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\newpage

\title{\ifnum\TeacherEdition=1 TEACHER VERSION  \else STUDENT VERSION \fi\\KINETICS - RATE OF CHEMICAL REACTION}




\markboth{Kinetics - Rate of Chemical Reaction}{Kinetics - Rate of Chemical Reaction}





\author{Brian Winkel,
Director SIMIODE\\
Chardon OH 44024 USA\\
\teacher{BrianWinkel@simiode.org}}


\makeStitlePDFLaTex


\begin{abstract}
We help students see the connection between college level chemistry course work and their differential equations coursework.  We do this through modeling kinetics, or rates of chemical reaction. We study zeroth, first, and second order reactions and offer many opportunities to model these chemical reactions with data, some of which comes from traditional introductory chemistry textbooks.  We ask students to verify their model through parameter estimation. We use Excel's Trendline addition to graphs/charts to select the models for the data and transformed data to take advantage of Trendline's set function choices and we also use Mathematica's direct nonlinear fitting capabilities.
\end{abstract}



\section*{SCENARIO DESCRIPTION\teacher{\footnote{In lieu of a formal COMMENTS section for the teacher after the SCENARIO DESCRIPTION section for students we intersperse this Teacher Version (and NOT the Student Version) with comments and pointers for the teacher in NB ({\it nota bene\/}) notes.  We refer to Mathematica as the computer algebra system (CAS) used in this material. However, any quality CAS, such as SAGE or Maple, will perform the same commands.}}}





Rate  or differential equations can model chemical reactions \cite{WinkelKinetics2011}. The rate of reaction is determined by the concentrations of the reactants. This is sometimes referred to as {\it The Law of Mass Action\/}.   For example, it is known that, ``The rate of decomposition is dependent on the temperature and concentration of the peroxide, as well as the pH and the presence of impurities and stabilizers.''\cite{WikipediaHydrogenPeroxide2007} Thus we consider reactions to be  dependent upon the concentration of a single reactant, say, $A$, where $[A]$ is the number of moles of $A$ present at time $t$.  The usual rates of reaction studied in the elementary texts are of the form,

\begin{equation}\label{eq:mthOrder}
\frac{d[A]}{dt} = - k [A]^m\, , \qquad {\rm with } \qquad  [A](0) = [A_0]\, . 
\end{equation}
with $k$ the {\it rate constant\/}  (presumed to be positive) and $m$ the {\it order\/} of the reaction, quite often an integer. Here $[A](0) = [A_0]$ is the initial amount of concentration of reactant. Usually the study at this level is restricted to $m = 0, 1, 2$, and these are called {\it zeroth}, {\it first}, and {\it second} order reactions, respectively.

\subsection*{ORDER OF REACTION - DEFINITIONS}

In this modeling scenario we are going to guide  more than we usually would in SIMIODE, for we are attempting to make a bridge between the notions of chemical kinetics found in the introductory chemistry course and the applications of differential equations. In practice chemists might have some idea of the order of a reaction, as defined below, from understanding of basic chemistry, but sometimes it is determining a mathematical model of a rate equation which helps them understand the nature of the reaction.  We proceed with the most common and elementary chemical reaction kinetics models.

\subsection*{Zeroth Order Reaction}



\begin{quote}

Zeroth order reactions are most often encountered when a substance such as a metal surface or an enzyme is required for the reaction to occur. For example, the decomposition reaction of nitrous oxide,		
$$2N_2O(g) \longrightarrow 2N_2(g) + O_2(g)\, , $$
occurs on a hot platinum surface. When the platinum surface is completely covered with $N_2O$ molecules, an increase in the concentration of $N_2O$ has no effect on the rate, since only those $N_2O$ molecules on the surface can react. Under these conditions the rate is a constant because it is controlled by what happens on the platinum surface rather than by the total concentration of $N_2O$, \ldots \cite[p.~657]{Zumdahl1992}.
\end{quote}

For a generic reaction we shall use $y(t) = [A] = [A(t)]$, where $[A]$ is the amount of reactant $A$. This is often given  in moles or moles/liter) with  time, $t$,  in sec.  Here we see the rate or differential equation is

$$\frac{dy}{dt} = - k y^0 = -k\, .$$
The solution to this is rather easy through direct integration, $y(t) = y(0)- k t$. Such reactions are self-evident from the data, for if we plot $y(t)$ vs.~$t$ we would see a linear function starting at $y = y(0)$ with a negative slope, $-k$. We can easily enter the data in Excel and use Trendline to pick off $k$. Since zeroth order reactions are quite rare we move on to first order reactions.

