How much faster is a reaction at #"313 K"# compared to at #"303 K"# if the activation energy is #"58 kJ/mol"#?
1 Answer
Feb 8, 2018
Should be about
The reaction follows Arrhenius kinetics:
#k = Ae^(-E_a//RT)# where
#k# is the rate constant at temperature#T# for the reaction with activation energy#E_a# ,#A# is the frequency factor, and#R# is the universal gas constant.
At two temperatures, the ratio of the rate constants becomes:
#color(blue)(k_2/k_1) = cancel(A/A)e^(-E_a//RT_2)/e^(-E_a//RT_1)#
#= "exp"(-E_a/R [1/T_2 - 1/T_1])#
#= "exp"(-"58 kJ/mol"/("0.008314472 kJ/mol"cdot"K") [1/"313 K" - 1/"303 K"])#
#= color(blue)(2.087)#
i.e.
#k_2 ~~ 2.087k_1# ,
and the reaction tends to be about twice as fast.
