Question

Many drugs decompose in blood by a first-order process. two tablets of aspirin supply 0.60 g of the active compound. after 30 min, this compound reaches a maximum concentration of 2 mg/100 ml of blood. if the half-life for its breakdown is 90 min, what is its concentration (in mg/100 ml) 4.0 h after it reaches its maximum concentration?

Answer 1

The concentration of drug after 4.5 hoursis $\boxed{{\text{0}}{\text{.25 mg/100 mL}}}$.

Further Explanation:

Radioactive decayis responsible to stabilizeunstable atomic nucleusaccompanied by the release of energy.

Half-life is the duration after which half of the original sample has been decayed and half is left behind. It is represented by ${t_{{\text{1/2}}}}$.

The relation between rate constant and half-life period for first-order reaction is as follows:

$k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}$                                                                                                    …… (1)

Here,

${t_{{\text{1/2}}}}$ is half-life period.

k is rate constant.

Substitute 90 min for ${t_{{\text{1/2}}}}$ in equation (1).

$\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}$  

Radioactive decay formula is as follows:

${\text{A}} = {{\text{A}}_0}{e^{ - kt}}$                                        …… (2)

Here

A is the concentration of sample after time t.

t is the time taken.

${{\text{A}}_0}$ is the initial concentration of sample.

k is the rate constant.

The time for the process is to be converted into min. The conversion factor for this is,

$1{\text{ hr}} = 60\min$  

Therefore time taken can be calculated as follows:

$\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}$  

Substitute 2 mg/100 mL for ${{\text{A}}_0}$ , $7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}$ for k and 270 min for t in equation (2).

$\begin{aligned}{\text{A}} &= \left( {\frac{{2{\text{ mg}}}}{{100{\text{ mL}}}}} \right){e^{ - \left( {7.7 \times {{10}^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}} \right)\left( {270{\text{ min}}} \right)}} \\&= \frac{{0.25{\text{ mg}}}}{{100{\text{ mL}}}} \\\end{aligned}$  

Learn more:

1. What nuclide will be produced in the given reaction? brainly.com/question/3433940
2. Calculate the nuclear binding energy: brainly.com/question/5822604

Answer details:

Grade: Senior School

Subject: Chemistry

Chapter: Radioactivity

Keywords: half-life, 0.25 mg/100 mL, 2 mg/100 mL, A, k, t, rate constant, 4.5 h, 270 min, radioactive decay formula.

Answer 2

The half-life of a first-order reaction is determined as follows: 

t½=ln2/k

From the equation, we can calculate the first-order rate constant:

k = (ln(2)) / t½ = 0.693 / 90 = 7.7 × 10⁻³

When we know the value of k we can then calculate concentration with the equation:

A₀ = 2 g/100 mL 

t = 2.5 h = 150min

A = A₀ × e^(-kt) =2 × e^(-7.7 × 10⁻³ × 150) = 0.63 g / 100ml

   = 6.3 × 10⁻⁴ mg / 100ml