Chapter 14:
Chemical Kinetics
Supplemental Instruction
IowaStateUniversity / Leader: / Chelsey
Course: / Chem 178
Instructor: / Greenbowe
Date: / Wed 1/26/11
  1. With the generic equation A + B + C  products, give: the one possible unimolecular rate law, the two possible bimolecular rate laws and the three possible termolecular rate laws.

Unimolecular: R = k[A]

Bimolecular:R = k [A]2 R = k [A][B]

Termolecular:R = k [A]3 R = k [A]2[B] R = k [A][B][C]

  1. The initial rate of the reaction A + B  C was measured for different starting concentrations of A and B. a) Determine the rate law for the reaction. b) Calculate the rate constant. c) What is the rate of the reaction when [A] = 0.050M and [B] = 0.100M?

Experiment #[A](M)[B](M)Initial Rate (M/s)

10.1000.2004.0x10-5

20.2000.10016.0x10-5

30.1000.1004.0x10-5

a)Rate = k [A]2

b)K = rate/[A]2 = (4.0x10-5M/s)/(0.100M)2 = 4.0x10-3 M-1s-1 one example of how to calculate the rate constant

c)Rate = (4.0x10-3M-1s-1) [0.050M]2 = 1.0x10-5 M/s

  1. The initial rate of the reaction 2NO + 2H2 N2 + 2H2O was measured for different starting concentrations of NO and H2. a) Determine the rate law for the reaction. b) Calculate the rate constant. c) What is the rate of the reaction when [NO] = 0.050M and [B] = 0.150M?

Experiment #[A](M)[B](M)Initial Rate (M/s)

10.100.101.23x10-3

20.200.102.46x10-3

30.100.204.92x10-3

a) Rate = k [NO]2[H2]

b)k = rate/[NO]2[H2] = (1.23x10-3M/s)/([0.10M]2[0.10M]) = 1.2 M-2s-1

c) Rate = (1.2 M-2s-1) [0.050M]2[0.150M]= 4.5x10-4M/s

  1. The integrated rate law for a first order reaction is ln[A]t – ln[A]0 = -kt. If you were to plot a ln[A]t vs. time graph, what would the slope of that graph equal to?

The opposite of the rate constant, -k

  1. Knowing that the integration of (d[x])/[x] = ln(x) and the integrated rate law of a first order reaction (given in #4), what would the integrated rate law be for a second order reaction knowing that the integration of (d[x])/[x2] = -1/x?

(1/[A]t) – (1/[A]0) = kt

  1. How would you find the rate constant of a second order reaction (rate = k[A]2) graphically if you were only given a table of the time and the concentration of the reactants?

You would convert the values of [A] to 1/[A] and then graph 1/[A] vs. time and find the slope which is equal to the rate constant, k.

  1. For the reaction 2H2O2 2H2O + O2, the concentration of H2O2 was measured over a period of 300 seconds. a) Determine the rate law for the reaction. (The integrated rate law of the reaction will give a linear line when plotted (ln[A]t or 1/[A]t) against time). Sketch the graph of ln[A]t vs. time and 1/[A]t vs. time for this reaction. b) Calculate the rate constant.

Time (s)[H2O2](M)

0.00.01000

50.00.00787

100.00.00649

200.00.00481

300.00.00380

a)Since the graph of 1/[H2O2] vs. time is linear, thereaction must be second order overall. Rate = k[H2O2]2

The sketch of ln[H2O2] vs. time should have a negative slope that is not quite linear (and concave up). The sketch of 1/[H2O2] should have a positive linear slope.

b)In calculating the slope you should find that k = 0.543 M-1s-1.