Problem 30 - Chemistry: Introducing Inorganic, Organic and Physical Chemistry

Question

The dissociation of propane in which a C-C bond breaks to form a methyl radical and an ethyl radical is a unimolecular reaction. \\[ \mathrm{C}_{3} \mathrm{H}_{8} \rightarrow \mathrm{CH}_{3}^{*}+\mathrm{C}_{2} \mathrm{H}_{5}^{*} \\] The rate of formation of \(\mathrm{CH}_{3}^{*}\) (and of \(\mathrm{C}_{2} \mathrm{H}_{5}\) ') is given by \(\frac{\mathrm{d}\left[\mathrm{CH} \mathrm{b}^{*}\right]}{\mathrm{d} t}=\mathrm{k}_{\text {overd }}\left[\mathrm{C}_{3} \mathrm{H}_{8}\right],\) where \(k_{\text {overd }}\) is the unimolecular rate constant for the reaction. (Sections 9.6 and 9.8 ) (a) Explain what is meant by the term 'unimolecular? The propane molecules obtain sufficient energy to dissociate by colliding with other molecules, \(M\), where \(M\) may be an unreactive gas such as nitrogen. The mechanism for this process can be written as where \(\mathrm{C}_{3} \mathrm{H}_{\mathrm{g}}^{*}\) is a propane molecule in a high energy state, which has sufficient energy to dissociate. (b) By applying the steady state approximation to \(\mathrm{C}_{3} \mathrm{H}_{8}^{*}\), derive an expression for \(\left[\mathrm{C}_{3} \mathrm{H}_{4}^{\prime}\right] .\) since the rate of formation of \(\mathrm{CH}_{3}^{*}\) is equal to \(k_{2}\left[\mathrm{C}_{3} \mathrm{H}_{8}^{*}\right],\) show that \\[ k_{\text {overal }}=\frac{k_{1} k_{2}[M]}{k_{-1}[M]+k_{2}} \\] (c) Show that: (i) when \([\mathrm{M}]\) is very large, \(k_{\text {overal }}=\frac{k_{1} k_{2}}{k_{-1}}\) (ii) when \([\mathrm{M}]\) is very small, \(k_{\text {oved }}=k_{1}[\mathrm{M}]\) (d) What are the rate-determining stages in the mechanism under each of the conditions in (c)(1) and (c)(ii)? Sketch a graph to show the dependence of \(k_{\text {overall }}\) on \([\mathrm{M}]\) (e) It is always better to find a linear expression to analyse experimental data. Show that \\[ \frac{1}{k_{\text {overal }}}=\frac{k_{-1}}{k_{\sqrt{2}}}+\frac{1}{k_{1}[M]} \\] A plot of \(1 / k_{\text {werall }}(y-\text { axis ) against } 1 /[\mathrm{M}] \text { ( } x\) -axis) is a straight line. What are the gradient and the intercept on the \(y\) -axis at \(1 /[\mathrm{M}]=0 ?\)

Step-by-Step Solution

Answer

A unimolecular reaction depends on a single reactant. Using the steady state approximation, the rate constant \(k_{\text{overal}}\) is derived. It behaves differently at high and low concentration of \([M]\), and a linear analysis gives a specific gradient and intercept.

Step 1: Define Unimolecular Reaction

A unimolecular reaction is a type of reaction where the rate is dependent on the concentration of a single molecule. In this case, the dissociation of propane into methyl and ethyl radicals occurs after the propane molecule gains sufficient energy through collisions. Here, the rate is proportional to the concentration of propane, \([\mathrm{C}_{3} \mathrm{H}_{8}]\).

