You can observe a lot just by watching.
—Yogi Berra, New York Yankees
For batch systems, the usual procedure is to collect concentration–time data, which we then use to determine the rate law. Table 71 gives a sevenstep procedure we will emphasize in analyzing reaction engineering data.
Data for homogeneous reactions is most often obtained in a batch reactor. After postulating a rate law in Step 1 and combining it with a mole balance in Step 2, we next use any or all of the methods in Step 5 to process the data and arrive at the reaction orders and specific reactionrate constants.
Analysis of heterogeneous reactions is shown in Step 6. For gas–solid heterogeneous reactions, we need to have an understanding of the reaction and possible mechanisms in order to postulate the rate law in Step 6B. After studying Chapter 10 on heterogeneous reactions, one will be able to postulate different rate laws and then use Polymath nonlinear regression to choose the “best” ratelaw and reactionratelaw parameters (see Example 103 on pages 494–496).
The procedure we will use to delineate the rate law and ratelaw parameters is given in Table 71.
TABLE 71 STEPS IN ANALYZING RATE DATA

Batch reactors are used primarily to determine ratelaw parameters for homogeneous reactions. This determination is usually achieved by measuring concentration as a function of time and then using either the integral, differential, or nonlinear regression method of data analysis to determine the reaction order, α, and specific reactionrate constant, k. If some reaction parameter other than concentration is monitored, such as pressure, the mole balance must be rewritten in terms of the measured variable (e.g., pressure, as shown in the example in Solved Problems on the CRE Web site at http://www.umich.edu/~elements/6e/07chap/pdf/excd51.pdf).
Process data in terms of the measured variable.
When a reaction is irreversible, it is possible in many cases to determine the reaction order α and the specific rate constant by either nonlinear regression or by numerically differentiating concentration versus time data. This latter method is most applicable when reaction conditions are such that the rate is essentially a function of the concentration of only one reactant; for example, if, for the decomposition reaction
A → Products
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}={\mathit{k}}_{\mathrm{A}}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}\hfill & \left(\mathrm{71}\right)\end{array}$
then the differential method may be used.
Assume that the rate law is of the form ${r}_{\text{A}}={k}_{\text{A}}{C}_{\text{A}}^{\alpha}$.
However, by utilizing the method of excess, it is also possible to determine the relationship between –r_{A} and the concentration of other reactants. That is, for the irreversible reaction
A + B → Products
with the rate law
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}={\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathit{C}}_{\mathrm{B}}^{\mathit{\beta}}& \left(\mathrm{72}\right)\end{array}$
where α and β are both unknown, the reaction could first be run in an excess of B so that C_{B} remains essentially unchanged during the course of the reaction (i.e., C_{B} ≈ C_{B0}) and
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathit{k}\mathrm{\prime}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\mathrm{73}\right)\end{array}$
where
$\mathit{k}\mathrm{\prime}={\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{B}}^{\mathit{\beta}}\approx {\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{B}\mathrm{0}}^{\mathit{\beta}}$
Method of excess
After determining α, the reaction is carried out in an excess of A, for which the rate law is approximated as
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathit{k}"{\mathit{C}}_{\mathrm{B}}^{\mathit{\beta}}& \left(\mathrm{74}\right)\end{array}$
where $k\mathrm{\prime}\mathrm{\prime}={\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}\approx {\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{A}\mathrm{0}}^{\mathit{\alpha}}$
Once α and β are determined, k_{A} can be calculated from the measurement of –r_{A} at known concentrations of A and B
$\begin{array}{cc}{\mathit{k}}_{\mathrm{A}}=\frac{\mathrm{}{\mathit{r}}_{\mathrm{A}}}{{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathit{C}}_{\mathrm{B}}^{\mathit{\beta}}}={\frac{\left(\mathrm{d}{\mathrm{m}}^{\mathrm{3}}\mathrm{/}\mathrm{m}\mathrm{o}\text{l}\right)}{\text{S}}}^{\alpha +\beta 1}& \left(\mathrm{75}\right)\end{array}$
Both α and β can be determined by using the method of excess, coupled with a differential analysis of data for batch systems.
One could also carry out many experiments using the methods of initial rates (http://www.umich.edu/~elements/6e/07chap/prof71.html) to find –r_{A0} as a function of C_{A0} and C_{B0} and then use regression to find k, α, and β as shown on the CRE Web site in Professional Reference Shelf R3 (http://www.umich.edu/~elements/6e/07chap/pdf/leastsquares.pdf).
The integral method is the quickest method to use to determine the rate law if the order turns out to zero, first, or second order. In the integral method, we guess the reaction order, α, in the combined batch reactor mole balance and rate law equation
$\begin{array}{cc}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathrm{}\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\mathrm{76}\right)\end{array}$
The integral method uses a trialanderror procedure to find the reaction order.
and integrate the differential equation to obtain the concentration as a function of time. If the order α we assumed is correct, then the appropriate plot (determined from this integration) of the concentration–time data should be linear. The integral method is used most often when the reaction order is known and it is desired to evaluate the specific reactionrate constant at different temperatures to determine the activation energy.
In the integral method of analysis of rate data, we are looking for the appropriate function of concentration corresponding to a particular rate law that is linear with time. We need to be familiar with the methods of obtaining these linear plots for reactions of zero, first, and second order.
It is important to know how to generate linear plots of functions of C_{A} versus t for zero, first, and secondorder reactions.
For the reaction
A → Products
carried out in a constantvolume batch reactor, the mole balance is
$\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}={\mathit{r}}_{\mathrm{A}}$
For a zeroorder reaction, r_{A} = –k, and the combined rate law and mole balance is
$\begin{array}{cc}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=k& \left(\mathrm{77}\right)\end{array}$
Integrating C_{A} = C_{A0} with at t = 0, we have
Zero order
A plot of the concentration of A as a function of time will be linear (Figure 71) with slope (–k) for a zeroorder reaction carried out in a constantvolume batch reactor.
If the reaction is first order (Figure 72), integration of the combined mole balance and the rate law
$\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathit{k}{\mathit{C}}_{\mathrm{A}}$
with the limits C_{A} = C_{A0} at t = 0 gives
First order
Consequently, we see that the slope of a plot of [ln (C_{A0}/C_{A})] as a function of time is indeed linear with slope k.
If the reaction is second order (Figure 73), then
$\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathrm{2}}$
#important
We need to know these plots for zero, first and secondorder reactions.
Integrating, with C_{A} = C_{A0} initially (i.e., t = 0), yields
$\begin{array}{c}\hline \frac{1}{{C}_{\text{A}}}\frac{1}{{C}_{\text{A0}}}=kt\\ \hline\end{array}\begin{array}{c}\left(7\text{10}\right)\end{array}$
Second order
We see that for a secondorder reaction a plot of (1/C_{A}) as a function of time should be linear with slope k.
In Figures 71, 72, and 73, we saw that when we plotted the appropriate function of concentration (i.e., C_{A}, ln C_{A}, or 1/C_{A}) versus time, the plots were linear, and we concluded that the reactions were zero, first, or second order, respectively. However, if the plots of concentration data versus time had turned out not to be linear, such as shown in Figure 74, we would say that the proposed reaction order did not fit the data. In the case of Figure 74, we would conclude that the reaction is not second order. After finding that the integral method for zero, first, and second orders also do not fit the data, one should use one of the other methods discussed in Table 71, such as the differential method of analysis or nonlinear regression.
The idea is to arrange the data so that a linear relationship is obtained.
It is important to restate that, given a reactionrate law, you should be able to quickly choose the appropriate function of concentration or conversion that yields a straight line when plotted against time or space time. The goodnessoffit of such a line may be assessed statistically by calculating the linear correlation coefficient, r^{2}, which should be as close to 1 as possible. The value of r^{2} is given in the output of Polymath’s nonlinear regression analysis.
The liquidphase reaction
Trityl (A) + Methanol (B) → Products (C)
was carried out in a batch reactor at 25°C in a solution of benzene and pyridine in an excess of methanol (${\mathrm{C}}_{\mathrm{B}\mathrm{0}}=\mathrm{0}\mathrm{.}\mathrm{5}\frac{\mathrm{m}\mathrm{o}\mathrm{1}}{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}$). (We need to point out that this batch reactor was purchased at the Sunday markets in Riça, Jofostan.) Pyridine reacts with HCl, which then precipitates as pyridine hydrochloride thereby making the reaction irreversible. The reaction is first order in methanol. The concentration of triphenyl methyl chloride (A) was measured as a function of time and is shown Table E71.1.
#BeCareful Watch for leaks!
TABLE E71.1 RAW DATA
t (min) 
0 
50 
100 
150 
200 
250 
300 
C_{A} (mol/dm^{3}) 
0.05 
0.038 
0.0306 
0.0256 
0.0222 
0.0195 
0.0174 
Use the integral method to confirm that the reaction is second order with regard to triphenyl methyl chloride.
Determine the pseudo rate constant k′ = kC_{B0}.
Assume the reaction is first order in methanol and determine the true specific reaction rate constant, k, for this over all thirdorder reaction.
Solution
We use the powerlaw model, Equation (72), along with information from the problem statement that the reaction is first order in methanol, (B), that is, β = 1 to obtain
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathit{k}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathrm{C}}_{\mathrm{B}}& \left(\text{E71.1}\right)\end{array}$
Excess methanol: The initial concentration of methanol (B) is 10 times that of trityl (A), so even if all A were consumed, 90% of B remains. Consequently, we will take the concentration of B as a constant and combine it with k to form
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathit{k}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathrm{C}}_{\mathrm{B}\mathrm{0}}=\mathit{k}\mathrm{\prime}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\text{E71.2}\right)\end{array}$
where k′ is the pseudo rate constant k′ = kC_{B}_{0} and k is the true rate constant. Substituting α = 2 and combining with the mole balance on a batch reactor, we obtain
$\begin{array}{cc}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathrm{k}\mathrm{\prime}{\mathrm{C}}_{\mathrm{A}}^{\mathrm{2}}& \left(\text{E71.3}\right)\end{array}$
Integrating with C_{A} = C_{A0} at t = 0
$\begin{array}{cc}\mathit{t}\text{}=\frac{\mathrm{1}}{\mathit{k}\mathrm{\u2019}}\mathrm{[}\frac{\mathrm{1}}{{\mathrm{C}}_{\mathrm{A}}}\mathrm{}\frac{\mathrm{1}}{{\mathrm{C}}_{\mathrm{A}\mathrm{0}}}\mathrm{]}& \left(\text{E71.4}\right)\end{array}$
Rearranging
from Equation (E71.5), we see that if the reaction is indeed second order then a plot of (1/C_{A}) versus t should be linear. Using the data in Table E71.1, we calculate (1/C_{A}) to construct Table E71.2.
TABLE E71.2 PROCESSED DATA
t (min) 
0 
50 
100 
150 
200 
250 
300 
C_{A} (mol/dm^{3}) 
0.05 
0.038 
0.0306 
0.0256 
0.0222 
0.0195 
0.0174 
1/C_{A} (dm^{3}/mol) 
20 
26.3 
32.7 
39.1 
45 
51.3 
57.5 
In a graphical solution, the data in Table E71.2 is used to construct a plot of (1/C_{A}) as a function of t, which will yield the pseudospecific reaction rate k′. This plot is shown in Figure E71.1. Again, Excel or Polymath could be used to find k′ from the data in Table E71.2.
We note in passing that if you used Excel or similar analysis, then the slope would most likely be part of your computer output.
