Question

Consider the collision-induced dissociation of \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) via the following mechanism: \\[ \begin{array}{l} \mathrm{N}_{2} \mathrm{O}_{5}(g)+\mathrm{N}_{2} \mathrm{O}_{5}(g) \frac{k_{1}}{\overline{k_{-1}}} \mathrm{N}_{2} \mathrm{O}_{5}(g)^{*}+\mathrm{N}_{2} \mathrm{O}_{5}(g) \\ \mathrm{N}_{2} \mathrm{O}_{5}(g) * \stackrel{k_{2}}{\longrightarrow} \mathrm{NO}_{2}(g)+\mathrm{NO}_{3}(g) \end{array} \\] The asterisk in the first reaction indicates that the reactan is activated through collision. Experimentally it is found that the reaction can be either first or second order in \(\mathrm{N}_{2} \mathrm{O}_{5}(g)\) depending on the concentration of this species. Derive a rate law expression for this reaction consistent with this observation.

Short answer

The rate law expression for the given reaction mechanism of N2O5 dissociation can be derived using the steady-state approximation for activated N2O5, and is given by: \(\textbf{Rate} = r_2 = k_2\left(\frac{k_1 [N2O5]^2}{(k_{-1} [N2O5]+k_2)}\right)\) This expression is consistent with the observation that the reaction can be either first or second order in N2O5, depending on its concentration.

Step-by-step solution

Elementary Reaction 1:

\(r_1 = k_1 [N2O5]^2 - k_{-1} [N2O5][N2O5^*]\)

Elementary Reaction 2:

\(r_2 = k_2[N2O5^*]\) Where: - \(r_1\) is the rate of elementary reaction 1 - \(r_2\) is the rate of elementary reaction 2 - \(k_1\) and \(k_{-1}\) are the forward and reverse rate constants for reaction 1 - \(k_2\) is the rate constant for reaction 2 - [N2O5] is the concentration of N2O5 - [N2O5^*] is the concentration of activated N2O5 #Step 2: Obtain the steady-state approximation for activated N2O5# Since the activated N2O5 is an unstable intermediate, it is reasonable to assume that its concentration stays constant during the reaction. This is known as the steady-state approximation:

Steady-state approximation:

\(\frac{d[N2O5^*]}{dt} \approx 0\) Applying the steady-state approximation, we can write:

Steady-state equation:

\(k_1 [N2O5]^2 - k_{-1} [N2O5][N2O5^*] = k_2[N2O5^*]\) #Step 3: Solve for activated N2O5 concentration# Rearrange the steady-state equation to get the concentration of activated N2O5 in terms of N2O5 concentration and rate constants:

Expression for activated N2O5 concentration:

\([N2O5^*] = \frac{k_1 [N2O5]^2}{(k_{-1} [N2O5] + k_2)}\) #Step 4: Write the overall rate law expression# Now, to obtain the overall rate law expression for the reaction, substitute the expression for [N2O5^*] (from step 3) into the rate expression for elementary reaction 2:

Rate law expression:

\(\textbf{Rate} = r_2 = k_2\left(\frac{k_1 [N2O5]^2}{(k_{-1} [N2O5]+k_2)}\right)\) This rate law expression is consistent with the observation that the reaction can be either first or second order in N2O5 depending on its concentration, as it contains both linear and quadratic concentration terms.

Key concepts

Steady-State Approximation

In chemical kinetics, the steady-state approximation is a valuable concept for understanding how reactions proceed, especially when intermediates are involved. This approximation assumes that the concentration of certain reactive intermediates, like the activated \(2O5^*\), remains relatively constant during the course of the reaction.

Why is this assumption helpful? Well, intermediates are often unstable and quickly convert into products or revert to reactants. By assuming their concentration doesn't change significantly, it simplifies the math required to describe complex reactions.

For this specific reaction, we utilized the steady-state approximation by setting \(\frac{d[N2O5^*]}{dt} \approx 0\) indicating the change in concentration of \([N2O5^*]\) over time is negligible. Applying this concept allows us to effectively analyze and derive the rate law expression.

Rate Law Expression

A rate law expression helps us understand how the rate of a reaction depends on the concentration of its reactants. It's derived by integrating different components of the reaction such as rate constants and concentrations of the reactants or intermediates.

In our process, the derived rate law expression was \(\textbf{Rate} = k_2\left(\frac{k_1 [N2O5]^2}{(k_{-1} [N2O5]+k_2)}\right)\). This particular expression reflects how the reaction rate depends on the concentration of \(2O5\), taking into account the forward and reverse paths of the activated complex.

Such an expression is crucial because it shows the relationship between the concentration of reactants and the speed of the reaction, highlighting the dependency on both quadratic and linear terms. It also illustrates how the reaction order can appear as first or second order depending on specific scenarios such as low or high concentrations of \(2O5\).

Elementary Reactions

An elementary reaction is a simple step within a complex reaction mechanism, describing a basic transformation of reactants to products. Each elementary step has its own unique rate law expression that relates to the specific concentrations of the reactants involved in that step.

For the dissociation of \(2O5\), we consider two elementary reactions:

• Elementary Reaction 1: \(r_1 = k_1 [N2O5]^2 - k_{-1} [N2O5][N2O5^*]\)
• Elementary Reaction 2: \(r_2 = k_2[N2O5^*]\)

These steps illustrate the transformation through an activated intermediate \(N2O5^*\), capturing both the forward and reverse directions.

Understanding elementary reactions is essential for unraveling how a mechanism works as they set the foundation for determining overall reaction rates, facilitating a clearer view on how the entire process unfolds.

Reaction Order

Reaction order is a critical aspect of chemical kinetics, indicating the power to which the concentration of a reactant is raised in the rate law expression. It helps predict how changes in reactant concentrations will affect the reaction rate.

For the given reaction involving \(2O5\), the experimentally observed reaction order is either first or second order. This variability in reaction order depends on the concentration of \(2O5\). When the concentration is low, it behaves as a first-order reaction, whereas at larger concentrations, the rate appears second order.

This change in reaction order is effectively explained by the derived rate law, which includes both linear and quadratic terms of \([N2O5]\). Understanding reaction order helps in predicting and controlling the rate of chemical reactions in practical applications.