U5.11 · CatalysisPractice · 催化

Without the catalyst the highest point (the transition state) is at 130; with it, the reaction takes two steps and the higher transition state is at 80. The reactants are at 20 in both, so the overall barrier falls from 110 to 60, a drop of 50 kJ/mol. Both curves have the same ends, so ΔH is unchanged.

NO, $\ce{SO2}$, and $\ce{O2}$ are all gases reacting in one phase: homogeneous catalysis. The other three use solid catalysts with gases or solutions, reacting on the solid surface: heterogeneous catalysis.

The catalyst’s new pathway serves both directions, lowering both activation energies by the same amount, so $k_\text{forward}$ and $k_\text{reverse}$ rise by the same factor and $K = k_\text{forward}/k_\text{reverse}$ is unchanged. A faster reaction is exactly what larger k values mean.

$\dfrac{k_\text{cat}}{k} = e^{\Delta E_a/RT}$ = $e^{\frac{20000}{8.314 \times 298}}$ = $e^{8.07}$ ≈ $3.2 \times 10^{3}$. Lowering Ea by just 20 kJ/mol makes k over three thousand times larger.

K is set by the energy difference between reactants and products (and the temperature). A catalyst leaves both ends alone and only lowers the hill between them, by the same amount in each direction, so both directions speed up equally: the equilibrium position stays put and is just reached sooner. The catalyst isn’t consumed either.