# 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"#?

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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")#