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The second-order reaction 2 Mn(CO)5 → Mn2(CO)10, has a rate constant equal to 3.0 × 109 M-1 s-1 at 25°C. If the initial concentration of Mn(CO)5 is 2.0 × 10-5 M, how long will it take for 90.% of the reactant to disappear?

Respuesta :

Answer:

t = 1.5 E-4 s

Explanation:

  • 2 Mn(CO)5 → Mn2(CO)10
  • - ra = K(Ca)∧α = - δCa/δt

∴ a: Mn(CO)5

∴ K = 3.0 E9 M-1 s-1................rate constant

∴ T = 25°C

∴ α = 2................ second-order

∴ Cao = 2.0 E-5 M................initial concentration

If  Ca = Cao - (Cao×0.90) ⇒ t = ?

⇒ Ca = 2.0 E-5 - ((2.0 E-5)(0.90)) = 2.0 E-6 M

⇒ - δCa/δt = KCa²

⇒ - ∫δCa/Ca² = K∫δt

⇒ [(1/Ca) - (1/Cao)] = K*t

⇒ t = [(1/Ca) - (1/Cao)] / K

⇒ t = [ (1/2 E-6) - (1/2 E-5)] / (3.0 E9)

⇒ t = 1.5 E-4 s

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