Problem 95 - Chemistry The Molecular Science

Question

Nitryl fluoride is an explosive compound that can be made by oxidizing nitrogen dioxide with fluorine: \(2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}(\mathrm{~g})\) Several kinetics experiments, all done at the same temperature and involving formation of nitryl fluoride, are summarized in this table: \begin{tabular}{cccccc} \hline & \multicolumn{2}{c} { Initial Concentration (mol/L) } & & \\ \cline { 2 - 4 } \cline { 6 } Experiment & {\(\left[\mathrm{NO}_{2}\right]\)} & {\(\left[\mathrm{F}_{2}\right]\)} & {\(\left[\mathrm{NO}_{2} F\right]\)} & & Initial Rate \(\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right)\) \\ \hline 1 & 0.0010 & 0.0050 & 0.0020 & & \(2.0 \times 10^{-4}\) \\ 2 & 0.0020 & 0.0050 & 0.0020 & & \(4.0 \times 10^{-4}\) \\ 3 & 0.0020 & 0.0020 & 0.0020 & & \(1.6 \times 10^{-4}\) \\ 4 & 0.0020 & 0.0020 & 0.0010 & & \(1.6 \times 10^{-4}\) \\ \hline \end{tabular} (a) Write the rate law for the reaction. (b) Determine what the order of the reaction is with respect to each reactant and each product. (c) Calculate the rate constant \(k\) and express it in appropriate units.

Step-by-Step Solution

Answer

Rate law: \(\text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}]\). Orders: 1 (\(\mathrm{NO}_{2}\)), 1 (\(\mathrm{F}_{2}\)). \(k = 40\) mol\(^{-1}\) L\(\,\)s\(^{-1}\).

Step 1: Identify the Reaction and Rate Law

The given reaction is \(2 \mathrm{NO}_{2}(\mathrm{~g})+\mathrm{F}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NO}_{2} \mathrm{~F}(\mathrm{~g})\). The general form of the rate law is \(\text{Rate} = k[\mathrm{NO}_{2}]^m[\mathrm{F}_{2}]^n\) where \(k\) is the rate constant, and \(m\) and \(n\) are the reaction orders with respect to \(\mathrm{NO}_{2}\) and \(\mathrm{F}_{2}\) respectively.

Step 2: Determine the Order with Respect to \(\mathrm{NO}_{2}\)

Compare experiments 1 and 2 where \([\mathrm{NO}_{2} ]\) changes and \([\mathrm{F}_{2} ]\) is constant. Doubling \([\mathrm{NO}_{2} ]\) from 0.0010 to 0.0020 mol/L doubles the rate from \(2.0 \times 10^{-4}\) to \(4.0 \times 10^{-4}\) mol L\(^{-1}\) s\(^{-1}\). This indicates that the reaction is first order with respect to \([\mathrm{NO}_{2} ]\) (\(m = 1\)).

Step 3: Determine the Order with Respect to \(\mathrm{F}_{2}\)

Compare experiments 2 and 3 where \([\mathrm{F}_{2} ]\) changes and \([\mathrm{NO}_{2} ]\) is constant. Halving \([\mathrm{F}_{2} ]\) from 0.0050 to 0.0020 mol/L reduces the rate from \(4.0 \times 10^{-4}\) to \(1.6 \times 10^{-4}\) mol L\(^{-1}\) s\(^{-1}\). This suggests a first order dependency with respect to \([\mathrm{F}_{2} ]\) (\(n = 1\)).

Step 4: Conclude Rate Law and Reaction Orders

From the experiments, the rate law is \(\text{Rate} = k[\mathrm{NO}_{2}]^1[\mathrm{F}_{2}]^1\) or simply \(\text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}]\). This indicates the reaction is first order with respect to both \(\mathrm{NO}_{2}\) and \(\mathrm{F}_{2}\).

Step 5: Calculate the Rate Constant \(k\)

Use the rate law from any experiment to calculate \(k\). From experiment 1: \(\text{Rate} = 2.0 \times 10^{-4} = k(0.0010)(0.0050)\). Solve for \(k\): \[k = \frac{2.0 \times 10^{-4}}{(0.0010)(0.0050)} = 40 \, \text{mol}^{-1} \text{L}\,\text{s}^{-1}\]

Step 6: State Final Answers

(a) The rate law is \(\text{Rate} = k[\mathrm{NO}_{2}][\mathrm{F}_{2}]\). (b) The reaction is first order in \(\mathrm{NO}_{2}\) and first order in \(\mathrm{F}_{2}\). (c) The rate constant \(k = 40 \) mol\(^{-1}\) L\(\,\)s\(^{-1}\).

Key Concepts

Rate Law

The rate law is an essential part of chemical kinetics. It describes how the speed of a reaction is related to the concentration of its reactants. For the reaction given, the rate law is expressed as \( \text{Rate} = k[\text{NO}_2]^m[\text{F}_2]^n \). Here, \( k \) is the rate constant, and \( m \) and \( n \) represent the reaction orders with respect to \( \text{NO}_2 \) and \( \text{F}_2 \), respectively.

The coefficients in the rate law may not always match the stoichiometric coefficients from the balanced equation. Instead, they are determined experimentally. Different experiments help identify these values by showing how changes in concentrations affect the rate. In this reaction, both \( m \) and \( n \) are found to be 1, indicating a first-order dependency on each reactant. This gives us a clear picture that doubling one reactant concentration also doubles the rate of reaction.

Reaction Order

Understanding reaction order is crucial for predicting how changes in concentration affect reaction rates. The order of reaction gives insight into the mechanism and complexity of the reaction. It is the sum of the powers of the concentration terms in the rate law expression. For this particular reaction, the rate law is \( \text{Rate} = k[\text{NO}_2][\text{F}_2] \) which makes it second order overall.

• First-order: If the reaction is first-order with respect to a reactant, the rate changes proportionally with the concentration of the reactant.
• Second-order: A second-order reaction could mean it's first-order in two different reactants or second-order in one reactant.

In this example, each reactant is individually first-order, making the overall reaction second-order. This classification helps in understanding the dynamics of the reaction and in solving rate equations.

Rate Constant

The rate constant \( k \) is a pivotal parameter in the rate law equation. It links the reaction rate with the concentrations of reactants. In simple terms, it gives you a way to quantify the intrinsic speed of the reaction under specific conditions.

The units of the rate constant differ depending on the total order of the reaction. For the given reaction with a second-order rate law \( \text{Rate} = k[\text{NO}_2][\text{F}_2] \), the units of \( k \) are \( \text{mol}^{-1} \text{L} \text{s}^{-1} \).

To calculate \( k \), you can use any experiment that provides initial concentrations and initial rate. Using experiment 1 in the exercise:
\[ k = \frac{2.0 \times 10^{-4}}{(0.0010)(0.0050)} = 40 \, \text{mol}^{-1} \text{L}\,\text{s}^{-1} \]

This measurement tells you how reactive \( \text{NO}_2 \) and \( \text{F}_2 \) are in forming \( \text{NO}_2 \text{F} \), providing valuable insight into the dynamics of the reaction process.