HELP WITH GEN CHEM 2, Please show work?

Let n be the number of half-lives.

(1/2)ⁿ = 0.250/2.00
(1/2)ⁿ = 1/8
(1/2)ⁿ = (1/2)³
Then, n = 3

Time taken = (20.0 seconds) × 3 = 60.0 seconds

you have to know the order of the reaction to solve this

from the rate equation
.. Rate =k * [CH3CN]
we can readily read the exponent on [CH3CN] to b "1"
.. rate = k * [CH3CN]¹
i.e
.. the order of the reaction is "1" or "first order" w.r.t. [CH3CN]

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next, from the "integrated" 1st order
.. ln[At] = -kt + ln[Ao]

rearranging
.. t = ( ln[Ao] - ln[At] ) / k

and from the relationship ln(A) = ln(b) = ln(a/b)
and from the concept of half life... t½... half life = time required for [At] to drop to ½ [Ao]
so that
.. t½ = ln( [Ao] / [At] ) / k = ln( [Ao] / ½[Ao] ) / k = ln(2) / k

or if we solve for "k"
.. k = ln(2) / t½

and now we can go back to the original integrated equation
.. ln[At] = - (ln(2) / t½) * t + ln[Ao]

and finally, we can rearrange and solve for t
.. t = (ln ([Ao] / [At])) * (t½ / ln(2))
.. t = ( ln(2/0.250) * (20.0s / ln(2))
.. t = ln(8) * 20.0s / ln(2)
.. t = ln(2³) / ln(2) * 20.0s
.. t = 3 * ln(2) / ln(2) * 20.0s.... .. ..since ln(a^b) = b * ln(a)
.. t = 3 * 20.0s
.. t = 60.0s

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notice how this would change if the reaction was zero or 2nd order in [CH3CN]

zero order
.. [At] = -kt + [Ao]
.. k = ([Ao] - [At]) / t
.. k = ([Ao] - ½[Ao]) / t½
.. k = [Ao] / (2*t½)
then
.. [At] = -([Ao] / (2*t½)) * t + [Ao]
and solving
.. t = 2 * ([Ao] - [At]) * t½ / [Ao] = 35s

2nd order..
.. 1/[At] = +kt + 1/[Ao]
.. k = (1/[At] - 1/[Ao]) / t
.. k = (2/[At] - 1/[Ao]) / t½
.. 1/[At] = + (1 / ([At] * t½)) * t + 1/[Ao]
and you get the picture I'm sure.

This looks like a basic concept half-life problem, i.e., what's the definition of half-life?
Set up a chart to visualize the concentration dividing in half every half-life time interval.
Time [CH3NC]
0s----- 2M
20s--- 1M
40s--- 0.5M
60s--- 0.25M
So it will take 3 half-lives or 1 min to reach the specified concentration.