📚 Equilibrium Calculations | 平衡常数计算
Equilibrium calculations bring quantitative precision to the qualitative idea of reversible reactions. For IB Chemistry students, mastering the mathematics of Kc, Kp, Q, and their link to Gibbs free energy is essential for understanding both industrial processes and biological systems. This article systematically covers the concepts, rules, and problem-solving techniques required for success in equilibrium questions, pairing every English explanation with its Chinese counterpart to support bilingual learning.
平衡计算为可逆反应的定性概念提供了定量的精确性。对于 IB 化学学生来说，掌握 Kc、Kp、Q 及其与吉布斯自由能关联的数学，是理解工业过程和生物系统的基础。本文系统地涵盖了平衡问题所需的概念、规则和解题技巧，并将每一段英文解释与其对应的中文说明配对，以支持双语学习。
1. Introduction to Dynamic Equilibrium | 动态平衡简介
Dynamic equilibrium occurs when a reversible reaction proceeds in both directions at the same rate, leaving macroscopic properties like concentration and pressure constant. It is a state of balance, but at the molecular level, the forward and reverse reactions continue indefinitely.
当可逆反应以相同的速率向两个方向进行，使得浓度、压强等宏观性质保持不变时，即达到动态平衡。这是一种平衡状态，但在分子水平上，正反应和逆反应仍在无限期地进行着。
The equilibrium can be approached from either direction. Whether starting with reactants only or products only, the system will eventually reach the same equilibrium composition if conditions are unchanged.
反应可以从任一方向趋近平衡。不论是仅从反应物开始还是仅从生成物开始，只要条件不变，体系最终都会达到相同的平衡组成。
2. The Equilibrium Constant Kc | 平衡常数 Kc
The equilibrium constant, denoted Kc, is a dimensionless or dimensional number that relates the concentrations of products and reactants at equilibrium at a given temperature. For a general reaction aA + bB ⇌ cC + dD, the expression is:
平衡常数，记作 Kc，是一个量纲或无量纲的数，用于关联给定温度下平衡时产物和反应物的浓度。对于一般反应 aA + bB ⇌ cC + dD，表达式为：
Kc = [C]c[D]d / ([A]a[B]b)
The square brackets denote concentration in mol dm-3. The stoichiometric coefficients become exponents in the expression. Kc is constant for a specific reaction at a fixed temperature; changing temperature alters Kc.
方括号表示以 mol dm-3 为单位的浓度。化学计量系数在表达式中成为指数。对于特定反应，在温度固定时 Kc 为常数；改变温度会改变 Kc 值。
3. Writing Kc Expressions: Rules | 书写 Kc 表达式的规则
Only species in the gaseous or aqueous phase appear in the Kc expression. Pure solids and pure liquids (including water as a solvent when it is in large excess) are omitted because their concentrations remain essentially constant.
只有处于气态或水溶液中的物种才会出现在 Kc 表达式中。纯固体和纯液体（包括作为溶剂大量存在的水）被省略，因为它们的浓度基本恒定。
For heterogeneous equilibria, this simplification is crucial. For example, in CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂]. Ignoring the solids greatly simplifies calculations.
对于多相平衡，这一简化至关重要。例如，在 CaCO₃(s) ⇌ CaO(s) + CO₂(g) 中，Kc = [CO₂]。忽略固体使得计算大大简化。
IB questions often test the ability to write correct expressions before doing any arithmetic. Always check the state symbols (s), (l), (g), (aq) given in the equation.
IB 考题经常在开始计算之前先检验学生书写正确表达式的能力。务必检查方程中给出的状态符号 (s)、(l)、(g)、(aq)。
4. ICE Tables: Initial-Change-Equilibrium | ICE 表格法：初始-变化-平衡
The ICE (Initial-Change-Equilibrium) table is the most powerful tool for equilibrium calculations. It organises the initial amounts, the stoichiometric changes, and the equilibrium concentrations in a systematic way.
