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Exp. 17: Kinetics: Determination of the order of a reaction

Exp. 17 – video (time: 36:06 minutes). 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.

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Exp. 17: Kinetics: Determination of the order of a reaction

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  1. Exp. 17 – video (time: 36:06 minutes) 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)

  2. Exp 17 experiment: I- 2 H2O2(aq) 2H2O (l) + O2(g) Decomposition of hydrogen peroxide catalyzed by iodide ion General expression for this reaction (rate law): Rate = k [H2O2]x [I-]y

  3. Rate = k [H2O2]x [I-]y 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 • A + A  A2 elementary (slow) • A2 + 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 = 1st order (linear) and rate changes same Change conc of reactant = 2nd order (square) and rate changes to the sq of change 20 x rate = rate 30 x rate = rate 21 x rate = 2rate 31 x rate = 3rate 22 x rate = 4rate 32 x rate = 9rate

  5. 3rd 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 23 x rate = 8 rate x + y + … = overall order

  6. Ex. A + B  P rate = k [A]x[B]y [A] [B]rate Exp 1 1M 1M 1M/s Two ways: 1.) inspection Exp 2 1M 2M 2M/s Exp 3 2M 1M 8M/s Compare exp1/exp2, [B] doubles and rate doubles, linear effect y= 1: 1st order Compare exp1/exp3, [A] doubles and rate is eight-fold; cube effect x= 3: 3rd order Rate = k [A]3 [B]

  7. Ex. A + B  P rate = k [A]x[B]y [A] [B]rate Exp 1 1M 1M 1M/s Exp 2 1M 2M 2M/s 2.) initial rate method Exp 3 2M 1M 8M/s Find x: exp3rate3 = k[A3]x [B3]y exp1 rate1 = k[A1]x [B1]y 8M/s = k[2M]x [1M]y 1M/s = k[1M]x [1M]y 8 = 2x log 8 = log 2x = 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 rate3 = k[A3]x [B3]y 8M/s = k[2M]3 [1M] 8 M/s = k (2M)3(1M) 8 M/s = k (8M3)(1M) M: 1 – 3 – 1 = – 3 M0.25 M0.50 M0.85 = M-1.10 1 /s = k M3 M: 0.25 – 0.50 – 0.85 = – 1.10 k = 1 M-3 s-1 rate = 1 M-3 s-1 [A]3 [B] overall order = 3 + 1 = 4th

  9. In experiment 17, rate of H2O2 decomposed will be determined by plotting volume of O2 gas generated vs. time. We will do 3 different experiments but only once each KI, mLH2O, mLH2O2, mLrate, mL/min Exp 1 10.00 15.00 5.00slope1 Exp 2 20.00 5.00 5.00slope2 Exp 3 10.00 10.00 10.00slope3 Note: each group only needs 50 mL of KI and H2O2 will be given out by the TA as needed. Y axisX 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 H2O2 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. Plot cumulative volume, mL vs time, min

  12. 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

  13. Slope of each line gives the rate for that experiment. Slope= rise = D y = D mL = y2-y1 = rate D mL run D x D time x2 – x1D 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 / H2O2 and total volume cancel out leaving the volume of solution the only variable.

  14. Conc of KI in experiment 1 (0.100 M KI) (10.00 mL) = conc KI1 30.00 mL Conc of KI in experiment 2 (0.100 M KI) (20.00 mL) = conc KI2 30.00 mL When compare exp 1 what happens? exp 2 Therefore, in this experiment D mL αD conc and the slope of the line equals the rate KI2 = 2 x KI1 because of vol

  15. Overall goal of experiment is to report the rate law expression for the decomposition of H2O2 : Rate = k (H2O2, mL)x (I-, mL)y Must give k with units, x, y, and overall order

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