The slope of the line is the specific reaction rate k′, which using the points at t = 0 and t = 200, k is calculated to be
$\begin{array}{cc}k\prime =\frac{\mathrm{\Delta}\mathrm{(}\frac{\mathrm{1}}{{\mathrm{C}}_{\mathrm{A}}}\mathrm{)}}{\mathrm{\Delta}\mathit{t}}=\frac{\mathrm{(}\mathrm{4}\mathrm{4}\mathrm{}\mathrm{2}\mathrm{0}\mathrm{)}\mathrm{(}\frac{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}{\mathrm{mol}}\mathrm{)}}{\mathrm{(}\mathrm{2}\mathrm{0}\mathrm{0}\mathrm{}\mathrm{0}\mathrm{)}\mathrm{min}}=0.12\frac{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}{\mathrm{mol}\mathrm{\cdot}\mathrm{m}\text{in}}& \left(\text{E71.6}\right)\end{array}$
We see from the Excel analysis and plot that the slope of the line is 0.12 dm^{3}/mol · min.
We now use Equation (E71.6), along with the initial concentration of methanol, to find the true rate constant, k.
$k=\frac{k\prime}{{C}_{\text{B0}}}=\frac{0.12}{0.5}\frac{{\text{dm}}^{3}/\text{mol/min}}{\text{mol/}{\text{dm}}^{3}}=0.24(\frac{{\text{dm}}^{3}}{\text{mol}}{)}^{2}/\mathrm{min}$
The rate law is
$\begin{array}{c}\hline {r}_{\text{A}}=[0.24{\left(\frac{{\text{dm}}^{3}}{\text{mol}}\right)}^{2}/\text{min}]{C}_{\text{A}}^{2}{C}_{\text{B}}\\ \hline\end{array}\begin{array}{c}\left(\text{E71.7}\right)\end{array}$
We note that the integral method tends to smooth the data.
Analysis: In this example, the reaction orders are known so that the integral method can be used to (1) verify the reaction is second order in trityl and (2) to find the specific pseudo reaction rate k′ = kC_{B0} for the case of excess methanol (B). Knowing k′ and C_{B0}, we then found the true rate constant k.
The differential method is an excellent technique to use when you want to look for inconsistencies in one’s experimental data. To outline the procedure used in the differential method of analysis, we consider a reaction carried out isothermally in a constantvolume batch reactor and the concentration of A, recorded as a function of time. By combining the mole balance with the rate law given by Equation (71), we obtain
$\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}={\mathit{k}}_{\mathrm{A}}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}$
Constantvolume batch reactor
After taking the natural logarithm of both sides of the above equation,
observe that the slope of a plot of [ln (–dC_{A}/dt)] as a function of (ln C_{A}) is the reaction order, α (see Figure 75).
Figure 75(a) shows a plot of [–(dC_{A}/dt)] versus [C_{A}] on loglog paper (or use Excel to make the plot) where the slope is equal to the reaction order α. The specific reaction rate, k_{A}, can be found by first choosing a concentration at a point or on the plot, say C_{Ap}, and then finding the corresponding value of [–(dC_{A}/dt)_{p}] on the line, as shown in Figure 75(b). The concentration point chosen, C_{Ap}, to find the derivative at C_{Ap}, need not be a data point; it just needs to be on the line. After raising C_{Ap} to the power α, we divide it into [–(dC_{A}/dt)_{p}] to determine k_{A}
Plot $\text{ln}\left(\frac{d{C}_{\text{A}}}{dt}\right)$ versus ln C_{A} to find α and k_{A}.
$\begin{array}{cc}{k}_{A}=\frac{(d{\text{C}}_{\text{A}}/dt{)}_{P}}{({\text{C}}_{\text{A}p}{)}^{\alpha}}& \left(\mathrm{712}\right)\end{array}$
To obtain the derivative (–dC_{A}/dt) used in this plot, we must differentiate the concentration–time data either numerically or graphically. Three methods to determine the derivative from data giving the concentration as a function of time are
Graphical differentiation
Numerical differentiation formulas
Differentiation of a polynomial fit to the data
Methods for finding $\frac{d{C}_{\text{A}}}{dt}$ from concentration–time data
We shall only discuss the graphical and numerical methods.
This method is very old (from slide rule days—“What’s a slide rule, Grandfather?”), when compared with the numerous software packages. So why do we use it? Because with this method, disparities in the data are easily seen. Consequently, it is advantageous to use this technique to analyze the data before planning the next set of experiments. As explained in Appendix A.2, the graphical method involves plotting (–ΔC_{A}/Δt) as a function of t and then using equalarea differentiation to obtain (–dC_{A}/dt). An illustrative example is also given in Appendix A.2.
#UsedBefore Laptops
In addition to the graphical technique used to differentiate the data, two other methods are commonly used: differentiation formulas and polynomial fitting.
See Appendix
Numerical differentiation formulas can be used when the data points in the independent variable are equally spaced, such as t_{1} – t_{0} = t_{2} – t_{1} = Δt. (See Table 72.)
TABLE 72 RAW DATA
Time (min) 
t_{0} 
t_{1} 
t_{2} 
t_{3} 
t_{4} 
t_{5} 
Concentration (mol/dm^{3}) 
C_{A0} 
C_{A1} 
C_{A2} 
C_{A3} 
C_{A4} 
C_{A5} 
Raw Data
The threepoint differentiation formulas^{1} shown in Table 73 can be used to calculate dC_{A}/dt.
^{1} B. Carnahan, H. A. Luther, and J. O. Wilkes, Applied Numerical Methods, New York: Wiley, 1969, p. 129.
TABLE 73 DIFFERENTIATION FORMULAS
Initial point: 
${\left(\frac{d{\text{C}}_{\text{A}}}{dt}\right)}_{{t}_{0}}=\frac{3{\text{C}}_{\text{A0}}+4{\text{C}}_{\text{A1}}{\text{C}}_{\text{A2}}}{2\mathrm{\Delta}t}$ 
(713) 
Interior points: 
$\begin{array}{c}(\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}{)}_{{\mathit{t}}_{\mathrm{i}}}=\frac{\mathrm{1}}{\mathrm{2}\mathrm{\Delta}\mathit{t}}\mathrm{[}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathrm{(}\mathit{i}+\mathrm{1}\mathrm{)}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathrm{(}\mathit{i}\mathrm{}\mathrm{1})}]\\ \lfloor \mathit{e}\mathrm{.}\mathit{g}\mathrm{.}\mathrm{,}\mathrm{}{\left(\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\right)}_{{t}_{3}}=\frac{\mathrm{1}}{\mathrm{2}\mathrm{\Delta}\mathit{t}}\mathrm{[}{\mathit{C}}_{\mathrm{A}\mathrm{4}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathrm{2}}\mathrm{]}\rfloor \end{array}$ 
(714) 
Last point: 
$(\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}{)}_{{\mathit{t}}_{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{2}\mathrm{\Delta}\mathit{t}}\mathrm{[}{C}_{\mathrm{A}\mathrm{3}}\mathrm{}\mathrm{4}{\mathit{C}}_{\mathrm{A}\mathrm{4}}+\mathrm{3}{\mathit{C}}_{\mathrm{A}\mathrm{5}}\mathrm{]}$ 
(715) 
Equations (713) and (715) are used for the first and last data points, respectively, while Equation (714) is used for all intermediate data points (see Step 5B in Example 72).
Now, using either the graphical method, differentiation formulas, or the polynomial derivative, Table 74 can be set up.
TABLE 74 PROCESSED DATA
Time 
t_{0} 
t_{1} 
t_{2} 
t_{3} 
Concentration 
C_{A0} 
C_{A1} 
C_{A2} 
C_{A3} 
Derivative 
$\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathrm{t}}{\mathrm{)}}_{\mathrm{0}}$ 
$\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathrm{t}}{\mathrm{)}}_{\mathrm{1}}$ 
$\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathrm{t}}{\mathrm{)}}_{\mathrm{2}}$ 
$\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathrm{t}}{\mathrm{)}}_{\mathrm{3}}$ 
Processed data
The reaction order can now be found from a plot of ln (–dC_{A}/dt) as a function of ln C_{A}, as shown in Figure 75(a), since
$\begin{array}{cc}\mathrm{ln}\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{)}=\mathrm{ln}\text{}{\mathit{k}}_{\mathrm{A}}+\mathit{\alpha}\text{}\mathrm{ln}\text{}{\mathit{C}}_{\mathrm{A}}& \left(\mathrm{716}\right)\end{array}$
Before solving the following example problems, review the steps to determine the reactionrate law from a set of data points (Table 71).
The reaction of triphenyl methyl chloride (trityl) (A) and methanol (B) discussed in Example 71 is now analyzed using the differential method.
$\begin{array}{cccccc}{\left({\text{C}}_{6}{\text{H}}_{5}\right)}_{3}\text{CCl}& +& {\text{CH}}_{3}\text{OH}& \to & {\begin{array}{c}\left({\text{C}}_{6}{\text{H}}_{5}\right)\end{array}}_{3}& {\text{COCH}}_{3}+\text{HCl}\\ \text{A}& +& \text{B}& \to & \text{C}\text{+}& \text{D}\end{array}$
The concentration–time data in Table E72.1 was obtained in a batch reactor.
TABLE E72.1 RAW DATA
Time (min) 
0 
50 
100 
150 
200 
250 
300 
Concentration of A (mol/dm^{3}) × 10^{3} (At t = 0, C_{A} = 0.05 M) 
50 
38 
30.6 
25.6 
22.2 
19.5 
17.4 
The initial concentration of methanol was 0.5 mol/dm^{3}.
Part (1) Determine the reaction order with respect to triphenyl methyl chloride.
Part (2) In a separate set of experiments, the reaction order wrt methanol was found to be first order. Determine the specific reactionrate constant.
Solution
Part (1) Find the reaction order with respect to trityl.
Step 1 Postulate a rate law.
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathrm{k}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathrm{C}}_{\mathrm{B}}^{\mathit{\beta}}& \left(\text{E72.1}\right)\end{array}$
Step 2 Process your data in terms of the measured variable, which in this case is C_{A}.
Step 3 Look for simplifications. Because the concentration of methanol is 10 times the initial concentration of triphenyl methyl chloride, its concentration is essentially constant
$\begin{array}{cc}{C}_{\text{B}}\cong {C}_{\text{B0}}& \left(\text{E72.2}\right)\end{array}$
Substituting for C_{B} in Equation (E72.1)
Step 4 Apply the CRE algorithm.
Mole Balance:
$\begin{array}{cc}\frac{\mathit{d}{\mathrm{N}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}={\mathit{r}}_{\mathrm{A}}\mathrm{V}& \left(\text{E72.4}\right)\end{array}$
Rate Law:
$\begin{array}{cc}\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathrm{k}\mathrm{\prime}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\text{E72.3}\right)\end{array}$
Stoichiometry: Liquid V = V_{0}
${\mathrm{C}}_{\mathrm{A}}=\frac{{\mathrm{N}}_{\mathrm{A}}}{{\text{V}}_{\mathrm{0}}}$
Combine: Mole balance, rate law, and stoichiometry
$\begin{array}{cc}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathrm{k}\mathrm{\prime}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\text{E72.5}\right)\end{array}$
Evaluate: Taking the natural log of both sides of Equation (E72.5)
$\begin{array}{cc}\mathrm{ln}\mathrm{[}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{]}=\mathrm{ln}\mathrm{k}\mathrm{\prime}+\mathit{\alpha}\mathrm{ln}{\mathrm{C}}_{\mathrm{A}}& \left(\text{E72.6}\right)\end{array}$
The slope of a plot of $\begin{array}{c}\mathrm{ln}\mathrm{[}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{]}\end{array}$ versus ln C_{A} will yield the reaction
order α with respect to triphenyl methyl chloride (A).