\subsection*{First Order Reaction}

Let us look at first order reactions in general:

$$\frac{dy}{dt} = - k y^1 = -k\, y\, \qquad {\rm with} \qquad y(0) = y_0\, .$$
Using the separation of variables technique we obtain
$$\frac{1}{y}\frac{dy}{dt} = -k\, ,$$
from which we can see the integrated form of our model
$$\ln(y) = -k\, t + c \, \quad {\rm where}\quad c = \ln(y(0))\, . $$
This gives us 
\begin{equation}
\ln(y) = -k\, t + \ln(y(0))\, .\label{eq:logged} 
\end{equation}
Now we can actually solve this differential equation for $y = y(t)$ and we do, but the chemist really is interested in determining the nature (order) of the reaction as well as the parameter $k$, called the {\it rate of the reaction\/}, and often stops at this point. Chemists refer to   (\ref{eq:logged}) as the {\it integrated form\/} of the rate law. They quite often use {\it this\/} form of the equation to do their parameter estimation and determination of the order of the reaction.  However, in mathematics we might wish to push further to a solution that reads ``$y(t$) is.'' Thus, after taking antilogarithms (or exponentiating both sides in ~(\ref{eq:logged}) we can produce a complete solution of the differential equation, 
\begin{equation}\label{eq:FirstOrderSolved}
y = y(t) = y(0)e^{-k\, t}\, .
\end{equation}







In the study of chemical reactions one of the simplest reactions is that of a decomposition of a substance, say hydrogen peroxide ($H_2\, O_2$). There is the phenomena of going to the medicine chest to find the hydrogen peroxide (and iodine!) to flush and clean a cut, only to find that what is in the bottle does not produce a white froth when applied to the cut as the medicine is supposed to do while it rids the cut of germs. The medicine is old and has lost its powers!  This is an example of the decomposition of $H_2\, O_2$ into water and oxygen ($2 H_2O_2 \longrightarrow 2 H_2O + O_2$) and we use the basic Law of Mass Action to conjecture a rate (differential) equation for $H_2O_2$. 
 This means for $[H_2O_2]$ moles/L of hydrogen peroxide:
$$\frac{d\, [H_2O_2]}{dt} = - k\cdot [H_2O_2]^m\, ,$$
for some number $m$. We seek to determine if this reaction is first order, i.e. if $m=1$.






\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|}\hline
Time $t$ (sec) & $[H_2O_2]$ (mol/L) \\ \hline\hline
0	&1.00	\\ \hline
120	&0.91	\\ \hline
300	&0.78	\\ \hline
600	&0.59	\\ \hline
1200	&0.37   \\ \hline
1800	&0.22	\\ \hline
2400	&0.13	\\ \hline
3000	&0.08	\\ \hline
3600	&0.05	\\ \hline
\end{tabular}
\vspace*{.4 true cm}

\parbox{6.3 true cm}{\caption{\label{table:H2O2} Collected data \cite[p.~682]{Zumdahl1992} on the reaction $\displaystyle{2H_2O_2(g) \longrightarrow	2 H_2O + O_2(g)}$.}}
\end{center}

