Kinetics: Reaction Rate Law and Calculations

In this module, we will conduct a detailed analysis of the kinetics of the reaction involving fluorine gas (F2) and nitrogen dioxide (NO2) and derive the order of the reaction using the provided initial rate data. The reaction is represented as follows:
F2 (g) + 2 NO2 (g) \rightarrow 2 NO_2F (g)

Rate Law Determination

To establish the comprehensive rate law for this reaction, we begin with the provided initial rate data from three different experiments that give insights into how the concentrations of the reactants affect the rate of reaction.

Given Initial Rate Data:

  • Experiment 1: [F2]₀ = 0.10 M, [NO2]₀ = 0.10 M, Initial Rate = 1.21 \times 10^{-3} M s⁻¹

  • Experiment 2: [F2]₀ = 0.20 M, [NO2]₀ = 0.20 M, Initial Rate = 4.79 \times 10^{-3} M s⁻¹

  • Experiment 3: [F2]₀ = 0.10 M, [NO2]₀ = 0.050 M, Initial Rate = 6.02 \times 10^{-4} M s⁻¹

a) Determining the Rate Law

To determine the order of the reaction concerning each reactant, we start with the general form of the rate law, expressed as:
Rate = k[F2]^m[NO2]^n
where m and n represent the orders of the reaction with respect to F2 and NO2 respectively. The task at hand involves analyzing the relationship between the concentration of reactants and the observed initial rates from different experiments.

Using Experiment 1 and Experiment 2:
Comparing experiments 1 and 2 allows us to understand how changes in concentrations affect the reaction rate:
\frac{Rate2}{Rate1} = \frac{k[F2]2^m[NO2]2^n}{k[F2]1^m[NO2]1^n}
Substituting the values yields:
\frac{4.79 \times 10^{-3}}{1.21 \times 10^{-3}} = \frac{[0.20]^m[0.20]^n}{[0.10]^m[0.10]^n}
This simplifies to:
3.96 = 2^m2^n
Assuming the sum of the orders m + n = 2 (based on the observation that doubling the concentrations results in quadruple the rate), a possible solution emerges as m = 1 and n = 1.

4. Confirm with Experiment 3:
Next, we confirm our findings by comparing experiments 1 and 3, using the respective values:
\frac{Rate3}{Rate1} = \frac{6.02 \times 10^{-4}}{1.21 \times 10^{-3}} = 0.497
Further simplifying gives:
\frac{[0.10]^m[0.050]^n}{[0.10]^m[0.10]^n} = \frac{[0.050]^n}{[0.10]^n} = (0.50)^n
Hence, we observe:
0.497 = 0.125^{n} indicating that n = 2.

6. Final Rate Law:

From this rigorous analysis, we conclude that the rate law for the reaction can be expressed as:
Rate = k[F2]^1[NO2]^2
This indicates that the reaction is first order with respect to fluorine and second order with respect to nitrogen dioxide, reflecting how their concentrations influence the overall reaction rate.

b) Finding the Rate Constant (k)

To determine the rate constant k, we can substitute one of the experimental datasets into our established rate law. Using the data from Experiment 1:
1.21 \times 10^{-3} = k[0.10]^1[0.10]^2
Thus, solving for k gives:
k = \frac{1.21 \times 10^{-3}}{(0.10)(0.01)} = \frac{1.21 \times 10^{-3}}{0.001} = 1.21 \text{ M}^{-2} \text{ s}^{-1}
This implies that the rate constant significantly influences the speed of the reaction under the conditions tested.

c) Integrated Rate Law for Given Conditions

For reaction conditions where [F2]₀ = 0.10 M and [NO2]₀ = 10.0 M, we aim to apply the integrated form of the rate law:
Since NO2 serves as an excess reactant, we can treat its concentration as a constant. Therefore, the reaction order can be simplified to first-order concerning F2, leading to the integrated rate law:
[F2] = [F2]0 - kt
Where k is the rate constant derived in part b.

d) Time Calculation for [F_2]

To calculate the duration necessary for [F_2] to diminish from 0.10 M to 0.025 M, we employ the integrated rate law:
0.025 = 0.10 - (1.21)(t)
Rearranging this equation for t leads to:
1.21t = 0.10 - 0.025 = 0.075
This gives:
t = \frac{0.075}{1.21} \approx 0.0620 \text{ seconds}

Summary

In summary, we have rigorously analyzed the reaction kinetics for the system involving F2 and NO2, calculated the rate law and the rate constant, and explored the behavior of the system over time. These insights are essential for predicting the outcomes of chemical reactions and optimizing conditions for chemical processes in various applications.