Step 2: Apply Steady State Approximation

We assume the intermediate, \([\mathrm{C}_{3} \mathrm{H}_{8}^*]\), reaches a steady state, meaning its rate of formation equals its rate of consumption. The reaction mechanism includes the formation of \([\mathrm{C}_{3} \mathrm{H}_{8}^*]\) and its dissociation: 1. \([\mathrm{C}_{3} \mathrm{H}_{8}] + [M] \xrightarrow{k_1} [\mathrm{C}_{3} \mathrm{H}_{8}^*] + [M]\) 2. \([\mathrm{C}_{3} \mathrm{H}_{8}^*] \xrightarrow{k_2} \mathrm{CH}_{3}^* + \mathrm{C}_{2} \mathrm{H}_{5}^*\)3. \([\mathrm{C}_{3} \mathrm{H}_{8}^*] + M \xrightarrow{k_{-1}} [\mathrm{C}_{3} \mathrm{H}_{8}] + M\). The steady-state approximation gives: \[ k_1 [\mathrm{C}_{3} \mathrm{H}_{8}][M] = k_{-1} [\mathrm{C}_{3} \mathrm{H}_{8}^*][M] + k_2 [\mathrm{C}_{3} \mathrm{H}_{8}^*] \]

Step 3: Derive Expression for Propane Intermediate

Rearrange the steady-state equation to solve for \([\mathrm{C}_{3} \mathrm{H}_{8}^*]\): \[[\mathrm{C}_{3} \mathrm{H}_{8}^*] = \frac{k_1 [\mathrm{C}_{3} \mathrm{H}_{8}][M]}{k_{-1} [M] + k_2}\]

Step 4: Derive Overall Rate Constant

Since the rate of formation of \(\mathrm{CH}_{3}^*\) is \(k_2 [\mathrm{C}_{3} \mathrm{H}_{8}^*]\), substitute the expression for \([\mathrm{C}_{3} \mathrm{H}_{8}^*]\):\[\frac{\mathrm{d} [\mathrm{CH}_{3}^*]}{\mathrm{d}t} = k_{\text{overal}} [\mathrm{C}_{3} \mathrm{H}_{8}] = k_2 \times \left( \frac{k_1 [\mathrm{C}_{3} \mathrm{H}_{8}][M]}{k_{-1}[M] + k_2} \right)\]Therefore, \[k_{\text{overal}} = \frac{k_1 k_2[M]}{k_{-1}[M] + k_2}\]

Step 5: Special Cases (High and Low \\[M\"])

(i) For \([\mathrm{M}]\) very large: \[ k_{\text{overal}} = \frac{k_1 k_2[M]}{k_{-1}[M] + k_2} \approx \frac{k_1 k_2}{k_{-1}} \] since \(k_2\) becomes negligible compared to \(k_{-1}[M]\).(ii) For \([\mathrm{M}]\) very small: \[ k_{\text{overal}} = \frac{k_1 k_2[M]}{k_{-1}[M] + k_2} \approx k_1[M] \] since \(k_{-1}[M]\) becomes negligible compare to \(k_2\).

Step 6: Determine Rate-Determining Steps

(i) With large \([M]\), the rate is determined by the formation and dissociation step controlled by kinetic constant \(k_{-1}\).(ii) With small \([M]\), the rate is mostly determined by the formation of the excited state mediated by \(k_1\).

Step 7: Graph Behavior of \[k_{\text{overall}}\]

Sketch a graph with \([M]\) on the x-axis and \(k_{\text{overall}}\) on the y-axis showing increases in \([M]\) leading to asymptotic behavior towards \(\frac{k_1 k_2}{k_{-1}}\) when \([M]\) is large and linear dependency \(k_1[M]\) when \([M]\) is small.

Step 8: Linear Analysis Expression

Given \(k_{\text{overal}} = \frac{k_1 k_2[M]}{k_{-1}[M] + k_2}\), the linear equation in terms of inverse can be written as: \[\frac{1}{k_{\text{overal}}} = \frac{k_{-1}}{k_2} + \frac{1}{k_1[M]}\]which is in the form \(y = mx + c\) where the y-intercept is \(\frac{k_{-1}}{k_2}\) and the gradient \(m\) is \(\frac{1}{k_1}\).

Step 9: Identify Gradient and Intercept

From the plot of \(\frac{1}{k_{\text{overall}}}\) against \(\frac{1}{[M]}\), the gradient is \(\frac{1}{k_1}\) and the y-intercept at \(\frac{1}{[M]} = 0\) is \(\frac{k_{-1}}{k_2}\).

