What a learner can do afterwards
- Sets the net rate of change of an intermediate to zero and eliminates it from the rate expression
- Derives the rate law for the same mechanism under the pre-equilibrium assumption and compares the two results
- States the condition on the rate constants under which each approximation holds
- Recovers a fractional or mixed order from a proposed mechanism and matches it against measured data
1 · Read
A mechanism lists the elementary steps a reaction really takes. A step that is slower than the rest is the rate-determining step, and the whole reaction can only go as fast as it allows. Steps often make an intermediate, a species produced in one step and used up in a later one.
Intermediates vanish so fast that you cannot measure their concentration. Your final rate law must therefore contain no intermediate concentration. Both approximations below exist to remove it. Never trust a mechanism whose rate law you have not matched against measured data, since mechanisms can predict fractional or mixed orders.
The steady-state approximation sets the net rate of change of the intermediate to zero. Write its formation rate minus its consumption rate, set the difference to zero, and solve for its concentration. Substitute that expression into the slow step rate and the intermediate is gone.
The pre-equilibrium approximation assumes the first step reaches equilibrium quickly, so you replace the intermediate with an equilibrium expression. In symbols, it needs the backward rate of the first step to dwarf the forward rate of the second. When the intermediate is eaten fast, the first step never balances, so steady state is the fair choice.
Both routes remove the intermediate from the rate law, and how fast it is consumed tells you which route is fair.
2 · Watch
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8 questions wait behind this lesson, each with its answer explained. Every answer feeds the sky: stars light as they are learned, and dim when it is time to come back.