Step 5 Find $\begin{array}{c}\mathrm{\left[}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{\right]}\end{array}$ as a function of C_{A} from concentration–time data.
Step 5A Graphical Method. #ForOldTimesSake. To see how this problem was solved in the 1960s, go to the Chapter 7 Expanded Material on the Web site.
Step 5B Finite Difference Method. As mentioned this technique helps identify disparities in the data, especially between successive measurements. We now show how to calculate (dC_{A}/dt) using the finite difference formulas (i.e., Equations (713)–(715)).
$\begin{array}{ccc}t=0\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{t=0}& =& \frac{3{C}_{\text{A0}}+4{C}_{\text{A1}}{C}_{\text{A2}}}{2\mathrm{\Delta}t}\hfill \\ & =& \frac{[3\left(50\right)+4\left(38\right)30.6]\times {10}^{3}}{100}\hfill \\ & =& \begin{array}{c}{\begin{array}{c}2.86\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\end{array}\cdot \text{min}\hfill \\ & & \begin{array}{c}\frac{d{C}_{\text{A}}}{dt}\times {10}^{4}=2.86{\text{mol/dm}}^{3}\end{array}\cdot \text{min}\hfill \\ t=50\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{1}& =& \frac{\begin{array}{c}{C}_{\text{A2}}{C}_{\text{A0}}\end{array}}{2\mathrm{\Delta}t}=\frac{(30.650)\times {10}^{3}}{100}\hfill \\ & =& {\begin{array}{c}1.94\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \\ t=100\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{2}& =& \frac{\begin{array}{c}{C}_{\text{A3}}{C}_{\text{A1}\text{}}\end{array}}{2\mathrm{\Delta}t}=\frac{(25.638)\times {10}^{3}}{100}\hfill \\ & =& {\begin{array}{c}1.24\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \\ t=150\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{3}& =& \frac{\begin{array}{c}{C}_{\text{A4}}{C}_{\text{A2}\text{}}\end{array}}{2\mathrm{\Delta}t}=\frac{(22.230.6)\times {10}^{3}}{100}\hfill \\ & =& {\begin{array}{c}0.84\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \\ t=200\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{4}& =& \frac{\begin{array}{c}{C}_{\text{A5}}{C}_{\text{A3}\text{}}\end{array}}{2\mathrm{\Delta}t}=\frac{(19.525.6)\times {10}^{3}}{100}\hfill \\ & =& {\begin{array}{c}0.61\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \\ t=250\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{5}& =& \left(\begin{array}{c}\frac{\begin{array}{c}{C}_{\text{A6}}{C}_{\text{A4}\text{}}\end{array}}{2\mathrm{\Delta}t}=\frac{(17.422.2)\times {10}^{3}}{100}\end{array}\right)\hfill \\ & =& {\begin{array}{c}0.48\times 10\end{array}}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \\ t=300\text{}{\left(\frac{d{C}_{\text{A}}}{dt}\right)}_{6}& =& \frac{{C}_{\text{A4}}4{C}_{\text{A5}}+3{C}_{\text{A6}}}{2\mathrm{\Delta}t}=\frac{[22.24\left(19.5\right)+3\left(17.4\right)]\times {10}^{3}}{100}\\ & =& 0.36\times {10}^{4}{\text{mol/dm}}^{3}\cdot \text{min}\hfill \end{array}$
These are the type of calculations you would do if in the event of a power failure or malfunctioning Excel regression spreadsheet.
I know these are tedious calculations, but someone’s gotta know how to do it.
We now enter the above values for (–dC_{A}/dt) in Table E72.2 and use it to plot columns 2 and $\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{\times}\mathrm{1}\mathrm{0}\mathrm{,}\mathrm{0}\mathrm{0}\mathrm{0}\mathrm{)}$ as a function of column 4 (C_{A} × 1000) on loglog paper, as shown in Figure E72.1. We could also substitute the parameter values in Table E72.2 into Excel to find α and k′. Note that most of the points for both methods fall virtually on top of one another. This table is, in a way, redundant because it is not necessary to always find (–dC_{A}/dt) by both techniques, graphical and finite difference.
TABLE E72.2 SUMMARY OF PROCESSED DATA
Graphical 
Finite Difference 

t (min) 
$\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{\times}\mathrm{1}\mathrm{0}\mathrm{0}\mathrm{00}$ 
$\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\mathrm{\times}\mathrm{1}\mathrm{0}\mathrm{0}\mathrm{00}$ 
C_{A} × 1000 
0 
3.0 
2.86 
50 
50 
1.86 
1.94 
38 
100 
1.20 
1.24 
30.6 
150 
0.80 
0.84 
25.6 
200 
0.68 
0.61 
22.2 
250 
0.54 
0.48 
19.5 
300 
0.42 
0.36 
17.4 
#HowToImpress YourEmployer
From Figure E72.1, we found the slope to be 1.99, so that the reaction is said to be second order (α = 2.0) with respect to triphenyl methyl chloride. Recall when the cycle lengths on the xaxis and the yaxis are the same on loglog paper, then we simply measure the slope with a ruler. However, with all the software on your computer it would be more impressive to your professor or employer if you typed the data into Excel or Polymath, or some other spreadsheet software, and then printed the results.
To evaluate k′, we can evaluate the derivative in Figure E72.1 at C_{A}_{p} = 20 × 10^{–3} mol/dm^{3}, which is
$\begin{array}{cc}\mathrm{(}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}{\mathrm{)}}_{\mathit{p}}\mathrm{=}\mathrm{0.5}\mathrm{\times}\mathrm{1}{\mathrm{0}}^{\mathrm{}\mathrm{4}}\text{mol/}{\text{dm}}^{3}\mathrm{\xb7}\text{min}& \left(\text{E72.7}\right)\end{array}$
then
$\begin{array}{cc}k\prime =\frac{{(\frac{d{C}_{\text{A}}}{dt})}_{\text{p}}}{{C}_{\text{A}p}^{2}}\hfill & \left(\text{E72.8}\right)\\ =\frac{0.5\times {10}^{4}{\text{mol/dm}}^{3}\cdot \text{min}}{(20\times {10}^{3}{\text{mol/dm}}^{3}{)}^{2}}=0.125\phantom{0}{\text{dm}}^{3}/\text{mol}\cdot \text{min}\hfill & \end{array}$
As will be shown in Section 75, we could also use nonlinear regression on Equation (E71.5) to find k′
The Excel graph shown in Figure E72.1 gives α = 1.99 and k′ = 0.13 dm^{3}/mol · min. We now set α = 2 and regress again to find k′ = 0.122 dm^{3}/mol · min.
ODE Regression. There are techniques and software becoming available whereby an ODE solver can be combined with a regression program to solve differential equations, such as
to find ${k}_{\text{A}}^{\prime}$ and α from concentration–time data.
Part (2) The reaction was said to be first order with respect to methanol, β = 1.
Assuming C_{B0} is constant at 0.5 mol/dm^{3} and solving for k yields^{2}
$\begin{array}{l}k\phantom{0}=\phantom{0}\frac{k\mathrm{\prime}}{{C}_{\mathrm{B}\mathrm{0}}}\phantom{0}=\phantom{0}\frac{\mathrm{0.1}\mathrm{2}\mathrm{2}\frac{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}{\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{\cdot}\mathrm{min}}}{\mathrm{0.5}\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}}\\ k\phantom{0}=\phantom{0}0.244({\text{dm}}^{3}/\text{mol}{)}^{2}/\text{min}\end{array}$
^{2} M. Hoepfner and D. K. Roper, “Describing temperature increases in plasmonresonant nanoparticle systems,” J. Therm. Anal. Calorim., 98(1), 197–202 (2009).
The rate law is
Analysis: In this example, the differential method of data analysis was used to find the reaction order with respect to trityl (α = 1.99) and the pseudo rate constant (k′ = 0.125 (dm^{3}/mol)/min). However, as we mentioned earlier it would probably be more impressive to your employer or professor if you typed the data into your computer and used a software package to find k. We discuss one such software package in Section 7.5. The reaction order was rounded up to α = 2 and the data was regressed again to obtain k′ = 0.122 (dm^{3}/mol)/min, again knowing k′ and C_{B0}, and the true rate constant is k = 0.244 (dm^{3}/mol)^{2}/min. We used two techniques to differentiate the concentration–time data to find $\left(\frac{d{C}_{\text{A}}}{dt}\right)$: the graphical technique and the finite difference technique.
By comparing the methods of analysis of the rate data presented in Examples 71 and 72, we note that the differential method tends to accentuate the uncertainties in the data, while the integral method tends to smooth the data, thereby disguising the uncertainties in it. In most analyses, it is imperative that the engineer know the limits and uncertainties in the data!! This prior knowledge is necessary to provide for a safety factor when scaling up a process from laboratory experiments to design either a pilot plant or fullscale industrial plant.
Integral method normally used to find k when order is known
In nonlinear regression analysis, we search for those parameter values that minimize the sum of the squares of the differences between the measured values and the calculated values for all the data points. Not only can nonlinear regression find the best estimates of parameter values, it can also be used to discriminate between different ratelaw models, such as the LangmuirHinshelwood models discussed in Chapter 10. Many software programs are available to find these parameter values so that all one has to do is enter the data. The Polymath software will be used to illustrate this technique. In order to carry out the search efficiently, in some cases one has to enter initial estimates of the parameter values close to the actual values. These estimates can be obtained using the linearleastsquares technique discussed on the CRE Web site Professional Reference Shelf R7.3 (http://www.umich.edu/~elements/6e/07chap/pdf/leastsquares.pdf).
Compare the calculated parameter value with the measured value.
We will now apply nonlinear regression to reactionrate data to determine the ratelaw parameters. Here, we make initial estimates of the parameter values (e.g., reaction order, specific rate constant) in order to calculate the concentration for each data point, C_{ic}, obtained by solving an integrated form of the combined mole balance and rate law. We then compare the measured concentration at that point, C_{im}, with the calculated value, C_{ic}, for the parameter values chosen. We make this comparison by calculating the sum of the squares of the differences at each point Σ(C_{im} – C_{ic})^{2}. We then continue to choose new parameter values and search for those values of the ratelaw parameters that will minimize the sum of the squared differences of the measured concentrations, C_{im}, and the calculated concentrations values, C_{ic}. That is, we want to find the ratelaw parameters for which the sum of the squares of the difference at all data points, that is, ∑(C_{im} – C_{ic})^{2} is a minimum. If we carried out N experiments, we would want to find the parameter values (e.g., E, activation energy, reaction orders) that minimize the quantity
${\mathit{\sigma}}^{\mathrm{2}}\mathrm{=}\frac{{\mathit{s}}^{\mathrm{2}}}{N\mathrm{}K}\phantom{0}=\phantom{0}\mathrm{\sum}_{i=1}^{N}\frac{\mathrm{(}{\mathit{C}}_{\mathit{i}\mathit{m}}\mathrm{}{\mathit{C}}_{\mathit{i}\mathit{c}}{\mathrm{)}}^{\mathrm{2}}}{N\mathrm{}K}\phantom{0000000}\left(\mathrm{717}\right)$
where
${s}^{\mathrm{2}}\mathrm{}=\phantom{0}\mathrm{\sum}_{i=1}^{i=N}{(\mathit{C}}_{\mathit{i}\mathit{m}}\mathrm{}{\mathit{C}}_{\mathit{i}\mathit{c}}{\mathrm{)}}^{\mathrm{2}}$ 

N 
= 
number of runs 
K 
= 
number of parameters to be determined 
C_{im} 
= 
measured concentration rate for run i 
C_{ic} 
= 
calculated concentration rate for run i 
One notes that if we minimize S^{2} in Equation (717), we minimize σ^{2}.
