Question

Nitryl fluoride can be made by treating nitrogen dioxide with fluorine: $$2 \mathrm{NO}_{2}(\mathrm{g})+\mathrm{F}_{2}(\mathrm{g}) \rightarrow 2 \mathrm{NO}_{2} \mathrm{F}(\mathrm{g})$$ Use the rate data in the table to do the following: (a) Write the rate equation for the reaction. (b) Indicate the order of reaction with respect to each component of the reaction. (c) Find the numerical value of the rate constant, \(k.\) $$\begin{array}{ccccl}\hline \text { Experiment } & {\left[\mathrm{NO}_{2}\right]} & {\left[\mathrm{F}_{2}\right]} & {\left[\mathrm{NO}_{2} \mathrm{F}\right]} & \left(\mathrm{mol} \mathrm{F}_{2} / \mathrm{L} \cdot \mathrm{s}\right) \\\\\hline 1 & 0.001 & 0.005 & 0.001 & 2.0 \times 10^{-4} \\\2 & 0.002 & 0.005 & 0.001 & 4.0 \times 10^{-4} \\\3 & 0.006 & 0.002 & 0.001 & 4.8 \times 10^{-4} \\\4 & 0.006 & 0.004 & 0.001 & 9.6 \times 10^{-4} \\\5 & 0.001 & 0.001 & 0.001 & 4.0 \times 10^{-5} \\\6 & 0.001 & 0.001 & 0.002 & 4.0 \times 10^{-5} \\\\\hline\end{array}$$

Short answer

1st order in NO2 and F2; rate constant k = 40 L/mol·s.

Step-by-step solution

Analyze Rate Law Form

The general rate law for the reaction is of the form \( ext{Rate} = k[ ext{NO}_2]^x [ ext{F}_2]^y \), where \( k \) is the rate constant, and \( x \) and \( y \) are the orders of the reaction with respect to \( ext{NO}_2 \) and \( ext{F}_2 \) respectively.

Determine Order with Respect to NO2

Compare experiments 1 and 2, where the concentration of \( ext{F}_2 \) is constant and the concentration of \( ext{NO}_2 \) is doubled:- Rate doubles from \( 2.0 imes 10^{-4} \) to \( 4.0 imes 10^{-4} \)- The reaction is first order with respect to \( ext{NO}_2 \) because doubling \( [ ext{NO}_2] \) doubles the rate.

Determine Order with Respect to F2

Compare experiments 1 and 5, where the concentration of \( ext{NO}_2 \) is constant and the concentration of \( ext{F}_2 \) changes:- When the concentration of \( ext{F}_2 \) decreases from 0.005 to 0.001, the rate decreases from \( 2.0 imes 10^{-4} \) to \( 4.0 imes 10^{-5} \), which is by a factor of 5.- The reaction is first order with respect to \( ext{F}_2 \) because decreasing \( [ ext{F}_2] \) by a factor of 5 decreases the rate by the same factor.

Write the Rate Law

Given that both \( ext{NO}_2 \) and \( ext{F}_2 \) are first order, the rate equation can be expressed as:\[ ext{Rate} = k [ ext{NO}_2][ ext{F}_2] \]

Calculate the Rate Constant k

Use experiment 1 to determine \( k \):- Rate \( = 2.0 \times 10^{-4} \, \text{mol/L} \cdot \text{s} \)- \( [ ext{NO}_2] = 0.001 \)- \( [ ext{F}_2] = 0.005 \)Plug values into the rate equation:\[ 2.0 \times 10^{-4} = k (0.001)(0.005) \]\[ k = \frac{2.0 \times 10^{-4}}{(0.001)(0.005)} = 40 \, \text{L/mol} \cdot \text{s} \]

Summarize the Orders and Rate Constant

The reaction is 1st order with respect to \( ext{NO}_2 \) and 1st order with respect to \( ext{F}_2 \), making it second order overall. The rate constant \( k \) is 40 L/mol·s.

Key concepts

Rate Law

The rate law is a mathematical expression that describes how the rate of a chemical reaction depends on the concentration of its reactants. It is usually given in the form:

• Rate = k [A]ⁿ [B]ᵐ

Here,

• Rate denotes the speed of the reaction.
• k stands for the rate constant.
• [A] and [B] represent the concentrations of the reactants.
• n and m are the orders of the reaction with respect to each reactant.

Understanding the rate law helps identify how changes in reactant concentrations influence the reaction rate. This information is crucial for controlling reactions in industrial and laboratory settings.
Often, determining the rate law involves experiments where concentrations are varied, and the effects on the reaction rate are observed.

Reaction Order

The reaction order indicates how the rate is affected by the concentration of a particular reactant. It is a key part of the rate law and can be a whole number or a fraction.
The overall reaction order is the sum of the orders with respect to each reactant.
In the exercise, experimentation revealed that:

• The reaction is first order with respect to NO₂, since doubling its concentration doubles the reaction rate.
• The reaction is also first order with respect to F₂, evident when changes in F₂ concentration directly change the rate.
• This makes the overall order of the reaction second order.

Reaction orders are not necessarily related to the coefficients in the balanced chemical equation; instead, they must be determined through experimentation.

Rate Constant

The rate constant, denoted as k, is a crucial factor in the rate law equation. It provides a proportional relationship between the reaction rate and the concentrations of reactants. The rate constant can vary with temperature, and its units depend on the overall order of the reaction.
For a second-order reaction, such as the one given in the exercise, the units of k are L/mol·s. This ensures that when concentrations are substituted into the rate equation, the units cancel out appropriately to give a rate in mol/L·s.
The rate constant is determined experimentally. Using data from experiment 1, the rate constant k was calculated as follows:

• Rate = 2.0 × 10⁻⁴ mol/L·s
• [NO₂] = 0.001 mol/L
• [F₂] = 0.005 mol/L
• Thus, k = (2.0 × 10⁻⁴) / (0.001 * 0.005) = 40 L/mol·s

This calculation shows how the rate constant quantifies the intrinsic properties of a chemical reaction.

Chemical Reaction

A chemical reaction involves the transformation of reactants into products. The rate of a reaction refers to how quickly this transformation occurs. In kinetics, understanding the reaction mechanism helps clarify how reactants are consumed and products are formed.
The exercise presents a specific reaction: 2 NO₂(g) + F₂(g) → 2 NO₂F(g). Here,

• NO₂ and F₂ are reactants.
• NO₂F is the product formed.

Chemical reactions can be influenced by various factors such as concentration, temperature, presence of a catalyst, and physical state of the reactants. In our example:

• Increasing the concentrations of NO₂ and F₂ increases the rate of reaction.
• The specific interplay of molecules and bonds defines the reaction pathway and outcome.

Understanding these mechanisms is vital for enhancing reactions in chemical laboratories and industrial processes.