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At 700 k, the rate constant for this reaction is 6.2 × 10–4 min–1. how many minutes are required for 10.0% of a sample of cyclopropane to isomerize to propene?

Respuesta :

jmonterrozar

Answer:

170 min

Step-by-step explanation:

We know that there is a kinetics of the first order, which has the following equation:

ln ([A] t / [A] o) = -k * t

Solving for t:

t = - ln ([A] t / [A] o) / k

How we want to know is the time required with 10%, so [A] t / [A] o = 100% - 10%, that is, 90% = 0.9. We also know that k = 6.2 * (10 ^ -4)

Knowing this, we replace:

t = - ln (0.90) /6.2 * (10 ^ -4)

t = 170 min

Which means it will take 170 minutes.