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

Question

anonymous · Answer

i'll give u an easy formula for this..
time taken for conc. of reactant to fall is, n*half life=initial conc/$$2^{n}$$

anonymous · Answer

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

anonymous · Answer

but remember this formula is only applicable for 1st order reactions...
or elase u may apply the general rate constant formulas.

aaronq · Answer

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

anonymous · Answer

what formula is A = Ao e^-kt ??

aaronq · Answer

general equation for exponential decay or growth

anonymous · Answer

oh that's right gahh sorry LOL