3. In pharmacology, first-order reactions are referred to as \"linear\" processe

Question

3. In pharmacology, first-order reactions are referred to as "linear" processes because the rate of elimination of the drug is proportional to the concentration of the drug in the blood plasma. Consider the elimination of the drug gentamicin, depicted by the graph. A 288 mg dose of gentamicin was taken such that the initial concentration (assuming 18.0 L volume of body is 16.0 mg/L. Linear y-axis a. What is the elimination half-life of the drug in the c 16 en 34.) plasma? g 10 g 8 b. How many hours does it take for the drug to reach a concentration of 3.50 mg/L? c. If a mg dose exactly every o 2 288 is taken 4 hours what is the concentration (mg/L of the drug in the plasma after 15.0 hours? time [h]

Explanation / Answer

For a first order reaction the integrated rate law is C = Co e-kt where C is the concentration at time t and Co is the initial concentration and k is the rate constant. It is logarithmically expressed as

k = (2.303/t) { log (a/a-x)} where a= Co and a-x = Ct

The initial concentration of the drug was given as 16mg/L Co= 16

and in the diagram it is given C = 16x e-o.34t which shows k = 0.34

For a first order reaction half time is related to k as t1/2 = 0.693/k

a)Thus half time for the elimination of drug = 0.693/0.34 =2.832h (on the graph time is in hrs.)

b)Time to reach concentration of 3.5 mg/L

using the rate equation   k = (2.303/t) { log (a/a-x)}

0.34 = 2.303/t [log 16/3.5]

Soving for t we get t= 4.47hrs.

c)

When the drug is taken every 4 hrous exactly, the concentration of drug changes every 4 hours.

step i) after first 4 hours

initial concentration a = 16mg/L k = 0.34 and t= 4 hrs

Thus 0.34 = 2.303/4 log 16/ Ct

solving we get Ct = concentration of drug remaining after 4 hours = 4.107mg/L

step ii) After 8 hrs

Now another dose of 288mg is added, which makes concentration of durg in plasma = 16 (newly added ) + 4.107 (already prsent from previous dose) = 20.107

Substituting this in the same formula similarly and calculating the remaining concentration of durg after 8hrs.

    k = (2.303/t) { log (a/a-x)}

0.34 = (2.303/4) log 20.107/Ct [Time should be taken 4 hrs only not 8 hrs, as we start after 4 hrs after adding fresh dose)

Solving for Ct we get Ct = 5.1622mg/L

step iii) After 12 hrs, added fresh 288mg that is 16mg/L is again added.

Thus the initial concentration at this point = 5.1622 + 16 = 21.1622

Substituting this value and calculating for Ct , we get Ct= 5.433mg/L

step iv) After 15 hrs.No fresh dose added

Thus initial concentration = remaining drug concentration = 5.433mg/L

Substituting this we get the remaining concentration of drug after 15hrs

0.34 = 2.303/3 Log (5.433/Ct)

Calculating Ct = 1.96mg/L

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