A-Level Chemistry Rate Equation 反应级数 k 计算
速率方程（rate equation）是 A-Level 化学 A2 阶段动力学部分的核心考点，也是历年真题中区分度最高的题目之一。许多同学能背出 rate = k[A][B] 的形式，却在"级数由什么决定""k 的单位怎么推""如何从数据反推反应级数"上反复丢分。这篇文章用中英双语，把速率方程的定义、反应级数的判断、实验测定方法、k 的单位推导和决速步骤一次讲透，最后配一道完整例题。
The rate equation is one of the most heavily tested ideas in A2 kinetics, and examiners use it to separate strong candidates from weak ones. Many students can quote rate = k[A][B] from memory, yet lose marks on three things: where the orders actually come from, how to derive the units of k, and how to work backwards from experimental data. This guide explains the definition, how to identify orders, the experimental methods, unit derivation, and the rate-determining step, then walks through a full worked example.
速率方程的定义:幂次由实验决定,与配平系数无关
对于一个反应 aA + bB 生成产物,速率方程写成 rate = k[A]^m [B]^n。其中 k 是速率常数（rate constant）,只受温度和催化剂影响;m 是反应物 A 的级数,n 是反应物 B 的级数,总反应级数 = m + n。最关键的规则是:级数 m、n 与方程式中的配平系数 a、b 没有任何必然联系,只能由实验数据测定,绝对不能看着方程式系数直接抄。
For a reaction aA + bB producing products, the rate equation is written as rate = k[A]^m[B]^n, where k is the rate constant, which depends only on temperature and catalysts, while m and n are the orders with respect to A and B. The overall order is m + n. The single most important rule is that m and n have no automatic connection to the stoichiometric coefficients a and b, so they must be found from experimental data and never copied from the balanced equation.
经典反例:过二硫酸根与碘离子的反应 S2O8^2- + 2I- -> 2SO4^2- + I2,方程式中 I- 的系数是 2,但实验测得的速率方程是 rate = k[S2O8^2-][I-],碘离子其实是一级。反过来,2NO + O2 -> 2NO2 实测 rate = k[NO]^2[O2],碰巧与系数一致。可见"看系数写级数"是动力学大题的第一大失分点。
A classic counter-example is the persulfate-iodide reaction S2O8^2- + 2I- -> 2SO4^2- + I2. The coefficient of I- is 2 in the equation, yet the measured rate equation is rate = k[S2O8^2-][I-]: iodide is first order. By contrast, 2NO + O2 -> 2NO2 gives rate = k[NO]^2[O2], which happens to match the coefficients. Writing orders straight from the equation is therefore the single most common way to lose marks in kinetics questions.
反应级数的含义:零级、一级、二级的浓度效应
级数回答的问题是:改变某一反应物浓度,速率会怎么变?有一个通用结论可以直接套用:若某反应物级数为 n,浓度变为原来的 x 倍,速率就变为 x^n 倍。于是:零级时速率与浓度无关,浓度翻倍速率不变;一级时浓度翻倍速率翻倍;二级时浓度翻倍速率变成 4 倍。看到"浓度加倍、速率变四倍"就能立刻锁定二级,这是选择题和简答题里最快的判断路径。
An order answers one question: how does changing a reactant concentration change the rate? A general rule saves time in every exam: if the order with respect to a reactant is n, multiplying its concentration by a factor x multiplies the rate by x^n. So for zero order the rate is independent of concentration; for first order, doubling the concentration doubles the rate; for second order, doubling it quadruples the rate. Whenever you see "concentration doubled, rate quadrupled", you can immediately identify second order.