To illustrate this technique, let’s consider the reaction
$\text{A}\to \text{Product}$
for which we want to learn the reaction order, α, and the specific reaction rate, k,
$\mathrm{}{\mathit{r}}_{\mathrm{A}}=\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}$
The reaction rate will be measured at a number of different concentrations. We now choose (guess) values of α and k, and calculate the concentration (C_{ic}) at each point, i, that the concentration was measured experimentally. We then subtract the calculated value (C_{ic}) from the measured value (C_{im}), square the result, and find the sum the squares for all the runs for the values of k and α that we have chosen.
This procedure is continued by further varying α and k until we find those values of k and α that minimize the sum of the squares. Many wellknown searching techniques are available to obtain the minimum value ${\sigma}_{\text{min}}^{2}$.^{3} Figure 76 shows a hypothetical plot of the sum of the squares as a function of the parameters α and k:
$\begin{array}{cc}{\mathit{\sigma}}^{\mathrm{2}}=\mathit{f}\mathrm{(}\mathit{k}\mathrm{,}\mathit{\alpha}\mathrm{)}& \phantom{00000000}\left(\mathrm{718}\right)\end{array}$
^{3} (a) University of Michigan colleagues B. Carnahan and J. O. Wilkes, Digital Computing and Numerical Methods, New York: Wiley, 1973, p. 405. (b) D. J. Wilde and C. S. Beightler, Foundations of Optimization, 2nd ed. Upper Saddle River, NJ: Pearson, 1979. (c) D. Miller and M. Frenklach, Int. J. Chem. Kinet., 15, 677 (1983).
Look at the top circle. We see that there are many combinations of α and k (e.g., α = 2.2, k = 4.8 or α = 1.8, k = 5.3) that will give a value of σ^{2} = 57. The same is true for σ^{2} = 1.85. We need to find the combination of α and k that gives the lowest value of σ^{2}.
In searching to find the parameter values that give the minimum of the sum of squares σ^{2}, one can use a number of optimization techniques or software packages. The computerized searching procedure begins by guessing parameter values and then calculating (C_{im} – C_{ic}) and then σ^{2} for these values. Next, a few sets of parameters are chosen around the initial guess, and σ^{2} is calculated for these sets as well. The search technique looks for the smallest value of σ^{2} in the vicinity of the initial guess and then proceeds along a trajectory in the direction of decreasing σ^{2} to choose the next set of different parameter values and determine the corresponding σ^{2} as shown as a contour plot in Figure 77. Starting at α = 1 and k = 1, the trajectory, shown by the arrows, is continually adjusted by choosing α and k. We see in Figure 77 that the trajectory leads us to a minimum value of σ^{2} = 0.045 (mol/dm^{3})^{2} at which point α = 2 and k = 5 (dm^{3}/mol).
If the equations are highly nonlinear, your initial guesses of = and k are very important. A number of software packages are available to carry out the procedure to determine the best estimates of the parameter values and the corresponding confidence limits. All we have to do is to (1) enter the experimental values into the computer, (2) specify the model, (3) enter the initial guesses of the parameters, then (4) push the “compute” button, and (5) view the best estimates of the parameter values along with 95% confidence limits. If the confidence limits for a given parameter are larger than the parameter itself, the parameter is probably not significant and should be dropped from the model as demonstrated in Chapter 10, Example 103. If all the model parameters are smaller than the confidence limits, that model should be eliminated and a new model equation chosen. The software is run again to analyze the new model.
We will now use nonlinear regression to determine the ratelaw parameters from concentration–time data obtained in batch experiments. We recall that the combined ratelaw stoichiometry mole balance for a constantvolume batch reactor is
$\begin{array}{cc}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}\phantom{0}=\phantom{0}\mathrm{}\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \phantom{000000000}\left(\mathrm{719}\right)\end{array}$
We now integrate Equation (76) for the case α ≠ 1 to give
${\mathit{C}}_{\mathrm{A}\mathrm{0}}^{\mathrm{1}\mathit{\alpha}}\mathrm{}{\mathit{C}}_{\mathrm{A}}^{\mathrm{1}\mathrm{}\mathit{\alpha}}\mathrm{=}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}\mathit{k}\mathit{t}$
Rearranging to obtain the concentration as a function of time, we obtain
Now we could use either Polymath or MATLAB to find the values of = and k that would minimize the sum of squares of the differences between the measured concentrations, C_{A}_{im}, and calculated concentrations, C_{A}_{ic}. That is, for N data points,
$\begin{array}{cc}{\mathit{s}}^{\mathrm{2}}\phantom{0}=\phantom{0}\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathit{i}\mathit{m}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathit{i}\mathit{c}}{\mathrm{)}}^{\mathrm{2}}\phantom{0}=\phantom{0}\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}\mathrm{[}{\mathit{C}}_{\mathrm{A}\mathit{i}\mathit{m}}\mathrm{}\mathrm{[}{\mathit{C}}_{\mathrm{A}\mathrm{0}}^{\mathrm{1}\mathit{\alpha}}\mathrm{}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}\mathit{k}{\mathit{t}}_{\mathit{i}}{\mathrm{]}}^{\mathrm{1}\mathrm{/}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}{\mathrm{]}}^{\mathrm{2}}& \phantom{00000}\left(\mathrm{721}\right)\end{array}$
we want the values of = and k that will make s^{2} a minimum.
If Polymath is used, one should use the absolute value, abs, for the term in brackets in Equation (721), that is,
$\begin{array}{cc}{\mathit{s}}^{\mathrm{2}}\phantom{0}=\phantom{0}\underset{i=1}{\overset{n}{\mathrm{\Sigma}}}{\mathrm{\left[}{\mathit{C}}_{\mathrm{A}\mathit{i}\mathit{m}}\mathrm{}{\mathrm{\left\{}\mathit{a}\mathit{b}\mathit{s}\mathrm{\left[}{\mathit{C}}_{\mathrm{A}\mathrm{0}}^{\mathrm{1}\mathit{\alpha}}\mathrm{}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}\mathit{k}{\mathit{t}}_{\mathit{i}}\mathrm{\right]}\mathrm{\right\}}}^{\mathrm{1}\mathrm{/}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}\mathrm{\right]}}^{\mathrm{2}}& \phantom{000000}\left(\mathrm{722}\right)\end{array}$
Another, and perhaps easier, way to solve for the parameter values is to use time rather than concentrations, rearranging Equation (720) to get
That is, we find the values of k and α that minimize
$\begin{array}{cc}{s}^{2}=\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}({t}_{im}{t}_{ic}{)}^{2}=\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}[{t}_{im}\frac{{C}_{\text{A0}}^{1\alpha}{C}_{\text{Ai}}^{1\alpha}}{k(1\alpha )}{]}^{2}& \left(\mathrm{724}\right)\end{array}$
Finally, a discussion of weighted least squares as applied to a firstorder reaction is provided in the Professional Reference Shelf R7.4 on the CRE Web site.
^{†} See tutorial at http://www.umich.edu/~elements/6e/07chap/live.html and http://www.umich.edu/~elements/6e/07chap/Example_73_Polymath_nonlinear_regression_tutorial.pdf.
We shall use the reaction and data in Examples E71 and E72 to illustrate how to use regression to find α and k′.
$\begin{array}{ccccccc}{\left({\text{C}}_{6}{\text{H}}_{5}\right)}_{3}\text{CCl}& +& {\text{CH}}_{3}\text{OH}& \to & {\left({\text{C}}_{6}{\text{H}}_{5}\right)}_{3}{\text{COCH}}_{3}& +& \text{HCl}\\ \text{A}& +& \text{B}& \to & \text{C}\text{+}& \text{D}& \end{array}$
The Polymath regression program is included on the CRE Web site. Recalling Equation (E72.5)
$\begin{array}{cc}\mathrm{}\frac{\mathit{d}{\mathrm{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=k\mathrm{\prime}{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \phantom{0000000000}\left(\text{E72.3}\right)\end{array}$
and integrating with the initial condition when t = 0 and C_{A} = C_{A0} for α ≠ 1.0
$\begin{array}{c}\hline t=\frac{1}{{k}^{\prime}}\frac{{\text{C}}_{\text{A0}}^{\left(1\alpha \right)}\text{}{\text{C}}_{\text{A}}^{\left(1\alpha \right)}}{\left(1\alpha \right)}\\ \hline\end{array}\begin{array}{c}\left(\text{E73.1}\right)\end{array}$
Given or assuming k′ and α, Equation (E72.5) can be solved (cf. Equation E73.1) to calculate the time t to reach a concentration C_{A}, or, alternatively we could calculate the concentration C_{A} at time t. 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 [σ^{2} = Σ(t_{im} – t_{ic})^{2}], or we could solve Equation (E73.1) for C_{A} and find α and k that minimize [σ^{2} = Σ(C_{Aim} – C_{Aic})^{2}].
We shall choose the former. So, substituting for the initial concentration C_{A0} = 0.05 mol/dm^{3} into Equation (E73.1)
$\begin{array}{cc}\begin{array}{c}t=\frac{1}{k\prime}\frac{(0.05{)}^{(1\alpha )}{C}_{\text{A}}^{(1\alpha )}}{(1\alpha )}\end{array}& \left(\text{E73.2}\right)\end{array}$
Tutorial with screen shots.
A brief tutorial on how to input data in Polymath on the CRE Web site (http://umich.edu/~elements/6e/07chap/Example_73_Polymath_nonlinear_regression_tutorial.pdf) shows screen shots of how to enter the raw data in Table E72.1 and how to carry out a nonlinear regression on Equation (E73.2).
For C_{A0} = 0.05 mol/dm^{3}, that is, Equation (E73.1) becomes
$\begin{array}{cc}{\mathit{t}}_{\mathit{c}}=\frac{1}{\mathit{k}\mathrm{\u2019}}\frac{\mathrm{(}\mathrm{0.05}{\mathrm{)}}^{\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}\mathrm{}{\mathit{C}}_{\mathrm{A}}^{\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}}{\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}& \left(\text{E73.3}\right)\end{array}$
We want to minimize s^{2} to give α and k′.
$\begin{array}{cc}{\mathit{s}}^{\mathrm{2}}=\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}\mathrm{(}{\mathit{t}}_{\mathit{i}\mathit{m}}\mathrm{}{\mathit{t}}_{\mathit{i}\mathit{c}}{\mathrm{)}}^{\mathrm{2}}=\underset{i=1}{\overset{N}{\mathrm{\Sigma}}}[{\mathit{t}}_{\mathit{i}\mathit{m}}\mathrm{}\frac{\mathrm{0}\mathrm{.}\mathrm{0}{\mathrm{5}}^{\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathit{i}\mathit{c}}^{\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}}{\mathit{k}\mathrm{\prime}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}{]}^{\mathrm{2}}& \left(\mathrm{725}\right)\end{array}$
We shall carry out a first regression to find approximate values of α and k. Next, we use a common procedure, which is to round α off to an integer and then carry out a second regression to find the best value of k after α has been set equal to an integer.
