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The rate constant for this second‑order reaction is 0.550 M−1⋅s−1 at 300∘C. A ⟶ productsHow long, in seconds, would it take for the concentration of A to decrease from 0.800 M to 0.380 M?

Respuesta :

princessesther2011

Answer:

5.29 seconds

Explanation:

The reaction follows a second-order

Rate = kC^2 = change in concentration/time

k = 0.550 M^-1.s^-1

C = 0.380 M

Co = 0.800 M

Change in concentration = Co - C = 0.8 - 0.38 = 0.42 M

0.55×0.38^2 = 0.42/t

t = 0.42/0.07942 = 5.29 seconds