# What is the half-life when the initial concentration is 0.51 M?

Jul 11, 2016

The identity of the substance is not known (so you can't look up the half-life itself), nor is the total amount of time passed known, so this question can only be answered in general, with hypothetical numbers.

The half-life equation is:

$\setminus m a t h b f \left(\left[A\right] = \frac{1}{{2}^{t \text{/"t_"1/2}}} {\left[A\right]}_{0}\right)$

Even though a first-order half-life (general-chemistry half-life decay) technically doesn't depend on the initial concentration of the substance, we would still have to know either:

• The final concentration OR the fraction of the substance leftover,
• AND the total time passed,

since the half-life equation contains the initial and final concentrations, as well as the total time passed and the half-life itself. The rate itself doesn't change due to concentration, though.

From here, you wouldn't need to know the identity of the substance, but we'd need to make up some numbers.

If we wanted, one option is to know that the final concentration is, say, $\text{0.1275 M}$.

Then we know what fraction of the substance was left after time $t$, and from that we can determine the half-life, ${t}_{\text{1/2}}$.

Again, we have no idea how much total time passed, but let's pretend $\text{1 hour}$ passed for kicks. :) That's the time $t$, and we are solving for ${t}_{\text{1/2}}$.

In this case, we have:

$\frac{1}{\left[A\right]} = {2}^{t \text{/"t_"1/2}} / \left({\left[A\right]}_{0}\right)$

$\frac{{\left[A\right]}_{0}}{\left[A\right]} = {2}^{t \text{/"t_"1/2}}$

Now to take the base 2 logarithm and get the exponent out.

${\log}_{2} \left(\frac{{\left[A\right]}_{0}}{\left[A\right]}\right) = \frac{t}{{t}_{\text{1/2}}}$

$\frac{1}{{\log}_{2} \left(\frac{{\left[A\right]}_{0}}{\left[A\right]}\right)} = {t}_{\text{1/2}} / t$

color(blue)(t_"1/2" = t/(log_2(([A]_0)/([A]))

Now simply plug in your numbers:

• $t = \text{1 hour}$
• ${\left[A\right]}_{0} = \text{0.51 M}$
• $\left[A\right] = \text{0.1275 M}$

Note that you could rewrite ${\log}_{2} \left(a\right)$ as $\frac{{\log}_{10} \left(a\right)}{{\log}_{10} \left(2\right)}$, where ${\log}_{10}$ is the log button on your calculator.