\vspace*{-.7 true cm}
\end{table}




We return to the task of determining order of the reaction and the parameter $k$. From  ~(\ref{eq:logged}) we can see that a first order reaction will produce a linear relationship between $\ln(y)$ and $t$. 
\begin{itemize}\item [1)] 
\begin{itemize}
\item [a)] Confirm that this is a first order reaction by logging the $[H_2O_2]$ (mol/L) column data in Table~\ref{table:H2O2} and plotting $\log [H_2O_2]$ vs $t$ to see a linear relationship.
\item [b)] Use Excel's Trendline to determine the slope of the line of best fit for the logged data from Table~\ref{table:H2O2}, i.e. the natural logarithm of $[H_2O_2]$ (mol/L) vs.~$t$.
\item [c)] Use your parameter $k$ obtained in (b) and the initial amount of $[H_2O_2]$ (mol/L) in (\ref{eq:FirstOrderSolved}) to determine what amount of  $[H_2O_2]$ (mol/L) the model predicts at times $t$ as offered in Table~\ref{table:H2O2}.
\item [d)] Compare these predictions with the observed data. Do you believe that the first order model does a good job of predicting?
\item [e)] Let us determine if this reaction is first order. To do this we shall assume $m=1$ and see how the resulting model compares to the data. We can use  Mathematica's {\tt FindMinimum} command on the sum of square errors between the observed values of $H_2O_2$  and the predicted model values (\ref{eq:FirstOrderSolved}) for the 9 time observations. Or we could use Excel's Solver command (with parameters as true variables in Solver) to do the same minimization of the sum of square errors. Both strategies can determine the best fitting  rate constant, $k$.  Plot the model prediction over the data and comment on what you see. Note: We can, of course, obtain good results using Excel's Trendline, but Excel's Solver is a broader tool which permits optimization, in this case minimization of the sum of square errors.
\end{itemize}

\end{itemize}


\subsection*{Second Order Reaction}



Now let us look at second order reactions in general:

$$\frac{dy}{dt} = - k y^2\, \qquad {\rm with} \qquad y(0) = y_0\, .$$
Using the separation of variables technique we obtain
$$\frac{1}{y^2}\frac{dy}{dt} = -k\, ,$$
from which we can integrate to see
$$\frac{1}{y}= k\, t + c \, \quad {\rm where}\quad c = \frac{1}{y(0)}\, . $$
This gives us 
\begin{equation}
\frac{1}{y} = k\, t + \frac{1}{y(0)}\, .\label{eq:second} 
\end{equation}
Now we can actually solve this differential equation for $y = y(t)$ and we do, but again the chemist really is interested in determining the nature (order) of the reaction and the parameter $k$ and will often stop at this point. A function for a solution to the rate or  differential equation can be found by  inverting both sides in (\ref{eq:second}),
\begin{equation}\label{eq:secondClosed}
y = y(t) =\frac{1}{k\, t + \frac{1}{y(0)}} = \frac{y(0)}{y(0) k\, t + 1}\, .
\end{equation}


\begin{itemize}
  \item [2)] Revisit the decomposition of hydrogen peroxide in Table \ref{table:H2O2} and show it is {\it not\/} a second order reaction.  Offer as complete a defense as you can - data fitting, plots, verbal argument, etc.
\end{itemize} 

We summarize this information in Table~\ref{table:OrderSummary} while in 
 Figure~\ref{fig:AllOrderPlots} we offer general plots to illustrate the $m=0,1,2,$ order reactions. In each case we give plots of the actual solution  and the integrated form of the solution.

\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|c|} \hline
Order of reaction & Differential Equation & Integrated Form & solution Form \\ \hline
0 &  $y'(t) = - k y(t)^0 = - k y(t)$  &$y(t) = y(0) - k t$ & $y(t) = y(0) - k t$ \\ \hline
1 &  $y'(t) = - k y(t)^{1} = - k y(t)$ & $\ln(y(t) = - k t + c$ & $y(t) = y(0) e^{-k t}$ \\ \hline
2 & $y'(t) = - k y(t)^{2} = - k y(t)^2$ & $\displaystyle{\frac{1}{y(t)} = k t + \frac{1}{y(0)}}$ & $\displaystyle{y(t) = \frac{y(0)}{y(0) k t + 1}}$ \\ \hline
\end{tabular}
\parbox{13.3 true cm}{\caption{\label{table:OrderSummary} Summary of $m = 0, 1, 2$ order kinetics differential equations with differential equation model, the integrated form of the solution through which chemists can obtain possibly a linear plot to confirm the order of the reaction, and a complete solution for a fully developed model.}}
\end{center}
\end{table}






\begin{figure}
\begin{center}
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsZerothOrderPlot}


\end{minipage} \ \hspace*{.3 true cm} \
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsLinearZerothOrderPlot}

\end{minipage}

\centerline{Plots of both solution and integrated form of solution for order  $m=0$.} 

\vspace*{.8 true cm}
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsFirstOrderPlot}

