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
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.
To calculate the time needed for the concentration of substance A to decrease to a certain value, we can use the half-life equation for a second-order reaction:
t = (1 / (k * [A₀])) * (1 / (final concentration - [A₀]))
Where:
- t is the time needed for the concentration to decrease
- k is the rate constant of the reaction
- [A₀] is the initial concentration of substance A
As given in the question, the half-life for the reaction is 55.8 s and the initial concentration of A is 0.69 M.
(a) To find the time needed for the concentration to decrease to 1/8 of its original value, we can substitute the values into the equation:
final concentration = 1/8 * [A₀]
t = (1 / (k * 0.69)) * (1 / (1/8 * 0.69))
t = (1 / (k * 0.69)) * (8 / 0.69)
t = 8 / (k * 0.69)
(b) Similarly, to find the time needed for the concentration to decrease to 1/9 of its original value:
final concentration = 1/9 * [A₀]
t = (1 / (k * 0.69)) * (1 / (1/9 * 0.69))
t = (1 / (k * 0.69)) * (9 / 0.69)
t = 9 / (k * 0.69)
To calculate the time needed for the concentration to decrease to the specified values, we would need to know the value of the rate constant (k) for the reaction.