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")#