[VISUAL] The data from a study of the decomposition of NO2(g) to form NO(g) and O2(g) are given in t — Kinetics Chemistry Question

STEPS:

1. Analyze the kinetic data provided in the study:
The table records the time ($t$), concentration of the reactant ($[NO_2]$), natural log of the concentration ($ln[NO_2]$), and the reciprocal of the concentration ($1/[NO_2]$) at regular $100\text{-second}$ intervals.

2. Recall the relationships between concentration functions and reaction order:
* For a zeroth-order reaction, a plot of concentration versus time is linear, indicating a constant rate of concentration decrease over time.
* For a first-order reaction, a plot of the natural logarithm of concentration ($ln[A]$) versus time is linear.
* For a second-order reaction, a plot of the reciprocal concentration ($\frac{1}{[A]}$) versus time is linear.

3. Test the data for first-order linearity:
Calculate the change in the $ln[NO_2]$ value over each $100\text{-second}$ interval using the table data:
* From $0$ to $100\text{ s}$: $-1.01 - (-0.693) = -0.317$
* From $100$ to $200\text{ s}$: $-1.25 - (-1.01) = -0.240$
Since the rate of change of $ln[NO_2]$ is not constant over equal time intervals, the reaction is not first-order.

4. Test the data for second-order linearity:
Calculate the change in the reciprocal concentration ($\frac{1}{[NO_2]}$) over each $100\text{-second}$ interval:
* From $0$ to $100\text{ s}$: $2.75 - 2.00 = \mathbf{+0.75\text{ L/mol}}$
* From $100$ to $200\text{ s}$: $3.50 - 2.75 = \mathbf{+0.75\text{ L/mol}}$
* From $200$ to $300\text{ s}$: $4.25 - 3.50 = \mathbf{+0.75\text{ L/mol}}$

5. Identify the reaction order and consistent rate law:
Because the reciprocal concentration, $\frac{1}{[NO_2]}$, increases by the same constant amount ($0.75\text{ L/mol}$) during every $100\text{-second}$ interval, a plot of $\frac{1}{[NO_2]}$ versus time yields a straight line with a constant positive slope equal to the rate constant $k$. According to the integrated rate laws, this linear relationship is consistent with a second-order reaction. Consequently, the differential rate law takes the form:
$$\mathbf{\text{Rate} = k [NO_2]^2}$$
This identifies Option B as the correct choice.

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WHY_OTHERS_WRONG:

• Option A is incorrect: This option represents a first-order rate law. If the decomposition were first-order, the natural log of concentration ($ln[NO_2]$) would decrease by a constant amount every $100\text{ seconds}$, which is refuted by the actual unequal differences in the data ($-0.317$ versus $-0.240$).
• Option C is incorrect: This option represents a rate law where the rate is inversely proportional to concentration. This is mathematically inconsistent with the linear increase in reciprocal concentration over time, which dictates a second-order rate dependence on the reactant concentration ($\text{Rate} = k [NO_2]^2$).
• Option D is incorrect: This option represents a rate law where the rate is inversely proportional to the square of concentration. The integrated rate law of a second-order reaction yields a linear relationship for the reciprocal concentration versus time ($\frac{1}{[NO_2]} = kt + \frac{1}{[NO_2]_0}$), which corresponds to a direct second-order rate law ($\text{Rate} = k [NO_2]^2$) rather than an inverse squared relationship.