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

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

Step-by-step 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* \text{2 h}\\\text{150 ng/mL} & = & 2k \text{ h}\\k & = & \frac{\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$^(-1)$}* \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 $\large \boxed{\textbf{0 ng/mL}}$}`