Provide me the second order equation of chemical kinetics?

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

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#