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Chemistry LibreTexts

Ea and Catalysts

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 iodide ions catalyze the decomposition of hydrogen peroxide \(\ce{H2O2}\),

\(\ce{2 H2O2 \rightarrow 2 H2O + O2}\)

Thus, by dissolving solid \(\ce{KI}\) in a solution of hydrogen peroxide, the formation of oxygen bubbles is 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 exemplified by the following examples.

Example 1

At 300 K, the activation energy, Ea, for the decomposition of \(\ce{H2O2}\) has been measured to be 75.3 kJ/mol. In the presence of a definite concentration of iodide ions, \(\ce{I-}\), 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

\(rate = k\ce{f([H2O2])}\)
\(rate\,' = k\,'\ce{f([H2O2])}\)

where \(\ce{f([H2O2])}\) is a function of the concentration of the reactant. Note that rate and rate' are rates of the reactions in the absence of and in the presence of iodide ions.

\(\begin{align}
k &= \ce{A e}^{-75300/(8.3145\times300)}\\ \\
k\,' &= \ce{A e}^{-56500/(8.3145\times300)}\\ \\
\dfrac{k}{k\,'} &= \dfrac{\ce e^{-75300/(8.3145\times300)}}{\ce e^{-56500/(8.3145\times300)}}\\ \\
    &= \ce e^{-(75300 - 56500)/(8.3145\times300)}\\ \\
    &= \ce e^{-18800/(8.3145\times300)}\\ \\
    &= 1877
\end{align}\)

Thus, rate' = 1877 times 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 particular concentration, the rate increases ten fold. 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

\(\begin{align}
\dfrac{\ce e^{-E\ce a/(8.3145\times300)}}{\ce e^{-19000/(8.3145\times300)}} &= \dfrac{k_{\ce{catalyst}}}{k}\\ \\
    &= 10\\ \\
\ce e^{-E\ce a/(8.3145\times300)} &= 10 \times \ce e^{-19000/(8.3145\times300)}\\ \\
    &= 0.00492\\ \\
\dfrac{-E\ce a}{(8.3145\times300)} &= \ln0.00492\\ \\
-E\ce a &= (8.3145\times300)\times(-5.315)\\ \\
E\ce a &= \mathrm{13257\: J/mol}\\ \\
   &= \mathrm{13.3\: kJ/mol}
\end{align}\)

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

Questions

  1. At 300 K, the activation energy, Ea for the decomposition of \(\ce{H2O2}\) has been measured to be 75.3 kJ/mol. In the presence of iodide ion, \(\ce{I-}\), 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 seconds are required to collect 0.1 mL if iodide is absent?

    Skill -
    Calculate the time required to accomplish a certain task when the rate is different.

  2. At 298 K, the activation energy, Ea for the decomposition of \(\ce{H2O2}\) has been measured to be 75.3 kJ/mol. In the presence of iodide ion, \(\ce{I-}\), 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 seconds are required to collect 100 mL if iodide is used as the catalyst?

  3. The activation energy of a reaction is 19.0 kJ/mol. When a catalyst is used at 300 K, the rate increases 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.

Solutions

  1. 20647 s = 5.74 hours
  2. 2.7E8 s = 3215 days or close to 10 years

    Discussion -
    Calculate the time required to collect when no catalyst is used.

  3. 7.51 kJ/mol

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