Problem 69 - Chemistry The Molecular Science

Question

Experiments show that the reaction of nitrogen dioxide with fluorine $$ 2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g}) $$ has the rate law $$ \text { Rate }=k\left[\mathrm{NO}_{2}\right]\left[\mathrm{F}_{2}\right] $$ and the reaction is thought to occur in two steps: $$ \begin{array}{l} \text { Step } 1: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \\ \text { Step } 2: \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \end{array} $$ (a) Show that the sum of this sequence of reactions gives the balanced equation for the overall reaction. (b) Which step is rate-determining?

Step-by-Step Solution

Answer

The sum of the reactions matches the overall balanced equation. Step 1 is the rate-determining step.

Step 1: Identify Overall Reaction

The overall reaction is given as \(2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g})\). This is the reaction obtained by adding both steps in the mechanism.

Step 2: Write the Two-Step Reactions

The two steps provided are: Step 1: \( \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \)Step 2: \( \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \)

Step 3: Add Step Reactions to Find Overall Reaction

Add Step 1 and Step 2 and cancel out the intermediate species \(\mathrm{F}(\mathrm{g})\). This results in: \[\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) + \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \longrightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) + \mathrm{FNO}_{2}(\mathrm{~g}) \]Cancelling \(\mathrm{F}(\mathrm{g})\) on both sides gives:\[2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{FNO}_{2}(\mathrm{~g})\]This confirms that the sum of the two steps matches the overall balanced reaction.

Step 4: Identify the Rate-Determining Step

The rate-determining step is the one that dictates the overall reaction kinetics, reflected in the rate law. The given rate law is \( \text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}] \).Comparing the rate law to the mechanism, Step 1: \(\mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g})\), involves both \([\mathrm{NO}_{2}]\) and \([\mathrm{F}_{2}]\) as reactants, matching the form of the rate law. Thus, Step 1 is the rate-determining step.

Key Concepts

Rate Laws

Rate laws describe how the speed of a chemical reaction depends on the concentrations of reactants. They are often written as:

• Rate = k[A][B],

where "Rate" is the rate of the reaction, "k" is the rate constant specific to the reaction, and [A] and [B] are the concentrations of the reactants.
In this particular exercise, the rate law is given as \( \text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}] \). This indicates that the reaction rate depends linearly on the concentration of \( \mathrm{NO}_{2} \) and the concentration of \( \mathrm{F}_{2} \).
This rate law is consistent with elementary reactions and implies that both \( \mathrm{NO}_{2} \) and \( \mathrm{F}_{2} \) must collide for the reaction to proceed.

Understanding rate laws helps predict how changing the concentration of a reactant affects the reaction speed. For instance, doubling the concentration of \( \mathrm{NO}_{2} \) or \( \mathrm{F}_{2} \) will double the reaction rate, assuming other conditions are constant.

Rate-Determining Step

In a multi-step reaction mechanism, the rate-determining step (RDS) is the slowest step. It limits the overall speed of the entire process, much like the narrowest part of a bottleneck regulates the flow.

The criteria for identifying the RDS are simple: it should match the reaction's experimental rate law. For the given reaction, the rate law \( \text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}] \) tells us that the RDS involves \( \mathrm{NO}_{2} \) and \( \mathrm{F}_{2} \) as reactants. Thus, this matches Step 1:

• \( \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \)

This is the step where \( \mathrm{NO}_{2} \) and \( \mathrm{F}_{2} \) react, implying it's the slowest and thus rate-determining.
Recognizing the RDS allows chemists to focus on improving reaction efficiencies and conditions by tackling the slowest part of the reaction.

Reaction Intermediates

Reaction intermediates are species that are formed in one step of a mechanism and consumed in another. They do not appear in the overall equation but are crucial to the mechanism's function.

In the exercise provided, \( \mathrm{F}(\mathrm{g}) \) is the intermediate. It arises from Step 1 and is consumed in Step 2:

• \( \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \rightarrow \mathrm{FNO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \)
• \( \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}(\mathrm{g}) \rightarrow \mathrm{FNO}_{2}(\mathrm{~g}) \)

Intermediates are often essential for the transformation to proceed and can sometimes be isolated if the reaction conditions are carefully controlled.
Studying reaction intermediates offers insights into how a reaction pathway proceeds, giving a deeper understanding of reaction mechanisms and aiding in designing more efficient chemical processes.