| English | 意味 |
|---|---|
| pre-equilibrium/priː ˌiːkwɪˈlɪbrɪəm/ | 予平衡 |
When the slow one comes second
- Sometimes the bottleneck sits in the middle, not the start.
- The slow step then needs an intermediate made earlier.
- But an intermediate cannot appear in the final rate law.
- A clever swap fixes this.
A fast equilibrium first
- The first step is a fast, reversible equilibrium.
- The second, slow step is the rate-determining one.
- Its rate law contains an intermediate from step one.
The pre-equilibrium method is used when the slow step...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
A fast equilibrium precedes the slow rate-determining step.
In this case the ____ step is not the first one, but the second.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The rate-determining (slow) step is second here.
Replace the intermediate
- Intermediates cannot appear in the final rate law.
- Use the fast equilibrium to express it in terms of reactants.
- Substitute, and the intermediate disappears.
An intermediate is allowed to remain in the final rate law.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
It must be substituted out using the fast equilibrium.
In fast $A \rightleftharpoons B$, slow $B + C \to D$, the intermediate $B$ is replaced using...
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
The equilibrium gives $[B]$ in terms of $[A]$, removing the intermediate.
The pre-equilibrium idea
- This is the pre-equilibrium 预平衡 approximation.
- The fast first step stays balanced while the slow step sips away the intermediate.
- The result is a rate law in real reactants only.
When the first step is fast
A fast equilibrium feeds a slow step - so the intermediate must be rewritten using the reactants.
Using a fast first equilibrium to replace an intermediate is the ____ approximation.
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
This is the pre-equilibrium approximation.
After substitution, the final rate law contains only reactants (and any catalyst).
難易度 1/5 · 100 コイン + 今日初回完了時に小ボーナス付与
No intermediates should remain in the final rate law.
Fast: $A \rightleftharpoons B$. Slow: $B + C \to D$.
- The slow-step rate is $k[B][C]$, but $B$ is an intermediate.
- The equilibrium gives $[B] \propto [A]$, so $\text{rate} = k'[A][C]$.
An intermediate must not appear in the final rate law -- always substitute it out using the fast equilibrium. This pre-equilibrium method applies when the first step is a fast, reversible step before the slow one. And the final rate law should contain only reactants (and perhaps a catalyst).
When a fast equilibrium precedes the slow step, its rate law holds an intermediate. The pre-equilibrium approximation uses the fast step to rewrite that intermediate in terms of real reactants, then substitutes it out -- leaving a rate law with no intermediates.