For the first order reaction A----> B+C carried out at 27°C if 3.8×10^-16 % of the reactant molecules exist in the activated state then what would be the activation energy of the reaction ?

1 Answer
Truong-Son N.
Mar 8, 2018

I got #1.0 xx 10^2# #"kJ/mol"# to two sig figs...

Consider the limit as #E_a -> 0#.

#lim_(E_a -> 0) Ae^(-E_a//RT) = A = k#

Thus, in this condition, the rate constant is equal to the frequency factor if the activation energy is zero. In that condition, #100%# of the molecules are "activated" and the reaction proceeds infinitely fast, i.e. #A# is the limiting rate constant for having #100%# activated molecules.

Thus, #k/A xx 100%# is the percent of molecules that can do the reaction.

#k/A xx 100% = 3.8 xx 10^(-16) %#

#=> k/A = 3.8 xx 10^(-18) = e^(-E_a//RT)#

This means

#color(blue)(E_a) = -RTln(3.8 xx 10^(-18))#

#= -"8.314472 J/mol"cdot"K" cdot (27+"273.15 K") cdot ln(3.8 xx 10^(-18))#

#=# #"100102 J/mol"#

#=# #color(blue)("100.102 kJ/mol")#

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