A-Level化学 化学平衡 Kc Kp 勒夏特列原理

A-Level化学 化学平衡 Kc Kp 勒夏特列原理

Chemical equilibrium is one of the most fundamental concepts in A-Level Chemistry, bridging the gap between reaction kinetics and thermodynamics. Understanding equilibrium allows chemists to predict reaction yields, optimise industrial processes such as the Haber and Contact processes, and explain natural phenomena at the molecular level. Mastery of this topic is essential for success in Paper 4 and Paper 5 calculations. 化学平衡是A-Level化学中最基础的概念之一，它将反应动力学和热力学联系起来。理解平衡使化学家能够预测反应产率、优化哈伯法和接触法等工业流程，并在分子层面解释自然现象。掌握这一主题对Paper 4和Paper 5的计算题至关重要。

What Is Dynamic Equilibrium

A system reaches dynamic equilibrium when the forward and backward reactions proceed at exactly the same rate, resulting in no net change in the concentrations of reactants and products. Crucially, the reactions do not stop： molecules continue to react in both directions. This is why we call it dynamic rather than static equilibrium. 当正向反应和逆向反应以完全相同的速率进行时，系统达到动态平衡，反应物和产物的浓度不发生净变化。关键是反应并没有停止：分子继续在两个方向上反应。这就是为什么我们称之为动态平衡而非静态平衡。

For equilibrium to be established, the system must be closed so that no matter escapes. Consider the Haber process： N2(g) + 3H2(g) ⇌ 2NH3(g). Once equilibrium is reached, nitrogen and hydrogen continue to combine to form ammonia at exactly the same rate that ammonia decomposes back into its constituent elements. 要建立平衡，系统必须是封闭的，使物质不会逸出。以哈伯法为例：N2(g) + 3H2(g) ⇌ 2NH3(g)。一旦达到平衡，氮气和氢气继续结合生成氨的速率，恰好等于氨分解回其组成元素的速率。

The Equilibrium Constant Kc

The equilibrium constant Kc is a ratio that expresses the relationship between the concentrations of products and reactants at equilibrium. For a general reaction aA + bB ⇌ cC + dD, the expression is： Kc = [C]^c [D]^d / [A]^a [B]^b, where the square brackets represent concentrations in mol/dm3. 平衡常数Kc是一个比值，表示平衡时产物浓度与反应物浓度之间的关系。对于一般反应aA + bB ⇌ cC + dD，表达式为：Kc = [C]^c [D]^d / [A]^a [B]^b，方括号表示以mol/dm3为单位的浓度。

Kc is temperature-dependent： changing the temperature changes the value of Kc, because the forward and backward reactions have different activation energies. A larger Kc value (Kc >> 1) indicates that the equilibrium position lies to the right, favouring products. A smaller Kc value (Kc << 1) indicates that the equilibrium favours reactants. Kc取决于温度：改变温度会改变Kc的值，因为正向和逆向反应具有不同的活化能。较大的Kc值（Kc >> 1）表明平衡位置偏右，有利于产物。较小的Kc值（Kc << 1）表明平衡有利于反应物。

A common exam question asks students to calculate Kc from equilibrium concentrations. The key steps are： calculate the equilibrium moles of all species, convert to concentrations by dividing by the volume, substitute into the Kc expression, and calculate the result. Remember that Kc has no units when the total number of moles on each side of the equation is equal. When the mole totals differ, Kc carries units such as mol/dm3 or (mol/dm3)^2. 常见的考试题目要求学生根据平衡浓度计算Kc。关键步骤是：计算所有物种的平衡摩尔数，除以体积转化为浓度，代入Kc表达式，计算结果。记住当方程两边总摩尔数相等时Kc没有单位。当摩尔总数不同时，Kc带有单位如mol/dm3或(mol/dm3)^2。

Equilibrium Constant Kp and Partial Pressures

For reactions involving gases, we often use Kp, the equilibrium constant expressed in terms of partial pressures. The partial pressure of a gas is the pressure that gas would exert if it occupied the container alone. It is calculated as： partial pressure = mole fraction × total pressure. 对于涉及气体的反应，我们通常使用Kp，即以分压表示的平衡常数。气体的分压是该气体单独占据容器时所施加的压力。计算方式为：分压 = 摩尔分数 × 总压。

