For the gas-phase reaction 2N2O5 → 4NO2 + O2 some rate constants are 105k/s—11.696.7324.9...

For the gas-phase reaction 2N2O5 → 4NO2 + O2 some rate constants are

(a) Use a spreadsheet to plot ln k versus 1/T and from the re­gression line find Ea and A. Use these values in the Arrhenius equation to find ki,calc at each temperature. Then calculate the percent error for each ki and calculate ∑i (ki,calc − ki)2. (b) Use the Solver to find Ea and A by minimizing ∑i (ki,calc − ki)2. To help the Solver, do the following: Take the initial Ea and A guesses as the values found in (a). In the Solver Options box, (1) check Use Automatic Scaling (this choice is appropriate when quantities involved in the optimization differ by several orders of magnitude); (2) check Central Derivatives (this gives a more accurate estimation of the derivatives used in the mini­mization); (3) change the Solver default Precision and Converg­ence values to values that are at least 105 times as small. Repeat the minimization several times, each time starting from differ­ent Ea and A guesses, and select the Ea and A pair that gives the lowest ∑i (ki,calc − ki)2. Compare ∑i (ki,calc − ki)2 with the value found in (a) and compare the percent errors in ki,calc with those in (a). Minimization of ∑i (ki,calc − ki)2 gives greater weight to the larger (higher-T) values of ki and gives a good fit to these values at the expense of fitting the smaller ki values.

The proper procedure to find Ea and A is to do several ki­netics runs at each T so that a standard deviation σi can be cal­culated for k at each T. Statistical weights ωi are calculated as  and the quantity ∑i ωi(ki,calc − ki)2 is minimized by using a program such as the Excel Solver. Alternatively, one can transform the Arrhenius equation to linear form by taking the log of both sides. When this linearization is done, the sta­tistical weights ωi must be adjusted to new values  For the transformation from k to ln k, it turns out that  If only a single measured rate constant is available at each T, a not unreasonable assumption is that each ki value has about the same percent error. This means that the standard de­viation σi (which is a measure of the typical error) is approxi­mately proportional to ki; σi = cki, where c is a constant. In this case, the weights for the linear plot become  , so all points of the linear plot have the same weight. The procedure of part (a) is then appropriate.