\end{minipage} \ \hspace*{.3 true cm} \
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsLinearFirstOrderPlot}

\end{minipage}

\centerline{Plots of both solution and integrated form of solution for order  $m=1$.} 

\vspace*{.8 true cm}
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsSecondOrderPlot}

\end{minipage} \ \hspace*{.3 true cm} \
\begin{minipage}[t]{7 true cm}

\includegraphics[width=6.9cm]{1-011-KineticsLinearSecondOrderPlot}

\end{minipage}

\centerline{Plots of both solution and integrated form of solution for order  $m=2$.} 

\vspace*{.3 true cm}
\parbox{7.9 true cm}{\caption{\label{fig:AllOrderPlots} Plots of both solution and integrated form of solution for orders (a) $m=0$, (b) $m=1$, and (c) $m=2$.}}
\end{center}
\end{figure}

\vspace*{-.9 true cm}
\subsection*{EXAMPLES OF DIFFERENT REACTION ORDERS}


\subsection*{Decomposition of $NO_2$}

Consider the following data (Table~\ref{table:NO2}) for the reaction describing the decomposition of $NO_2$, nitrogen oxide.


\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|} \hline
Time $t$ (sec) & $[NO_2]$ \\ \hline
0.5	&0.02	\\ \hline
1.0	&0.015	\\ \hline
1.5	&0.012	\\ \hline
2.0	&0.0087   \\ \hline
\end{tabular}
\vspace*{.4 true cm}

\parbox{6.3 true cm}{\caption{\label{table:NO2} Collected data \cite[p.~486]{KotzPurcell1987}   on the reaction $\displaystyle{2NO_2(g) \longrightarrow	2 NO (g) + O_2(g)}$.}}
\end{center}

\vspace*{-.6 true cm}

\end{table}




\begin{itemize}
\item [3)]
\begin{itemize}
\item [a)] First, show this reaction is {\it not\/} zeroth or first order.   Offer as complete a defense as you can - data fitting, plots, verbal argument, etc.
\item [b)] Using the data in Table~\ref{table:NO2}, transform the data according to (\ref{eq:second}) and determine the parameter $k$ in terms of best fit.  This is the way a chemist would do this problem.
\item  [c)] Consider the final solved form in (\ref{eq:secondClosed}) and perform a nonlinear minimization using Mathematica's {\tt FindMinimum} command on the sum of square errors between the observed values of $N_2O$  and the model (\ref{eq:secondClosed}) predicted values for the 4 time observations. Or we could use Excel's Solver command (with parameters as true variables in Solver) to do the same minimization of the sum of square errors. Both strategies can determine the best fitting  rate constant, $k$.  Plot the model prediction over the data and comment on what you see.
\item [d)] Use the parameter $k$ obtained in (b) or (c) and the initial amount of $[NO_2]$ (mol/L) in Table~\ref{table:NO2} to determine what amount of  $[NO_2]$ (mol/L) the model predicts at any given time $t$, say $t = 3, \, 5$, and 10. 
\item [e)] Comparing the predictions with the observed data, if you believe that the second order model does a good job of predicting the amount of $[NO_2]$ (mol/L) at time $t$ explain why. Otherwise explain why not.
\end{itemize}
\end{itemize}



\subsection*{Decomposition of $N_2O_5$}

\begin{itemize}
\item [4)] Consider the following data (Table~\ref{table:N2O5}) for the reaction describing the decomposition of $N_2O_5$, dinitrogen pentoxide.
\begin{equation}\label{eq:decomp}
2N_2O_5 \longrightarrow 4NO_2 + O_2
\end{equation}

\begin{table}
\begin{center}
\begin{tabular}{|r|l|} \hline
Time $t$ in s &$[N_2O_5](t)/M$ \\ \hline
0 & 0.310 \\ \hline
600 & 0.254 \\ \hline
1200 & 0.208 \\ \hline
1800 & 0.172 \\ \hline
2400 & 0.141  \\ \hline
3000 & 0.116  \\ \hline
3600 & 0.0964 \\ \hline
4200 & 0.0812 \\ \hline
4800 & 0.0669 \\ \hline
6000 & 0.0464 \\ \hline
\end{tabular}

