Problem 86 - Chemistry and Chemical Reactivity

Question

You want to study the hydrolysis of the beautiful green, cobalt-based complex called transdichlorobis-(ethylenediamine)cobalt(III) ion, 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 \(=\) \(\mathrm{H}_{2} \mathrm{NCH}_{2} \mathrm{CH}_{2} \mathrm{NH}_{2}\) trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) 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: \(\left.\quad \text { trans-ICo(en) }_{2} \mathrm{Cl}_{2}\right]^{+}(\mathrm{aq}) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ Fast: \(\quad\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow\) $$ \left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq}) $$ (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. green $$ \left.\underset{\text { red }}{\operatorname{Co}(\mathrm{en})_{2}}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}(\mathrm{aq})+\mathrm{Cl}^{-}(\mathrm{aq}) $$ 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. 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.

Step-by-Step Solution

Answer

(a) Rate law is \( \text{Rate} = k [\mathrm{Co}( ext{en})_{2}\mathrm{Cl}_{2}]^{+} \). (b) The reaction is first-order as the gray stage timing is independent of initial conditions. (c) Activation energy found from \( \ln k \) vs. \( 1/T \) plot using time to gray stage data.

Step 1: Analyze the Slow Step Mechanism

The mechanism suggests that the slow step is the breaking of the Co-Cl bond in the species \( \text{trans-}[\mathrm{Co}( ext{en})_{2}\mathrm{Cl}_{2}]^{+} \), leading to the formation of a five-coordinate intermediate \([\mathrm{Co}( ext{en})_{2}\mathrm{Cl}]^{2+} \), and releasing \( \mathrm{Cl}^{-} \). This suggests that the rate of reaction is dependent on the concentration of \( \text{trans-}[\mathrm{Co}( ext{en})_{2}\mathrm{Cl}_{2}]^{+} \).

Step 2: Determine Rate Law

Based on the slow step being the rate-determining step, the rate law for this reaction can be written as: \( \text{Rate} = k [\mathrm{Co}( ext{en})_{2}\mathrm{Cl}_{2}]^{+} \). This indicates it is a first-order reaction with respect to the green complex ion.

Step 3: Justification Based on Observations

The observation that the gray color (indicating the intermediate) is reached at the same time regardless of the initial concentration of the green starting material further supports that the reaction is first-order. In first-order reactions, the time taken for an intermediate concentration is constant irrespective of initial concentrations.

Step 4: Activation Energy Calculation Setup

Although the exact calculations require data, we can find the activation energy by plotting \( \ln k \) versus \( 1/T \). Since \( k \) is represented as \( -(1/t) \ln ([\mathrm{R}]/[\mathrm{R}]_0) \), where \([\mathrm{R}] \) is the concentration at time \( t \), the inverse of time taken to reach the gray color can give us a relative measure of \( k \).

Step 5: Interpretation and Data Use

If you have specific temperature and time data for the reaction to reach the gray stage, this can be used to plot \( \ln k \) against \( 1/T \). The slope of the line obtained from this plot equals \( -E_a/R \), allowing us to calculate the activation energy \( E_a \).

Key Concepts

Hydrolysis Reaction

Hydrolysis is an important type of chemical reaction, especially in coordination chemistry. In this reaction, a water molecule replaces another ligand in a complex ion. For our example, this involves a green cobalt-based complex ion, trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\), hydrolyzing to form the red complex ion \(\left[\mathrm{Co}(\mathrm{en})_{2}\left(\mathrm{H}_{2} \mathrm{O}\right) \mathrm{Cl}\right]^{2+}\).
The reaction mechanism consists of two steps. The first is a slow step where a chloride ion (\(\mathrm{Cl}^{-}\)) is released, resulting in a five-coordinate intermediate. This is followed by a fast step where water attacks the intermediate, forming the final product. This sequence illustrates how hydrolysis can lead to a change in ligand without altering the rest of the complex structure.
It's particularly fascinating because the slow initial step dictates the overall rate of reaction. Understanding these steps helps in predicting how coordination complexes will behave in solution, which is crucial in fields like catalysis and material science.

Rate Laws in Chemistry

Rate laws express how the rate of a reaction depends on the concentration of reactants. In coordination chemistry, rate laws are essential for understanding how different factors influence the speed of complex reactions.
In our case, the rate law is derived from the rate-determining slow step of the reaction. This step involves the breaking of the Co-Cl bond in the green complex ion, trans-\(\left[\mathrm{Co}(\mathrm{en})_{2} \mathrm{Cl}_{2}\right]^{+}\).
For this reason, the rate law can be expressed as: \[ \text{Rate} = k \left[ \mathrm{Co}(\mathrm{en})_{2}\mathrm{Cl}_{2}\right]^{+} \] This means the reaction is first-order with respect to the green complex ion. Hence, the rate of the reaction is solely dependent on the concentration of the green complex ion and not on the concentration of water or other ions in the solution.
By understanding and applying rate laws, chemists can predict how changes in conditions such as temperature or reactant concentration will affect reaction rates.

Activation Energy Calculation

Activation energy is a critical concept as it represents the minimum energy required for a reaction to occur. For our hydrolysis reaction, calculating the activation energy involves using the Arrhenius equation, which relates the rate constant \(k\) to the temperature \(T\):
The equation is: \[ k = A e^{-E_{a}/RT} \] where \( A \) is the pre-exponential factor, \( E_a \) is the activation energy, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin.
By plotting \( \ln k \) versus \( 1/T \), we can find the activation energy. The slope of this line is \(-E_a/R\), allowing us to solve for \(E_a\). In the case of our experiment, instead of measuring \(k\) directly, we use \(-(1/t) \ln ([\mathrm{R}]/[\mathrm{R}]_0)\) to relate the reaction time to the rate constant.
This approach gives a practical way to determine how temperature affects the speed of the reaction, and by extension, provides insight into the energy barriers present in chemical reactions.

First-order Reaction Kinetics

First-order reactions are characterized by a rate that depends linearly on the concentration of a single reactant. In first-order kinetics, the rate of reaction is directly proportional to the concentration of the reactant.
For our studied hydrolysis reaction, we know it follows first-order kinetics because the rate only depends on the concentration of the green complex ion, \(\left[ \mathrm{Co}(\mathrm{en})_{2}\mathrm{Cl}_{2}\right]^{+}\). A key observation supporting this conclusion is that the time to reach the intermediate gray color remains consistent regardless of initial concentrations.
This consistency is indicative of first-order kinetics, where reaction time constants are independent of initial reactant concentrations. Mathematically, the concentration of the reactant changes over time according to the formula: \[ [\mathrm{R}] = [\mathrm{R}]_0 e^{-kt} \] where \([\mathrm{R}]\) is the concentration at time \(t\), \([\mathrm{R}]_0\) is the initial concentration, and \(k\) is the rate constant.
Understanding these kinetics allows us to predict how the concentration of the reactant will decrease over time, providing important insight into the reaction dynamics and allowing for optimization of reaction conditions in the lab or industrial processes.