图像也是高频考点。零级反应:速率-浓度图是水平线,浓度-时间图是斜率恒定的直线。一级反应:速率-浓度图是过原点的直线,ln[A] 对时间作图得直线,且半衰期恒定,t1/2 = 0.693/k,与起始浓度无关,这是判断一级反应最漂亮的标志。二级反应:1/[A] 对时间作图得直线。考试时先看题目给的是哪张图,再对号入座。
Graphs are also favourite targets. For zero order the rate-concentration graph is a horizontal line and the concentration-time graph is a straight line with constant slope. For first order, the rate-concentration graph is a straight line through the origin, ln[A] plotted against time is linear, and the half-life is constant at t1/2 = 0.693/k, independent of the starting concentration, which is the cleanest signature of a first-order process. For second order, 1/[A] against time gives a straight line. Always check which graph the question provides before answering.
初速率法:三组数据反推级数
实验题最常考初速率法（initial rates method）:保持其他反应物浓度不变,只改变一种反应物的浓度,比较初速率的变化。因为每轮只动一个变量,级数就能被逐一分离出来。看这道经典小题:反应 A + B 生成产物,三组实验数据如下。第 1 组 [A] = 0.10 mol dm^-3、[B] = 0.10 mol dm^-3,初速率 2.0 x 10^-3 mol dm^-3 s^-1;第 2 组 [A] = 0.20、[B] = 0.10,初速率 8.0 x 10^-3;第 3 组 [A] = 0.10、[B] = 0.20,初速率仍为 2.0 x 10^-3。
The most common experimental design is the initial rates method: keep every concentration fixed except one, change that single reactant, and compare the initial rates. Because only one variable moves per run, the orders can be separated cleanly. Classic example: for a reaction A + B producing products, run 1 uses [A] = 0.10 mol dm^-3 and [B] = 0.10 mol dm^-3 with initial rate 2.0 x 10^-3 mol dm^-3 s^-1; run 2 uses [A] = 0.20 and [B] = 0.10 giving 8.0 x 10^-3; run 3 uses [A] = 0.10 and [B] = 0.20 giving 2.0 x 10^-3 again.
对比第 1、2 组:B 不变,A 的浓度翻倍,速率从 2.0 x 10^-3 变成 8.0 x 10^-3,正好是 4 倍,所以 A 是二级。对比第 1、3 组:A 不变,B 的浓度翻倍,速率完全没变,所以 B 是零级。于是速率方程写作 rate = k[A]^2[B]^0 = k[A]^2,总反应级数为 2。注意第 3 组里 B 明明参加了反应,级数却是零,再次证明级数与方程式无关,它只反映 B 的浓度变化是否影响决速步。
Comparing runs 1 and 2: B is fixed, [A] doubles, and the rate jumps from 2.0 x 10^-3 to 8.0 x 10^-3, a factor of four, so A is second order. Comparing runs 1 and 3: A is fixed, [B] doubles, and the rate does not change at all, so B is zero order. The rate equation is therefore rate = k[A]^2[B]^0 = k[A]^2, with overall order 2. Notice that B genuinely takes part in the reaction yet is zero order, which reinforces the rule that orders reflect only whether a concentration change affects the slow step, not the stoichiometry.
除初速率法外,连续监测法（continuous monitoring）同样重要:通过跟踪某物理量随时间的变化来作图定级数。常见手段包括:收集气体测量体积变化（如镁与酸的反应）、用色度计监测有色物质的颜色深浅、测溶液电导率跟踪离子浓度变化,以及碘钟这类时钟反应记录变色所需时间。这类题往往同时考查作图能力和"如何从图中得到初速率"（取 t 接近 0 时的切线斜率）。
Beyond initial rates, continuous monitoring is equally important: you track a physical property over time and use the graph to decide the order. Common techniques include collecting gas and measuring the volume change, using a colorimeter to follow the intensity of a coloured species, measuring conductivity to follow ion concentrations, and clock reactions such as the iodine clock that record the time taken for a colour change. These questions usually test graph skills and how to obtain an initial rate from the tangent at t close to zero.
