he rate constant for this zero‑order reaction is 0.0130 M ⋅ s - 1 at 300 ∘ C. A ⟶ products How long (in seconds) would it take for the concentration of A to decrease from 0.890 M to 0.280 M?

188 s

Explanation:

We are told the reaction is second order respect to A so we know the expression for the rate law is

rate = - Δ[A]/Δt = k[A]²

where the symbol Δ stands for change, [A] is the concentration of A, and k is the rate constant.

The integrated rate law for this equation from calculus is

1 / [A]t = kt + 1/[A]₀

where [A]t is the concentration of A at time t, k is the rate constant, and [A]₀ is the initial concentration.

Since we have all the information required to solve this equation lets plug our values

1 / 0.280 = 0.0130x t + 1 / 0.890

(1 / 0.280 - 1 / 0.890) M⁻¹ = 0.0130 M⁻¹ ·s⁻¹t

t = 188 s