Further Aspects of Equilibria | 化学平衡的进阶方面

📚 Further Aspects of Equilibria | 化学平衡的进阶方面

In A-level Chemistry, equilibrium is not just about predicting the direction of shift. A deeper quantitative understanding involves equilibrium constants K_c and K_p, their calculations from experimental data, the effect of temperature on their values, and applications to both homogeneous and heterogeneous systems. This article explores these further aspects, including acid-base equilibrium constants, to build a solid foundation for exam success.

在A-level化学中，平衡不仅仅是预测反应方向。更深层次的定量理解涉及平衡常数 K_c 和 K_p、从实验数据计算这些常数、温度对其数值的影响，以及对均相和多相系统平衡的应用。本文将探讨这些进阶内容，包括酸碱平衡常数，为考试成功打下坚实基础。

1. Review of Dynamic Equilibrium and Le Chatelier's Principle | 动态平衡与勒夏特列原理复习

At dynamic equilibrium, the rates of the forward and reverse reactions are equal, and the concentrations of reactants and products remain constant. Le Chatelier's principle states that if a system at equilibrium is subjected to a change in concentration, pressure or temperature, the position of equilibrium shifts to oppose that change.

在动态平衡中，正逆反应速率相等，反应物和产物的浓度保持不变。勒夏特列原理指出，如果平衡系统受到浓度、压力或温度的改变，平衡位置会移动以对抗这种变化。

However, qualitative predictions alone are insufficient for many applications. We need equilibrium constants to describe the exact composition of an equilibrium mixture and to understand how far a reaction goes under specific conditions.

然而，仅定性预测在许多应用中是不够的。我们需要平衡常数来描述平衡混合物的精确组成，并理解在特定条件下反应进行的程度。

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2. The Equilibrium Constant K_c | 平衡常数 K_c

For a homogeneous reaction occurring in solution or gas phase, the equilibrium constant in terms of concentration, K_c, is given by:

对于在溶液或气相中发生的均相反应，基于浓度的平衡常数 K_c 表达式为：

aA + bB ⇌ cC + dD    K_c = [C]^c[D]^d / [A]^a[B]^b

Square brackets denote equilibrium concentrations in mol dm⁻³. K_c is dimensionless only if the sum of powers in the numerator equals that in the denominator; however, units are often quoted for clarity.

方括号表示平衡浓度，单位为 mol dm⁻³。仅当分子中各幂次之和等于分母各幂次之和时，K_c 无量纲；但通常仍会注明单位以便清晰。

K_c is constant at a given temperature. Its magnitude gives insight into the position of equilibrium: K_c >> 1 implies products are favoured, while K_c << 1 means reactants dominate.

在给定温度下 K_c 为定值。其大小反映了平衡位置：K_c >> 1 表明产物占优势，而 K_c << 1 意味着反应物占主导。

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3. Calculating K_c from Experimental Data | 从实验数据计算 K_c

Experimental determination often involves measuring the equilibrium concentration of one species and using stoichiometry to find the others. Consider the esterification reaction:

实验测定通常涉及测量某一物种的平衡浓度，并利用化学计量关系求出其他浓度。考虑酯化反应：

CH₃COOH(l) + C₂H₅OH(l) ⇌ CH₃COOC₂H₅(l) + H₂O(l)

Example: 1.00 mol of ethanoic acid and 1.00 mol of ethanol were mixed. At equilibrium, 0.667 mol of ethyl ethanoate was formed in a total volume V. Determine K_c.

例题：将1.00 mol 乙酸与1.00 mol 乙醇混合。达到平衡时生成了0.667 mol 乙酸乙酯，总体积为 V。求 K_c。

Using an ICE table (Initial, Change, Equilibrium):

使用ICE表格（初始、变化、平衡）：

Species	Initial (mol)	Change (mol)	Equilibrium (mol)
CH₃COOH	1.00	–0.667	0.333
C₂H₅OH	1.00	–0.667	0.333
CH₃COOC₂H₅	0	+0.667	0.667
H₂O	0	+0.667	0.667

Equilibrium concentrations = mol/V. Because V cancels in the K_c expression (equal numbers of moles on each side), we can use moles directly:

平衡浓度 = 物质的量/V。由于 K_c 表达式中各组分物质的量相等，V 可约去，因此可直接使用物质的量：

K_c = (0.667 × 0.667) / (0.333 × 0.333) = 4.0 (no units)

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4. Gaseous Equilibria and K_p | 气相平衡与 K_p

For gas-phase reactions, it is often more convenient to use partial pressures. The equilibrium constant K_p is expressed in terms of the partial pressures of the gases, typically in atm or Pa.