TABLE E73.1 RESULTS OF 1ST REGRESSION
TABLE E73.2 RESULTS OF 2ND REGRESSION WITH α = 2
The results shown are
The first regression gives α = 2.04, as shown in Table E73.1. 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 E73.2) on k′ because the k′ given in Table E.73.1 is for α = 2.04. We now regress the equation
$\mathit{t}=\frac{\mathrm{1}}{\mathit{k}\mathrm{\u2019}}\mathrm{[}\frac{\mathrm{1}}{{\mathit{C}}_{\mathrm{A}}}\mathrm{}\frac{\mathrm{1}}{{\mathit{C}}_{\mathrm{A}\mathrm{0}}}\mathrm{]}$
The second regression gives k′ = 0.125 dm^{3}/mol · min. We now calculate k
$\mathit{k}=\frac{k\prime}{{C}_{\text{A0}}}=\frac{0.125\left({\text{dm}}^{3}/\text{mol}\right)}{0.5\left({\text{dm}}^{3}/\text{mol}\right)\text{min}}=\mathrm{0}\mathrm{.}\mathrm{2}\mathrm{5}{\left(\frac{\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}{\mathrm{m}\mathrm{o}\mathrm{l}}\right)}^{\mathrm{2}}\mathrm{/}\mathrm{min}$
Analysis: In this example, we showed how to use nonlinear 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 (dm^{3}/mol)/min giving a value of the true specific reaction rate of k = 0.25 (dm^{3}/mol)^{2}/min. We note that the reaction order is the same as that in Examples 71 and 72; however, the value of k is about 8% larger. The r^{2} and other statistics can be found in Polymath’s output.
One can also determine which model or equation best fits the experimental data by comparing the sums of the squares for each model and then choosing the equation with a smaller sum of squares and/or carrying out an Ftest. Alternatively, we can compare the residual plots for each model. These plots show the error associated with each data point, and one looks to see whether the error is randomly distributed or whether there is a trend in the error. When the error is randomly distributed, this is an additional indication that the correct rate law has been chosen. An example of model discrimination using nonlinear regression is given in Chapter 10, Example 103.
Data acquisition using the method of initial rates and a differential reactor is similar in that the rate of reaction is determined for a specified number of predetermined initial or entering reactant concentrations. A differential reactor (PBR) is normally used to determine the rate of reaction as a function of either concentration or partial pressure. It consists of a tube containing a very small amount of catalyst, usually arranged in the form of a thin wafer or disk. A typical arrangement is shown schematically in Figure 78. The criterion for a reactor being differential is that the conversion of the reactants in the bed is extremely small, as is the change in temperature and reactant concentration through the bed. As a result, the reactant concentration through the reactor is essentially constant and approximately equal to the inlet concentration. That is, the reactor is considered to be gradientless,^{4} and the reaction rate is considered spatially uniform within the bed.
^{4} B. Anderson, ed., Experimental Methods in Catalytic Research, San Diego, CA: Academic Press.
Most commonly used catalytic reactor to obtain experimental data
The differential reactor is relatively easy to construct at a low cost. Owing to the low conversion achieved in this reactor, the heat release per unit volume will be small (or can be made small by diluting the bed with inert solids) so that the reactor operates essentially in an isothermal manner. When operating this reactor, precautions must be taken so that the reactant gas or liquid does not bypass or channel through the packed catalyst, but instead flows uniformly across the catalyst. If the catalyst under investigation decays rapidly, as discussed in Chapter 10, the differential reactor is not a good choice because the reactionrate parameters at the start of a run will be different from those at the end of the run. In some cases, sampling and analysis of the product stream may be difficult for small conversions in multicomponent systems.
Limitations of the differential reactor
The differential reactor is operated at steady state.
For the reaction of species A going to product (P)
A → P
the volumetric flow rate through the catalyst bed is monitored, as are the entering and exiting concentrations (Figure 79). Therefore, if the weight of catalyst, ΔW, is known, the rate of reaction per unit mass of catalyst, ${\text{r}}_{\text{A}}^{\prime}$, can be calculated. Since the differential reactor is assumed to be gradientless, the design equation will be similar to the CSTR design equation. A steadystate mole balance on reactant A gives
$\begin{array}{ccccccc}\left[\begin{array}{c}\text{Flow}\\ \begin{array}{c}\begin{array}{c}\text{rate}\end{array}\\ \text{in}\end{array}\end{array}\right]& & \left[\begin{array}{c}\begin{array}{c}\text{Flow}\\ \text{rate}\end{array}\\ \text{out}\end{array}\right]& +& \left[\begin{array}{c}\text{Rateof}\\ \text{generation}\end{array}\right]& =& \left[\begin{array}{c}\text{Rateof}\\ \text{accumulation}\end{array}\right]\\ \left[{F}_{\text{A0}}\right]& & \left[{F}_{\text{A}e}\right]& +& \left[\left(\frac{\text{Rateofreaction}}{\text{Massofcatalyst}}\right)\left(\text{Massofcatalyst}\right)\right]& =& 0\\ \begin{array}{c}{F}_{\text{A0}}\end{array}& & \begin{array}{c}{F}_{\text{A}e}\end{array}& +& \left({r}_{\text{A}}^{\prime}\right)\left(\mathrm{\Delta}W\right)& =& 0\end{array}$
The subscript e refers to the exit of the reactor. Solving for ${r}_{\text{A}}^{\prime}$, we have
$\begin{array}{cc}{r}_{\text{A}}^{\prime}=\frac{{\mathrm{F}}_{\mathrm{A}\mathrm{0}}\mathrm{}{\mathrm{F}}_{\mathrm{A}\mathit{e}}}{\mathrm{\Delta}\mathit{W}}& \left(\mathrm{726}\right)\end{array}$
The mole balance equation can also be written in terms of concentration
Differential reactor design equation
or in terms of conversion or product flow rate F_{P}
The term F_{A0}X gives the rate of formation of product, F_{p}, when the stoichiometric coefficients of A and of P are identical. Adjustments to Equation (728) must be made when this is not the case.
For constant volumetric flow, Equation (728) reduces to
$\begin{array}{cc}\mathrm{}{r}_{\text{A}}^{\prime}=\frac{{\mathit{v}}_{\mathrm{0}}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathrm{0}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathrm{e}}\mathrm{)}}{\mathrm{\Delta}\mathit{W}}=\frac{{\mathit{v}}_{{\mathrm{0}}^{{\mathit{C}}_{\mathrm{P}}}}}{\mathrm{\Delta}\mathit{W}}& \left(\mathrm{729}\right)\end{array}$
Consequently, we see that the reaction rate, $\mathrm{}{r}_{\text{A}}^{\prime}$, can be determined by measuring the product concentration, C_{p}.
By using very little catalyst and large volumetric flow rates, the concentration difference, (C_{A0} – C_{Ae}), can be made quite small. The rate of reaction determined from Equation (729) can be obtained as a function of the reactant concentration in the catalyst bed, C_{Ab}
$\begin{array}{cc}\mathrm{}{r}_{\text{A}}^{\prime}=\mathrm{}{r}_{\text{A}}^{\prime}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathrm{b}}\mathrm{)}& \left(\mathrm{730}\right)\end{array}$
by varying the inlet concentration. One approximation of the concentration of A within the bed, C_{Ab}, would be the arithmetic mean of the inlet and outlet concentrations
$\begin{array}{cc}{\mathit{C}}_{\mathrm{A}\mathrm{b}}=\frac{{\mathit{C}}_{\mathrm{A}\mathrm{0}}+{\mathit{C}}_{\mathrm{A}\mathit{e}}}{\mathrm{2}}& \left(\mathrm{731}\right)\end{array}$
However, since very little reaction takes place within the bed, the bed concentration is essentially equal to the inlet concentration
C_{Ab} ≈ C_{A0}
so $\mathrm{}{r}_{\text{A}}^{\prime}$ is a function of C_{A0}
$\begin{array}{cc}\mathrm{}{r}_{\text{A}}^{\prime}=\mathrm{}{r}_{\text{A}}^{\prime}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathrm{0}}\mathrm{)}& \left(\mathrm{732}\right)\end{array}$
As with the method of initial rates (see the CRE Web site, http://www.umich.edu/~elements/6e/software/nonlinear_regression_tutorial.pdf), we find –r_{A0} as a function of C_{A0} and then use various numerical and graphical techniques to determine the appropriate algebraic equation for the rate law. When collecting data for fluid–solid reacting systems, care must be taken that we use high flow rates through the differential reactor and small catalyst particle sizes in order to avoid mass transfer–limitations. If data show the reaction to be first order with a low activation energy, say 8 kcal/mol, one should suspect the data are being collected in the mass transfer–limited regime. We will expand on mass transfer–limitations and how to avoid them in Chapter 10, and also in Chapters 14 and 15.
The formation of methane from carbon monoxide and hydrogen using a nickel catalyst was studied by Pursley.^{5} The reaction
3H_{2} + CO → CH_{4} + H_{2}O
^{5} J. A. Pursley, “An investigation of the reaction between carbon monoxide and hydrogen on a nickel catalyst above one atmosphere,” Ph.D. thesis, University of Michigan.
was carried out at 500°F in a differential reactor where the effluent concentration of methane was measured. The raw data is shown in Table E74.1.
TABLE E74.1 RAW DATA
Run 
P_{CO} (atm) 
P_{H2} (atm) 
C_{CH4} (mol/dm^{3}) 
1 
1 
1.0 
1.73 × 10^{–4} 
2 
1.8 
1.0 
4.40 × 10^{–4} 
3 
4.08 
1.0 
10.0 × 10^{–4} 
4 
1.0 
0.1 
1.65 × 10^{–4} 
5 
1.0 
0.5 
2.47 × 10^{–4} 
6 
1.0 
4.0 
1.75 × 10^{–4} 
P_{H2} is constant in Runs 1, 2, 3.
P_{CO} is constant in Runs 4, 5, 6.
The exit volumetric flow rate from a differential packed bed containing 10 g of catalyst was maintained at S_{0} = 300 dm^{3}/min for each run. The partial pressures of H_{2} and CO were determined at the entrance to the reactor, and the methane concentration was measured at the reactor exit.
Determine the rate law and ratelaw parameters. To better understand how we will determine the rate law and its parameters, we will break the solution into three parts, (a), (b), and (c), as shown below.
Relate the rate of reaction to the exit methane concentration. The reactionrate law is assumed to be the product of a function of the partial pressure of CO, f(CO), and a function of the partial pressure of H_{2}, g(H_{2}), that is,
$\begin{array}{cc}{r}_{{\text{CH}}_{4}}^{\prime}=\mathit{f}\mathrm{(}\mathrm{C}\mathrm{O}\mathrm{)}\mathrm{\cdot}\mathit{g}\mathrm{(}{\mathrm{H}}_{\mathrm{2}}\mathrm{)}& \left(\text{E74.1}\right)\end{array}$
Determine the ratelaw dependence on carbon monoxide, using the data generated in part (a). Assume that the functional dependence of ${r}_{{\text{CH}}_{4}}^{\prime}$ on P_{CO} is of the form
$\begin{array}{cc}{r}_{{\text{CH}}_{4}}^{\prime}\sim {\mathrm{P}}_{\mathrm{C}\mathrm{O}}^{\mathit{\alpha}}& \left(\text{E74.2}\right)\end{array}$
Generate a table of the reaction rate as a function of partial pressures of carbon monoxide and hydrogen, and then determine the ratelaw dependence on H_{2}.
Solution
(a) Calculate the Rates of Reaction. In this example the product composition, rather than the reactant concentration, is being monitored. The term $\left({r}_{\text{CO}}^{\prime}\right)$ can be written in terms of the flow rate of methane from the reaction
$\mathrm{}{r}_{\text{CO}}^{\prime}={r}_{{\text{CH}}_{4}}^{\prime}=\frac{{\mathrm{F}}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}}{\mathrm{\Delta}\mathit{W}}$
Substituting for F_{CH4} in terms of the volumetric flow rate and the concentration of methane gives
Since υ_{0}, C_{CH4}, and ΔW are known for each run, we can calculate the rate of reaction.
