Rate Laws

OUTCOME QUESTION(S): C12-3-10 RATE MECHANICS Vocabulary & Concepts Explain the concept of a reaction mechanism. Include: rate determining step, intermediates, and catalysts Determine the reaction orders and rate law of a chemical reaction from experimental data. Explain the scientific process connecting a chemical reaction to its experimental rate law, and to the prediction of an appropriate reaction mechanism. Include: connecting the rate law to the RDS Vocabulary & Concepts

What change in the rate is seen when [A] is tripled? Determining rate law (using initial rates method): Measure the effect of changes in concentration of one reactant on rate, while keeping the other reactant constant. What change in the rate is seen when [A] is tripled? Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 same Analyzing the changes across two trials Can be done with math or in your head 4 possible outcomes for the order (1ST, 2ND, 3RD, zero)

= x = 14 (omelets can be made) Using Proportionalities 12 6 28 x One day, Starving Steve used 12 eggs to make 6 omelets. If he uses 28 eggs the next time, how many omelets can he make? The change seen in the ratio of eggs used will be proportional to the change seen in the ratio of omelets made 12 6 = 28 x x = 14 (omelets can be made) Creating proportional relationships allows for the identification of an unknown part of one ratio…

= = Using proportions (part 1): H2O2 + HI → products same Rate = k[H2O2] [HI] x 1 y 0.01522 k [0.1]x  [0.2]y 0.00761 k [0.1]x  [0.1]y Rate2 k [H2O2]x  [HI]y Rate1 k [H2O2]x  [HI]y = The change in the ratio of initial rates will be proportional to the change in the ratio of rate laws 0.0152 0.2 y 0.0076 0.1 = Create proportions that put the bigger values over the smaller values to avoid decimals Notice: to simplify the math, the ratios of k and [H2O2] can be eliminated as they equal 1 y = (1) first order 2.0 = [2]y

= = Using proportions (part 2): H2O2 + HI → products same Rate = k[H2O2] [HI] 1 x 1 0.01523 k [0.2]x  [0.1]1 0.00761 k [0.1]x  [0.1]1 = Now chose two new comparison trials to isolate and solve for the order of H2O2 Look at the simplicity of the math after values are cancelled – could you do this in your head? The goal is to simplify the proportions to the point of “this change in concentrations - to what power - produces this change in rates?” 0.0152 0.2 x 0.0076 0.1 = x = (1) first order 2 = [2]x Rate = k[H2O2][HI]

= = Using proportions (part 1): A + B → products same Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 same Rate = k[A] [B] y x 2 18.02 k [0.3]x  [0.2]y 2.01 k [0.1]x  [0.2]y Rate2 k [A]x  [B]y Rate1 k [A]x  [B]y = Try to look at the two trials and figure out the order in your head first 18.0 0.3 x 2.0 0.1 = x = (2) second order 9.0 = [3]x

= = Using proportions (part 2): A + B → products Rate = k[A] [B] Trial [A] (mol/L) [B] (mol/L) Initial Rate (mol/Ls) 1 0.10 0.20 2.0 2 0.30 18.0 3 0.40 16.0 Rate = k[A] [B] 2 1 y 16.03 k [0.2]2  [0.4]y 2.01 k [0.1]2  [0.2]y = Now chose two new comparison trials to isolate and solve for the order of B 8.0 = 4  [2]y [A] cannot cancel since it doesn’t stay the same (no trials across which it does) – but we know the order (x) of A, so we solve for y 16.0 0.2 2  0.4 y 2.0 0.1  0.2 = 2.0 = [2]y y = (1) first order 8.0 = [2]2  [2]y Rate = k[A]2[B]

= = = One more try…. A + B → products Rate = k[A] [B] Rate = k[A][B] Tl [A]i mol/L [B]i Initial Rate (mol/Ls) 1 0.0100 0.0240 1.45 x 10−4 2 0.0120 7.25 x 10−5 3 0.0200 0.0480 5.80 x 10−4 Rate = k[A] [B] x 1 1 y Rate = k[A][B] Ratios can be “stacked” any way – chose 1 vs 2 (not 2 vs 1) here to avoid decimals y = (1) first order x = (1) first order Try to look at the two trials and figure out the order in your head first 1.45-41 k [0.01]x  [0.024]y 7.25-52 k [0.01]x  [0.012]y 5.80-43 k [0.02]x  [0.048]1 1.45-41 k [0.01]x  [0.024]1 4 = [2]x  2 = 2 = [2]x 1.45-4 0.024 y 7.25-5 0.012 = 5.80-4 0.02 x  0.048 1 1.45-4 0.01  0.024 = 2 = [2]y 4 = [2]x  [2]1

Rate = k[A]2[B] 2nd order + 1st order = 3rd order overall Overall order of reaction is the sum of the orders: x + y = overall reaction order 2nd order + 1st order = 3rd order overall This is just a minor addition – you will be asked to state the order for each reactant and the overall reaction order

3 A (g) + B (g) + 2 C (g) 2 D (g) + 3 E (g) a. Write the rate law for this reaction. b. Calculate the value of the rate constant (k). c. Calculate the missing rate for Trial #5. d. Calculate the missing [A]in Trial #6.

= = 1 2 same same same same same same Rate2 k [A]x  [B]y  [C]z same same same same same same a. Write the rate law for this reaction. All of these orders could be done in your head without the need for paper Rate2 k [A]x  [B]y  [C]z Rate1 k [A]x  [B]y  [C]z = x = (1) first order 0.40 0.20 x 0.20 0.10 = y = (2) second order Rate = k[A][B]2 z = (0) zero order

k (constant), we can use the data from any trial to rearrange b. Calculate the value of the rate constant (k). Rate = k[A][B]2 To find the value of k (constant), we can use the data from any trial to rearrange Don’t include units for k (too complex)

Rate = (200)(0.50 mol/L)(0.40 mol/L)2 c. Calculate the rate for Trial #5. To find a missing rate use the calculated k value and concentrations from that trial to solve Rate = k[A][B]2 (k= 200) Rate = (200)(0.50 mol/L)(0.40 mol/L)2 Rate = 16 mol/L· s

d. Calculate the concentration of A in Trial #6. Rate = k[A][B]2 (k= 200) For a missing concentration, use the calculated k value and rearrange to solve

Conclusion: Everything in the Rate Law must be determined experimentally: 1. Write a basic rate law with all reactants 2. Determine order for each reactant (1st/2nd/3rd/0) 3. Re-write the rate law with the determined order of reaction for each 4. Solve any problems (k, missing values…)

CAN YOU / HAVE YOU? C12-3-10 RATE MECHANICS Vocabulary & Concepts Explain the concept of a reaction mechanism. Include: rate determining step, intermediates, and catalysts Determine the reaction orders and rate law of a chemical reaction from experimental data. Explain the scientific process connecting a chemical reaction to its experimental rate law, and to the prediction of an appropriate reaction mechanism. Include: connecting the rate law to the RDS Vocabulary & Concepts