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Catalysts and Energy of Activation, Ea

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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?

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.

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.

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) = kcatalyst / 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

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

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