Collection and Analysis of Rate Data

Collection and Analysis of Rate Data Lec.10 week 11 and week 12

Determining the Order and Rate Constant fromExperimental Data • The first step in the analysis is to convert the experimental observations to plots of {reactant] versus time. • Sometimes it is easy to determine the order directly from the experimental data, though in general this does not happen.

Rate Laws • Zero Order Reaction • Overall reaction order = 0 • Rate = k[A]0 = k Expt [A] (M) Rate (M/s) 1 0.50 2.00 2 1.00 2.00 3 1.50 2.00

First Order Reaction • Overall reaction order = 1 • Rate = k[A] Expt [A] (M) Rate (M/s) 1 0.50 1.00 2 1.00 2.00 3 2.00 4.00

Rate Laws • Second Order Reaction • Overall reaction order = 2 • Rate = k[A]2 Expt [A] (M) Rate (M/s) 1 0.50 0.50 2 1.00 2.00 3 1.50 4.50

Rate Laws • Third Order Reaction • Overall reaction order = 3 • Rate = k[A]3 Expt [A] (M) Rate (M/s) 1 0.50 0.25 2 1.00 2.00 3 1.50 6.75

REMEMBER • Rate laws must be determined experimentally. • Determine the instantaneous reaction rate at the start of the reaction (i.e. at t = 0) for various concentrations of reactants. • You CANNOT determine the rate law by looking at the coefficients in the balanced chemical equation!

To determine the rate law from experimental data, • identify two experiments in which the concentration of one reactant has been changed while the concentration of the other reactant(s) has been held constant • determine how the reaction rate changed in response to the change in the concentration of that reactant. • Repeat this process using another set of data in which the concentration of the first reactant is held constant while the concentration of the other one is changed.

Example (1) • Find the orders of the following reactions by inspection, then calculate the rate constants.

Answer. • (a) Determination of n: • If you just take a look at the data you will find if the concentration up by a factor of two, the rate will up by a factor of two; concentration up by a factor of three, rate up by a factor of three. (it means that the reaction is first order). • (a) Determination of k: • Since first order reaction then • rate = k (conc. Of reactant)= k CA • Then k = rate/CA. • All data should be used and the average taken, giving.

(b) Determination of n: • concentration up by a factor of two, rate up by a factor of four, i.e. 22; • concentration up by a factor of four, rate up by a factor of 16, i.e. 42. • Then the reaction is second order. • (b) Determination of k: • rate = k[reactant]2 and k = rate/[reactant]2, • giving an average value of k = 1.88* 10-3 mol/ dm3 S-1.

(c) Determination of n: • altering the concentration, rate remains the same then the reaction is zero order. • (c ) Determination of k: • rate = k[reactant]0 =k • and the average is 5.0 *10-4 mol dm-3 h-1 • Pay particular attention to the units of k in these examples.

Example • The following data were obtained for the reaction A + B + C products: • (a) Write the rate law for the reaction. Explain your reasoning in arriving at your rate law. [Hint: Table 1 is useful here.] • (b) What is the overall order of the reaction? • (c) Determine the value of the rate constant. [Hint: Use your rate law from (a) and appropriate data.] • (d) Use the data to predict the reaction rate for experiment 5.

solution • In runs 1 and 2, [B]0 and [C]0 are constant; when [A]0 doubles (f = 2), r0increases by = 2 = 21: thus 1st order in [A]0. • In runs 1 and 3, [A]0 and [C]0 are constant; when [B]0 increases by the factor f = 2.416, v0 increases by a factor of = 5.839 = (2.416)2: thus 2nd order in [B]0. • In runs 3 and 4, [A]0 and [B]0 are constant; when [C]0 increases by a factor of f = 3, v0 increases by a factor of = 9 = 32: thus 2nd order in [C]0. (b) Fifth order overall. (c) k = 2.85 x 1012 L4 mol-4 s-1. (Use the rate law from (a) above and the data from any run to calculate k. (d) rate = v0 = 0.0113 mol L-1 s-1.

Example: The initial reaction rate of the reaction A + B C was measured for several different starting concentration of A and B. The following results were obtained. Determine the rate law for the reaction. Expt # [A] (M) [B] (M) Initial rate (M /s) 1 0.100 0.100 4.0 x 10-5 2 0.100 0.200 8.0 x 10-5 3 0.200 0.100 16.0 x 10-5

x 2 x 2 Expt # [A] (M) [B] (M) Initial rate (M /s) 1 0.100 0.100 4.0 x 10-5 2 0.100 0.200 8.0 x 10-5 3 0.200 0.100 16.0 x 10-5 Rate = k [A]m [B]n Compare experiments 1 and 2 to find n: [A] = constant [B] = doubles Rate doubles: 8.0 x 10-5 = 2.0 4.0 x 10-5 [2]n = 2.0 n = 1 Rate = k[A]m[B]1

x 2 x 4 Expt # [A] (M) [B] (M) Initial rate (M /s) 1 0.100 0.100 4.0 x 10-5 2 0.100 0.200 8.0 x 10-5 3 0.200 0.100 16.0 x 10-5 Rate = k [A]m [B]n Compare experiments 1 and 3 to find m: [A] = doubles [B] = constant Rate quadruples: 16.0 x 10-5 = 4.0 4.0 x 10-5 [2]m = 4.0 n = 2 Rate = k[A]2[B]

You can also solve this using algebra: Rate = k [A]m [B]n Compare experiments 1 and 3 to find m: Rate 2 = k [0.200 M]m [0.100 M]n = 16.0 x 10-5 =4.0 Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5 [0.200 M]m = 4.0 [0.100 M]m 2m = 4.0 only if m = 2 [2.00]m = 2.0 Rate = k[A]2[B]n

Rate Laws You can also solve this using algebra: Rate = k [A]m [B]n Compare experiments 1 and 2 to find n: Rate 2 = k [0.100 M]m [0.200 M]n = 8.0 x 10-5 = 2.0 Rate 1 k [0.100 M]m [0.100 M]n 4.0 x 10-5 [0.200 M]n = 2.0 [0.100 M]n 2n = 2.0 only if n = 1 [2.00]n = 2.0 Rate = k[A]2[B]

The Integrated Rate Equations • Rate equations are differential equations that in simple cases can be easily integrated using standard methods of calculus. For example, if the rate law has the form • the equation may be integrated (from t = 0 to t, and from [A]= [A]0to [A]t) for different reaction orders in [A](i.e., different values of n). The equations given in the table below are obtained for n = 0, 1, and 2. The useful feature of these equations is that they may be written in straight-line form, y = mx + b (where m = slope and b = y-intercept are constants).

Reaction Order Using the Integrated Rate Equations

1-A straightforward graphical methodfor finding the rate law

Collection and Analysis of Rate Data