Problem 12-103

Activation energy determination

To simplify:

• A reaction has a specific rate at room temperature (20 °C).
• Adding a catalyst accelerates the reaction (catalyst = decrease in activation energy: E_a).

The question is: To what temperature must the uncatalized reaction be heated to achieve the same rate as the catalyzed reaction?

Assumptions:

• The concentration of the reactants remains the same.
• The collision frequency factor (A in the Arrhenius equation) remains constant.

This problem is solved using the Arrhenius equation: k = A exp ^−E_a/RT

Solution 1:
You can combine two Arrhenius equations (two different conditions) and solve them.

If both k are identical then:   A exp ^{−(70000 J/mol)/RT} = A exp ^{−(43000 J/mol)/R(293 K)}

by canceling the constant "A", running "ln" on both sides and removing "R" (constant), the equation simplifies to:

70000 J/mol	=	42000 J/mol
T		293 K

Solution 2:
Another way to find the answer is to assign a fictitious number to the value "A," for example, 1 (constants),
and calculate k at 20 °C with the catalyst (E_a = 42.0 kJ/mol).

Then repeat the process with the same k value (and A = 1) to find the temperature of the uncatalyzed reaction with E_a = 70 kJ/mol.

Make sure to use kelvin for your temperature and joules (not kJ) for your E_a.
Obviously, your temperature should be higher...