Question

If an illegal drug is removed from the body according to a zero-order process and the concentration drops from 300 ng/mL to 150 ng/mL in two hours, what will the concentration be after two more hours?
a. 50 ng/mL
b. 0 ng/mL
c. No right choice.

Answer 1

Answer:

$\large \boxed{\text{b. 0 ng/mL}}$

Explanation:

The integrated rate law for a zero-order reaction is

A₀ - A = kt

1. Calculate the rate constant

$\begin{array}{rcl}\text{300 ng/mL - 150 ng/mL} & = & k\times \text{2 h}\\\text{150 ng/mL} & = & 2k \text{ h}\\k & = & \dfrac{\text{150 ng/mL} }{ \text{2 h} }\\\\& = & \textbf{75 ng \cdot mL ^{-1}\cdot h ^{-1} }\\\end{array}$

2. Calculate the new concentration

$\begin{array}{rcl}\text{A _{0} - A} & = & kt\\\text{300 ng \cdot mL ^{-1} - A} & = & \text{75 ng \cdot mL ^{-1}\cdot h ^{-1} }\times \text{4 h}\\\text{300 ng \cdot mL ^{-1} - A} & = & \text{300 ng \cdot mL ^{-1} }\\\text{A} & = &\text{300 ng/mL - 300 ng/mL}\\& = & \mathbf{0}\\\end{array}\\\text{The concentration after 4 h will be \large \boxed{\textbf{0 ng/mL}} }$