Question

\(\mathrm{N}_{2} \mathrm{O}_{5}\) decomposes according to the following equation: \(\mathrm{N}_{2} \mathrm{O}_{5} \rightarrow 2 \mathrm{NO}_{2}+(1 / 2) \mathrm{O}_{2}\). The rate expression \(=\) \(\left\\{-\mathrm{d}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\right\\} / \mathrm{dt}=\mathrm{k}\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\). The following mechanism has been proposed: \(\mathrm{N}_{2} \mathrm{O}_{5} \mathrm{k} \rightleftarrows \mathrm{NO}_{2}+\mathrm{NO}_{3}\) $$ \begin{aligned} &\mathrm{NO}_{2}+\mathrm{NO}_{3} \mathrm{k}_{1} \rightarrow \mathrm{NO}_{2}+\mathrm{O}_{2}+\mathrm{NO} \\ &\mathrm{NO}+\mathrm{NO}_{3} \mathrm{k}_{2} \rightarrow 2 \mathrm{NO}_{2} \end{aligned} $$ Show the rate of \(\mathrm{O}_{2}\) formation is directly proportional to \(\left[\mathrm{N}_{2} \mathrm{O}_{5}\right]\)

Short answer

The rate of O₂ formation can be shown to be directly proportional to [N₂O₅] by examining the proposed mechanisms and using the steady-state approximation for the intermediates. By substituting expressions for [NO₃] and [NO], we find that \(r(O_2) = C [N_2 O_5]\), where C is a constant, showing that \(r(O_2) \propto [N_2 O_5]\).

Step-by-step solution

Write the rate laws for each elementary reaction

The rate laws for each elementary reaction in the mechanism are as follows: Reaction 1: \[r_1 = k_1[N_2 O_5] - k_{-1}[NO_2][NO_3]\] Reaction 2: \[r_2 = k_2 [NO_2][NO_3]\] Reaction 3: \[r_3 = k_3 [NO][NO_3]\] where \(r_i\) is the rate of reaction i, and \(k_i\) is the rate constant for reaction i.

Determine the rate of O₂ formation

The rate of O₂ formation is equal to the rate of Reaction 2, which is: \[r(O_2) = r_2 = k_2[NO_2][NO_3]\] Now, we need to find a way to express this rate in terms of [N₂O₅].

Use the steady-state approximation

The steady-state approximation states that the rate of change of any intermediate is zero. Here, NO₂ and NO₃ are intermediates, so: \[\frac{d[NO_2]}{dt} = 0\] \[\frac{d[NO_3]}{dt} = 0\] Applying this approximation on the Reaction 1, and it gives us, \[k_1 [N_2 O_5] = k_{-1} [NO_2][NO_3]\] From which we get, \[[NO_3]= \frac{k_1}{k_{-1}} \frac{[N_2 O_5]}{[NO_2]}\] Next, apply steady-state approximation on intermediate NO: \[\frac{d[NO]}{dt} = k_{2}[NO_2][NO_3] - k_3 [NO][NO_3] = 0\] Now we substitute the expressions for [NO₃] we just obtained: \[[NO]= \frac{k_{2}[NO_2]}{k_3} \frac{[N_2 O_5]}{[NO_2]}\]

Express the rate of O₂ formation in terms of [N₂O₅]

Now, substitute [NO₃] and [NO] expressions in the rate of O₂ formation: \[r(O_2) = k_2 [NO_2] \cdot \frac{k_1}{k_{-1}} \frac{[N_2 O_5]}{[NO_2]} = C [N_2 O_5]\] where \(C\) is a constant that is not dependent on the concentration of N₂O₅. Now, we have shown that the rate of O₂ formation is directly proportional to [N₂O₅]: \[r(O_2) \propto [N_2 O_5]\]

Key concepts

Rate Expressions

Rate expressions are mathematical equations that describe how the concentration of a reactant or a product changes with time. These expressions are essential in chemical kinetics as they help us understand how fast a reaction proceeds. In general, a rate expression can be written as:

• For a reactant: \[ ext{Rate}_{ ext{reactant}} = -\frac{d[ ext{Reactant}]}{dt} \]
• For a product: \[ ext{Rate}_{ ext{product}} = \frac{d[ ext{Product}]}{dt} \]

The negative sign for reactants indicates that their concentrations decrease over time, while products increase. For the given exercise, the decomposition of \(N_2O_5\) is expressed as:\[ -\frac{d[N_2O_5]}{dt} = k[N_2O_5] \]where \(k\) is the rate constant. This expression shows the relationship between the rate of decomposition and the concentration of \(N_2O_5\). Keeping track of such expressions helps predict the outcome of a reaction at any given time.

Steady-State Approximation

The steady-state approximation is a useful assumption in reaction kinetics, especially when dealing with complex reaction mechanisms. It suggests that the concentration of an intermediate stays relatively constant over the course of the reaction. Because of this, the rate of formation and the rate of consumption of that intermediate are equal, resulting in a net rate of change of zero:\[ \frac{d[ ext{Intermediate}]}{dt} = 0 \]In our exercise, \(NO_2\) and \(NO_3\) are intermediates formed and consumed during the reaction sequence. Modeling their rates using this approximation lets us simplify the equations:

• \( \frac{d[NO_2]}{dt} = 0 \)
• \( \frac{d[NO_3]}{dt} = 0 \)

This approach is particularly helpful because it allows us to eliminate variables that are complex to measure or solve directly, thus simplifying the pathways into a manageable form that can be related back to the primary reactant, \([N_2O_5]\).

Elementary Reactions

Elementary reactions are the simplest form of a chemical reaction, representing a single mechanistic step. The complete pathway of a chemical reaction is often composed of a series of such elementary reactions that sum to the overall balanced equation. Each elementary reaction has its own unique rate expression known as an elementary rate law. These can be written directly based on the stoichiometry of the elementary step. For instance:

• Reaction 1: \( \mathrm{N}_2O_5 \rightarrow \mathrm{NO}_2 + \mathrm{NO}_3 \)

The rate law for this reaction is:\[ r_1 = k_1 [N_2O_5] \]Every elementary reaction in the proposed mechanism contributes to the overall behavior of the decomposition of \(N_2O_5\). Understanding these steps is crucial because they will dictate the rate and mechanism of the entire reaction process.

Reaction Mechanism

A reaction mechanism provides a step-by-step account of how a chemical reaction proceeds at the molecular level. It breaks down a complex reaction into elementary steps, detailing the sequence in which bonds break and form. Each step in the mechanism has its own kinetics and rate law that can contribute to the overall reaction rate. In our exercise, the proposed mechanism involves three main steps:

• \( \mathrm{N}_2\mathrm{O}_5 \;\rightleftarrows \;\mathrm{NO}_2 + \mathrm{NO}_3\)
• \( \mathrm{NO}_2 + \mathrm{NO}_3 \rightarrow \mathrm{NO}_2 + \mathrm{O}_2 + \mathrm{NO}\)
• \(\mathrm{NO} + \mathrm{NO}_3 \rightarrow 2\mathrm{NO}_2 \)

Mechanisms like these help in understanding why certain reaction pathways are favored over others and can explain observations like why the rate of \(O_2\) formation is contingent on \([N_2O_5]\). Knowing this helps in quantifying and modeling real-world chemical processes, influencing efforts like catalyst design and control of industrial chemical reactions.