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A different beta blocker is eliminated by a second-order process with a rate constant of 7.6x10-3 min-1. A patient is given 10. mg of the drug. What mass of the drug (in mg) remains in the body 5.0 hours after administration?

Respuesta :

Edufirst

Answer:

  • 0.42mg

Explanation:

A second order process for the elimination (disappearance) of a substance, B, follows the law:

        [tex]\dfrac{dB}{dt}=-kB^2[/tex]

Whose integrated form is:

        [tex]\dfrac{1}{B}=\dfrac{1}{B_0}+kt[/tex]

You are given:

  • B₀ = 10mg
  • k = 7.6 × 10⁻³ min⁻¹
  • t = 5.0hours = 300 min

Then, just subsititute and compute:

     

        [tex]\dfrac{1}{B}=\dfrac{1}{10mg}+7.6\times 10^{-3}mg^{-1}min^{-1}\times 300min[/tex]

(notice that the units of k should be mg⁻¹·min⁻¹)

        [tex]\dfrac{1}{B}=2.38mg^-1[/tex]

        [tex]B=0.42mg[/tex]

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