For run 1
${r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}=\left(\frac{\mathrm{3}\mathrm{0}\mathrm{0}\mathrm{d}{\mathrm{m}}^{\mathrm{3}}}{\mathrm{min}}\right)\frac{\mathrm{1.7}\mathrm{3}\mathrm{\times}\mathrm{1}{\mathrm{0}}^{\mathrm{}\mathrm{4}}}{\mathrm{1}\mathrm{0}\text{\hspace{0.17em}}\mathrm{g}\mathrm{c}\mathrm{a}\mathrm{t}}\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{/}\mathrm{d}{\mathrm{m}}^{\mathrm{3}}=\mathrm{5.2}\mathrm{\times}\mathrm{1}{\mathrm{0}}^{\mathrm{}\mathrm{3}}\frac{\mathrm{mol}\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}{\mathrm{g}\text{c}\mathrm{a}\mathrm{t}\mathrm{\times}\mathrm{min}}$
The rate for runs 2 through 6 can be calculated in a similar manner (Table E74.2).
TABLE E74.2 RAW AND CALCULATED DATA
Run 
P_{CO} (atm) 
P_{H2} (atm) 
C_{CH4} (mol/dm^{3}) 
${r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\left(\frac{{\text{molCH}}_{4}}{g\text{catmin}}\right)$ 
1 
1.0 
1.0 
1.73 ×10^{–4} 
5.2 × 10^{–3} 
2 
1.8 
1.0 
4.40 × 10^{–4} 
13.2 × 10^{–3} 
3 
4.08 
1.0 
10.0 ×10^{–4} 
30.0 × 10^{–3} 
4 
1.0 
0.1 
1.65 × 10^{–4} 
4.95 × 10^{–3} 
5 
1.0 
0.5 
2.47 × 10^{–4} 
7.42 × 10^{–3} 
6 
1.0 
4.0 
1.75 × 10^{–4} 
5.25 × 10^{–3} 
(b) Determining the RateLaw Dependence in CO. For constant hydrogen concentration (runs 1, 2, and 3), the rate law
${r}_{{\text{CH}}_{4}}^{\prime}=\stackrel{{k}^{\prime}}{\overbrace{kg\left({P}_{{\text{H}}_{2}}\right)}}{p}_{\text{CO}}^{\alpha}$
can be written as
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}=k\mathrm{\prime}{P}_{\mathrm{C}\mathrm{O}}^{\mathit{\alpha}}& \left(\text{E74.4}\right)\end{array}$
Taking the natural log of Equation (E74.4) gives us
$\mathrm{ln}\left({r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\right)=\mathrm{ln}{k}^{\prime}+\alpha \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{ln}\text{\hspace{0.17em}}{P}_{\text{CO}}$
We now plot ln $\left({r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\right)$ versus ln P_{CO} using runs 1, 2, and 3, for which H_{2} the concentration is constant, in Figure E74.1. We see from the Excel plot that α = 1.22.
Had we not been so pressed for time and included more points, we would have found that the reaction is essentially first order with α = 1, that is
$\begin{array}{cc}\mathrm{}{r}_{\text{CO}}^{\prime}=k\mathrm{\prime}{P}_{\mathrm{C}\mathrm{O}}& \left(\text{E74.5}\right)\end{array}$
If we had used more data (not given here) we would have found α = 1.
From the first three data points in Table E74.2 where the partial pressure of H_{2} is constant, we see the rate is linear in partial pressure of CO
${r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}=k\mathrm{\prime}{P}_{\mathrm{C}\mathrm{O}}\mathrm{\cdot}\mathit{g}\mathrm{(}{\mathrm{H}}_{\mathrm{2}}\mathrm{)}$
Now let’s look at the hydrogen dependence.
(c) Determining the RateLaw Dependence on H_{2}. It will not be intuitively obvious how to do this analysis until you have studied Chapter 10 where we have rate laws with concentrations or partial pressures in both the numerator and denominator of the rate law. From Table E74.2 it appears that the dependence of ${r}_{{\text{CH}}_{4}}^{\prime}$ on P_{H2} cannot be represented by a power law. Comparing run 4 with run 5 and then run 5 with run 6, we see that the reaction rate first increases with increasing partial pressure of hydrogen, and subsequently decreases with increasing P_{H2}. That is, there appears to be a concentration of hydrogen at which the rate is maximum. One set of rate laws that is consistent with these observations is:
At low H_{2} concentrations where ${r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}$ increases as P_{H2} increases, the rate law may be of the form
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\sim {P}_{{\mathrm{H}}_{\mathrm{2}}}^{{\mathit{\beta}}_{\mathrm{1}}}& \left(\text{E74.6}\right)\end{array}$
At high H_{2} concentrations where ${r}_{{\text{CH}}_{4}}^{\prime}$ decreases as P_{H2} increases, the rate law may be of the form
$\begin{array}{cc}{r}_{{\text{CH}}_{4}}^{\prime}\sim \frac{1}{{P}_{{\text{H}}_{2}}^{{\beta}_{2}}}& \left(\text{E74.7}\right)\end{array}$
Sketch of r′_{CH4} as a function of P_{H2}
We would like to find one rate law that is consistent with reactionrate data at both high and low hydrogen concentrations. After we have studied heterogeneous reactions in Chapter 10, we would recognize that Equations (E74.6) and (E74.7) can be combined into the form
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\sim \frac{{P}_{{\text{H}}_{2}}^{\text{}{\beta}_{1}}}{\mathrm{1}+\mathit{b}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{{\mathit{\beta}}_{\mathrm{2}}}}& \left(\text{E74.8}\right)\end{array}$
We will see in Chapter 10 that this combination and similar rate laws that have reactant concentrations (or partial pressures) in the numerator and denominator are common in heterogeneous catalysis.
Let’s see if the resulting rate law (E74.8) is qualitatively consistent with the rate observed.
For condition 1: At low P_{H2}, [b(P_{H2})^{β2} ≪ 1] and Equation (E74.8) reduces to
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\sim {P}_{{\mathrm{H}}_{\mathrm{2}}}^{{\mathit{\beta}}_{\mathrm{1}}}& \left(\text{E74.9}\right)\end{array}$
Equation (E74.9) is consistent with the trend in comparing runs 4 and 5.
For condition 2: At high P_{H2}, [b(P_{H2})^{β2} ≫ 1] and Equation (E74.8) reduces to
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}\sim \frac{\mathrm{(}{P}_{{\mathrm{H}}_{\mathrm{2}}}{\mathrm{)}}^{{\mathit{\beta}}_{\mathrm{1}}}}{\mathrm{(}{P}_{{\mathrm{H}}_{\mathrm{2}}}{\mathrm{)}}^{{\mathit{\beta}}_{\mathrm{2}}}}\sim \frac{\mathrm{1}}{\mathrm{(}{P}_{{\mathrm{H}}_{\mathrm{2}}}{\mathrm{)}}^{{\mathit{\beta}}_{\mathrm{2}}\mathrm{}{\mathit{\beta}}_{\mathrm{1}}}}& \left(\text{E74.10}\right)\end{array}$
where β_{2} > β_{1}. Equation (E74.10) is consistent with the trends in comparing runs 5 and 6.
Combining Equations (E74.8) and (E74.5)
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}=\frac{\mathit{a}{P}_{\mathrm{C}\mathrm{O}}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{{\mathit{\beta}}_{\mathrm{1}}}}{\mathrm{1}+\mathit{b}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{{\mathit{\beta}}_{\mathrm{2}}}}& \left(\text{E74.11}\right)\end{array}$
Typical form of the rate law for heterogeneous catalysis
We now use the Polymath regression program to find the parameter values a, b, β_{1}, and β_{2}. The results are shown in Table E74.3.
The corresponding rate law is
$\begin{array}{cc}{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}=\frac{0.025{P}_{\text{CO}}{P}_{{\text{H}}_{2}}^{0.61}}{\mathrm{1}+\mathrm{2.4}\mathrm{9}{P}_{{\mathrm{H}}_{\mathrm{2}}}}& \left(\text{E74.12}\right)\end{array}$
TABLE E74.3 FIRST REGRESSION
Polymath and Excel regression tutorials are given in the Chapter 7 Summary Notes on the CRE Web site.
We could use the rate law given by Equation (E74.12) as is, but there are only six data points, and we should be concerned about extrapolating the rate law over a wider range of partial pressures. We could take more data, and/or we could carry out a theoretical analysis of the type discussed in Chapter 10 for heterogeneous reactions. If we assume hydrogen undergoes dissociative adsorption on the catalyst surface (cf. Figure 1011), we would expect a dependence on the partial pressure of hydrogen to be to the ½ power. Because 0.61 is close to 0.5, we are going to regress the data again, setting β_{1} = ½ and β_{2} = 1.0. The results are shown in Table E74.4.
TABLE E74.4 SECOND REGRESSION
After studying the heterogeneous rate laws in Chapter 10, we will become more adept at postulating and analyzing this type of data for heterogeneous reactions.
Note: A complete tutorial of how to enter this example into Polymath can be found on the Web site at http://www.umich.edu/~elements/6e/software/nonlinear_regression_tutorial.pdf.
The rate law is now
$\begin{array}{c}\hline {r}_{{\text{CH}}_{4}}^{\prime}=\frac{{{0.018P}_{\text{CO}}P}_{{\text{H}}_{2}}^{1/2}}{{{1+1.49P}_{\text{H}}}_{2}}\\ \hline\end{array}$
where ${r}_{{\text{CH}}_{4}}^{\prime}$ is in (mol/gcat · s) and the partial pressures are in (atm).
We could also have set β_{1} = ½ and β_{2} = 1.0 and rearranged Equation (E74.11) in the form
$\begin{array}{cc}\frac{{P}_{\mathrm{C}\mathrm{O}}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{\mathrm{1}\mathrm{/}\mathrm{2}}}{{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}}=\frac{\mathrm{1}}{\mathit{a}}+\frac{\mathit{b}}{\mathit{a}}{P}_{{\mathrm{H}}_{\mathrm{2}}}& \left(\text{E74.13}\right)\end{array}$
Linearizing the rate law to determine the ratelaw parameters
A plot of ${P}_{\mathrm{C}\mathrm{O}}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{\mathrm{1}\mathrm{/}\mathrm{2}}/{r}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}^{\prime}$ as a function of P_{H2} should be a straight line with an intercept of 1/a and a slope of b/a. From the plot in Figure E74.2, we see that the rate law is indeed consistent with the ratelaw data.
Analysis: The reactionrate data in this example were obtained at steady state, and as a result, neither the integral method nor differential method of analysis can be used. One of the purposes of this example is to show how to reason out the form of the rate law and to then use regression to determine the ratelaw parameters. Once the parameters were obtained, we showed how to linearize the ratelaw (e.g., Equation (E74.13)) to generate a single plot of all the data, Figure (E74.2).
Four to six weeks in the lab can save you an hour in the library.
—G. C. Quarderer, Retired from Dow Chemical Co.
So far, this chapter has presented various methods of analyzing rate data. It is just as important to know in which circumstances to use each method as it is to know the mechanics of these methods. In the Expanded Material and in the Professional Reference Shelf (http://www.umich.edu/~elements/6e/07chap/prof75.html) on the CRE Web site, we give a thumbnail sketch of a heuristic to plan experiments to generate the data necessary for reactor design. However, for a more thorough discussion, the reader is referred to the books and articles by Box and Hunter.^{6}
^{6} G. E. P. Box, W. G. Hunter, and J. S. Hunter, Statistics for Experimenters: An Introduction to Design, Data Analysis, and Model Building, New York: Wiley, 1978.
