A kid has just been brought in to the hospital you work, he has taken some medic

Question

A kid has just been brought in to the hospital you work, he has taken some medication (theophylline) in tablet form. 2 hours before arriving at the hospital the child ingested 11, 100mg tablets. Theophylline isabsorbed into the bloodstream at a rate proportional to the amount present in the gastrointestinal tract (stomach and intestines). It is eliminated from the bloodstream at a rate proportional to the amount present in the bloodstream. Theophylline has an 5 hour absorption half life and an 5 hour elimination half life. A blood-level concentration of 100mg/L or more is seriously toxic. 200mg/L or more is fatal. The child has 2L of blood, and it is too late to induce vomiting.

Rate constants:

X]inital druf concentration = 1100mg /2L = 550mg/L

k1 (absorbtion half life rate)= t1/2 = 0.693/k1

0.693/5 hours = 0.1386 hours

k2 (elimination half life rate)= t1/2 = 0.693/k2

0.693/5 = 0.1386 hours

QUESTIONS:

One method of treating the child is to increase the rate at which theophylline is eliminated from the bloodstream by giving oral doses of charcoal (to absorb the drug). What would the half life of elimination need to be so the blood drug level never reaches a toxic level? (If the child in your case already has a toxic level before he reaches the hospital, what would the half life need to be so the blood drug level never reaches a fatal level?)

Note that since this treatment can only be given after the child reaches the hospital you will first have to determine the concentrations of X,Y and Z at 2 hours and then put them into the model as startingconcentrations.

New initial values:

[X]2 hrs =                     

[Y]2 hrs =                     

[Z]2 hrs =

k2 would need to be =

t1/2 would need to be =

Explanation / Answer

Problem.1) Provide New initial values:

ans. [X]2 hrs = 397.54mg

[Y]2 hrs = 152.46mg

[Z]2 hrs = 152.46mg

k2 would need to be = 0.87

t1/2 would need to be = 8hr

explaination: Rate of reaction dX/dT = Ka.[X]t. Thus apsorption rate is 76.23mg/L. SO After 2hr theophyllin in gastro intestinal track will be 550-(76.23*2)=397.54mg/ml. Apsorption in blood stream and elimination will be (76.23*2)=152.46mg. k2 must be lower so that theophylline concentration in blood stream less than 100mg/ml. k2 is depend on half time(t1/2). thus with t1/2= 8 hr theophylline concentration in blood stream will be 95mg/L and k2= 0.87.

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