\parbox{10 true cm}{\caption{\label{table:N2O5}   Data \cite{HenoldWalmsley1984} for the decomposition of $N_2O_5$  is given.}}
\end{center}\end{table}

\begin{itemize}
\item [a)] Plot the data.
  \item  [b)] From the plot  make a conjecture as to the order ($m=0,1, 2$) of the reaction.
  \item  [c)] Conduct a complete analysis, determining the order and the parameters. Plot  the data and  the model, being sure to defend what the order is and what the order is not vis-\'{a}-vis $m=0,1,2$ orders.
\end{itemize}

\teacher{\noindent NB:  To the Teacher.  We offer a complete analysis of this reaction in the resource Mathematica notebook (.nb and .pdf format) 1-011-T-Mma-Kinetics-TeacherVersion at the SIMIODE web site for this scenario. While the CAS used by any given teacher may differe from Mathematica we have offered sufficient comments in the notebook so as to permit a teacher to ``port'' the material to the CAS of choice. We also offer an Excel version of just the first order analysis in the Excel and pdf files 1-011-T-Excel-Kinetics-DecompositionOfN\_2O\_5-TeacherVersion.}
 \end{itemize}


\subsection*{Decomposition of $H_2O_2$}
\begin{itemize}
\item [5)] Consider the following data (Table~\ref{table:H2O2}) for the reaction describing the decomposition of $H_2O_2$, hydrogen peroxide.
\begin{equation}\label{eq:decomp}
2H_2O_2 \longrightarrow 2H_2O + O_2
\end{equation}

\begin{table}[h]
\begin{center}
\begin{tabular}{|r|l|} \hline
Time $t$ in s &$[H_2O_2]$ mol/L\\ \hline
0 & 1 \\ \hline
 120 &  0.91 \\ \hline
 300 &  0.78 \\ \hline
 600 &  0.59 \\ \hline
 1200 &  0.37 \\ \hline
 1800 &  0.22 \\ \hline
 2400 &  0.13 \\ \hline
 3000 &  0.082 \\ \hline
 3600 &  0.05 \\ \hline
\end{tabular}




\parbox{10 true cm}{\caption{\label{table:H2O2}   Data \cite[p.~682]{Zumdahl1992} for the decomposition of $H_2O_2$  is given.}}
\end{center}\end{table}

\begin{itemize}
\item [a)] Plot the data.
  \item  [b)] From the plot  make a conjecture as to the order ($m=0,1, 2$) of the reaction.
  \item  [c)] Conduct a complete analysis, determining the order and the parameters. Plot  the data and  the model, being sure to defend what the order is and what the order is not vis-\'{a}-vis $m=0,1,2$ orders.
\end{itemize}


\teacher{\noindent NB:  To the Teacher.  We offer a complete analysis of this reaction in the resource Mathematica notebook (.nb and .pdf format) 1-011-T-Mma-Kinetics-TeacherVersion at the SIMIODE web site for this scenario. We also offer an Excel version of just the first order analysis in the Excel and pdf files 1-011-T-Excel-Kinetics-DecompositionOfH\_2O\_2-TeacherVersion.}
\end{itemize}

\subsection*{Decomposition of $C_4H_6$}

\begin{itemize}
\item [6)] Consider the following data (Table~\ref{table:Buta}) for the reaction  of $C_4H_6$, butadiene, to form its dimer, a chemical structure formed from two similar sub-units.
\begin{equation}\label{eq:decomp}
2C_4H_6 (g) \longrightarrow C_8H_{12} (g)
\end{equation}

\begin{table}[h]
\begin{center}
\begin{tabular}{|r|l|} \hline
Time $t$ in s &$[C_4H_6]$ mol/L\\ \hline

0 & 0.01000 \\ \hline
1000 &  0.00625 \\ \hline
1800 &  0.00476 \\ \hline
2800 &  0.00370 \\ \hline
3600 &  0.00313 \\ \hline
4400 &  0.00270 \\ \hline
5200 &  0.00241 \\ \hline
6200 &  0.00208 \\ \hline
\end{tabular}




\parbox{11.5 true cm}{\caption{\label{table:Buta}   Data \cite[p.~654]{Zumdahl1992} for the reaction   of $C_4H_6$, butadiene, to form its dimer.}}
\end{center}
\vspace*{-.5 true cm}
\end{table}

\begin{itemize}
\item [a)] Plot the data.
  \item  [b)] From the plot  make a conjecture as to the order ($m=0,1, 2$) of the reaction.
  \item  [c)] Conduct a complete analysis, determining the order and the parameters. Plot  the data and  the model, being sure to defend what the order is and what the order is not vis-\'{a}-vis $m=0,1,2$ orders.

