1. Exp. 17: Kinetics: Determination of the order of a reaction Chemical Kinetics – is the study of rates of chemical reactions. The rate of a chemical reaction describes how fast a reaction proceeds. (basically: how quickly are reactants consumed and products produced) Exp. 17 – videoExp. 17 – video(time: 36:06 minutes)

2. Exp 17 experiment: I - 2 H 2 O 2 (aq)  2H 2 O (l) + O 2 (g) Decomposition of hydrogen peroxide catalyzed by iodide ion General expression for this reaction (rate law): Rate = k [H 2 O 2 ] x [I - ] y

3. Note: brackets typically refers to concentration in M. k is the specific rate constant and is related to a particular rxn and temperature x and y are referred to as the order of the reactant; it describes how the reactant concentration affects the rate of the reaction. Values are typically a positive integer but not always. orders determined experimentally not by stoichiometry of balanced equation 1)A + A  A 2 elementary (slow) 2)A 2 + B  C elementary (fast) 2A + B  C molecular

4. Order of a reactant is determined by the effect changing the reactant conc has on rate Change conc of reactant = 0 order (must be present) and rate remains same Change conc of reactant = 1 st order (linear) and rate changes same Change conc of reactant = 2 nd order (square) and rate changes to the sq of change 2 0 x rate = rate 3 0 x rate = rate 2 1 x rate = 2rate 3 1 x rate = 3rate 2 2 x rate = 4rate 3 2 x rate = 9rate

5. 3 rd order? Overall order of reaction equals the sum of all the orders. How do we determine the order? Collect data carefully with a well designed experiment 2 3 x rate = 8 rate x + y + … = overall order

6. Ex. A + B  Prate = k [A] x [B] y [A][B]rate Exp 11M1M1M/sTwo ways: 1.) inspection Exp 21M2M2M/s Exp 32M1M8M/s Compare exp1/exp2, [B] doubles and rate doubles, linear effect y= 1: 1 st order Compare exp1/exp3, [A] doubles and rate is eight-fold; cube effect x= 3: 3 rd order Rate = k [A] 3 [B]

7. Ex. A + B  Prate = k [A] x [B] y [A][B]rate Exp 11M1M1M/s Exp 21M2M2M/s2.) initial rate method Exp 32M1M8M/s Find x: exp3 rate 3 = k[A 3 ] x [B 3 ] y exp1 rate 1 = k[A 1 ] x [B 1 ] y 8M/s = k[2M] x [1M] y 1M/s = k[1M] x [1M] y 8 = 2 x log 8 = log 2 x = x log 2 x = log 8 = 3 log 2

8. Solving for k; we will use exp 3 data, but you can use any set rate 3 = k[A 3 ] x [B 3 ] y 8M/s = k[2M] 3 [1M] 8 M/s = k (2M) 3 (1M) 8 M/s = k (8M 3 )(1M) 1 /s = k M 3 k = 1 M -3 s -1 rate = 1 M -3 s -1 [A] 3 [B] overall order = 3 + 1 = 4 th M: 1 – 3 – 1 = – 3 M 0.25 M 0.50 M 0.85 M: 0.25 – 0.50 – 0.85 = – 1.10 = M -1.10

9. In experiment 17, rate of H 2 O 2 decomposed will be determined by plotting volume of O 2 gas generated vs. time. We will do 3 different experiments but only once each KI, mLH 2 O, mLH 2 O 2, mLrate, mL/min Exp 110.00 15.00 5.00 slope 1 Exp 220.005.005.00slope 2 Exp 310.0010.0010.00slope 3 Note: each group only needs 50 mL of KI and H 2 O 2 will be given out by the TA as needed. Y axis X axis

10. Pg 117 describes how the experiment will be conducted Important points: - levels of buret and drying tube must be equal for readings - you can dump excess water out - make sure all air bubbles are out - check for leaks - Add H 2 O 2 just before you are ready for exp to begin -wait 1 – 2 mL before call time “0 min” note: we are following the change in volume over a particular change in time; therefore, it doesn’t matter when call time zero.

11. time, min vol reading mL Cumulative vol., mL 0.002.000.00 1.004.102.10 2.006.154.15 3.008.306.30 4.0010.458.45 5.0012.3010.30 Plot cumulative volume, mL vs time, min

13. Graphing: -Must have a descriptive title -Label both axis with units -Large graph over majority of page, select axis increments which allows for this -Legend explaining data -Best line, not connecting the dots, and do not force through zero Graph 30 blocks available on x axis and 5 minutes of data: 5 min = 0.16 min  0.20 min 30 blocks block block

14. Slope of each line gives the rate for that experiment. Slope=rise=  y =  mL = y 2 -y 1 = rate  mL run  x  timex 2 – x 1  min Note: pick points on best line, not data points In this experiment, we are using volume instead of concentration in our rate unit. Notice in the experiment that the total volume is held constant to 30 mL in every experiment (volume of water changes to assure this). This means that because the way the experiment is designed that the original conc. of KI / H 2 O 2 and total volume cancel out leaving the volume of solution the only variable.

15. Conc of KI in experiment 1 (0.100 M KI) (10.00 mL) = conc KI 1 30.00 mL Conc of KI in experiment 2 (0.100 M KI) (20.00 mL) = conc KI 2 30.00 mL When compare exp 1what happens? exp 2 Therefore, in this experiment  mL α  conc and the slope of the line equals the rate KI 2 = 2 x KI 1 because of vol

16. Overall goal of experiment is to report the rate law expression for the decomposition of H 2 O 2 : Rate = k (H 2 O 2, mL) x (I -, mL) y Must give k with units, x, y, and overall order