If the initial concentration of "AB" in AB(g) -> A(g) + B(g) is "1.50 M", and k = "0.80 M"^(-1)cdot"s"^(-1), what amount of time passes until the concentration becomes 1//3 of what it began as in this second-order decomposition?

Aug 11, 2017

$\text{1.67 seconds}$.

We have

$A B \left(g\right) \to A \left(g\right) + B \left(g\right)$,

a second order decomposition or half-life with rate law

$r \left(t\right) = k {\left[A B\right]}^{2}$

and rate constant $k = {\text{0.80 M"^(-1)cdot"s}}^{- 1}$. The rate law can be equated with the instantaneous rate of disappearance of $A B$:

$r \left(t\right) = k {\left[A B\right]}^{2} = - \frac{1}{1} \frac{d \left[A B\right]}{\mathrm{dt}}$

By separation of variables, we can derive the integrated rate law for second order one-reactant processes.

$k \mathrm{dt} = - \frac{1}{{\left[A B\right]}^{2}} d \left[A B\right]$

${\int}_{0}^{t} k \mathrm{dt} = {\int}_{{\left[A B\right]}_{0}}^{\left[A B\right]} - \frac{1}{{\left[A B\right]}^{2}} d \left[A B\right]$

The integral of $- \frac{1}{x} ^ 2$ is $\frac{1}{x}$, so...

$k t = \frac{1}{\left[A B\right]} - \frac{1}{{\left[A B\right]}_{0}}$

Thus, the second-order integrated rate law is

$\textcolor{g r e e n}{\frac{1}{\left[A B\right]} = \frac{1}{{\left[A B\right]}_{0}} + k t}$,

the only one with a positive slope.

Since you know that its initial concentration is $\text{1.50 M}$ and that $\left[A B\right] = \frac{1}{3} {\left[A B\right]}_{0}$, we get for its "1/3rd life":

$\frac{1}{\frac{1}{3} {\left[A B\right]}_{0}} = \frac{1}{{\left[A B\right]}_{0}} + k {t}_{\text{1/3}}$

1/("0.50 M") = 1/("1.50 M") + ("0.80 M"^(-1)cdot"s"^(-1))t_"1/3"

color(blue)(t_"1/3") = (1/("0.50 M") - 1/("1.50 M"))/("0.80 M"^(-1)cdot"s"^(-1)) = (2 - 2/3)/(0.80) " s"

$=$ $\underline{\textcolor{b l u e}{\text{1.67 s}}}$