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4 March, 20:01

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?

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  1. 4 March, 20:25
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    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
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