Edexcel化学 动态平衡 Kc Kp 分压 勒沙特列
Edexcel A Level Chemistry 的官方 spec（9CH0，2015 年 9 月起施教，2017 年首次考试，现行版本为 2024 年 2 月发布的 Issue 3）把化学平衡分成 Topic 10 Equilibrium I 与 Topic 11 Equilibrium II 两个主题：前者要求为均相与非均相体系写出 Kc 表达式，后者要求在 atm 下用分压写出 Kp 并完成带单位的计算，还要解释温度如何通过改变平衡常数的取值来移动平衡位置。Paper 1（9CH0/01）与 Paper 3（9CH0/03）都会直接考到这两块内容。整章的公式只有两三条，失分却集中在三处：把纯固体与纯液体写进平衡常数、单位算错或漏写、解释温度影响时只说向吸热方向移动而不提平衡常数的变化。很多学生以为背住勒沙特列原理的三句话就能拿到这一章的分，阅卷时常见的答案整段缺了平衡常数变化这一步。下面按动态平衡的特征、Kc 表达式与单位、均相与非均相计算、浓度与压强的变化、温度对平衡常数的影响、催化剂的作用、Kp 与分压计算、工业条件取舍这八条线逐段拆开，每一段都配一个可以直接照算的数字例子。
Chemical equilibria sit in Topic 10 (Equilibrium I) and Topic 11 (Equilibrium II) of the Edexcel A Level Chemistry specification, coded 9CH0 with first teaching in September 2015, first assessment in 2017 and the current issue dated February 2024; both physical chemistry papers examine them. Topic 10 asks for Kc expressions covering homogeneous and heterogeneous systems, while Topic 11 moves the same logic onto partial pressures measured in atm and requires the effect of temperature to be explained through a change in the value of the constant. The marks rarely sit on the algebra; they sit on the detail, which is why pure solids and liquids left out of the expression, units worked out and carried through, and temperature tied to K rather than to a bare statement about the endothermic direction decide most of the grade.
可逆反应与动态平衡的特征 / Reversible Reactions and Dynamic Equilibrium
可逆反应在密闭容器里不会进行到底，当正反应与逆反应速率相等时体系进入动态平衡，各物质的浓度随时间保持不变。动态平衡的特征要写全：反应并未停止，两个方向仍在进行；体系必须密闭且温度恒定；浓度与分压不再变化，微观上正逆速率相等。Edexcel 在 Topic 10 里给出的实验体系是铁(III)离子与硫氰酸根离子生成血红色配离子，加入 Fe3+ 或 SCN- 后颜色加深，说明平衡向生成配合物的方向移动。另一个常用体系是 [Co(H2O)6]2+ 与 [CoCl4]2- 之间的平衡，粉红色的水合离子与蓝色的四面体离子随温度或氯离子浓度变化而互换。写平衡这个概念时不能写成反应已经停止，这类表述在阅卷时直接不得分。很多学生以为到达平衡就意味着反应物已经消耗完，实际上转化率由平衡常数的取值决定，与体系放了多久没有关系。实验室里判断是否到达平衡，看的是颜色或浓度不再随时间改变，而不是看容器里还剩多少反应物。
A reversible reaction in a closed vessel never runs to completion; equilibrium arrives when the forward and reverse rates become equal and the concentrations stop changing, even though both directions are still running underneath. Edexcel Topic 10 illustrates this with the blood-red iron(III) thiocyanate complex, which deepens when either Fe3+ or SCN- is added, and with the cobalt equilibrium between the pink hydrated ion and the blue tetrachloride ion that swaps sides as temperature or chloride concentration changes; describing that state as the reaction having stopped loses the mark outright. Conversion at equilibrium is fixed by the value of the equilibrium constant rather than by how long the mixture is left standing, so the same reaction can sit far to the left at one temperature and far to the right at another, and a practical judgement is made from colour or concentration that no longer drifts.