This chapter focused on obtaining and analyzing experimental data that is collected safely in both academia and industrial laboratories. To help promote safer laboratory practices, the Web site (http://umich.edu/~safeche/index.html), Process Safety Across the Chemical Engineering Curriculum, provides a number of videos on safety. These videos are classified into two types:
Video Snippets that are 1–5 minutes instructional videos
Case histories of academic accidents that resulted in fatalities or serious injuries and how they may have been prevented.
To access the Snippets Videos, go to the Safety Web site home page and click on Laboratory and Personal Safety (http://umich.edu/~safeche/lab_safety.html), which is under Course Specific Safety Modules. Upon clicking on Laboratory and Personal Safety, you will pull up the menu as shown here:
Safety Snippets Videos
The Sloppy Lab
Safety Gear
Evacuate
Videos of Case Histories
Experimenting with Danger
Guidelines
Active Attacker (https://www.dpss.umich.edu/content/preventioneducation/safetytips/activeattacker/)
UM Undergraduate Lab Safety Guidelines (PDF) (http://umich.edu/~safeche/assets/pdf/LabsSafetyGuidelines.pdf)
Introduction to Laboratory Safety for Graduate Students (PDF) (https://pubs.acs.org/doi/pdfplus/10.1021/acs.jchemed.8b00774)
ACS Safety Guidelines (https://www.acs.org/content/dam/acsorg/about/governance/committees/chemicalsafety/publications/acssafetyguidelinesacademic.pdf?logActivity=true)
To enhance your laboratory safety prowess, view each of the snippet videos and then write a onesentence takeaway lesson for each video.
Integral method:
Guess the reaction order and integrate the mole balance equation.
Calculate the resulting function of concentration for the data and plot it as a function of time. If the resulting plot is linear, you have probably guessed the correct reaction order.
If the plot is not linear, guess another order and repeat the procedure.
Differential method for constantvolume systems:
$\begin{array}{cc}\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}}=\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}& \left(\text{S71}\right)\end{array}$
Plot –ΔC_{A}/Δt as a function of t.
Determine –dC_{A}/dt from this plot.
Take the ln of both sides of Equation (S71) to get
$\begin{array}{cc}\mathrm{ln}(\mathrm{}\frac{\mathit{d}{\mathit{C}}_{\mathrm{A}}}{\mathit{d}\mathit{t}})=\mathrm{ln}\mathit{k}+\mathit{\alpha}\mathrm{ln}{\mathit{C}}_{\mathrm{A}}& \left(\text{S72}\right)\end{array}$
Plot ln(–dC_{A}/dt) versus ln_{C}A. The slope will be the reaction order α. We could use finitedifference formulas or software packages to evaluate (–dC_{A}/dt) as a function of time and concentration.
Nonlinear regression: Search for the parameters of the rate law that will minimize the sum of the squares of the difference between the measured rate of reaction and the rate of reaction calculated from the parameter values chosen. For N experimental runs and K parameters to be determined, use Polymath.
$\begin{array}{cc}{\mathit{\sigma}}^{\mathrm{2}}={\displaystyle \sum}_{\mathit{i}=\mathrm{1}}^{\mathrm{N}}\frac{\mathrm{}{P}_{\mathit{i}}\mathrm{(}\mathrm{m}\mathrm{e}\mathrm{a}\mathrm{s}\mathrm{u}\mathrm{r}\mathrm{e}\mathrm{d}\mathrm{)}\mathrm{}{P}_{\mathit{i}}\mathrm{(}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{c}\mathrm{u}\mathrm{l}\mathrm{a}\mathrm{t}\mathrm{e}\mathrm{d}\mathrm{)}{\mathrm{]}}^{\mathrm{2}}}{\mathrm{N}\mathrm{}\mathrm{K}}& \left(\text{S73}\right)\end{array}$
$\begin{array}{cc}{\mathit{s}}^{\mathrm{2}}={\displaystyle \sum}_{\mathit{i}=\mathrm{1}}^{\mathrm{N}}\mathrm{(}{\mathit{t}}_{\mathit{i}\mathit{m}}\mathrm{}{\mathit{t}}_{\mathit{i}\mathit{c}}{\mathrm{)}}^{\mathrm{2}}={\displaystyle \sum}_{\mathit{i}=\mathrm{1}}^{\mathrm{N}}[{\mathit{t}}_{\mathit{i}\mathit{m}}\mathrm{}\frac{{\mathit{C}}_{\mathrm{A}\mathrm{0}}^{\mathrm{1}\mathit{\alpha}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathit{i}}^{\mathrm{1}\mathrm{}\mathit{\alpha}}}{\mathit{k}\mathrm{(}\mathrm{1}\mathrm{}\mathit{\alpha}\mathrm{)}}{\mathrm{]}}^{\mathrm{2}}& \left(\text{S74}\right)\end{array}$
Caution: Be sure to avoid a false minimum in σ^{2} by varying your initial guess.
Modeling the differential reactor:
The rate of reaction is calculated from the equation
$\begin{array}{cc}\mathrm{}{r}_{\text{A}}^{\prime}=\frac{{\mathrm{F}}_{\mathrm{A}\mathrm{0}}\mathit{X}}{\mathrm{\Delta}\mathit{W}}=\frac{{\mathrm{F}}_{P}}{\mathrm{\Delta}\mathit{W}}=\frac{{\upsilon}_{\mathrm{0}}\mathrm{(}{\mathit{C}}_{\mathrm{A}\mathrm{0}}\mathrm{}{\mathit{C}}_{\mathrm{A}\mathit{e}}\mathrm{)}}{\mathrm{\Delta}\mathit{W}}=\frac{{\mathit{C}}_{P}{\upsilon}_{\mathrm{0}}}{\mathrm{\Delta}\mathit{W}}& \left(\text{S75}\right)\end{array}$
In calculating the reaction order, α
$\mathrm{}{r}_{\text{A}}^{\prime}=\text{\hspace{0.17em}}\mathit{k}{\mathit{C}}_{\mathrm{A}}^{\mathit{\alpha}}$
the concentration of A is evaluated either at the entrance conditions or at a mean value between C_{A0} and C_{Ae}. However, powerlaw models such as
$\begin{array}{cc}\mathrm{}{r}_{\text{A}}^{\prime}=k{\mathrm{C}}_{\mathrm{A}}^{\mathit{\alpha}}{\mathrm{C}}_{\mathrm{B}}^{\mathit{\beta}}& \left(\begin{array}{c}\text{S76}\end{array}\right)\end{array}$
are not the best way to describe heterogeneous reactionrate laws. Typically, they take the form
$\mathrm{}{r}_{\text{A}}^{\prime}=\frac{k{P}_{\mathrm{A}}{P}_{\mathrm{B}}}{\mathrm{1}+{\mathrm{K}}_{\mathrm{A}}{P}_{\mathrm{A}}+{\mathrm{K}}_{\mathrm{B}}{P}_{\mathrm{B}}}$
or a similar form, with the reactant partial pressures in the numerator and denominator of the rate law.
(http://umich.edu/~elements/6e/07chap/obj.html#/)
Interactive Computer Games (http://umich.edu/~elements/6e/icm/index.html)
The subscript to each of the problem numbers indicates the level of difficulty: A, least difficult; D, most difficult.
A = • B = ▪ C = ♦ D = ♦♦
Q71_{A} QBR (Question Before Reading). If you need to determine the rate law, what methods would you use to collect the data and how would you analyze it?
Q72_{A} i>clicker. Go to the Web site (http://www.umich.edu/~elements/6e/07chap/iclicker_ch7_q1.html) and view at least five i>clicker questions. Choose one that could be used as is, or a variation thereof, to be included on the next exam. You also could consider the opposite case: explaining why the question should not be on the next exam. In either case, explain your reasoning.
Q73_{A}
Listen to the audios on the CRE Web site. Pick one and say why it could be eliminated.
Create an original problem based on Chapter 7 material.
Design an experiment for the undergraduate laboratory that demonstrates the principles of chemical reaction engineering and will cost less than $500 in purchased parts to build. (From 1998 AIChE National Student Chapter Competition). Rules are provided on the CRE Web site.
K12 Experiment. Plant a number of seeds in different pots (corn works well). The plant and soil of each pot will be subjected to different conditions. Measure the height of the plant as a function of time and fertilizer concentration. Other variables might include lighting, pH, and room temperature. (Great grade school or high school science project.)
Q74_{B} Example 71. What is the error in assuming the concentration of species B is constant and what limits can you put on the calculated value of k? (i.e., k = 0.24 ±?)
Q75_{A} Example 73. Explain why the regression was carried out twice to find k′ and k.
Q76_{A} AWFOS–S. Which of the four Safety Snippets Videos most impressed you with getting across the safety aspect that the video was trying to highlight? Which of the four 1 to 2minute Snippets Videos was the most humorous while at the same time getting its point across?
Q77_{A} Go to the LearnChemE screencasts link for Chapter 7
(http://www.umich.edu/~elements/6e/07chap/learnchemevideos.html).
View the two screencasts on nonlinear regression and describe or list any differences between the two videos. Would you recommend these screencasts be assigned to next year’s class?
In the nonlinear regression screencast, what type of data do you need to regress for the reaction order?
P71_{A}
LEP Example Problem 73_{A}. The reaction A + B → C occurs in the liquid phase. The proposed empirical rate law is ${r}_{\text{A}}=\text{A exp(}E/RT){C}_{\text{A}}^{\alpha}{C}_{B}^{\beta}$. Look at the values of s^{2} as a function of α. At what value of α is s^{2} minimized?
Vary β keeping A fixed and describe what you find.
Vary A keeping β fixed and describe what you find.
Suggest a rate law that minimizes s^{2}.
Example 74: Nonlinear Regression
Follow the Polymath tutorial to regress the data given in this example to fit the rate law
${\mathit{r}}_{\mathrm{C}{\mathrm{H}}_{\mathrm{4}}}=\mathit{k}{P}_{\mathrm{C}\mathrm{O}}^{\mathit{\alpha}}{P}_{{\mathrm{H}}_{\mathrm{2}}}^{\mathit{\beta}}$
What is the difference in the correlation and sums of squares compared with those given in Example 74?
P72_{A}
Download the Interactive Computer Game (ICG) from the CRE Web site (http://www.umich.edu/~elements/6e/icm/ecology.html). Play the game and then record your performance number for the module that indicates your mastery of the material. Your professor has the key to decode your performance number.
ICM Ecology Performance # ________________.
Go to Professor Herz’s Reactor Lab on the Web at www.reactorlab.net. Do (a) one quiz, or (b) two quizzes from Division 1. When you first enter a lab, you see all input values and can vary them. In a lab, click on the Quiz button in the navigation bar to enter the quiz for that lab. In a quiz, you cannot see some of the input values: you need to find those with “???” hiding the values. In the quiz, perform experiments and analyze your data in order to determine the unknown values. See the bottom of the Example Quiz page at www.reactorlab.net for equations that relate E and k. Click on the “???” next to an input and supply your value. Your answer will be accepted if it is within ±20% of the correct value. Scoring is done with imaginary dollars to emphasize that you should design your experimental study rather than do random experiments. Each time you enter a quiz, new unknown values are assigned. To reenter an unfinished quiz at the same stage you left, click the [i] info button in the Directory for instructions. Turn in copies of your data, your analysis work, and the Budget Report.