\end{itemize}


\teacher{\noindent NB:  To the Teacher.  We offer a complete analysis of this reaction in the resource Mathematica notebook (.nb and .pdf format) 1-011-T-Mma-Kinetics-TeacherVersion at the SIMIODE web site for this scenario. }
\end{itemize}

\subsection*{Your Call}

\begin{itemize}
\item [7 )] Get an introductory chemistry text or check one out from the library and look up the section on chemical kinetics. Find a data set problem with directions to ascertain the order of the reaction and offer a write up  in this regard. 
\end{itemize}





\section*{SECOND ORDER REACTIONS OF THE TYPE $A + B \longrightarrow AB$}


From \cite[p~121]{Pennisi1972} we are advised,``In general, whenever one shows that a given chemical reaction is of the second order, one may assume the reaction to be bimolecular.''






Thus far we have considered $m^{\rm th}$ order reactions of the form (\ref{eq:mthOrder}). However, a more general definition of order describes situations like

\begin{equation}\label{eq:Basic}
 A + B  \longrightarrow AB \, . 
\end{equation}

We say a reaction is a general {\it   second order\/} reaction if the rate of the reaction is proportional to the product of the concentrations of the two reacting substances; here this is $A$ and $B$. 

Now, if we denote $C_A(t)$ and $C_B(t)$ as the concentrations of $A$ and $B$, say in moles per liter then  by the Law of Mass Action (which simply says the rate of the reaction is proportional to the product of the reactants available for the reaction)  (\ref{eq:Basic}) yields  the differential equation model (\ref{eq:AB},

\begin{equation}\label{eq:AB}
C_A'(t) = C_B'(t) =  k C_A\cdot C_B \, , \,\,\,  {\rm with} \,\,\, C_A(0) = A_0\,\,\,  {\rm and} \,\,\, C_B(0) = B_0\, .
\end{equation}

\noindent $C_A(0) = A_0$ and $C_B(0) = B_0$  are the number of moles per liter of $A$ and $B$ at the start of the reaction.    So if $y(t)$ represents the   number of moles per liter of $A$ or $B$ that have reacted in time $t$  , then $A_0 - y$  and $B_0 - y$ identify the concentrations of $A$ and $B$ remaining at time $t$. Now the differential equation (rate of reaction equation), (\ref{eq:AB}), actually can be rewritten as (\ref{eq:ABODE}),

\begin{equation}\label{eq:ABODE}
y'(t) =  k(A_0 - y)(B_0 - y)\, , y(0) = 0.
\end{equation}

\noindent Here $k>0$, as  the traditional way to offer rate constants  as positive numbers.

We have two cases to study:  Case I when the initial concentrations of $A$ and $B$  are  the same, i.e. $A_0 = B_0$ and Case II when the initial concentrations of $A$ and $B$  are not the same, i.e. $A_0 \neq B_0$.


\begin{itemize}\item [8)]
Find the solution to (\ref{eq:ABODE})  for Case I, using $A_0 = B_0$ with $y(0) = 0$ and $k = .0073$. Plot the solution over the time interval $t \in [0,600]$ s. What is the long term limit of $y(t)$. Explain why you are not surprised at this number.
\end{itemize}
 
\subsection*{Ethyl acetate and sodium hydroxide}

Consider the reaction of ethyl acetate with sodium hydroxide,
\begin{equation}\label{eq:Reaction}
\overbrace{CH_3COOC_2H_5}^{\rm ethyl\,\, acetate}\,\,\,\,+ \,\,\overbrace{NaOH}^{\rm sodium \,\, hydroxide} \longrightarrow  \,\,\,\,\overbrace{C_2H_5O_{11}}^{\rm ethyl\,\,  alcohol}\,\, + \,\,\,\overbrace{CH_3COONa}^{\rm sodium\,\,  acetate}\, 
\end{equation}

For ease in notation let $A = A(t)$ be the concentration of ${CH_3COOC_2H_5}$ in moles/liter at time $t$ in seconds and let $B = B(t)$ be the concentration of $NaOH$ in moles/liter at time $t$ in seconds. Notationally let us write $AB$ as the concentration of the product  $C_2H_5O_{11}$ in moles/liter and denote $CH_3COONa$ as a by product of the reaction.

