Reaction kinetics (A2)
A-Level Chemistry Topic 26 8:01 English narration · English + 中文 subtitles burned in
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Here is a real reaction: hydrogen peroxide reacting with iodide ions in acid.
这是一个真实的反应：过氧化氢在酸性条件下与碘离子反应。
Look at the balanced equation. It uses two iodide ions and two hydrogen ions every time, so doubling the acid should make it much faster.
看看这个配平方程式， 每次反应都用掉两个碘离子和两个氢离子，所以把酸的浓度加倍，反应应该快很多才对。
In the laboratory, doubling the acid changes the rate by nothing at all.
可是在实验室里，把酸加倍，速率完全不变。
A balanced equation cannot tell you the speed. Only an experiment can.
配平方程式并不能告诉你反应有多快， 只有实验才能。
Welcome to A2 reaction kinetics.
欢迎来到 A2 反应动力学。
We will write a rate equation, read the orders out of experimental data, find the rate constant and its units, use half-life, look inside a reaction at its steps, and see how catalysts really work.
我们将写出速率方程，从实验数据中读出反应级数， 求出速率常数和它的单位，使用半衰期，深入反应内部看它的分步， 并且看看催化剂到底是怎么工作的。
Let's begin.
让我们开始吧。
This is the rate equation. Rate equals k, times the concentration of A to the power m, times the concentration of B to the power n.
这就是速率方程：速率等于 k 乘以 A 的浓度的 m 次方，再乘以 B 的浓度的 n 次方。
Those powers are the orders: m is the order with respect to A, n the order with respect to B, and adding them gives the overall order.
这两个指数就是反应级数：m 是对 A 的级数，n 是对 B 的级数，把它们相加就得到总级数。
Each order is only zero, one or two.
每个级数只能是零、一或二。
The letter k is the rate constant.
字母 k 叫做速率常数。
And remember: the orders are not the numbers in the balanced equation.
记住：级数不是配平方程式里的系数。
What does each order mean in practice?
每个级数在实际中意味着什么？
Zero order: change the concentration and the rate does not move at all, so a graph of rate against concentration is flat.
零级：改变浓度，速率完全不变， 所以速率对浓度的图像是一条水平线。
First order: double the concentration and the rate doubles, giving a straight line through the origin.
一级：浓度加倍，速率也加倍， 图像是一条过原点的直线。
Second order: double the concentration and the rate goes up four times, and the graph curves upward.
二级：浓度加倍，速率变成四倍，图像是一条向上弯的曲线。
Learn those three shapes.
把这三种形状记牢。
Now the question the exam really asks: the initial rates method — four experiments, find the rate equation from how the initial rate changes.
现在来看考试真正会考的问题：初始速率法——四组实验，从初始速率如何变化求出速率方程。
Experiments two and three: only the iodide changes, it doubles, and the rate doubles, so first order in iodide.
先看实验二和实验三：只有碘离子在变，它加倍，速率也加倍，所以对碘离子是一级。
Experiments one and two: the iodide doubles, which alone would double the rate, but the rate did not change, so halving the peroxide must have halved it. First order in peroxide.
再看实验一和实验二：碘离子加倍，单靠它速率应该加倍，但速率没有变化， 所以过氧化氢减半一定把速率减半了，对过氧化氢也是一级。
Experiments three and four: the peroxide doubles and the rate doubles, exactly as expected, while the acid went up four times and did nothing.
最后看实验三和实验四：过氧化氢加倍，速率正好加倍，而酸增加到四倍却毫无作用， 所以对酸是零级。
The acid is zero order. So the rate equation is k, peroxide, iodide, and the overall order is two.
因此速率方程是 k 乘以过氧化氢浓度乘以碘离子浓度，总级数是二。
Next they ask for the rate constant, with its units.
接下来会让你求速率常数，并写出单位。
Put the numbers from experiment one into the rate equation and rearrange.
