# 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?

##### 1 Answer

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**

#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

#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/(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")#