The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half life of 46 minutes. Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 1.1 ug/mL. What will the concentration be 92 minutes later? Round your answer to 2 significant digits.
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half life of 46 minutes. Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 1.1 ug/mL. What will the concentration be 92 minutes later? Round your answer to 2 significant digits.
Chemistry
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ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
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![**Understanding Drug Elimination Kinetics: A Practical Example**
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half-life of 46 minutes.
Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 1.1 µg/mL. What will the concentration be 92 minutes later?
**Solution:**
Given data:
- Initial concentration (C₀) = 1.1 µg/mL
- Half-life (t₁/₂) = 46 minutes
- Time elapsed (t) = 92 minutes
We apply the first-order kinetics elimination formula:
\[ C = C₀ \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
Let's calculate the concentration (C) of the drug after 92 minutes:
\[ C = 1.1 \times \left(\frac{1}{2}\right)^{\frac{92}{46}} \]
\[ C = 1.1 \times \left(\frac{1}{2}\right)^{2} \]
\[ C = 1.1 \times \frac{1}{4} \]
\[ C = 1.1 \times 0.25 \]
\[ C = 0.275 \]
Rounding the answer to 2 significant digits:
\[ C \approx 0.28 \: \text{µg/mL} \]
Thus, the concentration of the drug in the bloodstream 92 minutes after injection will be approximately 0.28 µg/mL.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fd77309c6-927a-49f8-947f-1de5f0de81d4%2Fef722b35-4d65-4379-92a3-d53f6bf584d9%2F0lu5pet_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Drug Elimination Kinetics: A Practical Example**
The rate at which a certain drug is eliminated by the body follows first-order kinetics, with a half-life of 46 minutes.
Suppose in a particular patient the concentration of this drug in the bloodstream immediately after injection is 1.1 µg/mL. What will the concentration be 92 minutes later?
**Solution:**
Given data:
- Initial concentration (C₀) = 1.1 µg/mL
- Half-life (t₁/₂) = 46 minutes
- Time elapsed (t) = 92 minutes
We apply the first-order kinetics elimination formula:
\[ C = C₀ \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}} \]
Let's calculate the concentration (C) of the drug after 92 minutes:
\[ C = 1.1 \times \left(\frac{1}{2}\right)^{\frac{92}{46}} \]
\[ C = 1.1 \times \left(\frac{1}{2}\right)^{2} \]
\[ C = 1.1 \times \frac{1}{4} \]
\[ C = 1.1 \times 0.25 \]
\[ C = 0.275 \]
Rounding the answer to 2 significant digits:
\[ C \approx 0.28 \: \text{µg/mL} \]
Thus, the concentration of the drug in the bloodstream 92 minutes after injection will be approximately 0.28 µg/mL.
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