Question

A likely mechanism for the photolysis of acetaldehyde is \\[ \begin{array}{l} \mathrm{CH}_{3} \mathrm{CHO}(g)+h \nu \rightarrow \mathrm{CH}_{3} \cdot(g)+\mathrm{CHO} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \mathrm{CHO}(g)+\stackrel{k_{1}}{\longrightarrow} \mathrm{CH}_{4}(g)+\mathrm{CH}_{3} \mathrm{CO} \cdot(g) \\ \mathrm{CH}_{3} \mathrm{CO} \cdot(g) \stackrel{k_{2}}{\longrightarrow} \mathrm{CO}(g)+\mathrm{CH}_{3} \cdot(g) \\ \mathrm{CH}_{3} \cdot(g)+\mathrm{CH}_{3} \cdot(g) \stackrel{k_{3}}{\longrightarrow} \mathrm{C}_{2} \mathrm{H}_{6}(g) \end{array} \\] Derive the rate law expression for the formation of \(\mathrm{CO}(g)\) based on this mechanism.

Short answer

In the given reaction mechanism, we assumed the second step as the rate-determining step and derived the rate law expression for the formation of CO(g) as follows: \\[rate_{\mathrm{CO}} = \frac{k_{0}k_{1}[\mathrm{CH}_{3} \mathrm{CHO}]^2 - k_{1}k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]}{k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] + k_{3}[\mathrm{CH}_{3} \cdot]}\\] This expression is derived based on the assumption that the second step is the rate-determining step and using the steady-state approximation for the reaction intermediate concentration, [CH3 ·].

Step-by-step solution

Identify the rate-determining step

The rate-determining step is the slowest step in any reaction mechanism. In this case, it is not given which step is the slowest. Therefore, we will assume the second step to be the rate-determining step tentatively.

Write the rate law expression for the assumed rate-determining step

The rate of the second step can be written as -\\[rate = k_{1}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]\\]We know that the rate of the formation of CO(g) also depends on this step, so we can write -\\[rate_{\mathrm{CO}} = k_{1}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]\\]Now, we need to find an expression for [CH3 ·] to eliminate the intermediate from this equation.

Determine the relationships for the reaction intermediate

We can use the steady-state approximation to find the concentration of the reaction intermediate, [CH3 ·], which states that -\\[\frac{d[\mathrm{CH}_{3} \cdot]}{dt} = 0\\]From the reaction mechanism, we have 3 equations for the change in [CH3 ·] -\\[\frac{d[\mathrm{CH}_{3} \cdot]}{dt} = rate_{1} - rate_{2} + rate_{3}\\]Where,\\[rate_{1} = k_{0}[\mathrm{CH}_{3} \mathrm{CHO}]\\] \\[rate_{2} = k_{1}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]\\] \\[rate_{3} = -k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot] + k_{3}[\mathrm{CH}_{3} \cdot]^2\\] Now substitute these rate expressions into the steady-state equation and solve for [CH3 ·].

Solve for the reaction intermediate concentration

Substituting the rate expressions in the steady-state equation, we have -\\[0 = k_{0}[\mathrm{CH}_{3} \mathrm{CHO}] - k_{1}[\mathrm{CH}_{3} \cdot][\mathrm{CH}_{3} \mathrm{CHO}] - k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot] + k_{3}[\mathrm{CH}_{3} \cdot]^2\\]We can rearrange this equation to find an expression for [CH3 ·] in terms of reactants -\\[[\mathrm{CH}_{3} \cdot] = \frac{k_{0}[\mathrm{CH}_{3} \mathrm{CHO}] - k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot]}{k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] + k_{3}[\mathrm{CH}_{3} \cdot]}\\]Now we can substitute this expression back into the rate equation for CO(g).

Substitute the reaction intermediate concentration and simplify

Substituting the expression for [CH3 ·] in the rate equation for CO(g), we have -\\[rate_{\mathrm{CO}} = k_{1}\left(\frac{k_{0}[\mathrm{CH}_{3} \mathrm{CHO}] - k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot]}{k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] + k_{3}[\mathrm{CH}_{3} \cdot]}\right)[\mathrm{CH}_{3} \mathrm{CHO}]\\]After some simplification, the rate law expression for the formation of CO(g) becomes -\\[rate_{\mathrm{CO}} = \frac{k_{0}k_{1}[\mathrm{CH}_{3} \mathrm{CHO}]^2 - k_{1}k_{2}[\mathrm{CH}_{3} \mathrm{CO} \cdot][\mathrm{CH}_{3} \mathrm{CHO}]}{k_{1}[\mathrm{CH}_{3} \mathrm{CHO}] + k_{3}[\mathrm{CH}_{3} \cdot]}\\]This is the rate law expression for the formation of CO(g) based on the given mechanism and assuming the second step as the rate-determining step.

Key concepts

Photochemistry

Photochemistry is the branch of chemistry concerned with the chemical effects of light. It involves the reactions that occur when molecules absorb light and are excited to a higher energy state. In the provided exercise, photochemistry initiates the decomposition of acetaldehyde (CH₃CHO) when exposed to light, represented by \(hu\); this absorption causes the molecule to split into radicals. The light-induced reaction kicks off the entire mechanism by generating highly reactive species, making photochemistry a crucial part of understanding these reactions. Without the light-induced step, the subsequent steps of reactions would not occur, as the radicals necessary for progression are not generated. This concept helps us see the transformative power of photons in chemical reactions, making them a fundamental interest in both practical and theoretical chemistry applications.

Rate Law

The rate law provides an equation that relates the reaction rate with the concentrations of reactants and the rate constants. In the context of the given exercise, the rate law for the formation of carbon monoxide (CO) is based on the assumption that one of the steps is the rate-determining step. Typically, the rate-determining step is the slowest step of the reaction mechanism. Here, step two is tentatively assumed as the rate-determining step, leading to the rate expression \( k_{1}[ ext{CH}_3 ext{CHO}][ ext{CH}_3 ext{·}] \). The rate law provides important insights into how the concentration of intermediates or reactants affects the speed of the product formation. By understanding the rate law, chemists can predict how changes in concentration or conditions can affect the overall reaction process, which is pivotal in controlling and optimizing chemical processes.

Steady-State Approximation

The steady-state approximation is an essential concept for simplifying complex reaction mechanisms, especially those involving intermediates. It assumes that the concentration of intermediates remains constant over the time of the reaction, as their formation and consumption occur at the same rate. For the intermediate \([\text{CH}_3 ext{·}]\) in our example, the steady-state approximation implies \( \frac{d[\text{CH}_3 \cdot]}{dt} = 0 \).
This assumption allows simplifying the rate equations to find a manageable expression for the reaction rates in terms of stable reactants and products. By applying this steady-state concept, you can determine how intermediates contribute to the overall reaction without needing their exact concentrations at every moment, which can be mathematically and experimentally challenging.

Intermediates in Reactions

Intermediates are transient species that appear in the steps of a chemical reaction mechanism but are not present in the overall balanced equation. In the discussed mechanism, species like \(\text{CH}_3 \text{·}\) and \(\text{CH}_3 \text{CO} \cdot\) act as intermediates. They play a crucial role in the mechanism of a chemical reaction, acting as bridges between reactants and final products.
These intermediates are typically unstable and highly reactive, forming and disappearing quickly within the reaction. Understanding their behavior helps you comprehend the pathway of a chemical reaction better. They allow us to map out the transformation steps from reactants to products and are crucial in designing catalysts or new reaction processes by offering targets to stabilize or otherwise manipulate during chemical processes.