Question

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?

Answer 1

Answer:

• 0.42mg

Explanation:

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

        $\dfrac{dB}{dt}=-kB^2$

Whose integrated form is:

        $\dfrac{1}{B}=\dfrac{1}{B_0}+kt$

You are given:

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

Then, just subsititute and compute:

     

        $\dfrac{1}{B}=\dfrac{1}{10mg}+7.6\times 10^{-3}mg^{-1}min^{-1}\times 300min$

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

        $\dfrac{1}{B}=2.38mg^-1$

        $B=0.42mg$