📚 Calculations Involving the Rate Constant, k | 速率常数 k 的相关计算
For A-level Chemistry, the rate constant k is the proportionality constant in the rate equation. It allows chemists to calculate the rate of a reaction once the concentrations of reactants and the orders of reaction are known. The value of k is fixed for a given reaction at a constant temperature, but it changes if the temperature changes or if a catalyst is added.
在 A-level 化学中，速率常数 k 是速率方程中的比例常数。只要知道反应物浓度和反应级数，就可以用 k 计算反应速率。k 在温度不变的条件下对某个具体反应是常数，但温度改变或加入催化剂时，k 会发生变化。
rate = k [A]ᵐ [B]ⁿ
Here [A] and [B] are concentrations, while m and n are the orders of reaction with respect to A and B. The overall order is m + n.
其中 [A] 和 [B] 是反应物浓度，m 和 n 分别是对 A 和 B 的反应级数。总反应级数为 m + n。
1. Understanding the Role of k | 理解 k 的作用
The rate constant k is not affected by the concentrations of reactants. It is only affected by temperature and the presence of a catalyst. A larger k means a faster reaction under the same concentration conditions.
速率常数 k 不受反应物浓度影响，只受温度和催化剂影响。在相同浓度条件下，k 越大，反应速率越快。
For a given rate equation, the units of k must be worked out carefully because they depend on the overall order of reaction.
对于给定的速率方程，k 的单位必须根据总反应级数仔细推导，因为总级数不同，k 的单位也不同。
2. Units of k and Overall Order | 速率常数 k 的单位与总反应级数
Rate always has units of mol dm⁻³ s⁻¹. The units of k are found by rearranging the rate equation so that k is the subject. For a reaction with overall order n, the general unit expression is:
反应速率的单位始终是 mol dm⁻³ s⁻¹。将速率方程整理成 k 为被求项，就可以得到 k 的单位。对于总级数为 n 的反应，k 的单位通式为：
units of k = mol¹⁻ⁿ dm³ⁿ⁻³ s⁻¹
| Overall order | Units of k |
| 0 | mol dm⁻³ s⁻¹ |
| 1 | s⁻¹ |
| 2 | dm³ mol⁻¹ s⁻¹ |
| 3 | dm⁶ mol⁻² s⁻¹ |
You must be able to state these units or derive them from given data. Examiners frequently test whether you can match the units of k to the correct overall order.
你必须能够直接写出这些单位，或根据给定数据推导出来。考试中经常考查 k 的单位与总反应级数的对应关系。
3. Calculating k from a Single Initial Rate | 从单一初始速率计算 k
If the rate equation is already known, k can be calculated by substituting one set of concentration and rate data. For example, for the reaction 2NO + O₂ → 2NO₂, the rate equation is:
如果速率方程已知，就可以代入一组浓度和速率数据计算 k。例如，反应 2NO + O₂ → 2NO₂ 的速率方程为：
rate = k [NO]² [O₂]
If [NO] = 0.050 mol dm⁻³, [O₂] = 0.020 mol dm⁻³ and initial rate = 3.0 × 10⁻³ mol dm⁻³ s⁻¹, then:
若 [NO] = 0.050 mol dm⁻³，[O₂] = 0.020 mol dm⁻³，初始速率 = 3.0 × 10⁻³ mol dm⁻³ s⁻¹，则：
k = rate ÷ ([NO]² [O₂]) = 3.0 × 10⁻³ ÷ (0.050² × 0.020) = 60 dm⁶ mol⁻² s⁻¹
Always include the correct units with k because a number alone is not accepted in A-level answers.
k 必须带有正确单位，因为在 A-level 答案中只写数字是不给分的。
4. Using Initial Rate Tables to Find Orders and k | 利用初始速率表求反应级数和 k
When the rate equation is not given, you must use experimental initial rate data to determine the order with respect to each reactant first, then calculate k.
