Question

A compound decomposes with a half-life of 8.0 s and the half-life is independent of the concentration. How long does it take for the concentration to decrease to one-ninth of its initial value

Answer 1

Answer:

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.

Explanation:

The decomposition of the compound has an exponential behavior and process can be represented by this linear first-order differential equation:

$\frac{dc}{dt} = -\frac{1}{\tau}\cdot c(t)$

Where:

$\tau$ - Time constant, measured in seconds.

$c(t)$ - Concentration of the compound as a function of time.

The solution of the differential equation is:

$c(t) = c_{o} \cdot e^{-\frac{t}{\tau} }$

Where $c_{o}$ is the initial concentration of the compound.

The time is now cleared in the result obtained previously:

$\ln \frac{c(t)}{c_{o}} = -\frac{t}{\tau}$

$t = -\tau \cdot \ln \frac{c(t)}{c_{o}}$

Time constant as a function of half-life is:

$\tau = \frac{t_{1/2}}{\ln 2}$

Where $t_{1/2}$ is the half-life of the composite decomposition, measured in seconds.

If $t_{1/2} = 8\,s$, then:

$\tau = \frac{8\,s}{\ln 2}$

$\tau \approx 11.542\,s$

And lastly, given that $\frac{c(t)}{c_{o}} = \frac{1}{9}$ and $\tau \approx 11.542\,s$, the time taken for the concentration to decrease to one-ninth of its initial value is:

$t = -(11.542\,s)\cdot \ln\frac{1}{9}$

$t \approx 25.360\,s$

The concentration takes 25.360 seconds to decrease to one-ninth of its initial value.