Gen Chem 2Kinetics

Catalysis

Homogeneous, heterogeneous and enzyme catalysts open a new mechanism with a lower barrier.

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A catalyst speeds up a reaction without being used up. With multistep energy profiles we can say exactly how: it provides a different mechanism, whose highest barrier is lower.

Key idea

A catalyst opens a new pathway with a lower activation energy. It speeds up the forward and reverse reactions by the same factor, so it does not change $K$, $\Delta H$ or the amount of product at equilibrium; it only gets there sooner. Homogeneous catalysts are in the same phase as the reactants, heterogeneous ones are in a different phase (usually a solid surface), and enzymes are biological catalysts.

Why it works

Forward and reverse reactions pass over the same peak, in opposite directions. Lower the peak and both directions speed up equally. Because $K = k_\text{f}/k_\text{r}$, the ratio stays the same. Since the rate constant depends on $e^{-E_a/RT}$, even a modest drop in $E_a$ gives a big factor in rate.
Toggle the catalyst. The barrier drops for both directions, but the start and end energies (and so $\Delta H$) stay put.
With the catalyst on, a larger share of the same collisions has enough energy to react.

Worked examples

Worked example: A homogeneous catalyst

Iodide ion speeds up the decomposition of hydrogen peroxide:
Step 1 (slow): $\mathrm{H_2O_2 + I^- \to H_2O + IO^-}$
Step 2 (fast): $\mathrm{H_2O_2 + IO^- \to H_2O + O_2 + I^-}$
Find the overall reaction, the catalyst and the intermediate.
  1. Add and cancel: $\mathrm{2H_2O_2 \to 2H_2O + O_2}$.
  2. $\mathrm{I^-}$ is used in step 1 and remade in step 2: the catalyst. It is in solution with the peroxide, so it is homogeneous.
  3. $\mathrm{IO^-}$ is made and then used: the intermediate.

Answer: $\mathrm{2H_2O_2 \to 2H_2O + O_2}$; catalyst $\mathrm{I^-}$, intermediate $\mathrm{IO^-}$.

Worked example: How much faster?

A catalyst lowers $E_a$ by $20.0\ \text{kJ/mol}$ at $298\ \text{K}$. If $A$ is unchanged, by what factor does $k$ grow?
  1. $$\dfrac{k_\text{cat}}{k} = \dfrac{e^{-(E_a - 20\,000)/RT}}{e^{-E_a/RT}} = e^{20\,000/RT}$$
  2. $\dfrac{20\,000}{(8.314)(298)} = 8.07$, and $e^{8.07} = 3.2 \times 10^3$.

Answer: About 3200 times faster.

  • Heterogeneous: the platinum, palladium and rhodium in a car's catalytic converter turn CO and NO into $\mathrm{CO_2}$ and $\mathrm{N_2}$ on their surfaces; iron catalyzes the Haber process for ammonia.
  • Enzymes: catalase, in nearly every cell, breaks down hydrogen peroxide millions of times per second per enzyme molecule.

Common mistake

Thinking a catalyst raises the yield at equilibrium, or that it speeds up only the forward reaction. It lowers the barrier for both directions, so equilibrium comes sooner but lies in the same place. Check: a catalyst never appears in $K$ and never changes $\Delta H$.

Practice

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Sources and further reading

  • OpenStax Chemistry 2e, Section 12.7 Catalysis
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