# The concentration of a drug in the body is often expressed in units of milligrams per kilogram of body weight. The initial dose of a drug in an animal was 25.0 mg/kg body weight. After 2.00 hours, the concentration had dropped to 15.0 mg/kg body weight. If the drug is eliminated metabolically by a first-order process, what is the rate constant for the process in units of mn^(‒1) When we want to know the relationship between concentration (or amount) and time, we need to use the integrated rate law equations (see "Summary" section on link).

ln ([At]/[Ao]) = -kt

Since we know this is first order, we can use the appropriate integrated rate law (shown below) to find...

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When we want to know the relationship between concentration (or amount) and time, we need to use the integrated rate law equations (see "Summary" section on link).

ln ([At]/[Ao]) = -kt

Since we know this is first order, we can use the appropriate integrated rate law (shown below) to find the rate constant.

[A] represents the concentration of the substance at time zero or time t.  Even though it specifically is given as a concentration, we are really only worried about the ratio so we can use moles or grams in this formula as well.  We can put grams because the substance is the same so the ratio of grams will equal the ratio of the moles as well as the ratio of the concentration.

ln (15.0/25.0) = - k (2.00 hr)

Now we can solve for k to find 0.255.  For the units, we need to see that the time is in hours so the rate constant must include hours in its units (rather than minutes or seconds).  Also, the units of k must cancel with the units of time, so the units of k will be hr^-1 (i.e. inverse hours), so k = 0.255 hr^-1

When using the integrated rate law, we can solve for any one of the variables given the other three or solve for the ratio of the concentrations given k and t.

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