The reaction 2NO₂ → 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 integrated rate law for a first-order reaction is: ln[NO₂] = -kt + ln[NO₂]0 where [NO₂]0 is the initial concentration, k is the rate constant, t is time, and ln denotes the natural logarithm. We can rearrange this equation to solve for t: t = (ln[NO₂]0 - ln[NO₂])/k We are given the rate law and the value of k, so we can plug in the values: k = 1.40 × 10^-2 s^-1 [NO₂]0 = 1.00 M [NO₂] = 0.25 M (25.0% of the initial value) t = (ln[NO₂]0 - ln[NO₂])/k t = (ln[1.00] - ln[0.25])/1.40 × 10^-2 t = 213.7 s Therefore, the answer is B) 214 s.

The Reaction 2NO2 → 2NO + O2 Obeys the Rate

The Reaction 2NO₂ → 2NO + O₂ Obeys the Rate