# A certain reaction has an activation energy of 70.0 kJ/mol and a frequency factor of #A = 1.40 xx 10^12 "M"^(-1)"s"^(-1)#. What is the rate constant, #k#, of this reaction at #22.0 °"C"#?

##### 1 Answer

Jul 26, 2017

Given

#k = Ae^(-E_a//RT)# ,where the variables above are all known, except that

#T# is in#"K"# . This is a second order reaction (how do we know?).

There is no further rearrangement necessary; we already have the form of the equation we need.

#color(blue)(k) = (1.40 xx 10^(12) "M"^(-1)cdot"s"^(-1))e^(-"70.0 kJ/mol"//"0.008314472 kJ/mol"cdot"K" // "295.15 K")#

#= color(blue)(5.73 xx 10^(-1) "M"^(-1)cdot"s"^(-1))#

And we should recognize that exponentials have no units, i.e.