Reicher  \cite{Reicher1885} offers  experimental data in Table \ref{table:expData} where $t$ is time in seconds and the entries in the second row are the concentration of the product $AB$ in moles/liter at time $t$.  


\begin{table}
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|} \hline
$t$  & 0& 393& 699& 1010& 1265\\ \hline
$y(t)$& 0.0&0.0772& 0.1171& 0.1525& 0.1759\\ \hline
\end{tabular}
\parbox{12.3 true cm}{\caption{\label{table:expData} Data collected by Reicher \cite{Reicher1885} shows the  time ($t$) in seconds and amount of $AB$ produced in moles per liter ($y(t)$) at time $t$.}}
\end{center}
\end{table}

\begin{itemize}
\item [9)] In Table~\ref{table:expData}, we see there is initially $A_0 = 0.5638$ moles/liter concentration of $A$ and $B_0 = 0.3114$ moles/liter concentration of $B$. 
\begin{itemize}
\item [a)] Since it takes one mole/liter of each of $A$ and $B$ to produce one mole/liter of $AB$ build a mathematical model for $y'(t)$ the rate of change in $y(t)$ at time $t$. Defend every term in the model.
\end{itemize} 
		
If we  solve (using Mathematica's {\tt DSolve}) for $y(t)$ and place $A_0 = 0.5638$ and $B_0 = 0.3114$ into the solution we obtain the following function:

\begin{equation}\label{eq:ysol}
y(t) = \frac{0.149709 e^{0.3114 k\cdot  t}-0.149709
   e^{0.5638 k\cdot  t}}{0.265536 e^{0.3114 k\cdot 
   t}-0.480762 e^{0.5638 k\cdot  t}}\, .
\end{equation}

\item [b)] Construct the sum of square errors function (\ref{eq:SSEk}) for the difference between the model (\ref{eq:ysol}) predictions at time $t_i$ where $i = 1, 2, \ldots 5$ goes through the five observation data point we have in Table~\ref{table:expData} and the data on the concentration of product $AB$ offered in the second row of Table~\ref{table:expData}, i.e, $AB_i$, for $i = 1, 2, \ldots 5$:

\begin{equation}\label{eq:SSEk}
SSE(k) = \sum_{i=1}^5 (y(t_i) -   AB_i)^2 \, .
\end{equation}

\item [c)] Minimize $SSE(k)$ with respect to $k$ and obtain $k =0.00139862$ (using Mathematica's {\tt FindMinimum}) with an actual $SSE(0.00139862) = 0.0000134864$. Here,  create a very good fit. Indeed, we offer a plot of the data over the model in Figure~\ref{fig:1-011-KineticsDatak}.

\end{itemize}




\begin{figure}[!h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9cm]{1-011-KineticsDatak.eps}



\parbox{11.7 true cm}{\caption{\label{fig:1-011-KineticsDatak} Plot of the model over the data. We also include the horizontal line at value of $B_0 = 0.3114$ for this is the limiting value of the concentration of product $AB$ as it takes one mole/liter of each of $A$ and $B$ to produce one mole/liter of $AB$ and the concentration of $B$ is the smaller, and hence limiting, of the two concentrations.}}
\vspace*{-.3 true cm}
\end{figure}



\section*{SUBLIMATION OF CAMPHOR - \ldots SOMETHING COMPLETELY DIFFERENT} 

Camphor is a compound used to moth-proof clothing and possesses a strong odor, primarily because it sublimates from solid to gas.  In \cite{Kareem2000}  experiments were conducted to model the rate of sublimation and compare it to the collected data. An assortment of camphor balls was set aside and observations on the diameter of these spherical balls was taken over a period of 41 days.  We display the average diameter information for this set of camphor balls at various times in Table~\ref{table:camphor}.



