A second order reaction is 60% completed in 1 hour. What is the value of the rate constant, if the initial reactant concentration is #[A]_0# #"mol"cdot"L"^(-1)#?
1 Answer
#k = 3/(2[A]_0)# #"M"^(-1) cdot "hr"^(-1)#
A second order one-reactant reaction
#A -> B#
is given by the rate law:
#r(t) = k[A]^2 = -(d[A])/(dt)# where
#k# is the rate constant in#"M"^(-1)cdot"time"^(-1)# ,#[A]# is the molar concentation of reactant#A# , and#(d[A])/(dt)# is the change in concentation of#A# over a short time interval.
By separation of variables,
#kdt = -1/([A]^2)d[A]#
Integration gives:
#kint_(0)^(t)dt = int_([A]_0)^([A]) -1/([A]^2)d[A]#
#kt = 1/([A]) - 1/([A]_0)#
From here we COULD show that we obtained the integrated rate law for second order reactions:
#barul|stackrel(" ")(" "1/([A]) = kt + 1/([A]_0)" ")|#
If the reaction is
#kcdot("1 hr") = 1/(0.40[A]_0) - 1/([A]_0)#
#= 5/(2[A]_0) - 2/(2[A]_0)#
#= 3/(2[A]_0)#
As a result, the rate constant is given by:
#color(blue)(k = 3/(2[A]_0)color(white)(.)"M"^(-1)cdot"hr"^(-1))#
