Catalysts and Energy of Activation, Ea

Skills to develop

Redefine catalyst in terms of energy of activation
Calculate Ea when a catalyst is used from rate of reaction
If Ea is known, calculate the rate of reaction

Catalysts and Energy of Activation, Ea

For reactions that follow the Arrhenius rate law a catalyst can be re-defined as a substance that lowers the energy of activation Ea by providing a pathway (reaction mechanism), or transition state.

For example, it is well known that the iodide ions catalyze the decomposition of hydrogen peroxide H₂O₂,

2 H₂O₂ -> 2 H₂O + O₂ Thus, by dissolving solid KI in a solution of hydrogen peroxide, the formation of oxygen bubbles are accelerated. Of course, the reaction depends on concentrations of reactants and catalyst, but for a definite (or fixed) concentration, the relative reaction rates can be compared as examplified by the following examples.

Example 1

At 300 K, the activation energy, Ea for the decomposition of H₂O₂ has been measured to be 75.3 kJ/mol. In the presence of a definite concentration of iodide ions, I^-, the activation energy Ea has been estimated to be 56.5 kJ/mol. How much faster is the decomposition when the same concentration of iodide is present in the reaction?

Solution
If k and k_- represent the reaction rate constants of decomposition in the absence of iodide and in the presence of iodide ion (at a definite concentration) respectively, then

rate = kf([H₂O₂])
rate_- = k_-f([H₂O₂])
where f([H₂O₂]) is a function of the concentration of the reactant. Note that rate and rate_- are rates of the reactions in the presence of and in the presence of iodide ions. k = A exp(-75300/(8.3145*300)
k_- = A exp(-56500/(8.3145*300)
k / k_- = exp(-75300/(8.3145*300)) / exp(-56500/(8.3145*300))
= exp(-(75300 - 56500)/(8.3145*300))
= exp(-(18800)/(8.3145*300))
= 1877 Thus, rate_- = 1877 time rate, because the rate constant has increased 1877 times.

Discussion
Note that the presence of a catalyst allows the reaction to proceed at the same low temperature, but achieve a much faster rate of reaction.

Example 2

The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at a particcular concentration, the rate increase ten folds. Calculate the energy of activation when the catalyst is present.

Solution
This example differs from example 1 in that we know how much faster the reaction is and want to evaluate the activation energy. Let the activation energy in the presence of the catalyst be Ea, then

exp(-Ea/(8.3145*300)) / exp(-19000/(8.3145*300) = k_catalyst / k
= 10
exp(-Ea/(8.3145*300)) = 10 * exp(-19000/(8.3145*300)
= 0.00492
-Ea/(8.3145*300) = ln0.00492
-Ea = (8.3145*300)*(-5.315)
Ea = 13257 J/mol
= 13.3 kJ/mol

Discussion
The details of the calculation are given to illustrate the mathematic skills involved. Check out the units throughout the calculation please.

Confidence Building Questions

At 300 K, the activation energy, Ea for the decomposition of H₂O₂ has been measured to be 75.3 kJ/mol. In the presence of iodide ion, I^-, the activation energy Ea has been estimated to be 56.5 kJ/mol. If 11 s is required to collect 0.1 mL of oxygen when iodide is present, how many second is required to collect 0.1 mL if iodide is absent?

Skill -
Calculate the time required to accomplish certain task when the rate is different.
At 298 K, the activation energy, Ea for the decomposition of H₂O₂ has been measured to be 75.3 kJ/mol. In the presence of iodide ion, I^-, the activation energy Ea has been estimated to be 56.5 kJ/mol. When catalyzed by the enzyme catalase, the activation energy is 8.4 kJ/mol. If 1 s is required to collect 100 mL of oxygen when catalase is present, how many second is required to collect 100 mL if iodide is used as the catalyst?

Discussion -
Calculate the time required to collect when no catalyst is used.
The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at 300 K, the rate increase one hundred fold. Calculate the energy of activation when the catalyst is present.

Skill -
Calculate energy of activation when the rate of increase is known.