对于气相反应，使用分压通常更方便。平衡常数 K_p 用各气体的分压表示，单位常为 atm 或 Pa。

aA(g) + bB(g) ⇌ cC(g) + dD(g)    K_p = (P_C)^c(P_D)^d / (P_A)^a(P_B)^b

Partial pressure is related to mole fraction: P_A = x_A × P_total, where x_A = moles of A / total moles. K_p is also temperature-dependent and can have units, e.g., atm^Δn where Δn = (c+d) – (a+b).

分压与摩尔分数相关：P_A = x_A × P_总，其中 x_A = A的物质的量/总物质的量。K_p 也依赖于温度，并可能有单位，如 atm^Δn，其中 Δn = (c+d) – (a+b)。

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5. Calculating K_p from Mole Fractions | 从摩尔分数计算 K_p

To determine K_p, follow these steps: (1) calculate equilibrium moles; (2) find mole fractions; (3) calculate partial pressures using P_total; (4) substitute into the K_p expression.

要确定 K_p，遵循以下步骤：(1) 计算平衡物质的量；(2) 计算摩尔分数；(3) 利用总压计算分压；(4) 代入 K_p 表达式。

Example: N₂O₄(g) ⇌ 2NO₂(g). 1.00 mol N₂O₄ is placed in a vessel, and at equilibrium at 100 °C and total pressure 2.00 atm, 20% has dissociated. Find K_p.

例题：反应 N₂O₄(g) ⇌ 2NO₂(g)。将1.00 mol N₂O₄ 放入容器中，在100 °C、总压2.00 atm下达到平衡时，有20% 解离。求 K_p。

• uilibrium moles: N₂O₄ = 0.80 mol, NO₂ = 0.40 mol (since 0.20 mol N₂O₄ gives 0.40 mol NO₂).
• 平衡物质的量：N₂O₄ = 0.80 mol，NO₂ = 0.40 mol（因为0.20 mol N₂O₄ 生成0.40 mol NO₂）。
• Total moles = 1.20 mol.
• 总物质的量 = 1.20 mol。
• Mole fractions: x(N₂O₄) = 0.80/1.20 = 0.667, x(NO₂) = 0.40/1.20 = 0.333.
• 摩尔分数：x(N₂O₄) = 0.667，x(NO₂) = 0.333。
• Partial pressures: P(N₂O₄) = 0.667 × 2.00 = 1.334 atm, P(NO₂) = 0.333 × 2.00 = 0.666 atm.
• 分压：P(N₂O₄) = 1.334 atm，P(NO₂) = 0.666 atm。

K_p = (P_NO₂)² / P_N₂O₄ = (0.666)² / 1.334 = 0.333 atm

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6. Heterogeneous Equilibria | 多相平衡

When a reaction involves substances in different phases, the equilibrium constant expression omits pure solids and pure liquids because their concentrations (or activities) are constant and incorporated into the K value.

当反应涉及不同相的物质时，平衡常数表达式中省略纯固体和纯液体，因为它们的浓度（或活度）是常数，已包含在 K 值中。

Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g). Here, K_c = [CO₂] and K_p = P_CO₂. The solid phases do not appear.

示例：CaCO₃(s) ⇌ CaO(s) + CO₂(g)。此时，K_c = [CO₂]，K_p = P_CO₂，固体相不出现在表达式中。

This simplification is crucial when interpreting data: the decomposition pressure of CO₂ is a direct measure of the equilibrium constant at that temperature.

这种简化在解释数据时至关重要：CO₂ 的分解压力直接就是该温度下平衡常数的量度。

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7. Effect of Temperature on K | 温度对平衡常数的影响

Temperature is the only factor that changes the value of the equilibrium constant. According to Le Chatelier's principle and the van't Hoff equation:

温度是唯一能改变平衡常数值的因素。根据勒夏特列原理和范特霍夫方程：

• For an exothermic reaction (ΔH < 0), increasing temperature shifts equilibrium to the left, so K decreases.
• 对于放热反应（ΔH < 0），升温使平衡左移，因此 K 减小。
• For an endothermic reaction (ΔH > 0), increasing temperature shifts equilibrium to the right, so K increases.
• 对于吸热反应（ΔH > 0），升温使平衡右移，因此 K 增大。

This is why industrial processes like the Haber process are carried out at a compromise temperature: low temperature gives a higher K and yield (exothermic), but rate is too slow without a catalyst.