The mole fraction of a gas is the number of moles of that gas divided by the total number of moles in the mixture. For the Haber process at equilibrium, if the mixture contains 2 moles of NH3, 1 mole of N2, and 3 moles of H2 at a total pressure of 200 atm, the mole fraction of NH3 is 2/6 = 0.333, and its partial pressure is 0.333 × 200 = 66.7 atm. 气体的摩尔分数是该气体的摩尔数除以混合物中的总摩尔数。对于哈伯法在平衡时，如果混合物含有2 mol NH3、1 mol N2和3 mol H2，总压为200 atm，则NH3的摩尔分数为2/6 = 0.333，其分压为0.333 × 200 = 66.7 atm。

The Kp expression mirrors the Kc expression but uses partial pressures instead of concentrations. For aA(g) + bB(g) ⇌ cC(g) + dD(g)： Kp = (pC)^c (pD)^d / (pA)^a (pB)^b. Like Kc, Kp is temperature-dependent but is not affected by changes in total pressure or the presence of a catalyst. Kp表达式与Kc表达式类似，但使用分压代替浓度。对于aA(g) + bB(g) ⇌ cC(g) + dD(g)：Kp = (pC)^c (pD)^d / (pA)^a (pB)^b。与Kc一样，Kp取决于温度，但不受总压变化或催化剂存在的影响。

There is a direct mathematical relationship between Kc and Kp： Kp = Kc (RT)^Δn, where Δn is the change in the number of moles of gas (products minus reactants), R is the gas constant 8.31 J K^-1 mol^-1 when using pressure in Pa, and T is the absolute temperature in Kelvin. When Δn = 0, Kp = Kc because (RT)^0 = 1. This equation is essential for converting between the two constants in exam problems. Kc和Kp之间存在直接的数学关系：Kp = Kc (RT)^Δn，其中Δn是气体摩尔数的变化（产物减反应物），R是气体常数8.31 J K^-1 mol^-1（当压强以Pa为单位时），T是以开尔文为单位的绝对温度。当Δn = 0时，Kp = Kc，因为(RT)^0 = 1。该方程对于考试题目中两种常数之间的转换至关重要。

Le Chatelier’s Principle

Le Chatelier’s Principle states that if a system at equilibrium is subjected to a change in conditions, the position of equilibrium shifts to oppose the change and minimise its effect. This principle is a powerful predictive tool that helps chemists understand how concentration, pressure, temperature, and catalysts affect equilibrium systems. 勒夏特列原理指出，如果处于平衡状态的系统受到条件变化的影响，平衡位置会移动以对抗该变化并最小化其影响。这一原理是一个强大的预测工具，帮助化学家理解浓度、压力、温度和催化剂如何影响平衡系统。

Concentration changes： adding more reactant shifts equilibrium to the right (producing more product)； removing product also shifts equilibrium to the right. Conversely, adding product shifts equilibrium to the left. This is why industrial processes often continuously remove the desired product ： to push the equilibrium towards higher yields. 浓度变化：加入更多反应物使平衡向右移动（产生更多产物）；移除产物也使平衡向右移动。反之，加入产物使平衡向左移动。这就是为什么工业过程通常持续移除所需产物：推动平衡朝向更高的产率。

Pressure changes： for gaseous reactions, increasing pressure shifts equilibrium towards the side with fewer gas molecules, because this reduces the total pressure (opposing the change). In the Haber process, 4 moles of gas on the left become 2 moles on the right, so high pressure favours ammonia production. If the number of gas molecules is equal on both sides, pressure changes have no effect on the equilibrium position. 压力变化：对于气体反应，增加压力使平衡向气体分子较少的一侧移动，因为这会降低总压（对抗变化）。在哈伯法中，左边4 mol气体变为右边2 mol，因此高压有利于氨的生产。如果两侧气体分子数相等，压力变化对平衡位置没有影响。

Temperature changes： increasing temperature shifts equilibrium in the endothermic direction (the direction that absorbs heat), while decreasing temperature favours the exothermic direction. For the Haber process, the forward reaction is exothermic, so lowering the temperature would favour ammonia production. However, in practice, a compromise temperature of about 450°C is used because lower temperatures slow the reaction rate too much. 温度变化：升高温度使平衡向吸热方向移动（吸收热量的方向），而降低温度有利于放热方向。对于哈伯法，正向反应是放热的，因此降低温度有利于氨的生产。然而在实践中，使用约450°C的折中温度，因为较低温度会使反应速率过慢。

Catalysts and Equilibrium

A catalyst does not affect the position of equilibrium. It provides an alternative reaction pathway with a lower activation energy, speeding up both the forward and backward reactions equally by exactly the same amount. This is because a catalyst lowers the activation energy barrier for both directions identically, leaving the energy difference between reactants and products unchanged. A catalyst therefore helps a system reach equilibrium faster, but it does not change the equilibrium composition. Kc and Kp remain unchanged when a catalyst is added because the catalyst does not alter the relative energies of reactants and products. 催化剂不影响平衡位置。它提供具有较低活化能的替代反应路径，同等地加速正向和逆向反应。这是因为催化剂同等降低两个方向的活化能屏障，使反应物和产物之间的能量差保持不变。因此催化剂帮助系统更快达到平衡，但不改变平衡组成。加入催化剂时Kc和Kp保持不变，因为催化剂不改变反应物和产物的相对能量。

