Respuesta :
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
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
The concentration of drug after 4.5 hoursis [tex]\boxed{{\text{0}}{\text{.25 mg/100 mL}}}[/tex].
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 [tex]{t_{{\text{1/2}}}}[/tex].
The relation between rate constant and half-life period for first-order reaction is as follows:
[tex]k = \dfrac{{0.693}}{{{t_{{\text{1/2}}}}}}[/tex] …… (1)
Here,
[tex]{t_{{\text{1/2}}}}[/tex] is half-life period.
k is rate constant.
Substitute 90 min for [tex]{t_{{\text{1/2}}}}[/tex] in equation (1).
[tex]\begin{aligned}k &= \frac{{0.693}}{{90{\text{ min}}}} \\&= 7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}} \\\end{aligned}[/tex]
Radioactive decay formula is as follows:
[tex]{\text{A}} = {{\text{A}}_0}{e^{ - kt}}[/tex] …… (2)
Here
A is the concentration of sample after time t.
t is the time taken.
[tex]{{\text{A}}_0}[/tex] 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,
[tex]1{\text{ hr}} = 60\min[/tex]
Therefore time taken can be calculated as follows:
[tex]\begin{aligned}t&= \left( {4.5{\text{ hr}}} \right)\left( {\frac{{60{\text{ min}}}}{{1{\text{ hr}}}}} \right) \\&= 270{\text{ min}} \\\end{aligned}[/tex]
Substitute 2 mg/100 mL for [tex]{{\text{A}}_0}[/tex] , [tex]7.7 \times {10^{ - 3}}{\text{ mi}}{{\text{n}}^{ - 1}}[/tex] for k and 270 min for t in equation (2).
[tex]\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}[/tex]
Learn more:
- What nuclide will be produced in the given reaction? https://brainly.com/question/3433940
- Calculate the nuclear binding energy: https://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.
