Question

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.

Answer 1

ANSWER:

Given,

Rate of reaction = 3^rd order

k_r = 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 n^th 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).