\begin{itemize}\item [10)] Form a mathematical model for the sublimation of camphor balls and validate the model with the data in Table~\ref{table:camphor}.
\end{itemize} 


\begin{table}[!h]
\vspace*{.2 true cm}
\begin{center}
\begin{tabular}{|c|c|} \hline
Time (days) & Average Diameter (mm)\\ \hline
0.00 & 18.52 \\ \hline
20.50 & 14.56 \\ \hline
41.00 & 11.47 \\ \hline
\end{tabular}

\parbox{8.6 true cm}{\caption{\label{table:camphor} Data from \cite{Kareem2000} on the average diameter of camphor balls in ambient temperature over a 41 day period.}}
\end{center}
\vspace*{-.5 true cm}
\end{table}

In Table~\ref{table:camphorV} we see the data  displayed, but with the addition of the average volume of the camphor balls sampled at each time. 

\begin{table}[h]
\begin{center}
\begin{tabular}{|c|c|c|} \hline
Time (days) & Average Diameter (mm)& Average Volume (mm$^3$)\\ \hline
0.00 & 18.52 & 3325.99\\ \hline
20.50 & 14.56 &1616.15\\ \hline
41.00 & 11.47 &790.112\\ \hline
\end{tabular}
\parbox{8.6 true cm}{\caption{\label{table:camphorV} Data from \cite{Kareem2000} on the average diameter of camphor balls in ambient temperature over a 41 day period.}}
\end{center}
\end{table}


\teacher{\noindent NB: To Teachers. A very reasonable assumption is that the rate of sublimation of the camphor balls is proportional to average surface area of the balls sampled, so if $y(t)$ is the average volume of the balls sampled, then $y(t)^{2/3}$ is proportional to the surface area and hence the rate of change in the volume (using volume of sphere formula, $V = \frac{4}{3}\pi r^3$) of the average camphor balls is

\vspace*{-.2 true cm}

\begin{equation}\label{eq:camp}
y'(t) = - k y(t)^{2/3}\, . 
\end{equation}
}





\teacher{\noindent NB:  To the Teacher.  We offer a complete analysis of this reaction in the resource Mathematica notebook (.nb and .pdf format) 1-011-T-Mma-Kinetics-TeacherVersion at the SIMIODE web site for this scenario. 


We consider three   approaches for modeling $y(t)$ the average volume of the set of camphor balls:

\begin{itemize}
  \item [i)] $y'(t) = - k y(t)^{2/3}$, where we assume the sublimation of camphor is proportional to surface area of the balls hence proportional to volume raised to the 2/3 power. Explain why this follows. Here we get a sum of square errors of 12,650.2.
  \item [ii)] $y'(t) = - k y(t)^1$, where we assume the sublimation of camphor is proportional to volume of the balls hence proportional to volume raised to the first power. Here we get a sum of square errors of 11.8274.
  \item   [iii)] $y'(t) = - k y(t)^r$, where we assume the sublimation of camphor is proportional to volume of the balls raised to an unknown power, $r$.  Here we get $r = 1.01179$ a sum of square errors of $9.80621 \times 10^{-20}$.
\end{itemize}

\vspace*{-.4 true cm}

\begin{figure}[h]
\centering

\vspace*{-.1 true cm}
\includegraphics[width=9.5cm]{1-011-KineticsCamphorPlots}



\parbox{11. true cm}{\caption{\label{fig:1-011-KineticsCamphorPlots} Plots of the camphor ball data and the three models. Models (ii) and (iii) are almost perfect with $r$ equal to or almost 1, while (i) is off. }}
\vspace*{-.3 true cm}
\end{figure}
}

\teacher{\section*{NOTES FOR TEACHERS}

One of the referees suggested that this scenario gave too much away and did not really encourage students to build models, because the material prescribed one of three orders of reaction $m = 0, 1, 2$. This is an introduction to kinetics modeling and, as such, we decided to be more directive. We have found that chemistry courses, because they cannot count on calculus as a prerequisite, will often omit the section on chemical reaction kinetics, found in most texts. We intend this material to serve as a connector   between  elementary chemistry coursework and the study of differential equations.  Thus we are offering students in this scenario more guidance than one might want in a true modeling spirit.
}

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\end{document}



