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Thinking about a third-order reaction of the form 2A → P with kr = 6.50x10-4 dm6*mol-2*s-1,...

Thinking about a third-order reaction of the form 2A → P with kr = 6.50x10-4 dm6*mol-2*s-1, find the time required for the concentration of A to change to 0.015 mol*dm-3 from an initial concentration of 0.030 mol*dm-3.

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Answer #1

ANSWER:

Given,

Rate of reaction = 3rd order

kr = 6.500 x 10-4 dm6 mol-2 s-1

initial concentration, [A]o = 0.030 mol dm-3

final concentration, [A]t = 0.015 mol dm-3

For the nth order of reaction, rate law can be written as:

So, for third order (n=3) reaction, rate law is:

t = 2564076.9 sec

t = 712.24 hours

Hence, the time required for the concentration of A to change to 0.015 mol dm-3 from an initial concentration of 0.030 mol dm-3 is 2564076.9 sec (or, 712.24 hours).

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