把实验一的数据代入速率方程，然后变形求 k。
Two point four two times ten to the minus three, divided by zero point zero four five times zero point zero three, gives k equals one point seven nine.
二点四二乘以十的负三次方，除以零点零四五乘以零点零三，得到 k 等于一点七九。
Now the units — this is where marks disappear.
现在看单位——分就是在这里丢的。
Write the unit of every term and cancel.
把每一项的单位都写出来，然后约分。
Do that every time, because the units change with the overall order.
每次都要这样做，因为单位会随着总级数改变。
Once you have a rate equation, the exam loves to ask what happens if we change something.
有了速率方程，考试最爱问：如果改变某个条件会怎样？
Here the order in hydrogen is one, the order in nitrogen monoxide is two, so the overall order is three.
这里对氢气是一级，对一氧化氮是二级，所以总级数是三。
Halve the nitrogen monoxide: squared means a half times a half, so the rate falls to one quarter.
把一氧化氮浓度减半：因为是平方，等于二分之一乘二分之一，速率降到四分之一。
Triple both concentrations: three for the hydrogen times three squared for the monoxide gives twenty-seven times faster.
把两个浓度都变成三倍：氢气贡献三倍，一氧化氮贡献三的平方，速率变成二十七倍。
Just count the powers.
只要数一数指数就行。
Now the other way to spot first order: half-life.
判断一级反应还有另一种方法：半衰期。
The half-life is the time for the concentration to fall to half.
半衰期就是浓度下降到一半所需的时间。
For a first order reaction that time is always the same, whatever you start with — every step here is the same width.
对一级反应来说，不管起始浓度是多少，这个时间总是一样的—— 看这里每一段的宽度都相同。
So if two half-lives read off a graph match, the reaction is first order.
所以如果从图上读出的两个半衰期相等，这个反应就是一级。
Half-life also gives the rate constant: k is zero point six nine three divided by the half-life.
半衰期还能给出速率常数：k 等于零点六九三除以半衰期。
Careful: a constant half-life means first order, nothing else.
注意：半衰期恒定只说明是一级反应，别的都说明不了。
So where do these orders come from?
那么这些级数是从哪里来的呢？
Most reactions do not happen in one go.
大多数反应并不是一步完成的。
They happen in steps, and the slowest step is called the rate determining step.
它们分成若干步，其中最慢的一步叫做决速步骤。
Think of a crowd leaving through one narrow gate: the whole crowd moves at the speed of that gate.
想象一群人从一个窄门出去：整群人的速度由那道门决定。
Only the species used up to and including the slow step appear in the rate equation.
只有在决速步骤之前（包括决速步骤本身）用到的物种，才会出现在速率方程里。
A species made in one step and used up in a later step is an intermediate.
在某一步生成、又在后面某一步被消耗掉的物种，叫做中间体。
Let's use that.
我们来用一用。
Nitrogen monoxide reacts with hydrogen, and the measured rate equation is k, hydrogen, monoxide squared. Why can this not be one single step?
一氧化氮和氢气反应生成氮气和水，实验测得的速率方程是 k 乘以氢气浓度乘以一氧化氮浓度的平方。
One step would need four molecules to meet at exactly the same moment, which is far too unlikely.
为什么它不可能只有一步？ 一步反应需要四个分子在同一瞬间相遇，这太不可能了。
So here are three steps that work. Two monoxide molecules join into one bigger molecule. That molecule reacts with hydrogen, and this is the slow step.
所以可以这样分成三步： 两个一氧化氮分子先结合成一个较大的分子；这个分子再与氢气反应，这一步是慢步骤； 然后它生成的一氧化二氮再与氢气反应。
Then the dinitrogen oxide it made reacts with more hydrogen. Now count what is used up to the slow step: two monoxide and one hydrogen. That is exactly the rate equation.
现在数一数到慢步骤为止用掉了什么：两个一氧化氮和一个氢气， 这正好就是速率方程。
And the dinitrogen oxide is made in one step and used up in the next, so it is an intermediate.
