4.4: Catalysts and Energy of Activation
- Page ID
- 35692
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)- 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