Kc 表达式与单位 / The Kc Expression and Its Units
Kc 的表达式来自平衡时的浓度比：aA + bB 可逆生成 cC + dD 时，Kc 等于 [C] 的 c 次方乘 [D] 的 d 次方，除以 [A] 的 a 次方乘 [B] 的 b 次方，各浓度取平衡值并升到方程式里的系数次方。Edexcel 官方 spec 的 Topic 10 明确要求能为均相与非均相体系推导 Kc 表达式，纯固体与纯液体的浓度视为常数，写进表达式反而丢分。以碳酸钙分解为例，CaCO3(s) 可逆生成 CaO(s) 加 CO2(g)，Kc 就等于 [CO2]，两个固体项一个都不出现。氮气与氢气合成氨的 Kc 则是 [NH3]^2 除以 [N2] 乘 [H2]^3，单位是 (mol dm^-3)^-2。单位的算法是把分子与分母的浓度次方相减，写成 mol dm^-3 的幂次，Edexcel 的 mark scheme 把单位列为独立的一分，漏掉这一分与算错数值的代价一样。常见的错误是把平衡常数与反应速率混在一起比较，速率取决于活化能与浓度，Kc 只随温度改变。
The Kc expression is a ratio of equilibrium concentrations, so for the general reaction aA + bB in equilibrium with cC + dD it reads [C]^c [D]^d divided by [A]^a [B]^b, with every concentration raised to the coefficient in the equation. The Edexcel specification for Topic 10 requires the expression to be deduced for both homogeneous and heterogeneous systems, which means pure solids and pure liquids are treated as constants and left out; decomposing calcium carbonate gives Kc equal to [CO2] alone, because neither solid appears in the expression. The units come from subtracting the powers of concentration in the denominator from those in the numerator, written as powers of mol dm^-3, and the mark scheme awards the unit as a free-standing mark, so a missing unit costs exactly as much as a wrong number. Rate and equilibrium constant are different animals: the rate depends on activation energy and concentration, whereas Kc responds only to temperature.
Kc 计算与均相非均相体系 / Worked Kc Calculations
计算题先列三行数据：初始量、变化量与平衡量，再把平衡一行代入表达式。以 700 K 时氢气与碘蒸气生成碘化氢为例，平衡浓度 [H2] 与 [I2] 都是 0.0050 mol dm^-3，[HI] 为 0.0900 mol dm^-3，Kc 等于 0.0900 的平方除以 0.0050 乘 0.0050，结果是 324，浓度的二次方上下约掉，Kc 没有单位。再看四氧化二氮分解为二氧化氮的体系，平衡时 [N2O4] 为 0.0080 mol dm^-3，[NO2] 为 0.0040 mol dm^-3，Kc 等于 0.0040 的平方除以 0.0080，得到 2.0 x 10^-3 mol dm^-3，分子二次方减分母一次方，剩下一个浓度的量纲。非均相体系的题眼永远在固体上，碳与水蒸气反应生成一氧化碳和氢气时，Kc 只写 [CO] 乘 [H2] 除以 [H2O]，碳的浓度不进表达式。有效数字按题干给出的最少位数保留，Edexcel 通常接受两到三位，数值与单位要分开核对。错得最多的是算完变化量直接代入平衡表达式，忘记把变化量乘上方程式的系数。
A calculation question is best handled by writing three rows of data, starting amounts, changes and equilibrium amounts, and then substituting only the equilibrium row. For hydrogen reacting with iodine vapour at 700 K, an equilibrium mixture with [H2] and [I2] at 0.0050 mol dm^-3 and [HI] at 0.0900 mol dm^-3 gives Kc equal to 0.0900 squared divided by 0.0050 times 0.0050, which is 324 with no units at all because the powers of concentration cancel. Dinitrogen tetroxide breaking down into nitrogen dioxide behaves differently: 0.0080 mol dm^-3 against 0.0040 mol dm^-3 yields 2.0 x 10^-3 mol dm^-3, since two powers in the numerator and one in the denominator leave a single concentration behind. Heterogeneous systems hide their marks in the solid term, as with carbon and steam, where Kc holds only the two gaseous products over the steam term. The arithmetic usually fails in one place: the change is calculated correctly and then inserted without multiplying by the coefficient in the equation.
浓度与压强对平衡位置的影响 / Concentration and Pressure Changes
改变浓度会移动平衡位置，却不改变平衡常数的取值，这一点是判断答案是否完整的标准。向体系中加入某一种反应物，浓度商在瞬间小于 Kc，正反应速率大于逆反应速率，平衡右移直到浓度商重新等于 Kc。把产物从体系中移走的效果相同，合成氨的工业流程里用冷凝把氨液化分出，正是靠这一步不断把平衡拉向产物一侧。压强的变化只对有气体参与且两边气体摩尔数不相等的体系有效，增大压强时平衡移向气体摩尔数较少的一侧，因为这一侧的分压上升得更慢。氮气与氢气合成氨是四摩尔气体生成两摩尔气体，增大压强有利于氨的生成；而氢气与碘蒸气生成碘化氢两边都是两摩尔气体，加压对平衡位置没有影响。体积减半而温度不变时，各组分浓度翻倍，浓度商会偏离 Kc，体系随即移动以抵消这一偏离。学生最常把压强的影响一句带过，没有指出两边气体摩尔数的比较，这类答案只能拿到一半的分数。
Changing a concentration shifts the position of equilibrium without touching the value of the constant, and a complete answer says so explicitly. Adding a reactant makes the reaction quotient smaller than Kc for an instant, so the forward rate exceeds the reverse rate and the system moves right until the quotient matches Kc again; removing a product does the same job, which is exactly why the industrial ammonia process liquefies and drains off the ammonia as it forms. Pressure only matters when the number of gas moles differs between the two sides, and an increase pushes the system towards the side with fewer gas moles because that side gains pressure more slowly. Nitrogen and hydrogen combining into ammonia turns four moles of gas into two, so high pressure helps, whereas hydrogen with iodine vapour keeps two moles on each side and therefore ignores pressure altogether. Halving the volume at constant temperature doubles every concentration, and the system then moves to bring the disturbed quotient back to Kc.