Key Concepts

Unimolecular Reactions

Unimolecular reactions are quite fascinating as they involve the transformation of a single molecule into different products. This type of reaction typically depends on the concentration of just one reactant molecule. In the context of the propane dissociation, the reaction transforms one propane molecule into methyl and ethyl radicals. The interesting part is that the energy for this transformation is generally acquired through interactions and collisions with other molecules.
To get specific, when propane molecules collide with other molecules, they gain energy. For example, this energy can come from an unreactive gas like nitrogen. Once a propane molecule has the necessary energy level, it can transition to an excited state. In this energized form, the propane is ready to break its C-C bond, leading to the unimolecular dissociation into radicals. Thus, the rate of the reaction is a direct reflection of the concentration of the excited propane molecules and is characterized by the unimolecular rate constant.

• Single molecule transforms into products.
• Rate depends on the excited state of the molecule.
• Collisions with other molecules provide the needed energy.

Steady-State Approximation

The steady-state approximation is a crucial concept in chemical kinetics, especially when dealing with reactions involving intermediates. It helps to simplify complex reaction mechanisms. In this scenario, the intermediate is the energized propane molecule, \([\mathrm{C}_{3} \mathrm{H}_{8}^{*}]\).
Under the steady-state approximation, it is assumed that the concentration of this intermediate does not change with time. This means that the rate at which it is formed equals the rate at which it is consumed.
Applying this approximation to the dissociation of propane, we derive a relation that allows us to express the concentration of the excited intermediate in terms of other steady-state parameters.
Here, the reaction sequence includes:

• Formation of the excited state by colliding with \[M\].
• Dissociation of the excited state into radicals.
• Deactivation of the excited state back to the stable propane form.

By using steady-state approximation, it leads us to a key equation that simplifies how we view these intermediate reactions: \([\mathrm{C}_{3} \mathrm{H}_{8}^{*}] = \frac{k_1 [\mathrm{C}_{3} \mathrm{H}_{8}][M]}{k_{-1} [M] + k_2}\). This allows us to concentrate on the major reactants and products, while the behavior of the intermediate is neatly categorized.

Rate-Determining Step

In chemical kinetics, finding the rate-determining step is like discovering the bottleneck in a process. This step controls the overall rate of the reaction, especially in complex mechanisms where several steps are involved. In the propane dissociation, depending on the concentration of \[M\], this critical step can change.
For high \[M\], we deal with rapid collisions leading to a formation-dominated sequence. The rate is primarily dictated by the dissociation to radicals, as the deactivation is much faster. This puts the emphasis on the rate constant \(k_{-1}\), making dissociation the rate-determining step.
In contrast, a low concentration of \[M\] shifts the focus to the rate-limiting collision process, as there are fewer encounters to energize the propane. Here, the rate-determining step is marked by the formation of the excited state, driven by \(k_1\).

• High \[M\]: Dissociation is slower, \(k_{-1}\) governs.
• Low \[M\]: Formation is slower, \(k_1\) governs.

The rate-determining step thus depends heavily on the concentration of \[M\], impacting which step governs the speed of the overall reaction.

Rate Constant

The rate constant is a fundamental part of chemical kinetics, acting as a proportionality factor that connects the reaction rate with reactant concentrations through their order. In propane's unimolecular dissociation, we have different constants for each mechanism step, like \(k_1\), \(k_2\), and \(k_{-1}\).
These constants are temperature-dependent and specific to each reaction's step. They help determine the reaction speed and can vary significantly with changes in reaction conditions, such as temperature and pressure.
The overall rate constant, \(k_{\text{overall}}\), becomes a crux in simplifying and understanding reaction velocity. It combines individual steps and constants, providing a clearer picture of how quickly a reaction progresses under specific conditions. This is influenced by the concentration of \[M\].

• Determines reaction speed.
• Specific to individual reaction steps.
• Affected by temperature and pressure.

Understanding the rate constant and its components helps explain and predict how a chemical reaction will behave, making it indispensable for analyzing reaction kinetic data and for plotting the progress over time.