J.R. S. answered 02/11/18
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Disappearance of oil in the first 20 min when initial [oil] = 0.5M = 0.25 M/20 min = 0.0125 M/min
Disappearance of oil in the next 40 min when initial [oil] =0.25 M = 0.125/40 min = 0.003125 M/min
Doubling [ ] from 0.25 to 0.5 results in a 4 fold increase in rate, thus the reaction is 2nd order
For second order, the integrated rate equation is 1/[A] = 1/[A]o + kt.
From rate = k[oil]2, you can solve for k using any of the rates and corresponding [ ] in the given problem. After finding k, substitute it in the integrated rate equation using [A]o = 1 and [A] 0.03 and solve for t.