Exhibit 13-15 Consider the decomposition of NO 2 at 300 ° C as shown below to answer the following question(s). This reaction has been experimentally determined to follow second order kinetics with a rate constant of 0.543 M - 1 s - 1 .
2 NO 2 (g)→2 NO (g) + O 2 (g)
Assume an initial concentration of NO 2 in a closed vessel of 5.00 M.
-Refer to Exhibit 13-15. What is the half-life for this second order reaction?

Question

Quizplus · Accepted Answer

The half-life for a second-order reaction is given by the formula $t_{1/2} = \frac{1}{k[A]_0}$, where $k$ is the rate constant and $[A]_0$ is the initial concentration. Substituting $k = 0.543 \, M^{-1}s^{-1}$ and $[A]_0 = 5.00 \, M$, we get $t_{1/2} = \frac{1}{0.543 \times 5.00} = 0.368 \, sec$.

Exhibit 13-15 Consider the Decomposition of NO2 at 300 °

Exhibit 13-15 Consider the Decomposition of NO₂ at 300 °