Data for, and answers to, the following problems are found in the folder "Ch. 14 (Nonlinear Regression)" in the "Problems & Solutions" folder on the CD.

1. First Order Reaction. The absorbance vs. time data in Table 14-1 was recorded for a chemical reaction. The reaction was believed to follow a firstorder exponential decay:

t, sec |
Aobsd |
t, sec |
Aobsd |

0 |
0.002000 |
10 |
0.000077 |

1 |
0.001441 |
11 |
0.000051 |

2 |
0.001070 |
12 |
0.000036 |

3 |
0.000739 |
13 |
0.000026 |

4 |
0.000542 |
14 |
0.000021 |

5 |
0.000367 |
15 |
0.000014 |

6 |
0.000263 |
16 |
0.000010 |

7 |
0.000200 |
17 |
0.000007 |

8 |
0.000140 |
18 |
0.000005 |

9 |
0.000100 |

Determine the rate constant k using the Solver.

2. Logistic Curve I. The data in Table 14.2 can be described by a simple logistic curve

Determine the constant a using the Solver.

X |
y |
x |
y |

-8 |
0.0150 |
1 |
0.6198 |

-7 |
0.0338 |
2 |
0.7292 |

-6 |
0.0468 |
3 |
0.8177 |

-5 |
0.0712 |
4 |
0.8843 |

-4 |
0.1152 |
5 |
0.9206 |

-3 |
0.1850 |
6 |
0.9547 |

-2 |
0.2716 |
7 |
0.9706 |

-1 |
0.3775 |
8 |
0.9863 |

0 |
0.4972 |
10 |
0.6198 |

3. Logistic Curve II. The logistic function a j y =---+ d

1 + e x takes into account offsets on the x-axis and the j-axis. Using the data in Table 14-3, determine the constants a, b, c and d using the Solver.

x |
y |

-5 |
9.99 |

-3 |
9.96 |

-1 |
10.06 |

0 |
10.08 |

1 |
10.29 |

2 |
10.48 |

3 |
10.73 |

4 |
10.84 |

5 |
11.00 |

7 |
11.00 |

9 |
11.03 |

10 |
11.05 |

4. Autocatalytic Reaction. The data in Table 14-4 describes the time course of an autocatalytic reaction with two pathways: an uncatalyzed path (A -» B )

and an autocatalytic path (A-»B). [A]0 = 0.0200 mol L . The rate law (the differential equation) is

Use any method from Chapter 10 to simulate the [B] = F(t) data, then use the Solver to obtain k0 and k,.

t, sec |
[Bl, mol L~1 |
t, sec |
[Bl, mol L~1 |

0 |
0.0000 |
550 |
0.0149 |

50 |
0.0002 |
600 |
0.0161 |

100 |
0.0000 |
650 |
0.0175 |

150 |
0.0008 |
700 |
0.0190 |

200 |
0.0009 |
750 |
0.0188 |

250 |
0.0024 |
800 |
0.0196 |

300 |
0.0034 |
850 |
0.0198 |

350 |
0.0052 |
900 |
0.0201 |

400 |
0.0077 |
950 |
0.0199 |

450 |
0.0094 |
1000 |
0.0203 |

500 |
0.0127 |

5. van Deemter Equation. Gas chromatography is an analytical technique that permits the separation and quantitation of complex mixtures. The mixture flows through a chromatographic column in a stream of carrier gas (usually helium), where the components separate and are detected. In the analysis of a sample of gasoline, for example, the components are separated based on their volatility, the lowest-boiling emerging from the separation column first. The degree of separation can be treated mathematically in the same way as for fractional distillation: a column can be considered to have a number of theoretical plates, just as a distillation tower in a refinery has actual "plates" for the separation of different petroleum products (naphtha, gasoline, diesel fuel, etc.). For gas chromatography, separation efficiency is usually expressed in terms of HETP (Height Equivalent to a Theoretical Plate), the column length divided by the number of theoretical plates. Separation efficiency is a function of the carrier gas flow rate v, as shown in the following figure. There is an optimum flow rate that provides the smallest HETP; too fast and there is not sufficient time for equilibration, too slow and gaseous diffusion allows the components to re-mix. The van Deemter Equation describes the relationship between HETP and carrier gas flow rate:

HETP = A+ B/v + Cv where v = carrier gas flow velocity. The data in Table 14-5 (also on the CD) shows measurements of HETP for a gas chromatographic column, using different flow rates.

v, cm/sec |
HETP, cm |

0.9 |
0.64 |

1.5 |
0.51 |

3.0 |
0.42 |

4.2 |
0.47 |

5.6 |
0.55 |

7.0 |
0.63 |

8.0 |
0.69 |

9.0 |
0.75 |

Use the Solver to obtain the least-squares coefficients A, B and C for the van Deemter equation.

