A first-order reaction has a half life of 35.7 seconds. How long will it take for this reaction to be 84.4% complete?

Answer 1

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Half life formula can be written as

N = N₀ (½)^(t/t½)

N/N₀ = (½)^(t/t½)

ln (N/N₀) = ln (½)^(t/t½)

ln (N/N₀) = (t/t½) ln(½)

(t/t½) = ln (N/N₀) / ln(½)

t = t½ [ln (N/N₀) / ln(½)]   ...(1)

Given t½ = 35.7 s

Balance percentage of reaction = 100% - 84.4% = 15.6%

[N/N₀] = 15.6 %

[N/N₀] = 0.156

plug in (1)

t = 35.7 [ln (0.156) / ln(½)]t = 95.7 s

It will take about 95.7 s for the reaction to be 84.4% complete

Answer 2

since it's first order

.. ln[At] = -kt + ln[Ao]

.. kt = ln[Ao] - ln[At]= ln([Ao] / [At]).. . . .recall = ln(a) - ln(b) = ln(a/b)

.. t = ln([Ao] / [A]) / k

and

.. k = ln(2) / half life

subbing

.. t = (ln([Ao] / [At]) / ln(2)) * half life

we want to know when [At] = 0.156 * [Ao] (that's 84.4% complete)

so that

.. t = (ln([Ao] / 0.156[Ao]) / ln(2)) * 35.7 sec

.. t = (ln(1 / 0.156)) / ln(2)) * 35.7 sec

.. t = (ln(1) - ln(0.156)) / ln(2)) * 35.7 sec.. .. . .recall ln(a/b) = ln(a) - ln(b)

.. t = -ln(0.156) / ln(2) * 35.7 sec.. .. .. .. .. . . . recall ln(1) = 0

and now we're ready to calc. 

.. t = 95.7 sec