1. Kinetics Problems

2. Finding Reaction Orders by Experiment - I Given the reaction: O 2 (g) + 2 NO (g) 2 NO 2 (g) The rate law for this reaction in general terms is:Rate = k [O 2 ] m [NO] n To find the reaction orders, we run a series of experiments, each with a different set of reactant concentrations, and determine the initial reaction rates. Experiment Initial Reactant Concentrations (mol/l) Initial Rate O 2 NO (mol/L·s) 1 1.10 x 10 -2 1.30 x 10 -2 3.21 x 10 -3 2 2.20 x 10 -2 1.30 x 10 -2 6.40 x 10 -3 3 1.10 x 10 -2 2.60 x 10 -2 12.8 x 10 -3 4 3.30 x 10 -2 1.30 x 10 -2 9.60 x 10 -3 5 1.10 x 10 -2 3.90 x 10 -2 28.8 x 10 -3 By using the data from experiments 1&2 we can calculate the values of the coefficients m & n for the rate equation:

3. Determining Reaction Orders by Experiment - II Rate 2 [O 2 ] 2 m [O 2 ] 2 Rate 1 [O 2 ] 1 m [O 2 ] 1 = m The oxygen concentration changes for these two experiments but NO is the same and the constant k is the same, they will cancel out. Using the concentrations from the preceding table and the rate: 6.40 x 10 -3 mol/L s 2.20 x 10 -2 mol/L 3.21 x 10 -3 mol/L s 1.10 x 10 -2 mol/L m = 1.99 = (2.00) m m = 1 The reaction is first order in O 2, when O 2 doubles, the rate doubles. To find the order of NO we will compare experiments 3 & 1 in which oxygen is constant, but NO doubles, and the rate between the two experiments goes up substantially.

4. Determining Reaction Orders by Experiment - III Rate 3 k[O 2 ] 3 m [NO] 3 n Rate 1 k[O 2 ] 1 m [NO] 1 n = As before, k and for this one, the concentration of oxygen is not varying, and will cancel out, leaving only the rate and the concentration of No to consider. Rate 3 [NO] 3 Rate 1 [NO] 1 = Substituting in the values from the table, we get the following: 12.8 x 10 -3 mol/L s 2.60 x 10 -2 mol/L = 3.21 x 10 -3 mol/L s 1.30 x 10 -2 mol/L 3.99 = (2.00) n n = 2 The reaction is second order in NO: when [NO] doubles, the rate quadruples, Thus the rate law is: Rate = k [O 2 ][NO] 2

5. Determining the Reactant Concentration at a Given Time! - I Problem: The conversion of cyclopropane to propene occurs with a first order rate constant equal to 1.7 x 10 -2 hr -1. (a) If the initial concentration of cyclopropane is 1.200 M, what is the concentration after 20.00 hrs? (b) What fraction of cyclopropane has decomposed? Plan: (a) To find the concentration of cyclopropane at time t, [C 3 H 6 ] t. the problem tells us it is a first-order reaction so we use the integrated rate law: We know k,t and [C 3 H 6 ] 0, so we can solve for [C 3 H 6 ] t. (b) The fraction decomposed is the concentration that has decomposed divided by the initial concentration: ln [C 3 H 6 ] 0 - ln [C 3 H 6 ] t = kt Fraction decomposed = [C 3 H 6 ] 0 - [C 3 H 6 ] t [C 3 H 6 ] 0

6. Determining the Reactant Concentration at a Given Time - II Solution: (a) Rearranging the integrated rate expression and solving for ln [C 3 H 6 ] t : ln [C 3 H 6 ] t = ln[C 3 H 6 ] 0 - kt ln [C 3 H 6 ] t = ln(1.200 mol/L) -(1.7 x 10 -2 hr -1 )(20.00 hr) Ln[C 3 H 6 ] t = 0.18232 - 0.03400 = 0.14832 [C 3 H 6 ] t = 1.159 mol/L (b) Finding the fraction that has decomposed after 20.00 hrs: Fraction decomposed = [C 3 H 6 ] 0 - [C 3 H 6 ] t [C 3 H 6 ] 0 Fraction decomposed = = 0.03417 (1.200 mol/L) (1.200 mol/L) - (1.159 mol/L)

7. Determining Molecularity and Rate Laws for Elementary Steps Problem: The following two reactions are for the stepwise neutralization of sulfuric acid by gaseous ammonia: Plan: We find the overall equation from the sum of the simple steps. The molecularity of each step equals the total number of reactant particles. We write the rate law for each step using the the molecularities as reactant orders. Solution: (a) Writing the overall equation: (1) NH 3 (g) + H 2 SO 4 (g) NH 4 + (g) + HSO 4 - (g) (2) NH 3 (g) + HSO 4 - (g) NH 4 + (g) + SO 4 -2 (g) (a) Write the overall balanced equation. (b) Determine the molecularity of each step. (c) Write the rate law for each step. NH 3 (g) + H 2 SO 4 (g) NH 4 + (g) + HSO 4 - (g) NH 3 (g) + HSO 4 - (g) NH 4 + (g) + SO 4 -2 (g) 2 NH 3 (g) + H 2 SO 4 (g) 2 NH 4 + (g) + SO 4 -2 (g)

8. Determining Molecularity and Rate Laws for Elementary Steps Solution Cont. (b) Determining the molecularity of each step: (1) Step #1 has two reactants, ammonia and sulfuric acid, and is therefore bimolecular (2) Step#2 has two reactants, ammonia and hydrogen sulfate, and is therefore bimolecular (c) Writing the rate laws for the elementary reactions: (1) Rate 1 = k 1 [NH 3 ][H 2 SO 4 ] (2) Rate 2 = k 2 [NH 3 ][HSO 4 - ] The overall reaction and rate is: 2 NH 3 (g) + H 2 SO 4 (g) 2 NH 4 + (g) + SO 4 -2 Rate = k[NH 3 ] 2 [H 2 SO 4 ]

9. For the reaction: A + 2B  2C, the following data were collected: [A][B]initial rate 0.20 M0.10 M0.300 mol/L min 0.50 M0.30 M0.360 mol/L min 0.80 M0.30 M1.44 mol/L min React 1

10. a)What is the expected initial rate if [A]0 = 0.50 M, and [B]0 = 0.80 M? b)Assuming the initial rate is given in terms of production of C, what is the initial rate of change in concentration A in the first experiment? d)Assuming the initial rate is given in terms of production of C, what is the initial rate of change in concentration B in the first experiment? React 1

11. How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? React 2

12. For the reaction: A + 2B  2C, the following data were collected: [A][B]initial rate 0.20 M0.10 M0.300 mol/L min 0.50 M0.30 M0.360 mol/L min 0.80 M0.30 M1.44 mol/L min React 1

13. a)What is the expected initial rate if [A]0 = 0.50 M, and [B]0 = 0.80 M? b)Assuming the initial rate is given in terms of production of C, what is the initial rate of change in concentration A in the first experiment? d)Assuming the initial rate is given in terms of production of C, what is the initial rate of change in concentration B in the first experiment? React 1

14. How do exponents (orders) in rate laws compare to coefficients in balanced equations? Why? React 2