Finding the Rate Law From a Mechanism
Introduced| English | Meaning |
|---|---|
| rate-determining step/reɪt dɪˈtɜːmɪnɪŋ step/ | rate-determining step |
Predicting speed from the steps
- Given a proposed sequence, can you predict the speed equation?
- If you know which step drags, you almost can.
- The slow step hands you the orders directly.
- It is a satisfying bit of detective work.
Start with the slow step
- The rate law comes from the rate-determining step 决速步.
- Use its reactants and their coefficients as the orders.
- A slow $A + B \to \dots$ gives $\text{rate} = k[A][B]$.
The rate law is written from the reactants of the...
The slow (rate-determining) step's reactants set the rate law.
When it is the first step
- If the slow step is first, its reactants are simple starting materials.
- Read the rate law directly from them, with no extra work.
- This is the most common exam case.
When the slow step is the first step, you can read the rate law directly from its reactants.
Its reactants are starting materials, so no substitution is needed.
A quick recipe
- Find the slowest step in the mechanism.
- Write $\text{rate} = k \times$ its reactants, each raised to its coefficient.
- Check that it matches the experimental rate law.
Order the steps to find a rate law from a mechanism.
Find the bottleneck, write its rate law, then verify.
Mechanism: $\text{NO}_2 + \text{NO}_2 \to \text{NO}_3 + \text{NO}$ (slow), then a fast step.
- The slow step has two $\text{NO}_2$ molecules.
- So $\text{rate} = k[\text{NO}_2]^2$.
Which step sets the rate law?
Sort the parts of a mechanism by whether they control the observed rate.
If the slow first step is $\text{NO}_2 + \text{NO}_2 \to \text{products}$, the rate law is...
Two $\text{NO}_2$ react in the slow step, so it is second order.
Coefficients can be used as orders for an elementary step but not the overall equation.
Only elementary steps allow coefficients as orders.
If an ____ appears in the slow step's rate law, you must replace it.
An intermediate must be substituted using an earlier fast equilibrium.
This direct method works cleanly only when the slow step is first (or uses only starting materials). If an intermediate appears in the slow step's rate law, you must replace it -- that is the next lesson. And coefficients-as-orders applies to the elementary step, never the overall equation.
To find a rate law from a mechanism, locate the rate-determining step and write $\text{rate} = k \times$ its reactants raised to their coefficients. When the slow step is first, you read the law straight off, like $\text{rate} = k[\text{NO}_2]^2$. Check it against experiment.