hi. a first-order reaction has a half life of 26.4 seconds. how long will it take for the concentration of the reactant in the reaction to fall to 1/8 of its initial value? please help me omg

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hi. a first-order reaction has a half life of 26.4 seconds. how long will it take for the concentration of the reactant in the reaction to fall to 1/8 of its initial value? please help me omg

Chemistry
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i'll give u an easy formula for this.. time taken for conc. of reactant to fall is, n*half life=initial conc/\[2^{n}\]
since here it is given 1/8 of its initial value so accordin to the formula, 3*26.4=initial conc/\[2^{3}\] thus time taken is=3*26.4=79.2secs
but remember this formula is only applicable for 1st order reactions... or elase u may apply the general rate constant formulas.

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first order reaction so use: \[t _{1/2}=\frac{ \ln2 }{ k }\] find k then, use: \[A=A _{0}e ^{-kt}\] assume Ao= initial = 1 then A = 1/8 solve for t, time, which will be in seconds
what formula is A = Ao e^-kt ??
general equation for exponential decay or growth
oh that's right gahh sorry LOL

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