ICE（初始-变化-平衡）表格是平衡计算中最有力的工具。它系统地整合了初始物质的量、根据化学计量关系发生的变化以及平衡浓度。
Set up the table with rows for each species. Under Initial, place known initial concentrations. In Change, use -x for reactants, +x for products, multiplied by stoichiometric coefficients. Equilibrium row sums Initial + Change.
为每种物质设置行。在初始一行填入已知的起始浓度。在变化一行，反应物用 -x，生成物用 +x，并乘以化学计量系数。平衡一行等于初始与变化之和。
For example, for H₂(g) + I₂(g) ⇌ -x, -x, +2x; Equilibrium: 1.0-x, 1.0-x, 2x. Substitute into Kc expression and solve for x.
例如，对于 H₂(g) + I₂(g) ⇌ 2HI(g)，若初始 [H₂] = 1.0，[I₂] = 1.0，[HI] = 0，则变化为 -x、-x、+2x；平衡浓度为 1.0-x、1.0-x、2x。代入 Kc 表达式并求解 x。
5. Calculating Kc from Equilibrium Data | 从平衡数据计算 Kc
Once equilibrium concentrations are known or found via ICE, Kc is calculated by substituting into the expression. Always check whether the problem gives initial and equilibrium amounts in moles or concentrations, and whether the volume is 1.0 dm³.
一旦已知平衡浓度或通过 ICE 表求得，将数值代入表达式即可计算 Kc。务必检查题目给出的是物质的量还是浓度，以及体积是否为 1.0 dm³。
If amounts are given in moles, divide by the volume (in dm³) to obtain concentrations. Common IB trap: providing moles in a 500 cm³ flask without converting to dm³. Always convert cm³ to dm³ by dividing by 1000.
如果给出的是物质的量，需除以体积（以 dm³ 为单位）获得浓度。IB 常设陷阱：给出 500 cm³ 烧瓶中的物质的量却不要求转换单位。切记将 cm³ 除以 1000 转换为 dm³。
Kc may have units depending on the change in the number of moles of gas in the reaction, Δn. However, IB often treats Kc as dimensionless for simplicity, but students should be aware of units to check consistency.
根据反应中气体物质的量的变化 Δn，Kc 可能具有单位。不过 IB 通常为简化起见将 Kc 视为无量纲，但学生应了解单位以确保一致性。
6. Gas Equilibria and Kp | 气体平衡与 Kp
For reactions involving gases, the equilibrium constant can be expressed in terms of partial pressures, denoted Kp. Partial pressure is the pressure a gas would exert if it occupied the container alone, and is related to mole fraction and total pressure.
对于涉及气体的反应，平衡常数可以用分压表示，记作 Kp。分压是气体单独占据容器时的压强，它与摩尔分数和总压强有关。
total moles of gas. Then partial pressure = mole fraction × total pressure. So pA = (nA/ntotal) × Ptotal.
气体的摩尔分数 = 该气体的物质的量 / 气体的总物质的量。分压 = 摩尔分数 × 总压强。即 pA = (nA/ntotal) × Ptotal。
Kp expression mirrors Kc but uses partial pressures instead of concentrations. For aA + bB ⇌ cC + dD: Kp = (pCc × pDd) / (pAa × pBb). Kp is also temperature-dependent.
Kp 表达式与 Kc 类似，只是用分压代替浓度。对于 aA + bB ⇌ cC + dD：Kp = (pCc × pDd) / (pAa × pBb)。Kp 同样依赖于温度。
7. Relationship Between Kp and Kc | Kp 与 Kc 的关系
as equation: p = (n/V)RT = [concentration]RT. For the reaction stoichiometry, this yields:
Kp 和 Kc 通过理想气体方程式关联：p = (n/V)RT = [浓度]RT。根据反应计量关系，可得：
Kp = Kc (RT)Δn
moles of gas. R is the gas constant, and T is temperature in Kelvin. This equation is particularly useful when converting between Kp and Kc in IB problems.
