Question

the following graph shows data from an experiment in which no₂(g) decomposes as represented by the equation.\\(\ce{no2(g) \
ightarrow no(g) + \frac{1}{2}o2(g)}\\)\
the graph could be used to support which of the following statements?\
a the half - life of the reaction is constant.\
b the magnitude of the rate constant, \\(k\\), is approximately 10.\
c the unit of the rate constant, \\(k\\), is \\(\mathrm{s}^{-1}\\).\
d the rate law for the reaction is \\(\text{rate} = k\ce{no2}^2\\)?

Explanation:

Step1: Analyze the graph type

The graph is of \( \frac{1}{[\text{NO}_2]} \) vs. time, which is a straight line. For a second - order reaction, the integrated rate law is \( \frac{1}{[\text{A}]}=kt + \frac{1}{[\text{A}]_0} \), and the plot of \( \frac{1}{[\text{A}]} \) vs. \( t \) is linear. So this reaction is second - order with respect to \( \text{NO}_2 \).

Step2: Evaluate Option A

For a second - order reaction, the half - life formula is \( t_{1/2}=\frac{1}{k[\text{A}]_0} \). Since \( [\text{A}]_0 \) changes, the half - life of a second - order reaction is not constant. So Option A is incorrect.

Step3: Evaluate Option B

From the integrated rate law for second - order reaction \( \frac{1}{[\text{NO}_2]}=kt+\frac{1}{[\text{NO}_2]_0} \). The slope of the line \( \frac{1}{[\text{NO}_2]} \) vs. \( t \) is equal to \( k \). Let's take two points from the graph. At \( t = 0 \), \( \frac{1}{[\text{NO}_2]}=\frac{1}{100} \) (assuming the y - axis is \( \frac{1}{[\text{NO}_2]} \) with units such that at \( t = 0 \), the value is \( \frac{1}{100} \) (let's say the concentration units are such that \( [\text{NO}_2]_0=\frac{1}{100} \) inverse concentration units) and at \( t = 100 \) s, \( \frac{1}{[\text{NO}_2]}=\frac{1}{50} \) (wait, no, looking at the graph, the y - axis is \( \frac{1}{[\text{NO}_2]} \) with values 50, 100, 150, 200, 250, 300? Wait, maybe I misread. Wait, the y - axis is labeled \( \frac{1}{[\text{NO}_2]} \) and at \( t = 0 \), the value is 100 (so \( \frac{1}{[\text{NO}_2]_0}=100 \)), at \( t = 100 \) s, the value is 150? Wait, no, the x - axis is time (s) and y - axis is \( \frac{1}{[\text{NO}_2]} \). Let's take two points: \( (t_1, y_1)=(0, 100) \) and \( (t_2, y_2)=(100, 150) \). The slope \( k=\frac{y_2 - y_1}{t_2 - t_1}=\frac{150 - 100}{100-0}=\frac{50}{100}=0.5 \), not approximately 10. So Option B is incorrect.

Step4: Evaluate Option C

For a second - order reaction, the rate law is \( \text{rate}=k[\text{NO}_2]^2 \). The units of rate are \( \text{concentration}/\text{time} \) (e.g., \( \text{M/s} \)), and the units of \( [\text{NO}_2]^2 \) are \( \text{M}^2 \). So \( k=\frac{\text{rate}}{[\text{NO}_2]^2} \), the units of \( k \) are \( \text{M}^{-1}\text{s}^{-1} \), not \( \text{s}^{-1} \). So Option C is incorrect.

Step5: Evaluate Option D

Since the plot of \( \frac{1}{[\text{NO}_2]} \) vs. \( t \) is linear, the reaction follows the second - order integrated rate law, which means the rate law is \( \text{rate}=k[\text{NO}_2]^2 \). So Option D is correct.

Answer:

D. The rate law for the reaction is \( \text{rate}=k[\text{NO}_2]^2 \)