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

• 0.42mg

Step-by-step explanation:

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

`(dB)/(dt)=-kB^2`

Whose integrated form is:

`(1)/(B)=(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:

`(1)/(B)=(1)/(10mg)+7.6* 10^(-3)mg^(-1)min^(-1)* 300min`

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

`(1)/(B)=2.38mg^-1`

`B=0.42mg`