-3
$\begingroup$

I need to calculate the half-life in hours and I got the decay constant in min, how do I do? Which formula do I use? Is it just ln2/decay constant in minutes?

$\endgroup$
4
  • 2
    $\begingroup$ Don't you think converting from minutes to hours is not "chemistry"? $\endgroup$ – Nisarg Bhavsar Apr 24 at 18:03
  • 1
    $\begingroup$ Ať what age are children taught how many minutes is 1 hour ? $\endgroup$ – Poutnik Apr 24 at 18:07
  • $\begingroup$ Thank you for the kind answers :)))) my question was more about which formula to use. $\endgroup$ – Jermaine Apr 24 at 18:09
  • $\begingroup$ The formula is always the same, just units of halftime and time constant differ. Or, if you insist to use mixed units, you have to involve explicit unit conversion factors. It is similar if you use mixed units for relation of distance, speed and time. $\endgroup$ – Poutnik Apr 24 at 19:03
3
$\begingroup$

If you know the decay constant ($\lambda$), you use:

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

The unit of $\lambda$ is $\pu{s-1}, \pu{min-1}, \pu{h-1}, \pu{d-1},$ or $\pu{a-1}$, while that of $\ln 2$ is unitless. Consequently, $t_{1/2}$ can be calculated as $\pu{s}, \pu{min}, \pu{h}, \pu{d},$ or $\pu{a}$, respectively. Since you have $\lambda$ in $\pu{min-1}$, what you have to do is converted it to $\pu{h-1}$ and calculate $t_{1/2}$ in $\pu{h}$ or calculate $t_{1/2}$ in $\pu{min}$ and converted the answer to $\pu{h}$.

For example, suppose your $\lambda = \pu{0.005 min-1}$.

$$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{\ln 2}{\pu{0.005 min-1}} = \frac{0.693}{\pu{0.005 min-1}}= \pu{139 min} = \pu{2.31 h}$$

Or,

$$\lambda = \pu{0.005 min} = \pu{0.005 min-1} \times \frac{\pu{60 min}}{\pu{1 h}} = \pu{0.30 h-1}$$

Thus, $$t_{1/2} = \frac{\ln 2}{\lambda} = \frac{\ln 2}{\pu{0.30 h-1}} = \frac{0.693}{\pu{0.30 h-1}}= \pu{2.31 h}$$

$\endgroup$
0

Not the answer you're looking for? Browse other questions tagged or ask your own question.