其中 Δn = (c+d) - (a+b)，即气体物质的量的变化。R 是气体常数，T 是开尔文温度。在 IB 题目中，当需要在 Kp 和 Kc 之间进行换算时，该方程尤其有用。
p = Kc, and both are dimensionless. This simplifies calculation for reactions like H₂ + I₂ ⇌ 2HI. Always calculate Δn carefully.
请注意，如果 Δn = 0，Kp = Kc，并且两者均无量纲。对于像 H₂ + I₂ ⇌ 2HI 这样的反应，这简化了计算。务必仔细计算 Δn。
8. Reaction Quotient Q: Predicting Direction | 反应商 Q：预测反应方向
The reaction quotient, Q, has the same mathematical form as Kc or Kp but uses concentrations or pressures at any point, not necessarily at equilibrium. Comparing Q with K tells us the direction the reaction must shift to reach equilibrium.
反应商 Q 与 Kc 或 Kp 具有相同的数学形式，但它使用的是任意时刻（不一定是平衡时）的浓度或压强。将 Q 与 K 进行比较，可以告诉我们反应必须向哪个方向移动才能达到平衡。
shift to the left (more reactants) to decrease Q. If Q = K, the system is at equilibrium.
若 Q < K，体系将向右移动（生成更多产物）以使 Q 增大至 K。若 Q > K，则向左移动（生成更多反应物）以使 Q 减小。若 Q = K，体系处于平衡状态。
Q is especially handy when predicting the effect of sudden changes, such as adding a reactant or changing volume/pressure. By recalculating Q and comparing with K, you can correctly predict the shift direction without memorising Le Chatelier's principle qualitatively.
在预测突然变化的效应时，如添加反应物或改变体积/压强，Q 就特别方便。通过重新计算 Q 并与 K 进行比较，可以正确预测移动方向，而不必定性记忆勒夏特列原理。
9. Effect of Temperature on K (Le Chatelier and van't Hoff) | 温度对 K 的影响（勒夏特列与范特霍夫）
Temperature is the only factor that changes the value of the equilibrium constant. For an exothermic reaction (ΔH < 0), increasing temperature decreases K because heat is a product. For an endothermic reaction (ΔH > 0), increasing temperature increases K.
温度是唯一能改变平衡常数值的因素。对于放热反应（ΔH < 0），升高温度会降低 K，因为热是产物；对于吸热反应（ΔH > 0），升高温度会增大 K。
The quantitative relationship is given by the van' suming ΔH° remains constant over the temperature range. This equation can be used to calculate K at a new temperature.
定量关系由范特霍夫方程给出。其简化形式为：ln(K₂/K₁) = -(ΔH°/R)(1/T₂ - 1/T₁)，并假设 ΔH° 在温度范围内恒定。该方程可用于计算新温度下的 K。
IB questions may ask you to predict and explain the sign of ΔH given K values at two temperatures. A larger K at higher temperature indicates an endothermic forward reaction.
IB 考题可能会要求根据两个温度下的 K 值预测并解释 ΔH 的符号。高温下 K 更大，表明正反应是吸热的。
10. Equilibrium and Gibbs Free Energy | 平衡与吉布斯自由能
determination of the equilibrium constant from standard free energy data, and vice versa.
平衡与热力学之间通过方程 ΔG° = -RT ln K 存在深刻的联系。这使得我们可以从标准自由能数据确定平衡常数，反之亦然。
A positive ΔG° gives K < 1, indicating a reactant-favoured reaction. When ΔG° = 0, K = 1.
ΔG° 为负对应 K > 1，表明在标准条件下反应倾向于产物；ΔG° 为正则 K < 1，反应倾向于反应物。当 ΔG° = 0 时，K = 1。
is equation is powerful for linking thermodynamic spontaneity and equilibrium composition.
