We want to study the hydrolysis of the beautiful green, cobalt-based complex
called trans-dichlorobis(ethylenediamine) cobalt(III) ion,
(Check your book to see figure)
In this hydrolysis reaction, the green complex ion
trans\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\) forms the
red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2}
\mathrm{O}\right) \mathrm{Cl}\right]^{2+}\) as a \(\mathrm{Cl}^{-}\) ion is
replaced with a water molecule on the \(\mathrm{Co}^{3+}\) ion (en
\(=\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) ).
The reaction progress is followed by observing the color of the solution. The
original solution is green, and the final solution is red, but at some
intermediate stage when both the reactant and product are present, the
solution is gray.
Changes in color with time as \(\mathrm{Cl}^{-}\) ion is replaced by
\(\mathrm{H}_{2} \mathrm{O}\) in a cobalt(III) complex. The shape in the middle
of the beaker is a vortex that arises because the solutions are being stirred
using a magnetic stirring bar in the bottom of the beaker.
Reactions such as this have been studied extensively, and experiments suggest
that the initial, slow step in the reaction is the breaking of the Co-Cl bond
to give a five-coordinate intermediate. The intermediate is then attacked
rapidly by water.
Slow:
$$\begin{aligned}\operatorname{trans}-\left[\mathrm{Co}(\mathrm{en})_{2}
\mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) & \rightarrow
\\\&\left[\mathrm{Co}(\mathrm{en})_{2}
\mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq})\end{aligned}$$
Fast:
$$\begin{aligned}\left[\mathrm{Co}(\mathrm{en})_{2}
\mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{aq}) &
\rightarrow \\\&\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2}
\mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})\end{aligned}$$
(a) Based on the reaction mechanism, what is the predicted rate law?
(b) As the reaction proceeds, the color changes from green to red with an
intermediate stage where the color is gray. The gray color is reached at the
same time, no matter what the concentration of the green starting material (at
the same temperature). How does this show the reaction is first order in the
green form? Explain.
(c) The activation energy for a reaction can be found by plotting In \(k\)
versus \(1 / T .\) However, here we do not need to measure \(k\) directly.
Instead, because \(k=-(1 / t) \ln \left([\mathrm{R}] /[\mathrm{R}]_{0}\right),\)
the time needed to achieve the gray color is a measure of \(k .\) Use the data
below to find the activation energy.
$$\begin{array}{cc}\text { Temperature }^{\circ} \mathrm{C} & \text { (for the
Same Initial Concentration) } \\\\\hline 56 & 156 \mathrm{s} \\\60 & 114
\mathrm{s} \\\65 & 88 \mathrm{s} \\\75 & 47 \mathrm{s} \\\\\hline\end{array}$$