Worked Example： Calculating Kc

Consider the reaction： H2(g) + I2(g) ⇌ 2HI(g). At equilibrium at 700 K, a 2.0 dm3 flask contains 0.40 mol H2, 0.40 mol I2, and 2.40 mol HI. To calculate Kc： first find equilibrium concentrations：[H2] = 0.40/2.0 = 0.20 mol/dm3, [I2] = 0.40/2.0 = 0.20 mol/dm3, [HI] = 2.40/2.0 = 1.20 mol/dm3. Then： Kc = [HI]^2 / ([H2][I2]) = (1.20)^2 / (0.20 × 0.20) = 1.44 / 0.04 = 36.0. Since there are 2 moles of gas on each side, Kc has no units. The large Kc value confirms that the equilibrium strongly favours products. 考虑反应：H2(g) + I2(g) ⇌ 2HI(g)。在700 K时平衡，一个2.0 dm3的烧瓶中含有0.40 mol H2、0.40 mol I2和2.40 mol HI。计算Kc：首先求平衡浓度：[H2] = 0.40/2.0 = 0.20 mol/dm3, [I2] = 0.40/2.0 = 0.20 mol/dm3, [HI] = 2.40/2.0 = 1.20 mol/dm3。则：Kc = [HI]^2 / ([H2][I2]) = (1.20)^2 / (0.20 × 0.20) = 1.44 / 0.04 = 36.0。由于两侧各有2 mol气体，Kc没有单位。较大的Kc值证实平衡强烈有利于产物。

The industrial synthesis of methanol provides a classic example of equilibrium optimisation. The reaction CO(g) + 2H2(g) ⇌ CH3OH(g) is exothermic (ΔH = -91 kJ mol^-1). According to Le Chatelier’s Principle, high pressure favours the forward reaction (3 moles produce 1 mole), while low temperature favours exothermic product formation. In practice, manufacturers use a copper-zinc oxide catalyst at 250°C and 50-100 atm ： a compromise optimising both yield and rate. 甲醇的工业合成为平衡优化提供了经典案例。反应CO(g) + 2H2(g) ⇌ CH3OH(g)是放热的（ΔH = -91 kJ mol^-1）。根据勒夏特列原理，高压有利于正向反应（3 mol产生1 mol），而低温有利于放热产物形成。实践中制造商使用铜锌氧化物催化剂在250°C和50-100 atm下操作：这是优化产率和速率的折中方案。

Common Exam Pitfalls

Students often confuse the rate of reaction with the position of equilibrium. A catalyst increases the rate at which equilibrium is reached but does not shift the equilibrium position. Similarly, increasing temperature always increases reaction rate, but its effect on equilibrium position depends on whether the forward reaction is exothermic or endothermic. 学生经常混淆反应速率和平衡位置。催化剂增加达到平衡的速率，但不改变平衡位置。同样，升高温度总是增加反应速率，但其对平衡位置的影响取决于正向反应是放热还是吸热。

Another common mistake is forgetting that pure solids and liquids are omitted from Kc and Kp expressions because their concentrations are effectively constant. Only aqueous and gaseous species appear in the equilibrium expression. For example, in the reaction CaCO3(s) ⇌ CaO(s) + CO2(g), Kc = [CO2] only, because the two solids have constant concentrations. 另一个常见错误是忘记纯固体和液体因浓度基本恒定而从Kc和Kp表达式中省略。只有水溶液和气态物种出现在平衡表达式中。例如，在反应CaCO3(s) ⇌ CaO(s) + CO2(g)中，Kc = [CO2]，因为两种固体具有恒定的浓度。

Finally, examiners frequently test the distinction between “equilibrium position shifts” and “Kc changes”. Only temperature changes alter the value of Kc. Concentration and pressure changes shift the equilibrium position but leave Kc unchanged. This is because Kc is a constant at a given temperature ： the system adjusts concentrations to restore the same ratio after a disturbance. 最后，考官经常测试”平衡位置移动”和”Kc变化”之间的区别。只有温度变化改变Kc的值。浓度和压力变化使平衡位置移动但Kc保持不变。这是因为Kc在给定温度下是常数：系统在干扰后调整浓度以恢复相同的比值。