此外，当反应不在标准条件下时，ΔG = ΔG° + RT ln Q。在平衡时（ΔG = 0，Q = K），此式简化为 ΔG° = -RT ln K。该方程在联系热力学自发性与平衡组成方面作用强大。
11. Worked Examples: Putting It All Together | 综合计算示例
Example 1: Kc from given equilibrium concentrations. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), at equilibrium [N₂] = 0.50, [H₂] = 0.10, [NH₃] = 0.20 mol dm-3. Calculate Kc.
示例 1：根据给定的平衡浓度计算 Kc。对于 N₂(g) + 3H₂(g) ⇌ 2NH₃(g)，平衡浓度 [N₂] = 0.50，[H₂] = 0.10，[NH₃] = 0.20 mol dm-3，计算 Kc。
) = (0.20)² / (0.50 × (0.10)³) = 0.04 / (0.50 × 0.001) = 0.04 / 0.0005 = 80 (unit: dm⁶ mol-2 or dimensionless per IB context).
解答： ]² / ([N₂][H₂]³) = (0.20)² / (0.50 × (0.10)³) = 0.04 / (0.50 × 0.001) = 0.04 / 0.0005 = 80（单位：dm⁶ mol-2 或根据 IB 习惯视为无量纲）。
Example 2: Using ICE table. H₂(g) + I₂(g) ⇌ 2HI(g) Kc = 50 at 700 K. If initial [H₂] = [I₂] = 0.50 M, find equilibrium [HI].
示例 2：利用 ICE 表格。H₂(g) + I₂(g) ⇌ 2HI(g)，Kc = 50（700 K）。若初始 [H₂] = [I₂] = 0.50 M，求平衡时 [HI]。
2x; eqm 0.50-x, 0.50-x, 2x. Kc = (2x)² / (0.50-x)² = 50. Taking square root: 2x/(0.50-x) = √50 x = 7.07(0.50-x) → 2x = 3.535 - 7.07x → 9.07x = 3.535 → x = 0.39. Then [HI] = 2x = 0.78 M.
建立 ICE 表：初始 H₂ 0.50，I₂ 0.50，HI 0；变化 -x，-x，+2x；平衡 0.50-x，0.50-x，2x。Kc = (2x)² / (0.50-x)² = 50。两边开平方：2x/(0.50-x) = √50 ≈ 7.07。求解得 x = 0.39，则 [HI] = 2x = 0.78 M。
12. Common Pitfalls and Tips | 常见错误与提示
Pitfall 1: Forgetting to convert mass or moles to concentration. Always divide moles by volume in dm³. If volume is not 1 dm³, the ICE table still requires concentrations, not just moles.
陷阱 1：忘记将质量或物质的量转换为浓度。务必用物质的量除以体积（dm³）。如果体积不是 1 dm³，ICE 表格仍需使用浓度，而非仅为物质的量。
Pitfall 2: Including solids or pure liquids in K expressions. Remember that only (g) and (aq) species belong there.
陷阱 2：在 K 表达式中包含了固体或纯液体。请记住只有 (g) 和 (aq) 物种才应写入。
Pitfall 3: Using Kp when given Kc or vice versa without converting via Kp = Kc(RT)Δn. Always check the units and the phase of reactants and products.
陷阱 3：在已知 Kc 时直接使用 Kp，或相反，而没有通过 Kp = Kc(RT)Δn 进行转换。务必检查反应物和产物的相态及单位。
Tip: When solving quadratic equations from ICE tables, if Kc is very large or very small, you may use approximation (neglecting x) to simplify. However, always check the 5% rule or verify the approximation after solving.
提示：在解由 ICE 表格产生的二次方程时，如果 Kc 很大或很小，可使用近似（忽略 x）简化计算。但解出后务必用 5% 规则检验近似是否有理。
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