The half-life for the second-order reaction of a substance A is 55.8 s when concentration of initial A = 0.69 M. Calculate the time needed for the concentration of A to decrease to the following values.
(a) 1/8 of its original value
(b) 1/9 of its original value
3 answers
i tried multiplying 7/8 and 8/9 by .69 each to find the respective initial concentrations of A to then use for the half life equation t=1/ k*[ A ]
use t= (1/ [ A]) - (1/ [A ]o) over k
t1/2 = 1/k[Ao]
1/(A) - 1/(Ao) = kt
I think these should get both a and b.
1/(A) - 1/(Ao) = kt
I think these should get both a and b.
1 answer