这就是为什么哈伯法等工业过程在折中温度下进行：低温可得到较高的 K 值和产率（放热反应），但若无催化剂速率过慢。

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8. Effect of Concentration, Pressure and Catalysts | 浓度、压力和催化剂的影响

Changes in concentration or pressure shift the position of equilibrium but do not alter the value of K_c or K_p, provided temperature remains constant. The system adjusts to restore the same ratio of products to reactants at equilibrium.

浓度或压力的变化会移动平衡位置，但不会改变 K_c 或 K_p 的值，只要温度保持恒定。系统会进行调整，以使产物与反应物的比值恢复至相同的平衡常数。

Catalysts speed up both forward and reverse reactions equally, so they reduce the time to reach equilibrium but have no effect on the equilibrium constant or the equilibrium yield.

催化剂同等程度地加快正逆反应，因此它们缩短达到平衡的时间，但对平衡常数和平衡产率没有影响。

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9. Acid–Base Equilibria: K_a and K_b | 酸碱平衡：K_a 和 K_b

The concept of equilibrium constants extends naturally to acid–base chemistry. For a weak acid HA in water:

平衡常数的概念自然延伸到酸碱化学。对于弱酸 HA 在水溶液中：

HA(aq) + H₂O(l) ⇌ H₃O⁺(aq) + A⁻(aq)    K_a = [H₃O⁺][A⁻] / [HA]

K_a is the acid dissociation constant; a larger K_a indicates a stronger acid. Similarly, for a weak base B:

K_a 是酸解离常数；K_a 越大，酸性越强。类似地，对于弱碱 B：

B(aq) + H₂O(l) ⇌ BH⁺(aq) + OH⁻(aq)    K_b = [BH⁺][OH⁻] / [B]

These equilibrium constants allow calculation of pH, pOH, and the concentrations of species in solution, linking quantitative equilibria to acid–base titrations and buffers.

这些平衡常数可用于计算 pH、pOH 以及溶液中各组分的浓度，将定量平衡与酸碱滴定和缓冲溶液联系起来。

The ionic product of water K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ at 298 K, and the relationship pK_a + pK_b = 14 holds for conjugate acid–base pairs in aqueous solution.

水的离子积 K_w = [H₃O⁺][OH⁻] = 1.0 × 10⁻¹⁴ mol² dm⁻⁶ (298 K)，且在水溶液中共轭酸碱对满足 pK_a + pK_b = 14。

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10. Summary and Exam Tips | 总结与备考建议

When tackling equilibrium problems, remember: always write the balanced equation; use ICE tables for clarity; convert to equilibrium concentrations or partial pressures correctly; and note the units of K. Temperature changes are the only way to alter K, while concentration, pressure, and catalyst changes only shift the position. In acid–base equilibria, apply the same principles using K_a, K_b and K_w.

在解决平衡问题时，记住：务必写出配平方程式；利用 ICE 表格使过程清晰；正确转换为平衡浓度或分压；并注意 K 的单位。温度变化是唯一能改变 K 的方式，而浓度、压力和催化剂的改变仅移动平衡位置。在酸碱平衡中，运用 K_a、K_b 和 K_w 的原理与之相同。

Practise a variety of K_c and K_p calculations, including those where the reaction vessel volume is unknown, and where inert gases are added at constant pressure. For acids and bases, be confident converting between K_a, pK_a, and pH.

练习各类 K_c 和 K_p 的计算，包括容器体积未知以及恒压下加入惰性气体的情形。对于酸碱部分，要熟练掌握 K_a、pK_a 与 pH 之间的换算。

Mastering these further aspects of equilibria will not only secure high marks but also provide a deeper appreciation of the quantitative nature of chemical systems.

掌握这些化学平衡的进阶内容，不仅能在考试中取得高分，还能更深刻地理解化学体系的定量本质。

Published by TutorHao | Chemistry Revision Series | aleveler.com

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