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\begin{document}

{\bf Question}

The reaction scheme $A
\overset{k^{12}}{\underset{k_{21}}{\rightleftharpoons}}B
\overset{k^{23}}{\underset{k_{32}}{\rightleftharpoons}}C$ is
described by the rate equations:



\begin{eqnarray*}  \frac{dA}{dt} &amp; = &amp; -k_{12}A +k_{21}B \\ \frac{dB}{dt}
&amp; = &amp; k_{12}-(k_{21}+k_{23})B + k_{32}C \\ \frac{dC}{dt}  &amp;= &amp;
k_{23}B - k_{32}C
\end{eqnarray*}

Solve these equations when $k_{12} = k_{23} = k_{32}= 1, k_{21}
 =2$ and
\begin{description}
\item[(i)]
$A(0) = p, B(0) = C(0) = 0$;
\item[(ii)]
$B(0) = p, A(0) = C(0) = 0$.
\end{description}
What is the equilibrium concentrations in each case?

{\bf Answer} \vspace{.2in}

In matrix form the equations become: $\left(
\begin{array}{c} \frac{dA}{dt} \\ \frac{dB}{dt} \\ \frac{dC}{dt}
\end{array} \right) = \left( \begin{array}{rrr} -1 &amp; 2 &amp; 0 \\ 1 &amp;
-3 &amp; 1 \\ 0 &amp; 1 &amp; -1 \end{array} \right) \left( \begin{array}{c}
A\\ B \\ C
\end{array} \right)$

\begin{eqnarray*}
 \left| \begin{array}{ccc} {-1 - \lambda} &amp; 2 &amp; 0 \\ 1 &amp; {-3 -
 \lambda}&amp; 1  \\0 &amp; 1 &amp; {-1 - \lambda} \\ \end{array} \right|&amp; = &amp; -(1 + \lambda) \left|
 \begin{array}{cc} {-3 - \lambda} &amp; 1 \\ 1 &amp; {-1 - \lambda} \\
\end{array} \right| - 2 \left| \begin{array}{cc} 1 &amp; 1 \\ 0 &amp; {-1 - \lambda} \\
\end{array} \right|  +0 \\ &amp; = &amp; (-1 - \lambda)[(1 + \lambda)(3 + \lambda) - 1] +2(1+ \lambda) \\ &amp; = &amp;
\ -\lambda (1+ \lambda)(\lambda +4)  = 0
\\ {\rm so \ \ \ } \lambda &amp; = &amp; 0, -1, -4
\end{eqnarray*}

$ \underline {\lambda = 0}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} -1 &amp; 2 &amp; 0 \\ 1 &amp; -3 &amp; 1 \\ 0 &amp; 1 &amp; -1\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {-x + 2y} &amp; = &amp; 0 \\ {x - 3y + z} &amp;
= &amp; 0 \\ {y - z} &amp; = &amp; 0 \\ \end{array} \right\}
\begin{array}{lrcl}  {\rm let\ \ } &amp; y &amp; = &amp; \alpha \\ {\rm so\ \ } &amp; x &amp;
= &amp; 2\alpha \\ {} &amp; z &amp; = &amp; \alpha \end{array}$

Suitable eigenvector $\left( \begin{array}{c} 2\alpha \\ \alpha \\
\alpha \\ \end{array} \right)$

$ \underline {\lambda = -1}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} 0 &amp; 2 &amp; 0 \\ 1 &amp; -2 &amp; 0 \\ 0 &amp; 1 &amp; 0\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {2y} &amp; = &amp; 0  \\ {x -2y + z} &amp; = &amp;
0\\ {y} &amp; = &amp; 0 \end{array} \right\} \begin{array}{lrcl} {} &amp; y &amp;
= &amp; 0 \\ {\rm let\ \ } &amp; x &amp; = &amp; \beta \\ {\rm so\ \ } &amp; z &amp; = &amp;
-\beta \\  \end{array}$

Suitable eigenvector $\left( \begin{array}{c} \beta \\ 0 \\ -\beta
\\ \end{array} \right)$

$ \underline {\lambda = -4}$ \hspace{.2in} Solve $\left(
\begin{array}{rrr} 3 &amp; 2 &amp; 0 \\ 1 &amp; 1 &amp; 1 \\ 0 &amp; 1 &amp; 3\\ \end{array} \right)
\left( \begin{array}{r} x \\ y\\ z\\ \end{array} \right) = \left(
\begin{array}{r} 0 \\ 0 \\0 \\ \end{array} \right)$

or $ \left. \begin{array}{rcl} {3x + 2y} &amp; = &amp; 0 \\ {x + y +z} &amp; =
&amp; 0 \\ y + 3z &amp; = &amp; 0 \\ \end{array} \right\}
\begin{array}{lrcl} {\rm let\ \ } &amp; x &amp; = &amp; 2\gamma \\ {\rm so\ \ }
&amp; y &amp; = &amp; -3\gamma\\ {} &amp; z &amp; = &amp; \gamma \end{array}$

