Writing the Rate Law · 书写速率方程
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| rate law/reɪt lɔː/ | 速率定律 | sù lǜ dìng lǜ |
| rate constant/reɪt ˈkɒnstənt/ | 速率常数 | sù lǜ cháng shù |
| order/ˈɔːdə/ | 反应级数 | fǎn yìng jí shù |
The formula for speed
- Double a reactant and the speed might double -- or quadruple.
- Each reaction has its own sensitivity to concentration.
- A short equation captures exactly how speed depends on amounts.
- But you can only find it by experiment, not by guessing.
速度的公式
- 把某反应物加倍,速度可能翻倍——也可能变四倍。
- 每个反应对浓度都有自己的敏感度。
- 一个简短的方程精确抓住速度如何依赖于量。
- 但你只能通过实验找到它,而不能靠猜。
The rate law
- The rate law 速率定律 links speed to concentrations:
- Here $k$ is the rate constant 速率常数, and $m$ and $n$ are the orders.
速率定律
- 速率定律把速度与浓度联系起来:
- 这里 $k$ 是速率常数,$m$ 和 $n$ 是级数。
The rate constant $k$ changes with temperature. · 速率常数$k$随温度变化。
$k$ depends on temperature but not on concentration. · $k$ 取决于温度而不取决于浓度。
Reaction orders
- The order 反应级数 tells how strongly a concentration affects the rate.
- First order: double $[A]$ and the rate doubles. Second order: double and it quadruples.
- Orders come from experiments, not from the equation's coefficients.
反应级数
- 反应级数告诉你某浓度对速率影响有多强。
- 一级:$[A]$ 加倍,速率翻倍。二级:加倍,变四倍。
- 级数来自实验,而不是方程的系数。
For · 支持 $\text{rate} = k[A]^2$, doubling $[A]$ multiplies the rate by... · 对于 $\text{rate} = k[A]^2$，将 $[A]$ 加倍会使速率乘以...
Second order means $2^2 = 4$ times the rate. · 二级反应意味着速率乘以 $2^2 = 4$。
Reaction orders are determined by... · 反应级数由...决定
Orders must be found experimentally, not from coefficients. · 反应级数必须通过实验确定，而不能直接从系数得出。
A reaction is first order in $[A]$. Tripling $[A]$ changes the rate by a factor of... · 某反应物对$[A]$呈一级反应。将$[A]$增大三倍，反应速率变为原来的...
First order means the rate scales directly, so 3 times. · 一级反应意味着速率与浓度成正比，因此是3倍。
Overall order
- Add the individual orders to get the overall order ($m + n$).
- It sums up the reaction's total sensitivity to concentration.
- The units of $k$ depend on that overall order.
总级数
- 把各个级数相加得到总级数($m + n$)。
- 它汇总了反应对浓度的总敏感度。
- $k$ 的单位取决于那个总级数。
For · 支持 $\text{rate} = k[A]^2[B]^1$, what is the overall order? · 对于$\text{rate} = k[A]^2[B]^1$，总反应级数是多少？
Overall order $= 2 + 1 = 3$. · 总反应级数为$= 2 + 1 = 3$。
For $\text{rate} = k[A]^2[B]$, what happens if you double $[A]$?
- $[A]$ is second order, so doubling it multiplies the rate by $2^2 = 4$.
- The rate quadruples.
对于 $\text{rate} = k[A]^2[B]$,如果把 $[A]$ 加倍会怎样?
- $[A]$ 是二级,所以加倍会把速率乘以 $2^2 = 4$。
- 速率变四倍。
Find the reaction order · 确定反应级数
Use how the rate responds to concentration to find the order in A. · 利用速率对浓度的响应来确定 A 的级数。
The rate law normally contains only the ____ concentrations, not the products. · 速率方程通常只包含____的浓度，而不包含产物。
Products usually do not appear in the rate law. · 产物通常不出现在速率方程中。
Orders are found by experiment, not read off the balanced equation's coefficients -- a very common trap. Only reactants normally appear in the rate law; products usually do not. And the rate constant $k$ changes with temperature but not with concentration.
级数由实验确定,而不是从配平方程的系数读出——这是很常见的陷阱。通常只有反应物出现在速率定律里;产物一般不出现。而且速率常数 $k$ 随温度变化,但不随浓度变化。
The rate law $\text{rate} = k[A]^m[B]^n$ links speed to concentrations through the rate constant $k$ and the orders $m$, $n$. Orders come from experiment, not coefficients, and their sum is the overall order. A concentration's order sets how strongly it changes the rate.
速率定律 $\text{rate} = k[A]^m[B]^n$ 通过速率常数 $k$ 和反应级数 $m$、$n$ 把速度与浓度联系起来。级数来自实验而非系数,其和是总级数。某浓度的级数决定它对速率影响多强。