The reaction of nitrogen dioxide with fluorine generates nitryl fluoride:

2NO2(g) + F2(g) --> 2NO2F(g)

To determine the rate law for this reaction, a chemist performed several initial-rate experiments using different initial concentration, with the details tabulated below.

Experiment (#) [NO2] (M)
[F2] (M) [NO2F] (M)
Initial Rate (M/s)
1 1.0 x 10-3 5.0 x 10-3 0.10 x 10-4 2.0 x 10-4
2 2.0 x 10-3 5.0 x 10-3 0.10 x 10-4 4.0 x 10-4
3 2.0 x 10-3 5.0 x 10-3 1.00 x 10-3 4.0 x 10-4
4 2.0 x 10-3 7.5 x 10-3 0.10 x 10-4 6.0 x 10-4
Determine the rate constant for this reaction.

Based on data given, Rate = k[NO2][F2]^1.2

Pick any row and substitute concentration and rate values. Solve for k. I got 115.4 M^-1.s^-1

Well, let's break it down! The rate law for a reaction can be determined by comparing the initial rates of different experiments and how they vary with changes in concentration.

From experiment 1 to experiment 2, the only change in concentration is the [NO2] value, which doubles. However, the initial rate also doubles. This suggests that the rate is directly proportional to the concentration of NO2. We can represent this as:

Rate = k[NO2]^x

Next, let's compare experiment 2 and 3. This time, the [NO2F] value changes, but the initial rate remains the same. So, the rate doesn't depend on the concentration of NO2F. We can conclude that the exponent, x, for NO2F is 0:

Rate = k[NO2]^1[NO2F]^0

Simplifying this:

Rate = k[NO2]

Finally, let's compare experiment 3 and 4. Both [NO2] and [F2] change, but the initial rate remains the same. This indicates that the rate doesn't vary with the concentration of F2. Thus, the exponent for F2 is also 0:

Rate = k[NO2]^1[F2]^0

Simplifying this even further:

Rate = k[NO2]

From all our experiments, we can conclude that the rate law for the reaction is Rate = k[NO2]. So now, we need to determine the rate constant, k.

Let's use the data from experiment 1:

[NO2] = 1.0 x 10^-3 M
[NO2F] = 0.10 x 10^-4 M
Rate = 2.0 x 10^-4 M/s

Using the rate law equation with these values:

2.0 x 10^-4 M/s = k(1.0 x 10^-3 M)

Simplifying:

k = (2.0 x 10^-4 M/s) / (1.0 x 10^-3 M)

k = 2.0 x 10^-4 s^-1

So, the rate constant for this reaction is 2.0 x 10^-4 s^-1.

To determine the rate law for the reaction, we need to analyze the data from the experiments and see how the initial concentrations of reactants affect the initial rate of the reaction.

Let's analyze the experiments:

Experiment 1:
[NO2] = 1.0 x 10^(-3) M
[F2] = 5.0 x 10^(-3) M
[NO2F] = 0.10 x 10^(-4) M
Initial Rate = 2.0 x 10^(-4) M/s

Experiment 2:
[NO2] = 2.0 x 10^(-3) M
[F2] = 5.0 x 10^(-3) M
[NO2F] = 0.10 x 10^(-4) M
Initial Rate = 4.0 x 10^(-4) M/s

Experiment 3:
[NO2] = 2.0 x 10^(-3) M
[F2] = 5.0 x 10^(-3) M
[NO2F] = 1.00 x 10^(-3) M
Initial Rate = 4.0 x 10^(-4) M/s

Experiment 4:
[NO2] = 2.0 x 10^(-3) M
[F2] = 7.5 x 10^(-3) M
[NO2F] = 0.10 x 10^(-4) M
Initial Rate = 6.0 x 10^(-4) M/s

To determine the rate law, we compare Experiments 1 and 2:

The concentration of F2 is constant while the concentration of NO2 doubles. The initial rate also doubles. This indicates a direct relationship between the concentration of NO2 and the initial rate.

Therefore, the rate law can be written as: Rate = k[NO2]^n[F2]^m

Next, we compare Experiments 2 and 3:

The concentration of NO2 is constant, while the concentration of NO2F doubles. The initial rate remains the same. This indicates that the concentration of NO2F does not affect the initial rate.

Therefore, m = 0.

The rate law can now be written as: Rate = k[NO2]^n[F2]^0

Simplifying, we get: Rate = k[NO2]^n

Next, we compare Experiments 1 and 4:

The concentration of F2 increases by a factor of 1.5, while the concentration of NO2 and NO2F are constant. The initial rate increases by a factor of 3.

This indicates that the reaction is first order with respect to F2.

Therefore, n = 1.

The rate law can now be written as: Rate = k[NO2]^1[F2]^1

Simplifying further, we get: Rate = k[NO2][F2]

Now, let's use one of the experiments (Experiment 1) to determine the value of the rate constant (k).

[NO2] = 1.0 x 10^(-3) M
[F2] = 5.0 x 10^(-3) M
Initial Rate = 2.0 x 10^(-4) M/s

Plugging in these values, we have:

2.0 x 10^(-4) M/s = k(1.0 x 10^(-3) M)(5.0 x 10^(-3) M)

Simplifying this equation, we find:

k = (2.0 x 10^(-4) M/s) / (1.0 x 10^(-3) M)(5.0 x 10^(-3) M)

k = 4.0 x 10^(-2) M^(-1) s^(-1)

Therefore, the rate constant for this reaction is 4.0 x 10^(-2) M^(-1) s^(-1).

To determine the rate law for the reaction, we need to analyze the experimental data and find the relationship between the initial concentrations and the initial rates.

From the given data, we can see that experiments 1 and 2 have the same initial concentrations of NO2 and F2 but yield different initial rates. This suggests that the concentration of NO2 might be the determining factor in the rate of the reaction.

Comparing experiments 1 and 3, we see that the concentration of NO2 remains the same, but the concentration of NO2F is increased by a factor of 10. However, the initial rate remains the same. This suggests that the concentration of NO2F does not affect the rate of the reaction.

Comparing experiments 2 and 4, we see that the concentration of F2 remains the same, but the concentration of NO2 is doubled. The initial rate is also doubled. This suggests that the rate is directly proportional to the concentration of NO2.

Based on these observations, we can propose the following rate law:

Rate = k[NO2]^1[F2]^0[NO2F]^0

Since the reaction rate is only dependent on the concentration of NO2, we can simplify the rate law to:

Rate = k[NO2]^1

Now, we can use any of the given experiments to calculate the rate constant (k). Let's use experiment 1:

[NO2] = 1.0 x 10^(-3) M
Initial Rate = 2.0 x 10^(-4) M/s

Using the rate law equation:

2.0 x 10^(-4) M/s = k(1.0 x 10^(-3) M)^1

Solving for k:

k = 2.0 x 10^(-4) M/s / (1.0 x 10^(-3) M)

k = 0.2 s^(-1)

Therefore, the rate constant for this reaction is 0.2 s^(-1).