A reaction rate doubles when the temperature increases from 0 degrees Celsius to 10 degrees Celsius, maintaining everything else equal. What is the activation energy for this reaction? (R = 8.314 J/mol K)

Answer 1

Hey Walter, for this we use the arrhenius equation:
In(K2/K1)= -Ea/R (1/T2-1/T1)
As it says the reaction rate doubles. so for instance if you take K1= 1 then K2 will equal to 2. Now plug this into In(K2/K1)= -Ea/R (1/T2-1/T1)
Now after we do In(K2/K1)= 0.693
0.693=-Ea/R (1/T2 - 1/T1). Now we have to get -Ea by itself.
0.693 X (8.314)/ (1/T1-1/T2)= -Ea
5.763/ (1/T2 -1/T1) = -Ea
5.763/ (1/(273+10) - 1/(273 + 0)= -Ea
5.763/ [ (3.53x10^-3) - (3.66x10^-3) ] = -Ea
Now just simplify the equation and you should end up with...
5.763/ -1.3x10^-4 = -Ea
-44330.76 = -Ea
44330.76 J/mol = Ea
:) hope that helps.