如果题目没有给出速率方程，你必须先利用初始速率实验数据确定各反应物的级数，再计算 k。
| Experiment | [A] / mol dm⁻³ | [B] / mol dm⁻³ | Initial rate / mol dm⁻³ s⁻¹ |
| 1 | 0.10 | 0.10 | 2.0 × 10⁻³ |
| 2 | 0.20 | 0.10 | 4.0 × 10⁻³ |
| 3 | 0.20 | 0.20 | 1.6 × 10⁻² |
Comparing experiments 1 and 2: [A] doubles while [B] is constant, and the rate doubles. This shows first order with respect to A. Comparing experiments 2 and 3: [B] doubles while [A] is constant, and the rate increases by a factor of 4. This shows second order with respect to B.
对比实验 1 和 2：保持 [B] 不变，[A] 加倍，速率加倍，说明对 A 是一级。对比实验 2 和 3：保持 [A] 不变，[B] 加倍，速率变为 4 倍，说明对 B 是二级。
The rate equation is therefore rate = k [A][B]². Using experiment 1, k = 2.0 × 10⁻³ ÷ (0.10 × 0.10²) = 2.0 dm⁶ mol⁻² s⁻¹.
因此速率方程为 rate = k [A][B]²。使用实验 1 的数据，k = 2.0 × 10⁻³ ÷ (0.10 × 0.10²) = 2.0 dm⁶ mol⁻² s⁻¹。
5. First-Order Integrated Equation and k | 一级反应的积分式与 k
For a first-order reaction A → products, the concentration of A at time t is related to the initial concentration by the integrated rate equation:
对于一级反应 A → products，反应物 A 在时间 t 的浓度与初始浓度之间的关系可用积分速率方程表示：
ln [A]ₜ = ln [A]₀ − kt
This equation can be rearranged to calculate k from concentration-time data. If [A] falls from 0.80 mol dm⁻³ to 0.20 mol dm⁻³ in 50 s, then:
该方程可变形后用浓度-时间数据计算 k。若 [A] 在 50 s 内从 0.80 mol dm⁻³ 降至 0.20 mol dm⁻³，则：
k = ln([A]₀ ÷ [A]ₜ) ÷ t = ln(0.80 ÷ 0.20) ÷ 50 = ln 4 ÷ 50 = 0.0277 s⁻¹
This method is especially useful when a set of concentrations at different times is given, rather than initial rates.
当题目给出不同时间的浓度而不是初始速率时，这种方法特别有用。
6. Half-Life Method for First-Order k | 用半衰期求一级反应 k
For a first-order reaction, the half-life t½ is independent of concentration and is related to k by:
对于一级反应，半衰期 t½ 与浓度无关，并与 k 有以下关系：
t½ = ln 2 ÷ k = 0.693 ÷ k
If the half-life of a first-order reaction is 120 s, then k = 0.693 ÷ 120 = 5.78 × 10⁻³ s⁻¹. Conversely, if k is known, the half-life can be calculated directly.
如果一级反应的半衰期为 120 s，则 k = 0.693 ÷ 120 = 5.78 × 10⁻³ s⁻¹。反之，已知 k 也可以直接计算半衰期。
7. The Arrhenius Equation and k | 阿伦尼乌斯方程与 k
The rate constant k depends on temperature. The Arrhenius equation shows this relationship:
速率常数 k 随温度变化。阿伦尼乌斯方程表达了这一关系：
k = A e^(−Eₐ ÷ RT)
In this equation, A is the pre-exponential factor, Eₐ is the activation energy, R is the gas constant, and T is the absolute temperature in kelvin.
式中 A 是指前因子，Eₐ 是活化能，R 是气体常数，T 是开尔文温度。
Taking natural logarithms gives the linear form:
对两边取自然对数，得到线性形式：
ln k = ln A − Eₐ ÷ (RT)
This version is used to calculate k at different temperatures or to determine Eₐ graphically.
