Posted by chemist on May 20, 2004, at 12:04:21
In reply to Thank You Chemist ! (nm) » chemist, posted by chess on May 20, 2004, at 1:59:37
it occurred to me that i should clarify the half-life business a little...there have been statements (mine, others) that have been made along the lines of ``if the elimination half-life is 30 hours, then if you took 1 mg klonopin, at 30 hours there would be 0.5 mg in your system, and at 60 hours 0.25 mg, and so forth.'' this is not entirely accurate. assuming that the elimination can be described by first-order kinetics, we need to know k, the rate constant, which will be in units of inverse time. going through the math, you end up with the usual expression (i am assuming the activity coefficient for klonopin - K - to be unity, thus [K] == {K}) for t_{1/2} = (ln 2)/k. this yields (assuming t_{1/2} == 30 hours) k = 0.023 hr^{-1}. for the time at which one quarter of your initial dose remains, t_{1/4} = (ln 4)/k = 60.26 hours. the decay is exponential, and my colleagues would berate me if they ever found out that i had not clarified....thus, t_{1/10} = 100.11 hours, and t_{1/1000} = 300.34 hours, which means that 1 microgram of klonopin - which is 3 orders of magnitude of the lowest detection limit i am aware of for the metabolite (i.e., part per trillion) - will be present at 1.8 weeks after the initial 1 mg dose....all the best, chemist
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thread:347588
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