📚 IB Chemistry: Complete Guide to Chemical Equilibrium Calculations | IB化学：化学平衡计算题型全解析
Chemical equilibrium calculations are one of the most frequently tested areas in IB Chemistry HL and SL papers. From writing Kc and Kp expressions to performing ICE table calculations, this guide systematically breaks down every question type you may encounter, with step-by-step strategies and common pitfall warnings.
化学平衡计算是IB化学HL和SL试卷中出现频率最高的考点之一。从书写Kc和Kp表达式到使用ICE表格进行计算，本指南系统拆解你可能遇到的每一类题型，并给出逐步解题策略与常见误区警示。
1. Writing Equilibrium Constant Expressions | 平衡常数表达式的书写
Every equilibrium calculation begins with correctly writing the equilibrium constant expression. For a general reaction aA + bB ⇌ cC + dD, the equilibrium constant Kc is expressed as the product of product concentrations raised to their stoichiometric coefficients, divided by the product of reactant concentrations raised to their coefficients.
一切平衡计算都以正确书写平衡常数表达式为起点。对于一般反应 aA + bB ⇌ cC + dD，平衡常数 Kc 表示为产物浓度按其化学计量系数幂次方的乘积，除以反应物浓度按其化学计量系数幂次方的乘积。
Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ)
Only aqueous (aq) and gaseous (g) species appear in the expression; pure solids (s) and pure liquids (l) are omitted.
只有水溶液（aq）和气态（g）物质出现在表达式中；纯固体（s）和纯液体（l）被省略。
For gaseous equilibria, Kp uses partial pressures in atm or kPa instead of concentrations.
对于气相平衡，Kp 使用以 atm 或 kPa 表示的分压，而非浓度。
The coefficients in the balanced equation become the exponents in the expression — this is non-negotiable.
平衡方程式中的计量系数成为表达式中的指数——这是不可妥协的规则。
2. The ICE Table Method | ICE表格法
The ICE (Initial, Change, Equilibrium) table is the most powerful tool for solving equilibrium concentration problems. It organizes the initial amounts, the change that occurs as the system reaches equilibrium, and the final equilibrium amounts.
ICE（Initial-初始, Change-变化, Equilibrium-平衡）表格是求解平衡浓度问题最强大的工具。它系统整理了系统的初始量、趋向平衡时发生的改变量以及最终平衡量。
Step 1: Write the balanced equation. Step 2: Fill in the ICE row by row. Step 3: Substitute equilibrium values into K expression.
For example, for the reaction N₂(g) + 3H₂(g) ⇌ ount of N₂ consumed at equilibrium. Then the change row is -x for N₂, -3x for H₂, and +2x for NH₃. The equilibrium row becomes 1.0 - x, 3.0 - 3x, and 2x respectively.
例如，对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，若初始 [N₂] = 1.0 mol dm⁻³、[H₂] = 3.0 mol dm⁻³，设 x 为平衡时 N₂ 的消耗量，则变化行中 N₂ 为 -x，H₂ 为 -3x，NH₃ 为 +2x。平衡行分别变为 1.0 - x、3.0 - 3x 和 2x。
Assign x based on the species with a coefficient of 1 whenever possible to simplify algebra.
尽量选择化学计量系数为1的物质来设 x，以简化代数运算。
The change row must follow the stoichiometric ratio exactly: for every x of N₂ consumed, 3x of H₂ is consumed and 2x of NH₃ is produced.
变化行必须严格遵循化学计量比：每消耗 x 的 N₂，同时消耗 3x 的 H₂，生成 2x 的 NH₃。
3. Calculating Kc from Equilibrium Concentrations | 由平衡浓度求Kc
This is the most direct question type. You are given all equilibrium concentrations and asked to calculate the value of Kc with units. The key is to substitute correctly and then handle units carefully.
这是最直接的题型。题目给出所有平衡浓度，要求计算 Kc 的值及单位。关键在于正确代入，并谨慎处理单位。
Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ) with units (mol dm⁻³)ᶜ⁺ᵈ⁻ᵃ⁻ᵇ
For the reaction H₂(g) + I₂(g) ⇌ c = (1.40)² / (0.20 × 0.20) = 49. The units are (mol dm⁻³)² / (mol dm⁻³)² = dimensionless.
