At a particular temperature,N₂O₅ decomposes according to a first-order rate law with a half-life of 3.0 s.If the initial concentration of N₂O₅ is 1.0 $\times$ 10¹⁶ molecules/cm³,what will be the concentration in molecules/cm³ after 16.1 s?

Question

Quizplus · Accepted Answer

The half-life of the reaction is given as 3.0 s. Therefore, the rate constant (k) can be calculated as: t1/2 = ln(2) / k => k = ln(2) / t1/2 => k = ln(2) / 3.0 = 0.231 s^-1 Using the first-order rate law equation: ln[Nt/No] = -kt where Nt is the concentration at time t and No is the initial concentration. Substituting t = 16.1 s, No = 1.0 x 10^10 molecules/cm3, and k = 0.231 s^-1, we get: ln[Nt/1.0 x 10^10] = -0.231 x 16.1 => ln[Nt/1.0 x 10^10] = -3.722 Taking the antilog of both sides, we get: [Nt/1.0 x 10^10] = e^-3.722 = 1.52 x 10^-2 Multiplying both sides by the initial concentration, we get: Nt = 1.52 x 10^-2 x 1.0 x 10^10 = 2.4 x 10^8 molecules/cm3 Therefore, the concentration of N₂O₅ after 16.1 s is 2.4 x 10^8 molecules/cm3, which is closest to option A.

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At a Particular Temperature,N₂O₅ Decomposes According to a First-Order Rate $\times$