而一氧化二氮在前一步生成、在下一步被消耗，所以它是中间体。
One more thing about k.
关于 k 还有一点。
The rate constant is only constant while the temperature stays the same.
只有温度不变时，速率常数才是常数。
Heat the mixture, and a much larger fraction of the molecules can pass the activation energy — the curve you met at AS.
给混合物加热，能越过活化能的分子比例就会大得多——就是你在 AS 学过的那条曲线。
Nothing in the rate equation changes except k, and k gets bigger, so the reaction speeds up.
速率方程里除了 k 本身没有任何东西改变，而 k 变大了，所以反应变快。
In the exam say both halves: raising the temperature increases the rate constant, and therefore increases the rate.
考试时两半都要写：升高温度会增大速率常数，因而增大反应速率。
Now catalysts.
现在讲催化剂。
The A2 question is always the same: state the type, then explain.
A2 的问法总是一样：说出类型，再解释原因。
Homogeneous means the catalyst is in the same physical state as the reactants — iron ions dissolved among other ions in solution.
均相是指催化剂与反应物处于相同的物理状态——比如溶液中的铁离子混在其他离子之中。
Heterogeneous means a different state — solid manganese dioxide dropped into hydrogen peroxide solution.
多相是指处于不同的状态——比如把固体二氧化锰放进过氧化氢溶液里。
It is one mark, but you only earn it if you write both halves.
这只有一分，但你必须两半都写出来才能拿到。
How does a solid catalyst actually work?
固体催化剂到底是怎么工作的？
Three stages, and the mark scheme wants all three.
分三个阶段，评分标准三个都要。
First, adsorption: the reactant molecules stick onto the catalyst surface.
第一，吸附：反应物分子附着在催化剂表面上。
Second, the bonds inside them are weakened, and they are held close together, so they react easily.
第二，分子内部的键被削弱，同时它们被固定在靠近且方向合适的位置，所以更容易反应。
Third, desorption: the products leave the surface and free it again.
第三，脱附：产物离开表面，把位置腾出来。
Iron does this in the Haber process; platinum, palladium and rhodium do it in a catalytic converter.
哈伯法中的铁就是这样工作的；催化转化器里的铂、钯和铑也是如此。
A homogeneous catalyst works differently: it joins in.
均相催化剂的工作方式不同：它会参与进去。
Iodide ions react with peroxydisulfate ions, but both are negative, so they repel, the activation energy is high and the reaction is slow.
碘离子与过二硫酸根离子反应， 但两者都带负电，会互相排斥，活化能很高，反应很慢。
Add iron three ions.
现在加入三价铁离子。
In the first step, iron three takes electrons from the iodide and becomes iron two.
第一步，三价铁从碘离子那里夺走电子，自己变成二价铁。
In the second step, the peroxydisulfate turns the iron two back into iron three.
第二步，过二硫酸根又把二价铁氧化回三价铁。
Now every step is a positive ion meeting a negative one, which is easy.
这样每一步都是一个正离子遇到一个负离子，都很容易发生。
The catalyst was used in one step and reformed in a later step, so it comes back unchanged.
催化剂在一步中被消耗，又在后面一步被重新生成，所以它原样回来了。
Three marks students give away.
三个学生常送掉的分。
First, never read the orders off the balanced equation; they come only from data, and say which experiments you compared.
第一，绝不要从配平方程式里读级数； 级数只能来自实验数据，而且你必须写出你比较了哪两组实验。
Second, always give the units of k — write them, cancel them, and the mark is yours.
第二，一定要写出 k 的单位——把单位写出来、约分，这一分就是你的。
Third, know the difference between an intermediate and a catalyst: an intermediate is made first and used up later; a catalyst is used up first and made again later.
第三，弄清中间体和催化剂的区别：中间体是先生成、后被消耗； 催化剂是先被消耗、后又生成。
Get those right and this topic is yours.
做对这三点，这个专题就是你的了。