Chemistry Calculate the half-life of the antibiotic Ofloxacin in this patient

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The discussion focuses on calculating the half-life of the antibiotic ofloxacin in a 70 kg patient treated with ciprofloxacin. The initial half-life is calculated as approximately 4.158 hours based on the patient's volume of distribution and clearance. After the patient's condition worsens, the volume of distribution increases to 125 liters, but the calculated half-life remains unchanged at 4.158 hours. Several participants point out potential errors in the original calculations and clarify terminology related to the medication and its pharmacokinetics. The conversation emphasizes the importance of accurate units and understanding the relationship between volume of distribution and clearance in determining half-life.
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Homework Statement
half life time
Relevant Equations
๐‘‡1/2 =(๐‘™๐‘›2) ๐‘ฅ ๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘‘๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ÿ๐‘–๐‘๐‘ข๐‘ก๐‘–๐‘œ๐‘›/ clarence.
A 70 kg patient is treated with ciprofloxacin (2 x 500 mg / d) for an infection
pulmonary. Ciprofloxacin is an antibiotic eliminated primarily by the kidneys, its
apparent volume of distribution is 3 [1 / kg) and its clearance is 0.5 (Wh x kg).
Calculate the half-life of ofloxacin in this patient. Later, his situation
degrades: it produces edema and the volume of distribution of ofloxacin increases to 125 liters.
After calculating the new half-life of ofloxacin in this patient,
Choose
the correct statement among the following 5:
Please choose an answer:
A. The half-life is extended by 60 min. NS.
B. The half-life is shortened by 30 min. NS.
C. The half-life is lengthened by 4h10 min. NS.
D. The half-life time remains unchanged.
E. The half-life is extended by 100 min. NS.to calculate the first half-life I did:

T1/2 =(ln(2) x volume distribution)/ clearance

v=70Kg x 3L/KG =210L
clearance 0.5L/H KG x70KG=35

T1/2= (ln(2) x 210)/ 35 = 4.158h

to calculate the second half-life:

k= ln(2) / T1/2
K= ln2/4.158 =0.166

clearance= K. V
clearance= 0.166 X 125= 20.83

T1/2= (ln (2) x125)/ 20.83 =4.158h

I get that the half-life time remains the same; I don't see where my error is. you can help me?
 
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I am layman in medicine and chemistry. As I have scarce knowledge on your system mechanism half time you say reminds me exponential decay as radioactive materials.
 
havenly said:
Homework Statement:: half life time
Relevant Equations:: ๐‘‡1/2 =(๐‘™๐‘›2) ๐‘ฅ ๐‘‰๐‘œ๐‘™๐‘ข๐‘š๐‘’ ๐‘‘๐‘’ ๐‘‘๐‘–๐‘ ๐‘ก๐‘Ÿ๐‘–๐‘๐‘ข๐‘ก๐‘–๐‘œ๐‘›/ clarence.

A 70 kg patient is treated with ciprofloxacin (2 x 500 mg / d) for an infection
pulmonary. Ciprofloxacin is an antibiotic eliminated primarily by the kidneys, its
apparent volume of distribution is 3 [1 / kg) and its clearance is 0.5 (Wh x kg).
Calculate the half-life of ofloxacin in this patient. Later, his situation
degrades: it produces edema and the volume of distribution of ofloxacin increases to 125 liters.
After calculating the new half-life of ofloxacin in this patient,
Choose
the correct statement among the following 5:
Please choose an answer:
A. The half-life is extended by 60 min. NS.
B. The half-life is shortened by 30 min. NS.
C. The half-life is lengthened by 4h10 min. NS.
D. The half-life time remains unchanged.
E. The half-life is extended by 100 min. NS.to calculate the first half-life I did:

T1/2 =(ln(2) x volume distribution)/ clearance

v=70Kg x 3L/KG =210L
clearance 0.5L/H KG x70KG=35

T1/2= (ln(2) x 210)/ 35 = 4.158h

to calculate the second half-life:

k= ln(2) / T1/2
K= ln2/4.158 =0.166

clearance= K. V
clearance= 0.166 X 125= 20.83

T1/2= (ln (2) x125)/ 20.83 =4.158h

I get that the half-life time remains the same; I don't see where my error is. you can help me?
Hi @havenly. Welcome to PF.

Iโ€™m sure someone here can help, but there are some problems with your question. Using my limited knowledge, here is a list of the problems I spotted. If you can fix them, you are more likely to get help.

1. Does (2 x 500 mg / d) mean a single dose of 1000mg? If not, what does it mean?

2. You have given the apparent volume of distribution as โ€œ3 [1 / kg) โ€œ. This makes no sense. Apparent volume of distribution should be in units of litres (l). I guess you mean โ€˜apparent volume of distribution per kg of body massโ€ (= 3 l/kg). Is that what you mean?

3. You said โ€œclearance is 0.5 (Wh x kg)โ€ but that makes no sense, Clearance is typically measured in units of l/h. Do you mean โ€œclearance per kg of body massโ€ which has units of l/(h.kg)?

4. You changed โ€œciprofloxacinโ€ to โ€œofloxacinโ€ halfway through the question. They are different antibiotics. Is that a mistake?

5. The question says new volume of distribution increases to 125 litres. Do you means 125 litres or 125 litres/kg?. I.e. is 125 in the same units as the initial value (3 l/kg)?

6. When the volume of distribution increases to 125 litres, does the clearance stay the same or change?

7. What do the letters โ€˜NSโ€™ stand for, in the list of alternative answers?

Iโ€™m guessing your question has been translated from another language (French?) which could be the reason for some of the problems.
 
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