Question

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?

Answer 1

`"1.67 seconds"`.

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We have

`AB(g) -> A(g) + B(g)`,

a second order decomposition or half-life with rate law

`r(t) = k[AB]^2`

and rate constant `k = "0.80 M"^(-1)cdot"s"^(-1)`. The rate law can be equated with the instantaneous rate of disappearance of `AB`:

`r(t) = k[AB]^2 = -1/1 (d[AB])/(dt)`

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

`kdt = -1/([AB]^2)d[AB]`

`int_(0)^(t) kdt = int_([AB]_0)^([AB])-1/([AB]^2)d[AB]`

The integral of `-1/x^2` is `1/x`, so...

`kt = 1/([AB]) - 1/([AB]_0)`

Thus, the second-order integrated rate law is

`color(green)(1/([AB]) = 1/([AB]_0) + kt)`,

the only one with a positive slope.

Since you know that its initial concentration is `"1.50 M"` and that `[AB] = 1/3[AB]_0`, we get for its "1/3rd life":

`1/(1/3[AB]_0) = 1/([AB]_0) + kt_"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"`

`=` `ulcolor(blue)("1.67 s")`