这个形式用于计算不同温度下的 k，或通过作图求活化能 Eₐ。
8. Calculating k at a New Temperature | 计算新温度下的 k
When two rate constants at two temperatures are involved, use the two-point Arrhenius equation:
当涉及两个温度下的速率常数时，使用阿伦尼乌斯方程的两点式：
ln(k₂ ÷ k₁) = (Eₐ ÷ R) × (1 ÷ T₁ − 1 ÷ T₂)
For example, Eₐ = 55.0 kJ mol⁻¹, k₁ = 2.5 × 10⁻³ s⁻¹ at T₁ = 300 K. To find k₂ at 320 K, first convert Eₐ to J mol⁻¹: 55.0 × 10³ J mol⁻¹. Then:
例如，Eₐ = 55.0 kJ mol⁻¹，k₁ = 2.5 × 10⁻³ s⁻¹，T₁ = 300 K。求 320 K 时的 k₂，先要把 Eₐ 换算成 J mol⁻¹：55.0 × 10³ J mol⁻¹。然后：
ln(k₂ ÷ k₁) = (55 000 ÷ 8.31) × (1 ÷ 300 − 1 ÷ 320) = 1.379
k₂ = 2.5 × 10⁻³ × e^(1.379) = 9.93 × 10⁻³ s⁻¹
Remember that temperature must always be in kelvin and activation energy in J mol⁻¹ when R = 8.31 J K⁻¹ mol⁻¹ is used.
使用 R = 8.31 J K⁻¹ mol⁻¹ 时，温度必须用开尔文，活化能必须用 J mol⁻¹。
9. Graphical Determination of k and Eₐ | 图形法求 k 和活化能
For a first-order reaction, plotting ln [A]ₜ against time t gives a straight line with slope equal to −k. This allows k to be calculated from the gradient without knowing individual rates.
对于一级反应，以 ln [A]ₜ 对时间 t 作图得到一条直线，斜率为 −k。这样无需知道各个时刻的反应速率，就可以从斜率求出 k。
gradient = −k
For the Arrhenius equation, plotting ln k against 1 ÷ T gives a straight line with slope equal to −Eₐ ÷ R. The intercept is ln A.
对于阿伦尼乌斯方程，以 ln k 对 1 ÷ T 作图得到一条直线，斜率为 −Eₐ ÷ R，截距为 ln A。
gradient = −Eₐ ÷ R
These graphical methods are common in Cambridge A-level practical and written questions.
这些图形法在剑桥 A-level 实验和笔试题目中很常见。
10. Common Pitfalls and Exam Tips | 常见错误与应试技巧
Always match units of k to the overall order. A second-order reaction should have k in dm³ mol⁻¹ s⁻¹, not s⁻¹.
务必让 k 的单位与总反应级数匹配。二级反应的 k 单位应为 dm³ mol⁻¹ s⁻¹，而不是 s⁻¹。
When using initial rates, compare experiments where only one concentration changes. If both concentrations change at once, you cannot isolate the order directly.
使用初始速率时，要比较只有一个浓度发生变化的实验。如果两个浓度同时改变，就不能直接确定单个反应物的级数。
Do not confuse k with the reaction rate. k is constant at constant temperature, while the rate changes as concentrations change.
不要把 k 与反应速率混淆。恒温条件下 k 是常数，而速率会随浓度变化而变化。
In Arrhenius calculations, temperature must be in kelvin and Eₐ must be in J mol⁻¹ if R = 8.31 J K⁻¹ mol⁻¹. Also remember that a catalyst increases k by lowering Eₐ, not by changing concentration.
在阿伦尼乌斯计算中，温度必须用开尔文，若 R = 8.31 J K⁻¹ mol⁻¹，Eₐ 必须用 J mol⁻¹。还要记住催化剂通过降低 Eₐ 提高 k，而不是通过改变浓度。
Published by TutorHao | Chemistry Revision Series | aleveler.com
更多咨询请联系16621398022（同微信）
屏轩国际教育cambridge primary/secondary checkpoint, cat4, ukiset,ukcat,igcse,alevel,PAT,STEP,MAT, ibdp,ap,ssat,sat,sat2课程辅导，国外大学本科硕士研究生博士课程论文辅导