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
Truong-Son N.
Apr 11, 2018

#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 #60%# completed in #"1 hr"#, #0.60[A]_0# was lost from #[A]_0#, giving #[A] = 0.40[A]_0# after #t = "1 hr"#. Thus,

#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))#

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