-07 80.04.29E-06 9.85E-08 100.02.32E-06

17. Rates depend on concentrations From the table for Crystal Violet we can plot rate vs concentration: Looks pretty linear. The rate law is: rate = k [CV + ] with: k = 2.68*10 -2 s -1 (the rate constant). Another rate law determination

18. It is not always this easy Here we could get a set of rates from a single experiment. Often this cannot be done: you have to stop the reaction to analyze the results. Start reaction, stop it after a short interval, analyze. Do this for a number of initial concentrations  obtain a number of initial rates. Need lots of experiments to get accurate curves!

19. Using initial rates Initial concConc after 1 sInitial rate 5.00E-054.89E-051.14E-06 4.00E-053.91E-059.10E-07 3.00E-052.93E-056.82E-07 2.50E-052.44E-055.68E-07 2.00E-051.95E-054.55E-07 1.50E-051.47E-053.41E-07 1.00E-059.77E-062.27E-07

20. More complicated rate laws 2 NO + O 2  2 NO 2 expt[NO][O 2 ]initial rate 10.020.010.028 20.02 0.057 30.020.040.114 40.040.020.227 50.010.020.014

21. More complicated rate laws 2 NO + O 2  2 NO 2 expt[NO][O 2 ]initial rate 10.020.010.028 20.02 0.057 30.020.040.114 40.040.020.227 50.010.020.014 linear in [O 2 ] rate  [O 2 ]

22. More complicated rate laws 2 NO + O 2  2 NO 2 expt[NO][O 2 ]initial rate 10.020.010.028 20.02 0.057 30.020.040.114 40.040.020.227 50.010.020.014 definitely not linear in [NO] ! quadratic ???

23. More complicated rate laws 2 NO + O 2  2 NO 2 expt[NO][O 2 ]initial rate 10.020.010.028 20.02 0.057 30.020.040.114 40.040.020.227 50.010.020.014 quadratic (second-order) in [NO] rate  [NO] 2

24. Putting it all together 2 NO + O 2  2 NO 2 So we have rate  [O 2 ]([NO] constant) rate  [NO] 2 ([O 2 ] constant) Combining these gives rate = k [O 2 ][NO] 2 with k = 7069 L 2 mol -2 s -1 Reaction is: first-order in O 2 second-order in NO third-order overall expt[NO][O 2 ]initial rateest k 10.020.010.0287000 20.02 0.0577125 30.020.040.1147125 40.040.020.2277094 50.010.020.0147000 est k = rate/([O 2 ][NO] 2 )

25. What do we mean by "the rate" ? 2 NO + O 2  2 NO 2 For every single molecule of O 2, two molecules of NO are consumed and two molecules of NO 2 are produced. "The rate" of the reaction is defined as contains every component divided by its coefficient in the balanced reaction equation.

26. Stoichiometry and rate laws Rate laws can not be deduced from the balanced reaction equation! They must be measured experimentally. The results are not always intuitive. NO 2 + CO  NO + CO 2 rate = k [NO 2 ] 2 Consequences of a rate law

27. Stoichiometry and rate laws You can even have non-integer orders: trans-2-butenecis-2-butene rate(cis  trans) = k [cis-2-butene][I 2 ] ½ first-order in cis-2-butene half-order in I 2 A more complicated rate law