P73_{A} Skiing Hand Warmers.^{†} Once a year, Professor Dr. Sven Köttlov loves to go skiing with his students in the mountains of Jofostan. Before going, he retires to his basement in his small but adequate university housing to make hand warmers for everyone.
^{†} B. Opegdra, A.M.R.P. Journal of Chemical Elevation on Time, D01:10.1021/acs.jchemed.8600012. J. Gordon and K. Chancy, J. Chem. Educ., 82(7), 1065 (July 2005).
Here he mixes iron (steel wool), sodium chloride, and a few other proprietary ingredients, then seals them in an airtight Ziploc^{®} compartment. When the warmers are needed, the bag is broken and the iron is exposed to air generating heat by the exothermic reaction
$\begin{array}{c}\text{4Fe}\left(\text{s}\right)+3{\text{O}}_{2}\left(\text{g}\right)\text{}\rightleftarrows \text{}{\text{2Fe}}_{2}{\text{O}}_{3}\end{array}$
To determine the reaction kinetics, a steel wool sample was weighed, cleaned, and put in a sealed container where the percent of oxygen was measured as a function of time and the following data was recorded:
t(min) 
0 
3 
5 
8 
15 
20 
25 
[% O_{2}] 
21 
15 
12 
10 
5 
3 
2 
For a specified steel wool weight and surface area,the proposed relationship for the percent oxygen in the bag as a function of time is
$\begin{array}{cc}[\%{\text{O}}_{2}]=3\text{exp}(~kt)& \left(\text{P73.1}\right)\end{array}$
Assuming the surface area of steel wool is approximately constant and the bag is sealed, use the CRE algorithm to derive Equation (P73.1).
Compare the rate law and data. By inspection, without carrying out any data analysis, can you determine immediately that the rate law is correct or not? Explain.
Use the data to determine the rate law and ratelaw parameters.
The data shows the reaction is virtually complete after 25 minutes, while the students will be skiing for 5–7 hours. Professor Köttlov had finger nail polish removed from his nails with acetic acid immediately before performing the experiment. It is known that acid accelerates the reaction. Can you take this information into account? If so, how would you do it?
P74_{A} When arterial blood enters a tissue capillary, it exchanges oxygen and carbon dioxide with its environment, as shown in this diagram.
The kinetics of this deoxygenation of hemoglobin in blood was studied with the aid of a tubular reactor by Nakamura and Staub (J. Physiol., 173, 161).
${\text{HbO}}_{2}\underset{{k}_{1}}{\overset{{k}_{1}}{\rightleftarrows}}\text{Hb}+{\text{O}}_{2}$
Although this is a reversible reaction, measurements were made in the initial phases of the decomposition so that the reverse reaction could be neglected. Consider a system similar to the one used by Nakamura and Staub: the solution enters a tubular reactor (0.158 cm in diameter) that has oxygen electrodes placed at 5cm intervals down the tube. The solution flow rate into the reactor is 19.6 cm^{3}/s with C_{A0} = 2.33 × 10^{–6} mol/cm^{3}.
Electrode position 
1 
2 
3 
4 
5 
6 
7 
Percent decomposition of HbO_{2} 
0.00 
1.93 
3.82 
5.68 
7.48 
9.25 
11.00 
Using the method of differential analysis of rate data, determine the reaction order and the forward specific reactionrate constant k_{1} for the deoxygenation of hemoglobin.
Repeat using regression.
P75_{A} The irreversible isomerization
$\text{A}\to \text{B}$
was carried out in a constantvolume batch reactor and the following concentration–time data were obtained:
t (min) 
0 
5 
8 
10 
12 
15 
17.5 
20 
C_{A} (mol/dm^{3}) 
4.0 
2.25 
1.45 
1.0 
0.65 
0.25 
0.06 
0.008 
Determine the reaction order = and the specific reaction rate k in appropriate units.
P76_{B} OEQ (Old Exam Question). The liquidphase irreversible reaction
A → B + C
is carried out in a CSTR. To learn the rate law, the volumetric flow rate, υ_{0}, (hence τ = V/υ_{0}) is varied and the effluent concentrations of species A are recorded as a function of the space time τ. Pure A enters the reactor at a concentration of 2 mol/dm^{3}. Steadystate conditions exist when the measurements are recorded.
Run 
1 
2 
3 
4 
5 
τ (min) 
15 
38 
100 
300 
1200 
C_{A} (mol/dm^{3}) 
1.5 
1.25 
1.0 
0.75 
0.5 
Determine the reaction order and specific reactionrate constant.
What experiments would you do to determine the activation energy?
If you were to take more data, where would you place the measurements (e.g., τ)?
It is believed that the technician may have made a dilution factorof10 error in one of the concentration measurements. What do you think?
Note: All measurements were taken at steadystate conditions. Be careful; this problem is sneaky.
P77_{A} The reaction
A → B + C
was carried out in a constantvolume batch reactor where the following concentration measurements were recorded as a function of time.
t (min) 
0 
5 
9 
15 
22 
30 
40 
60 
C_{A} (mol/dm^{3}) 
2 
1.6 
1.35 
1.1 
0.87 
0.70 
0.53 
0.35 
Use nonlinear least squares (i.e., regression) and one other method to determine the reaction order, α, and the specific reaction rate, k.
Nicolas Bellini wants to know, if you were to take more data, where would you place the points? Why?
Prof. Dr. Sven Köttlov from Jofostan University always asks his students, if you were to repeat this experiment to determine the kinetics, what would you do differently? Would you run at a higher, lower, or the same temperature? Take different data points? Explain what you think he is expecting them to say and why.
It is believed that the technician made a dilution error in the concentration measured at 60 min. What do you think? How do your answers compare using regression (Polymath or other software) with those obtained by graphical methods?
P78_{A} OEQ (Old Exam Question). The following data were reported (from C. N. Hinshelwood and P. J. Ackey, Proc. R. Soc. [Lond]., A115, 215) for a gasphase constantvolume decomposition of dimethyl ether at 504°C in a batch reactor. Initially, only (CH_{3})_{2}O was present.
Time (s) 
390 
777 
1195 
3155 
∞ 
Total Pressure (mmHg) 
408 
488 
562 
799 
931 
Why do you think the total pressure measurement at t = 0 is missing? Can you estimate it?
Assuming that the reaction
(CH_{3})_{2}O → CH_{4} + H_{2} + CO
is irreversible and goes virtually to completion, determine the reaction order and specific reaction rate k. (Ans: k = 0.00048 min^{–1})
What experimental conditions would you suggest if you were to obtain more data?
How would the data and your answers change if the reaction were run at a higher temperature? A lower temperature?
P79_{A} OEQ (Old Exam Question). In order to study the photochemical decay of aqueous bromine in bright sunlight, a small quantity of liquid bromine was dissolved in water contained in a glass battery jar and placed in direct sunlight. The following data were obtained at 25°C:
Time (min) 
10 
20 
30 
40 
50 
60 
ppm Br_{2} 
2.45 
1.74 
1.23 
0.88 
0.62 
0.44 
Determine whether the reaction rate is zero, first, or second order in bromine, and calculate the reactionrate constant in units of your choice.
Assuming identical exposure conditions, calculate the required hourly rate of injection of bromine (in pounds per hour) into a sunlit body of water, 25000 gal in volume, in order to maintain a sterilizing level of bromine of 1.0 ppm. (Ans: 0.43 lb/h)
Apply to this problem one or more of the six ideas discussed in Table P4 in the Complete PrefaceIntroduction on the Web site (http://www.umich.edu/~elements/6e/toc/PrefaceComplete.pdf).
Note: ppm = parts of bromine per million parts of brominated water by weight. In dilute aqueous solutions, 1 ppm ≡ 1 milligram per liter. (From California Professional Engineers’ Exam.)
P710_{C} The reactions of ozone were studied in the presence of alkenes (from R. Atkinson et al., Int. J. Chem. Kinet., 15(8), 721). The data in Table P710_{C} are for one of the alkenes studied, cis2butene. The reaction was carried out isothermally at 297 K. Determine the rate law and the values of the ratelaw parameters.
TABLE P710_{C} RATE AS A FUNCTION OF OZONE AND BUTENE CONCENTRATIONS
Run 
Ozone Rate 
Ozone Concentration 
Butene Concentration 
1 
1.5 
0.01 
10^{–12} 
2 
3.2 
0.02 
10^{–11} 
3 
3.5 
0.015 
10^{–10} 
4 
5.0 
0.005 
10^{–9} 
5 
8.8 
0.001 
10^{–8} 
6 
4.7^{*} 
0.018 
10^{–9} 
^{*}Hint: Ozone also decomposes by collision with the wall.
P711_{A} OEQ (Old Exam Question). Tests were run on a small experimental reactor used for decomposing nitrogen oxides in an automobile exhaust stream. In one series of tests, a nitrogen stream containing various concentrations of NO_{2} was fed to a reactor, and the kinetic data obtained are shown in Figure P711_{A}. Each point represents one complete run. The reactor operates essentially as an isothermal backmix reactor (CSTR). What can you deduce about the apparent order of the reaction over the temperature range studied?
The plot gives the fractional decomposition of NO_{2} fed versus the ratio of reactor volume V (in cm^{3}) to the NO_{2} feed rate, F_{NO2.0} (g mol/h), at different feed concentrations of NO_{2} (in parts per million by weight). Determine as many ratelaw parameters as you can.
P712_{A} The thermal decomposition of isopropyl isocyanate was studied in a differential packedbed reactor. From the data in Table P712_{A}, determine the reactionratelaw parameters.
TABLE P712_{A} RAW DATA^{†}
Run 
Rate 
Concentration 
Temperature 
1 
4.9 × 10^{–4} 
0.2 
700 
2 
1.1 × 10^{–4} 
0.02 
750 
3 
2.4 × 10^{–3} 
0.05 
800 
4 
2.2 × 10^{–2} 
0.08 
850 
5 
1.18 × 10^{–1} 
0.1 
900 
6 
1.82 × 10^{–2} 
0.06 
950 
^{†} Jofostan Journal of Chemical Engineering, Vol. 15, page 743 (1995).
A wide variety of techniques for measuring the concentrations of the reacting species may or may not be found in
THORNTON W. BURGESS, Mr. Toad and Danny the Meadow Mouse Take a Walk. New York: Dover Publications, Inc., 1915.
H. SCOTT FOGLER AND STEVEN E. LEBLANC, with BENJAMIN RIZZO, Strategies for Creative Problem Solving, 3rd ed. Upper Saddle River, NJ: Pearson, 2014.
CHESTER L. KARRASS, In Business As in Life, You Don’t Get What You Deserve, You Get What You Negotiate. Hill, CA: Stanford Street Press, 1996.
J. W. ROBINSON, Undergraduate Instrumental Analysis, 5th ed. New York: Marcel Dekker, 1995.
DOUGLAS A. SKOOG, F. JAMES HOLLER, and TIMOTHY A. NIEMAN, Principles of Instrumental Analysis, 5th ed. Philadelphia: Saunders College Publishers, Harcourt Brace College Publishers, 1998.
The design of laboratory catalytic reactors for obtaining rate data is presented in
H. F. RASE, Chemical Reactor Design for Process Plants, Vol. 1. New York: Wiley, 1983, Chap. 5.
The sequential design of experiments and parameter estimation is covered in
G. E. P. BOX, J. S. HUNTER, and W. G. HUNTER, Statistics for Experimenters: Design, Innovation, and Discovery, 2nd ed. Hoboken, NJ: Wiley, 2005.