速率常数 k:单位推导与温度影响
速率常数 k 的单位由总反应级数决定,推导原则一句话:速率的单位固定是 mol dm^-3 s^-1,把它除以浓度的单位幂次即可。总级数为 1 时,k = rate 除以 [A],单位是 s^-1;总级数为 2 时,k = rate 除以 [A]^2,单位是 mol^-1 dm^3 s^-1;总级数为 3 时,单位是 mol^-2 dm^6 s^-1。通用公式:(mol dm^-3)^(1-n) s^-1,其中 n 是总级数。先定总级数再写单位,顺序别反。
The units of the rate constant k are fixed by the overall order. The principle is simple: the rate always has units mol dm^-3 s^-1, and you divide by the units of the concentrations raised to their powers. For overall order 1, k = rate divided by [A], giving s^-1. For overall order 2, k = rate divided by [A]^2, giving mol^-1 dm^3 s^-1. For overall order 3, the units are mol^-2 dm^6 s^-1. The general expression is (mol dm^-3)^(1-n) s^-1 where n is the overall order. Work out the overall order first, then write the units.
k 的数值只随温度与催化剂改变。温度每升高约 10 K,k 大约变为原来的 2 到 4 倍,这正是 Arrhenius 方程 k = Ae^(-Ea/RT) 的直观体现:温度升高使更多分子越过活化能垒。催化剂通过降低活化能 Ea 同样显著增大 k。真题里常这样设问:同一反应、同一浓度下速率变快了,但速率方程形式不变,原因就是温度升高或加入催化剂导致 k 增大。反过来,看到"速率方程没变",第一个要想到的就是 k 变了。
The value of k changes only with temperature and catalysts. Raising the temperature by roughly 10 K typically multiplies k by a factor of two to four, exactly what the Arrhenius equation k = Ae^(-Ea/RT) predicts: more molecules acquire enough energy to cross the activation barrier. Catalysts increase k dramatically by lowering the activation energy Ea. A favourite exam question gives the same reaction and concentrations with a faster rate but an unchanged rate equation, and the answer is simply that a higher temperature or a catalyst has increased k. Whenever the rate equation is unchanged, suspect that k has changed.
决速步骤:速率方程如何揭示反应机理
绝大多数反应是多步进行的,整体速率由最慢的一步,即决速步骤（rate-determining step,简称 RDS）控制。规律如下:凡是出现在决速步中的反应物,以及决速步之前快速平衡中产生并能进入决速步的物种,都会出现在速率方程里;决速步之后的任何步骤都不影响速率方程。因此,一个速率方程就是一张机理的指纹,可以用它来检验或排除假说机理。
Most reactions proceed through several steps, and the overall rate is controlled by the slowest one, the rate-determining step (RDS). The rule is: any reactant that appears in the RDS, plus any species produced in a fast equilibrium before the RDS that feeds into it, will appear in the rate equation, while steps after the RDS have no effect on it. A rate equation is therefore a fingerprint of the mechanism, and you can use it to test or eliminate proposed mechanisms.
经典案例:反应 NO2 + CO -> NO + CO2 若是一步完成的碰撞,速率方程应同时包含 NO2 和 CO;但实验测得 rate = k[NO2]^2,与 CO 无关,说明机理只能是:第一步 2NO2 -> NO3 + NO（慢）,第二步 NO3 + CO -> NO2 + CO2（快）。慢步里根本没有 CO,所以 CO 的浓度不影响速率。解题时先看速率方程中出现了哪些物种,再让机理中的慢步包含这些物种,这就是机理题的标准打法。
A classic example: if NO2 + CO -> NO + CO2 happened in one collision step, the rate equation would contain both NO2 and CO. But experiment gives rate = k[NO2]^2, independent of CO, so the mechanism must be: step 1, 2NO2 -> NO3 + NO, slow; step 2, NO3 + CO -> NO2 + CO2, fast. CO never appears in the slow step, so its concentration cannot affect the rate. When solving mechanism questions, list the species that appear in the rate equation, then make sure the slow step of your proposed mechanism contains exactly those species.
