Calculate the new activation energy at #37^@ "C"# if the catalyzed reaction is #200# times as fast, and the uncatalyzed activation energy was #"100 kJ/mol"#?
The equation is
#ln(k_("cat")/(k_"uncat")) = -1/(RT) xx [E_(a,"cat") - E_(a,"uncat")]#
The equation is
1 Answer
#"86.34 kJ/mol"# .
Remember to convert
I would start from the original form of the Arrhenius equation, just to verify first that the given equation is correct.
You know that you are at the same temperature twice, but you have two different rate constants and two different activation energies.
So:
#k_(uncat) = Ae^(-E_(a,uncat)"/"RT)#
#k_(cat) = Ae^(-E_(a,cat)"/"RT)#
Assuming the pre-exponential factor is the same and only the mechanism changes,
#(k_(cat))/(k_(uncat)) = (cancel(A)e^(-E_(a,cat)"/"RT))/(cancel(A)e^(-E_(a,uncat)"/"RT))#
#= e^(-E_(a,cat)"/"RT + E_(a,uncat)"/"RT)#
Take the
#bb(ln((k_(cat))/(k_(uncat)))) = -E_(a,cat)/(RT) + E_(a,uncat)/(RT)#
#= bb(-1/(RT)xx[E_(a,cat) - E_(a,uncat)])#
Okay, your equation is correct. Now, if you know that the reaction rate increased by a factor of
#A -> B#
in two scenarios is
#r_(uncat)(t) = k_(uncat)[A]# ,
#r_(cat)(t) = k_(cat)[A]# ,where
#(r_(cat))/(r_(uncat)) = 200# .
Therefore, with the same reactants and products, divide these rates to get:
#(r_(cat))/(r_(uncat)) = (k_(cat))/(k_(uncat))#
So, the ratio of the rates is the ratio of the rate constants at the same temperature.
That means you can plug in the ratio of the two rates into the Arrhenius equation:
#=> ln(200) = -1/(8.314472 cancel"J""/mol"cdotcancel"K" xx "1 kJ"/(1000 cancel"J") xx (37 + 273.15 cancel"K"))#
#xx [E_(a,cat) - "100 kJ"/"mol"]#
Simplify the expression by evaluating some of it:
#ln(200) = -"0.3878 mol/kJ" xx (E_(a,cat) - "100 kJ"/"mol")#
Solve for the catalyzed activation energy:
#color(blue)(E_(a,cat)) = ln(200)/(-"0.3878 mol/kJ") + "100 kJ/mol"#
#=# #color(blue)("86.34 kJ/mol")#