6. NMR Titration. The protonation constants K\ and K2 of a diprotic acid H2A were determined by NMR titration. (Protonation constants, for example,

are used in this example because they simplify the equilibrium expressions The chemical shift 8 of a hydrogen near the acidic sites was measured at a number of pH values over the range pH 1 to pH 11. The data are shown in the following Figure (data table and figure are on the CD that accompanies this book).

Figure 14-12. NMR titration.

At any pH value there are three acid-base species in solution: H2A, HA-and A2"; the observed chemical shift is given by the expression

Figure 14-12. NMR titration.

At any pH value there are three acid-base species in solution: H2A, HA-and A2"; the observed chemical shift is given by the expression

where ay is the fraction of the species in the form containing j acidic hydrogens and is the chemical shift of the species. The a values can be calculated using the expressions below:

Use the Solver to determine K\, K2, So, 5\ and &i.

7. 2-D Regression. Using the Power vs. Speed and Throttle setting data in problem 13-6, find the coefficients for the polynomial fitting equation

8. Deconvolution of a Spectrum I. Use the data in Table 14-6 (also found on the CD in the worksheet "Deconvolution I") to deconvolute the spectrum. Close examination of the spectrum will reveal that it consists of four bands. Use a Gaussian band shape, i.e.,

Acalc ~ max eXP , where Aca\c is the calculated absorbance at a given wavelength, Amaii is the absorbance at A.max, jc is the wavelength or frequency (nm or cm-1), /j. is the x at ^max and s is an adjustable parameter related to, but not necessarily equal to, the standard deviation of the Gaussian distribution or to the bandwidth at half-height of the spectrum.

X, nm |
Absorbance |
X, nm |
Absorbance |
X, nm |
Absorbance |

350 |
0.032 |
420 |
0.860 |
490 |
0.373 |

360 |
0.055 |
430 |
1.050 |
500 |
0.222 |

370 |
0.097 |
440 |
1.146 |
510 |
0.127 |

380 |
0.163 |
450 |
1.120 |
520 |
0.071 |

390 |
0.279 |
460 |
0.995 |
530 |
0.040 |

400 |
0.429 |
470 |
0.790 |
540 |
0.024 |

410 |
0.645 |
480 |
0.569 |
550 |
0.012 |

9. Deconvolution of a Spectrum II. Use the data in the worksheet "Deconvolution II" to deconvolute the spectrum of K3[Mn(CN)6] in 2M KCN, shown in Figure 14-13. Use a Gaussian band shape. It should be clear from the figure that the spectrum contains multiple bands, perhaps five or more.

Figure 14-13. Spectrum of K3[Mn(CN)6].

200 250 300 350 400

Wavelength, nm

Figure 14-13. Spectrum of K3[Mn(CN)6].

10. Spectrum of a Mixture. The UV-visible spectra of pure solutions of cobalt2+, nickel2+ and copper2+ salts, and of a mixture of the three, are given on the CD-ROM over the wavelength range 350-820 nm. Instead of using absorbance readings at only three wavelengths to calculate the concentrations of the three salts in the mixture (as was done in problem 9-4), use the data at all 236 wavelength data points to calculate the three concentrations. Use the relationship A = sbc, where e, the molar absorptivity, is a dimensionless constant for a particular species at a particular wavelength, b is the light path length (1.00 cm in this experiment) and c is the molar concentration. For the mixture, ^obsd = sCoCCo + £NjCNi + £c»CCu at each wavelength.

Use the Solver Statistics macro to obtain the standard deviations of the three concentrations.

11. Multiple-Wavelength Regression. Dissociation of the second hydrogen ion of Tiron (l,2-dihydroxybenzene-3,5-disulfonate, H2L) does not begin until the pH is raised above 10. The pKa2 of Tiron was determined spectrophotometrically by recording the spectrum at constant Tiron concentration and varying pH. The spectra are shown in the following figure; the absorbance readings (from 226 nm to 360 nm in 2-nm increments) at each pH value are tabulated on the CD that accompanies this text.

Figure 14-14. Spectra of Tiron at pH values between 10 and 12.

Wavelength, nm

Figure 14-14. Spectra of Tiron at pH values between 10 and 12.

The equilibrium reaction being measured is (charges omitted for clarity)

The dissociation of H2L to HL~ is complete at pH values of 10 and higher, and can be neglected. The concentrations of L and HL are given by the following expressions:

where LT is the total concentration of Tiron in the solution. The absorbance at a given wavelength is the sum of the contributions of the two species, that is,

where e is the molar absorptivity of the species, a constant at a given wavelength.

Calculate the Ka value and the sL and Ehl values at each wavelength, in one global minimization. (Excel's Solver can handle up to 200 changing cells, so we are pushing the limit here.) You will need to calculate the sum-of-squares-of-residuals for each wavelength, and minimize the "grand total" for all wavelengths. The Solver may have trouble "digesting" all this data. If so, use the Solver with data at a single wavelength to get the values of Ka, sL and eHl, then use these as starting value for a global minimization.

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