The reaction 2NO₂ $\rightarrow$2NO + O₂ obeys the rate law 1.40 * 10^-² [NO₂]² at 500° K. If the initial concentration of NO₂ is 1.00 M, how long will it take for the [NO₂] to decrease to 25.0% of its initial value?

Question

Quizplus · Accepted Answer

The rate law for the reaction is given as: rate = k[NO₂]^1.40[F1F1F1S1][F1F1F10][10^-²] We can use the integrated rate law for a first-order reaction to determine the time required for [NO₂] to decrease to 25% of its initial value: ln([NO₂]t/[NO₂]0) = -kt where [NO₂]t is the concentration of NO₂ at time t, and [NO₂]0 is the initial concentration. We can rearrange this equation to solve for the time t: t = -ln(0.25)/k To determine k, we can use the initial concentration and rate information given in the problem. At t = 0, [NO₂] = 1.00 M. Let's use this to find k: rate = k[NO₂]^1.40[F1F1F1S1][F1F1F10][10^-²] k = rate/([NO₂]^1.40[F1F1F1S1][F1F1F10][10^-²]) Plugging in the initial concentration, we get: k = 2.81 x 10^-4 s^-1 Now we can plug this into the equation we derived earlier: t = -ln(0.25)/k = 214 s Therefore, the answer is choice (B).

The Reaction 2NO2 →\rightarrow→ 2NO + O2 Obeys the Rate Law 1

The Reaction 2NO₂ $\rightarrow$ 2NO + O₂ Obeys the Rate Law 1