Concentration-Time Data

We will now use nonlinear regression to determine the rate law parameters from concentration–time data obtained in batch experiments. We recall that the combined rate law-stoichiometry-mole balance for a constant-volume batch reactor is

7-6

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We now integrate Equation (7-6) to give

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Rearranging to obtain the concentration as a function of time, we obtain

7-15

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Now we could use Polymath or MATLAB to find the values of α and k that would minimize the sum of squares of the differences between the measured and calculated concentrations. That is, for N data points,

7-16

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we want the values of α and k that will make s2 a minimum.

If Polymath is used, one should use the absolute value for the term in brackets in Equation (7-16), that is,

7-17

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Another way to solve for the parameter values is to use time rather than concentrations:

7-18

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That is, we find the values of k and α that minimize

7-19

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Finally, a discussion of weighted least squares as applied to a first-order reaction is provided in the Professional Reference Shelf R7.4 on the DVD-ROM.

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Example 7-3. Use of Regression to Find the Rate Law Parameters

We shall use the reaction and data in Examples E7-1 and E7-2 to illustrate how to use regression to find α and k′.

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The Polymath regression program is included on the DVD-ROM. Recalling Equation (E5-1.5)

E7-2.5

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and integrating with the initial condition when t = 0 and CA = CA0 for α ≠ 1.0

E7-3.1

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We can proceed two ways from this point, both of which will give the same result. We can search for the combination α and k that minimizes [σ2 = Σ(timtic)2], or we could solve Equation (E7-4.3) for CA and find α and k that minimize [σ2 = Σ(CSimCAic)2]. We shall choose the former.

Substituting for the initial concentration CA0 = 0.05 mol/dm3

E7-3.2

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The Polymath tutorial on the DVD-ROM shows screen shots of how to enter the raw data in Table E7-2.1 and to carry out a nonlinear regression on Equation (E7-3.2). For CA0 = 0.05 mol/dm3, that is, Equation (E7-3.1) becomes

E7-3.3

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We want to minimize s2 to give α and k′.

7-19

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The result of the first and second Polymath regressions are shown in Tables E7-3.1 and E7-3.2.

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The first regression gives α = 2.04, as shown in Table E7-3.1. We shall round off α to make the reaction second order, (i.e., α = 2.00). Now having fixed α at 2.0, we must do another regression (cf. Table E7-3.2) on k′ because the k′ given in Table E.7-3.1 is for α = 2.04. We now regress the equation

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The second regression gives k′ = 0.125 dm3/mol · min. We now calculate k

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Analysis: In this example we showed how to use non-linear regression to find k′ and α. The first regression gave α = 2.04 which we rounded to 2.00 and then regressed again for the best value of k′ for α = 2.0 which was k′ = 0.125 (dm3/mol)/min giving a value of the true specific reaction rate of k = 0.25 (mol/dm3)2/min. We note that the reaction order is the same as that in Examples 7-1 and 7-2; however, the value of k is about 8% larger. The r2 and other statistics are in Polymath’s output.

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