Suitable eigenvector $\left( \begin{array}{c} 2\gamma \\ -3\gamma
\\ \gamma \\ \end{array} \right)$

Hence the general solution is \begin{eqnarray*} \left(
\begin{array}{c} A \\ B\\ C
\end{array} \right) &amp; = &amp; \alpha \left( \begin{array}{c} 2 \\ 1 \\ 1
\end{array} \right)e^{0t} + \beta \left( \begin{array}{c} 1 \\ 0
\\ -1 \end{array} \right)e^{-t} + \gamma \left( \begin{array}{c} 2 \\
-3 \\ 1 \end{array} \right)e^{-t} \\ &amp; = &amp; \alpha \left(
\begin{array}{c} 2 \\ 1 \\ 1
\end{array} \right) + \beta \left( \begin{array}{c} 1 \\ 0
\\ -1 \end{array} \right)e^{-t} + \gamma \left( \begin{array}{c} 2 \\
-3 \\ 1 \end{array} \right)e^{-t} \end{eqnarray*}
\begin{description}
\item[(i)]
if A(0) = p and B(0) = C(0) = 0 then

$\left. \begin{array}{lrcccl} (1) &amp; A(0) &amp; = &amp; p &amp; = &amp; 2\alpha +
\beta + 2\gamma \\ (2)&amp; B(0) &amp; = &amp; 0 &amp; = &amp; \alpha - 3\gamma \\(3)
&amp; C(0) &amp; = &amp; 0 &amp; = &amp; \alpha - \beta + \alpha \end{array} \right\}$
$\begin{array}{l} (2) \Rightarrow \alpha = 3\gamma {\rm \ \ }(*)\\
{\rm sub\ (*)\ into\ (3)\ } \Rightarrow \beta = 4\gamma \\ {\rm
sub\ (*)\ into\ (1)\ } \\ \Rightarrow p = 6\gamma + 4\gamma +
2\gamma = 12\gamma \\ \Rightarrow \gamma = \frac{p}{12}
\end{array}$

Hence $\gamma = \frac{p}{12}, \alpha = \frac{3p}{12} =
\frac{p}{4}$ and $\beta = \frac{4p}{12} = \frac{p}{3}$. This gives
the particular solution $\displaystyle \left( \begin{array}{c} A
\\B
\\ C
\end{array} \right) = \frac{p}{4} \left( \begin{array}{c} 2 \\ 1 \\
1 \end{array} \right) + \frac{p}{3}\left( \begin{array}{c} 1 \\ 0
\\ -1 \end{array} \right)e^{-t} + \frac{p}{12}\left( \begin{array}{c} 2 \\ -3 \\
1 \end{array} \right)e^{-4t}.$

\item[(ii)]
if B(0) = q and A(0) = C(0) = 0 then

$\left. \begin{array}{lrcccl} (1) &amp; A(0) &amp; = &amp; 0 &amp; = &amp; 2\alpha +
\beta + 2\gamma \\ (2)&amp; B(0) &amp; = &amp; q &amp; = &amp; \alpha - 3\gamma \\(3)
&amp; C(0) &amp; = &amp; 0 &amp; = &amp; \alpha - \beta + \alpha \end{array} \right\}$
$\begin{array}{l} (1)-(2) \Rightarrow 3\beta = 0 \Rightarrow \beta
= 0 {\rm \ \ }(*)\\ {\rm sub\ (*)\ into\ (3)\ } \Rightarrow \gamma
= -\alpha \\ {\rm sub\ \ into\ (2)\ } \\ \Rightarrow \alpha =
\frac{q}{4} \\ \Rightarrow \gamma = -\frac{q}{4}
\end{array}$

This give the particular solution $\left( \begin{array}{c} A \\B
\\ C \end{array} \right) ={\displaystyle \frac{q}{4}} \left[ \left( \begin{array}{c} 2 \\ 1 \\
1 \end{array} \right) -\left( \begin{array}{c} 2 \\ -3
\\ 1 \end{array} \right)e^{-4t} \right].$
\end{description}
Note:$\lim_{t \to \infty} e^{-t} = 0$ and $\lim_{t \to \infty}
e^{-4t} = 0$ Hence as t increases, the exponential part of each
solution ``dies away".  The equilibrium concentration is the limit
of the concentration taken as $t \rightarrow \infty$.

Cases(i) Equilibrium concentration $\left( \begin{array}{c} A \\ B
\\ C \end{array} \right) = {\displaystyle \frac{p}{4}}\left( \begin{array}{c} 2 \\ 1
\\ 1 \end{array} \right)$

Cases(ii) Equilibrium concentration $\left( \begin{array}{c} A \\
B \\ C \end{array} \right) = {\displaystyle \frac{q}{4}}\left(
\begin{array}{c} 2
\\ 1 \\ 1 \end{array} \right)$

\end{document}
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