For an uncatalyzed reaction, the activation energy was #"76 kJ/mol"#. Upon being catalyzed, the activation energy was lowered by #"20 kJ/mol"#. How much faster is the reaction?
1 Answer
If you want any real answer, you have to supply your temperature...
You can find the ratio of catalyzed to uncatalyzed rate constants.
The Arrhenius equation relates rate constants
#k = Ae^(-E_a//RT)# where
#R# is the universal gas constant and#T# is the temperature in#"K"# .
For two reactions with different activation energies, we then have (assuming the frequency factor stays the same):
#k_(cat)/k = (e^(-E_a,cat//RT))/(e^(-E_a//RT))#
#= e^((-E_a,cat + E_a)//RT)#
#= e^((E_a-E_a,cat)/(RT))#
Since the ratio of the rate constants is the ratio of the rates, the reaction is this much faster when catalyzed:
#color(blue)(r_(cat)/(r) = k_(cat)/k) = e^(("76 kJ/mol" - "56 kJ/mol")/("0.008314472 kJ/mol"cdot"K" cdot T)#
#= color(blue)(e^(2405.44//T))#
If you want a real answer, you'll have to supply a temperature...
