1/T in 1/K In K 0.0035 5.65 0.0034 5.68 0.00337 5.69 0.00331 5.71 0.00321 5.74 y= 312.45x XXXXXXXXXX 11 A.) Calculate the Activation Energy (E a ) for the reaction. E a should be in kJ/mol. This means...

y= 312.45x + 6.7432

11 A.) Calculate the Activation Energy (E_a
) for the reaction.  E_a
should be in kJ/mol.  This means you should use R in kJ/mol as well (8.314 x 10^-3
kJ/mol). Show all work.

B.) Calculate the preexponential factor (A) from the trendline equation.  Show all your work.

C.) As studied in class, the exponential factor in the Arrhenius Equation, , equals
the fraction of collisions that have the needed activation energy.

D.) Use the E_a
you obtained in this experiment (#11a) to calculate this fraction for the
highest temperature
run – (approx.) 38.0^o
Remember that “e to the x” is the inverse of the ln function on some calculators.  Also remember that, this being a fraction,
you should obtain a number between zero and 1, in this case quite a bit closer to zero!
 Since E_a
is in kJ/mol, use the Gas Constant, R, as 8.314 x 10^-3
kJ/mol K.  The temperature should be in K.

e.) Calculate this fraction at the
lowest temperature
of the experiments, approx. 11.0^o

f.) Does the ratio of these fractions reflect the ratio of the k’s that were found? In other words, does the dependence of the fraction of collisions with the needed E_a
on temperature completely explain the change in rate constant with temperature?

g.) Find the ratio of the rate of reaction with the catalyst to the rate of reaction without the catalyst.

Comment on what this ratio tells you about the effect of a catalyst on a reaction rate.

h.) Use your result for
E_a

and the fact that

H
for this reaction is -350. kJ to sketch a
reaction profile
diagram for this system. (This is a diagram similar to those given on page 1 of this handout.) Show the effect of the catalyst on the reaction profile.