对于反应 H₂(g) + I₂(g) ⇌ 2HI(g)，若平衡时 [H₂] = 0.20 mol dm⁻³、[I₂] = 0.20 mol dm⁻³、[HI] = 1.40 mol dm⁻³，则 Kc = (1.40)² / (0.20 × 0.20) = 49。单位为 (mol dm⁻³)² / (mol dm⁻³)²，即无单位。
Always check the stoichiometric coefficients: squaring the HI concentration is a common point of error.
始终检查化学计量系数：对 HI 浓度取平方是常见的易错点。
fference in number of gaseous moles: only when Δn = c + d - a - b = 0 will Kc be unitless.
注意气态物质的摩尔数之差：只有当 Δn = c + d - a - b = 0 时，Kc 才是无单位的。
4. Solving for Unknown Concentrations Using Kc | 利用Kc求未知浓度
When Kc is known and initial concentrations are given, you must set up an ICE table and solve a quadratic equation. This is the classic 7-mark IB exam question.
当 Kc 已知且给出初始浓度时，你需要建立 ICE 表格并解一元二次方程。这是IB考试中经典的7分题型。
Consider the esterification reaction: CH₃COOH(aq) + C₂H₅OH(aq) ⇌ thanol are mixed, and Kc = 4.0, let x be the equilibrium concentration of the ester. Then:
考虑酯化反应：CH₃COOH(aq) + C₂H₅OH(aq) ⇌ CH₃COOC₂H₅(aq) + H₂O(l)。由于水是纯液体，在 Kc 表达式中省略。若各取 1.0 mol 乙酸和乙醇混合，且 Kc = 4.0，设 x 为平衡时酯的浓度，则：
4.0 = x² / (1.0 - x)²
l dm⁻³. Notice that taking square roots cleverly avoids solving a full quadratic equation.
两边开平方得 x / (1.0 - x) = 2.0，故 x = 0.67 mol dm⁻³。注意：两边开平方可巧妙地避免解完整的二次方程。
When the numerator and denominator have the same sum of exponents, taking square roots is a powerful simplification.
当分子与分母的指数之和相同时，两边同时开方是极有效的简化手段。
If quadratic solving is unavoidable, always choose the root that gives positive equilibrium concentrations.
若不可避免地需要解二次方程，始终选择能给出正值平衡浓度的根。
5. The Reaction Quotient Q and Direction of Shift | 反应商Q与反应方向判断
The reaction quotient Q has the same form as Kc but is calculated using concentrations that are not necessarily at equilibrium. Comparing Q with Kc tells you which direction the reaction will proceed.
反应商 Q 与 Kc 形式相同，但使用不一定是平衡状态的浓度进行计算。比较 Q 与 Kc 的大小可判断反应进行的方向。
If Q < Kc: forward direction dominates (→). If Q > Kc: reverse direction dominates (←). If Q = Kc: at equilibrium.
For the reaction N₂(g) + 3H₂(g) ⇌ .10 × (0.10)³) = 10,000. Since Q > Kc, the reaction shifts left, producing more N₂ and H₂.
对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，已知 Kc = 500。设某一时刻 [N₂] = 0.10 mol dm⁻³、[H₂] = 0.10 mol dm⁻³、[NH₃] = 1.0 mol dm⁻³，则 Q = (1.0)² / (0.10 × (0.10)³) = 10,000。因为 Q > Kc，反应向左移动，生成更多的 N₂ 和 H₂。
Q is the single most important tool for predicting shifts when a system is disturbed — far more reliable than Le Chatelier intuition alone.
Q 是判断系统受扰动后移动方向最重要的工具——远比仅仅依靠勒夏特列原理的直觉更可靠。
In gas reactions, if concentrations change due to volume changes, recalculate Q before concluding.
在气体反应中，如果浓度因体积变化而改变，务必先重新计算 Q 再下结论。
6. Mole Fraction and Partial Pressure | 摩尔分数与分压
ns, and partial pressures. The partial pressure of a gas = mole fraction × total pressure.
对于气相平衡，你必须熟练掌握摩尔数、摩尔分数和分压之间的相互换算。某气体的分压 = 摩尔分数 × 总压。
P(A) = (nₐ / ntotal) × Ptotal
For the reaction 2SO₂(g) + O₂(g) ⇌ 0.20 × 3.0 = 0.60 atm, P(O₂) = 0.30 × 3.0 = 0.90 atm, and P(SO₃) = 0.50 × 3.0 = 1.50 atm.
