Question

Provide me the second order equation of chemical kinetics?

Answer 1

Well, it depends on what about it you want... you could just read your book...

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A second-order rate law for a one-reactant process is given by

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

for the reaction

`aA -> bB`

where:

• `r(t)` is the rate of reaction in molarity per unit time.
• `a` is the stoichiometric coefficient of `A`.
• `[A]` is the molar concentration of `A`.
• `(d[A])/(dt)` is the rate of change in concentration of `A` over time. It's negative for reactant `A` as it disappears during the course of the reaction.

A useful equation to derive is the second-order integrated rate law. From the starting rate law, separation of variables gives:

`akdt = -1/([A]^2)d[A]`

Integration of the left side from time zero to time `t` and the right side from initial concentration `[A]_0` to current concentration `[A]` gives:

`ak int_(0)^(t) dt = int_([A]_0)^([A])-1/([A]^2)d[A]`

The integral of `-1/x^2` gives `1/x`, so:

`akt = 1/([A]) - 1/([A]_0)`

Thus, the second-order integrated rate law is:

`bb barul|stackrel(" ")(" "1/([A]) = akt + 1/([A]_0)" ")|`

Typically for simplicity we take `a = 1`, so the version you'll see in most textbooks is:

`color(blue)(bb barul|stackrel(" ")(" "1/([A]) = kt + 1/([A]_0)" ")|)`

Lastly, for the second-order half-life of such a reactant `A` is found when `[A] = 1/2[A]_0`. As a result,

`2/([A]_0) = kt_("1/2") + 1/([A]_0)`

Solving for the half-life,

`1/([A]_0) = kt_("1/2")`

`=> color(blue)(bb barul|stackrel(" ")(" "t_("1/2") = 1/(k[A]_0)" ")|)`

Of course, there exist other reaction orders, for which the equations WILL differ. The following assumes the coefficient `a = 1`.

ZERO ORDER

`[A] = -kt + [A]_0`

`t_"1/2" = ([A]_0)/(2k)`

FIRST ORDER

`ln[A] = -kt + ln[A]_0`

`t_"1/2" = (ln2)/k`

SECOND ORDER

`1/([A]) = kt + 1/([A]_0)`

`t_"1/2" = 1/(k[A]_0)`

THIRD ORDER

You usually don't have to use this, but here it is.

`1/(2[A]^2) = kt + 1/(2[A]_0^2`

`t_"1/2" = 3/(2k[A]_0^2`