Ans. First order kinetics-
ln ([A]t / [A]0) = -kt -equation 1
where, [A]t = Final concentration
[A]0 = Initial concentration
k = rate constant
t = time of reaction
Also, Half-life of first order reaction, t½ = 0.693 / k
Or, k = 0.693 / t½ - equation 2
Combining equation 1 and 2 –
ln ([A]t / [A]0) = -(0.693 / t½) t
or, ln [A]t – ln [A]0 = -(0.693 / t½) t - equation 3
Given-
Initial [drug], [A]0 = 20 mg/ dL
Time, t = 5 x (half-life) = 5 x 2 hrs = 10 hrs
Now,
Putting the values in equation 3-
ln [A]t – ln (20) = -(0.693 / 2 hr ) x 10 hr
Or, ln [A]t – 2.9957 = - 3.465
Or, 2.303 log [A]t = - 3.465 + 2.9957 = -0.4693
Or, log [A]t = -0.4693 / 2.303 = -0.203778
Or, [A]t = antilog (- 0.203778) = 0.625
Therefore, final [Drug] in patient’s body, [A]t = 0.625 mg/ dL
We want to calculate the average concentration of a relatively short acting drug in a patient....
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