The half-life of the first order reaction, A → products, is 33.8 s. What was the original concentration, [A]O in mol/L, if after 2.5 minutes, the concentration of A is 0.0796 mol/L? Round your answer to three significant figures.

I think the formular is 1/(A)t = kt + 1/(A)0

So I tried 1/33.8s = (0.0796)

so it = 1/0.0796.

I know this is not correct, but I think I'm using the right formula. I looked back at my notes and nothing is really making any sense, but this formula.


For a first order reaction $\ce{A ->[k]P}$, the rate constant is given by

$$k[A] = - \frac{d[A]}{dt} $$

So, at any time $t$, the concentation $[A]_t$ is

$$[A]_t = [A]_0 \cdot e^{-kt}$$

At the half-life time $t_{1/2}$, $[A]_t = \frac{1}{2}[A]_0$, isn't it?

Would you agree that we can calculate $k$ using $k = \frac{ln 2}{t_{1/2}}\ $?

Knowing $k$ and $[A]_t$, calculation of $[A]_0$ doesn't seem impossible now ;).


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