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?

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