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Catalysts and activation energy

Choose the presence or absence of a catalyst and compare the barriers and speeds. Without catalyst, Ea=50, ΔH=−20 kJ/mol. The start and end energies of the two paths are the same.

    C15 · Catalysts and activation energyReady
    Choose the presence or absence of a catalyst and compare the barriers and speeds. Without catalyst, Ea=50, ΔH=−20 kJ/mol. The start and end energies of the two paths are the same. You can also experiment with the numbers and controls below.

    Chemistry · Reaction rate · Equilibrium / Principles and exploration

    Catalysts and activation energy — principles and investigation

    Catalysts provide a pathway with a lower activation energy barrier that the reactants must overcome. Compare the reaction rates and fraction of particles that can cross the barrier by varying the catalytic path and temperature. The speed at which a reaction reaches equilibrium and the composition at equilibrium are two different issues.

    How does it work?

    The fraction of particles above the barrier increases, making the reaction faster. K does not change at the same temperature.

    Key relationship

    k catalyst/k non-catalyst = exp[(50−Ea)×1000/RT]

    Try this investigation

    1. Choose the presence or absence of a catalyst and compare the barriers and speeds. Without catalyst, Ea=50, ΔH=−20 kJ/mol. The start and end energies of the two paths are the same.
    2. Record the catalytic pathway without catalyst and Ea 40 kJ/mol at 25 °C.
    3. Change Ea or temperature one by one in the catalytic path. The equilibrium constant does not change with the presence or absence of a catalyst.

    Interpreting your results

    Which quantities does a catalyst leave unchanged at the same temperature?

    Catalysts lower the barriers to forward and reverse reactions. The reactant and product energies and ΔH are the same, so K is the same.

    Assumptions and limitations

    It is a hypothetical energy path and ΔH=−20 kJ/mol. Assume the same frequency factor A=10⁷ s⁻¹. We simplified the Boltzmann energy distribution into a single energy degree of freedom model where p(E)=exp(−E/RT)/(RT). The actual three-dimensional translational energy distribution, catalyst surface, adsorption, and multi-step reactions are omitted. The rate curve only covers the initial forward reaction A consumption and does not calculate the equilibrium composition.