对于反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)，若平衡时总压为 3.0 atm，摩尔分数分别为 X(SO₂) = 0.20、X(O₂) = 0.30、X(SO₃) = 0.50，则分压分别为 P(SO₂) = 0.20 × 3.0 = 0.60 atm、P(O₂) = 0.30 × 3.0 = 0.90 atm、P(SO₃) = 0.50 × 3.0 = 1.50 atm。
Mole fractions always sum to 1; use this as a check.
摩尔分数之和恒等于1，可用作检验。
When initial moles and total pressure are given but equilibrium moles are unknown, set up the ICE table in moles first, then convert to partial pressures at the end.
当已知初始摩尔数和总压但未知平衡摩尔数时，先以摩尔数为单位建立 ICE 表格，最后再转换为分压。
7. Relating Kp and Kc | Kp与Kc的换算
The relationship between Kp and Kc for a gas-phase reaction depends on Δn, the change in the number of moles of gas. This is a favourite HL exam question.
气相反应的 Kp 与 Kc 之间的关系取决于 Δn，即气体摩尔数的变化量。这是HL考试偏爱的考点。
Kp = Kc(RT)^Δn
emperature in Kelvin, and Δn = (moles of gaseous products) - (moles of gaseous reactants).
其中 R = 0.0821 L atm mol⁻¹ K⁻¹ 或 8.314 J mol⁻¹ K⁻¹，T 为热力学温度（开尔文），Δn = (气态产物的摩尔数) - (气态反应物的摩尔数)。
For the reaction N₂(g) + 3H₂(g) ⇌ = Kc(RT)⁻² = Kc / (RT)². The conversion factor clearly depends on both temperature and Δn.
对于反应 N₂(g) + 3H₂(g) ⇌ 2NH₃(g) 在 400 K 时，Δn = 2 - 4 = -2。因此 Kp = Kc(RT)⁻² = Kc / (RT)²。换算因子显然依赖于温度和 Δn 两者。
When Δn = 0, Kp = Kc — no conversion is needed.
当 Δn = 0 时，Kp = Kc——无需换算。
pressure is in atm and concentration in mol dm⁻³, using R = 0.0821 maintains consistency.
密切关注单位：若压强以 atm、浓度以 mol dm⁻³ 为单位，使用 R = 0.0821 可保持一致性。
8. Temperature Dependence of K | 平衡常数对温度的依赖
Only temperature can change the value of the equilibrium constant itself. Adding or removing substances, changing pressure, or using a catalyst does not alter K.
只有温度才能改变平衡常数本身的数值。增加或移除物质、改变压强或使用催化剂均不会改变 K。
Endothermic reactions: increasing T increases K. Exothermic reactions: increasing T decreases K.
Consider the endothermic reaction N₂O₄(g) ⇌ 2NO₂(g) (ΔH > 0). Raising the temperature supplies energy that drives the endothermic forward reaction, so K increases. Conversely, for the exothermic reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) (ΔH < 0), raising the temperature decreases K, reducing the equilibrium yield of SO₃.
考虑吸热反应 N₂O₄(g) ⇌ 2NO₂(g)（ΔH > 0）。升高温度提供了驱动正向吸热反应的能量，因此 K 增大。反之，对于放热反应 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)（ΔH < 0），升高温度会使 K 减小，从而降低 SO₃ 的平衡产率。
The van' is: ln(K₂/K₁) = -ΔH°/R × (1/T₂ - 1/T₁), which you may encounter in HL extension questions.
范特霍夫方程将其定量化：ln(K₂/K₁) = -ΔH°/R × (1/T₂ - 1/T₁)，这在HL拓展题中可能遇到。
Mentally verify: does your K value change in the direction predicted by Le Chatelier's principle? If not, revisit your reasoning.
心算验证：你的 K 值变化方向是否与勒夏特列原理预测的一致？若不一致，请重新审视推理过程。
9. Multiple Equilibria and Combined K Values | 多重平衡与K值组合
When a reaction can be expressed as the sum of two or more elementary reactions, the equilibrium constant for the overall reaction is the product of the individual K values.
当一个反应可以被表示成两个或多个基本反应之和时，总反应的平衡常数等于各分步 K 值的乘积。
If reaction 3 = reaction 1 + reaction 2, then K₃ = K₁ × K₂.
