Finding the Rate Law From a Mechanism · 从机理推导速率方程
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| rate-determining step/reɪt dɪˈtɜːmɪnɪŋ step/ | 决速步 | jué sù bù |
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.
从步骤预测速度
- 给定一个提出的序列,你能预测速度方程吗?
- 如果你知道哪一步拖后腿,你几乎就能。
- 慢步骤直接把级数递给你。
- 这是一件令人满足的侦探工作。
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 $\text{rate} = k[A][B]$.
从慢步骤开始
- 速率方程来自决速步。
- 用它的反应物和它们的系数作为级数。
- 慢的 $A + B \to \dots$ 给出 $\text{rate} = k[A][B]$。
The rate law is written from the reactants of the... · 速率方程是根据...的反应物书写的
The slow (rate-determining) step's reactants set the rate law. · 慢的（决速）步骤的反应物决定了速率方程。
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.
当它是第一步时
- 如果慢步骤在最前,它的反应物是简单的起始物。
- 直接从它们读出速率方程,无需额外工作。
- 这是最常见的考试情形。
When the slow step is the first step, you can read the rate law directly from its reactants. · 当慢步骤是第一步时，可以直接从其反应物读出速率方程。
Its reactants are starting materials, so no substitution is needed. · 它的反应物是起始原料，因此不需要替换。
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.
一个快速配方
- 找出机理中最慢的步骤。
- 写 $\text{rate} = k \times$ 它的反应物,各自升到自己的系数次方。
- 检查它是否与实验速率方程相符。
Order the steps to find a rate law from a mechanism. · 排序步骤以从机理中找到速率方程。
Find the bottleneck, write its rate law, then verify. · 找到瓶颈，写出其速率方程，然后验证。
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 $\text{rate} = k[\text{NO}_2]^2$.
机理:$\text{NO}_2 + \text{NO}_2 \to \text{NO}_3 + \text{NO}$(慢),然后一个快步骤。
- 慢步骤有两个 $\text{NO}_2$ 分子。
- 所以 $\text{rate} = k[\text{NO}_2]^2$。
Which step sets the rate law? · 哪一步决定了速率方程？
Sort the parts of a mechanism by whether they control the observed rate. · 根据各部分是否控制观测到的速率来整理机理的各个部分。
If the slow first step is $\text{NO}_2 + \text{NO}_2 \to \text{products}$, the rate law is... · 如果慢的第一步是$\text{NO}_2 + \text{NO}_2 \to \text{products}$，速率方程为...
Two $\text{NO}_2$ react in the slow step, so it is second order. · 两个$\text{NO}_2$在慢步骤中反应，因此是二级反应。
Coefficients can be used as orders for an elementary step but not the overall equation. · 系数可作为基元步骤的级数，但不能作为总方程式的级数。
Only elementary steps allow coefficients as orders. · 只有基元步骤允许使用系数作为级数。
If an ____ appears in the slow step's rate law, you must replace it. · 如果____出现在慢步骤的速率方程中，则必须将其替换。
An intermediate must be substituted using an earlier fast equilibrium. · 必须利用早期的快速平衡替换中间体。
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.
这个直接方法只有在慢步骤在最前(或只用起始物)时才干净利落。如果慢步骤的速率方程里出现中间体,你必须替换它——那是下一课。而且系数当级数只适用于基元步骤,绝不适用于总方程。
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.
要从机理求速率方程,找到决速步,写 $\text{rate} = k \times$ 它的反应物升到各自系数次方。当慢步骤在最前时,你直接读出方程,如 $\text{rate} = k[\text{NO}_2]^2$。再与实验对照。