4.4: Catalysts and Energy of Activation
- Page ID
- 35692
- 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.
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 \(\ce{H2O2}\),
\[\ce{2 H2O2 \rightarrow 2 H2O + O2} \nonumber\]
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.
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.
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
- 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. - 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?
- 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
- 20647 s = 5.74 hours
- 2.7E8 s = 3215 days or close to 10 years
Discussion -
Calculate the time required to collect when no catalyst is used. - 7.51 kJ/mol