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 remaining concentration after 60 seconds?

Question

Quizplus · Accepted Answer

Given that the reaction follows second-order kinetics, we can use the second-order rate law equation: $\frac{1}{[A]_t} = kt + \frac{1}{[A]_0}$, where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant. Plugging in the given values ($k = 0.543 \, \text{M}^{-1}\text{s}^{-1}$, $[A]_0 = 5.00 \, \text{M}$, $t = 60 \, \text{s}$), we can solve for $[A]_t$, which results in a concentration significantly different from zero, indicating that choice D is correct.

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

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