完整例题:写速率方程、算 k 并推导单位
综合题现在登场。反应 X + 2Y 生成产物,实验数据如下:第 1 组 [X] = 0.10 mol dm^-3、[Y] = 0.20 mol dm^-3,初速率 5.0 x 10^-4 mol dm^-3 s^-1;第 2 组 [X] = 0.20、[Y] = 0.20,初速率 1.0 x 10^-3;第 3 组 [X] = 0.10、[Y] = 0.40,初速率 2.0 x 10^-3。第一步,比较第 1、2 组:Y 不变,X 翻倍,速率翻倍,所以 X 是一级。第二步,比较第 1、3 组:X 不变,Y 翻倍,速率变成 4 倍,所以 Y 是二级。
Now a full question. For a reaction X + 2Y producing products, run 1 has [X] = 0.10 mol dm^-3 and [Y] = 0.20 mol dm^-3 with initial rate 5.0 x 10^-4 mol dm^-3 s^-1; run 2 has [X] = 0.20 and [Y] = 0.20 with rate 1.0 x 10^-3; run 3 has [X] = 0.10 and [Y] = 0.40 with rate 2.0 x 10^-3. Step 1: comparing runs 1 and 2, Y is fixed, [X] doubles and the rate doubles, so X is first order. Step 2: comparing runs 1 and 3, X is fixed, [Y] doubles and the rate quadruples, so Y is second order.
第三步写出速率方程:rate = k[X][Y]^2,总级数为 1 + 2 = 3。第四步代第 1 组数据求 k:k = rate 除以 ([X][Y]^2) = 5.0 x 10^-4 除以 (0.10 x 0.20^2) = 5.0 x 10^-4 除以 4.0 x 10^-3 = 0.125。第五步,总级数为 3,所以 k 的单位是 mol^-2 dm^6 s^-1,完整答案 k = 0.125 mol^-2 dm^6 s^-1。最后可验证:把 k 代回第 2 组,rate = 0.125 x 0.20 x 0.20^2 = 1.0 x 10^-3,与实验值吻合,说明推导无误。
Step 3: the rate equation is rate = k[X][Y]^2 with overall order 1 + 2 = 3. Step 4: substitute run 1 to find k: k = rate divided by ([X][Y]^2) = 5.0 x 10^-4 divided by (0.10 x 0.20^2) = 5.0 x 10^-4 divided by 4.0 x 10^-3 = 0.125. Step 5: with overall order 3, the units of k are mol^-2 dm^6 s^-1, so the full answer is k = 0.125 mol^-2 dm^6 s^-1. As a check, substitute k back into run 2: rate = 0.125 x 0.20 x 0.20^2 = 1.0 x 10^-3, matching the experimental value and confirming the derivation.
最后盘点常见失分点:把配平系数当作反应级数;比较两组数据时忘记保持第三个浓度不变,导致级数判断错误;k 的单位只写 s^-1 而不看总级数;初速率取错切线导致斜率偏差;以及把 k 当成会随浓度改变的量。备考建议:把近五年真题里的动力学数据题集中练一遍,每道题都坚持"先分离变量、再写方程、最后验算单位"三步走,正确率会明显提升。
Finally, the common mark traps: copying stoichiometric coefficients as orders; comparing two runs without holding the third concentration constant, which corrupts the order; writing s^-1 as the unit of k without considering the overall order; taking the wrong tangent when extracting an initial rate; and treating k as if it depended on concentration. For revision, work through every kinetics data question in the past five years of past papers, and always follow the three-step routine: isolate the variable, write the equation, then verify the units.
海外直订AQA GCSE Chemistry Student Book AQA GCSE化学学生手册¥588.0📱 去京东购买京东精选
A-Level / GCSE / IB / KS2-KS3 全科辅导,一对一或小班均可
16621398022 同微信
Visit aleveler.com
或截图后用「扫一扫 → 相册」选择本图