1/K). If a reaction is multiplied by a coefficient n, the new K is Kⁿ. For example, for 2A ⇌ B with K₁ = 4.0, the reverse reaction B ⇌ 2A has K = 1/4.0 = 0.25, and the half-reaction A ⇌ ½B has K = √4.0 = 2.0.
若反应被逆向书写，新的 K 为原来 K 的倒数（1/K）。若反应系数乘以 n，新的 K 为 Kⁿ。例如，对于 2A ⇌ B 且 K₁ = 4.0，逆反应 B ⇌ 2A 的 K = 1/4.0 = 0.25；而半反应 A ⇌ ½B 的 K = √4.0 = 2.0。
These rules follow directly from the law of mass action — no memorization beyond understanding the algebra of exponents.
这些规则直接来源于质量作用定律——理解了指数运算，无需死记硬背。
In multi-step equilibria, identify the common species and cancel them carefully, just as in Hess's law calculations.
在多步平衡中，仔细识别共同物质并相消，原理与赫斯定律计算相同。
10. Common Pitfalls and Exam Tips | 常见易错点与应试技巧
Students frequently lose marks in equilibrium calculations due to a handful of recurring errors. Being aware of these before the exam is half the battle won.
学生在平衡计算中常因几个反复出现的错误而失分。考前意识到这些错误就已成功了一半。
| Pitfall | Correct Approach |
| Writing H₂O(l) in Kc expression | Omit pure liquids and solids |
| Forgetting to square/cube exponents | Match exponents to stoichiometric coefficients exactly |
| Using initial concentrations as equilibrium concentrations | Always complete the ICE table first |
| Choosing the wrong quadratic root | Eliminate roots giving negative concentrations |
| Confusing Q and K in shift predictions | Use the Q vs K comparison, not memorised rules alone |
In the exam, always write down the K expression first before substituting values. This earns you method marks even if your final arithmetic is incorrect. Keep an eye on significant figures: IB usually requires answers to 2 or 3 significant figures.
考试中，务必先写出 K 的表达式再代入数值。即使最终计算有误，这也能为你赢得方法步骤分。注意有效数字：IB通常要求答案保留2至3位有效数字。
11. Worked Example: Full 7-Mark Question | 真题演练：完整7分题
t us work through a complete IB-style question together. Given the reaction CO(g) + H₂O(g) ⇌ in temperature, if 2.00 mol of CO and 2.00 mol of H₂O are placed in a 1.00 dm³ container, calculate the equilibrium concentration of each species.
让我们完整演练一道IB风格题目。已知反应 CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)，某温度下 Kc = 5.00。若将 2.00 mol 的 CO 和 2.00 mol 的 H₂O 置于 1.00 dm³ 容器中，求各物种的平衡浓度。
Step 1 | ICE Table:
| Conc. (mol dm⁻³) | CO | H₂O | CO₂ | H₂ |
| Initial | 2.00 | 2.00 | 0 | 0 |
| Change | -x | -x | +x | +x |
| Equilibrium | 2.00 - x | 2.00 - x | x | x |
Step 2 | Write Kc expression and substitute:
Kc = [CO₂][H₂] / ([CO][H₂O]) = x² / (2.00 - x)² = 5.00
Step 3 | Take square root: x / (2.00 - x) = √5.00 = 2.236, giving x = 2.236(2.00 - x), so x(1 + 2.236) = 4.472, x = 1.382 mol dm⁻³.
步骤3 | 两边开平方： x / (2.00 - x) = √5.00 = 2.236，得 x = 2.236(2.00 - x)，于是 x(1 + 2.236) = 4.472，x = 1.382 mol dm⁻³。
Step 4 | State equilibrium concentrations: 618 mol dm⁻³; [CO₂] = [H₂] = 1.382 mol dm⁻³. Verification: Kc = (1.382)² / (0.618)² = 5.00 ✓
步骤4 | 写出平衡浓度： [CO] = [H₂O] = 2.00 - 1.382 = 0.618 mol dm⁻³；[CO₂] = [H₂] = 1.382 mol dm⁻³。验算：Kc = (1.382)² / (0.618)² = 5.00 ✓
Always verify your answer by substituting back into the K expression — this catches algebraic errors instantly.
始终将答案代回 K 表达式验算——这能立刻发现代数错误。
Show the ICE table and K expression in full; IB examiners award marks for each correct step, not just the final answer.
完整呈现 ICE 表格和 K 表达式；IB考官按正确步骤给分，而非仅看最终答案。
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