温度与平衡常数的取值 / Temperature and the Value of K
温度是唯一能改变平衡常数取值的条件，Edexcel 的 Topic 11 把它写成了明确要求：温度对平衡位置的影响要用平衡常数取值的变化来解释。正向放热的反应升温时 Kc 与 Kp 都变小，产物在平衡混合物中的比例下降，平衡位置随之左移；正向吸热的反应升温时平衡常数变大，平衡右移。氮气与氢气合成氨的正向反应放热约 92 kJ mol^-1，温度升高会让 Kp 减小，工厂因此在产率与速率之间做取舍。氢气与碘蒸气生成碘化氢的正向反应吸热，升温时 Kc 反而变大，这与合成氨的方向正好相反。阅卷时只写向吸热方向移动而不提 K 的变化，这一分通常拿不到，因为题目问的正是平衡常数的变化。很多学生以为勒沙特列原理本身就是一个完整的解释，其实它只是对结果的概括，定量层面的依据仍然落在 Kc 与 Kp 的取值上。
Temperature is the only variable that changes the value of the equilibrium constant, and the Edexcel Topic 11 requirements spell this out by asking for the effect of temperature on the position of equilibrium to be explained through a change in the value of Kc or Kp. When the forward reaction is exothermic, raising the temperature lowers both Kc and Kp, the proportion of product in the equilibrium mixture falls and the position shifts left; an endothermic forward reaction raises the constant and pushes the position right. Ammonia synthesis releases roughly 92 kJ mol^-1, so a higher temperature lowers Kp and the plant has to trade yield against speed, whereas hydrogen with iodine vapour takes heat in and shows a larger Kc as the temperature rises. An answer that only states a shift in the endothermic direction without mentioning the constant misses the mark, because the change in the constant is precisely what the question is testing.
催化剂与达到平衡的时间 / Catalysts and the Time Taken to Reach Equilibrium
催化剂改变的是达到平衡所需的时间，不改变平衡位置，也不改变平衡常数的取值。它通过提供一条活化能更低的反应路径同时加快正反应与逆反应，两个方向的速率按相同倍数提高，因此正逆速率相等的那个状态出现得更早，而平衡混合物的组成完全一样。Edexcel 的 Topic 11 要求知道平衡常数的取值不受浓度、压强或催化剂的影响，这句话在选择题与解释题里都出现过。哈伯法使用铁催化剂，接触法使用五氧化二钒，两者都属于让工业过程在更低温度下仍然有可接受速率的做法，产率本身仍由温度与压强决定。实验里加催化剂后颜色变化的到达时间缩短，颜色深浅的最终读数不变，这是最直观的判断依据。常见的错误是把催化剂写成能提高产率或使平衡右移，这类答案与前面的温度影响混在一起写，阅卷时两处一起扣分。
A catalyst changes how quickly equilibrium is reached rather than where it settles, leaving both the position and the value of the constant untouched. It opens a route with a lower activation energy and speeds up the forward and reverse reactions by the same factor, so the state where the two rates match simply arrives earlier and the composition of the equilibrium mixture stays identical. The Edexcel Topic 11 requirements include the point that the value of the constant is unaffected by changes in concentration or pressure or by the addition of a catalyst, and this idea appears in both multiple choice and explanation questions. Iron serves the Haber process and vanadium(V) oxide the Contact process, letting each plant run at a temperature low enough for a workable rate while temperature and pressure still govern the yield, and in the laboratory the colour change simply arrives sooner with the final shade unchanged.
Kp 摩尔分数与分压计算 / Kp, Mole Fractions and Partial Pressures
气体平衡用分压表示，分压等于该气体的摩尔分数乘体系总压，摩尔分数是该气体的物质的量除以体系中气体总物质的量。Edexcel Topic 11 要求用 atm 为单位的分压写出 Kp，并计算数值与单位。以二氧化硫氧化为例，2SO2(g) + O2(g) 可逆生成 2SO3(g)，平衡时若摩尔分数分别为 0.20、0.20 与 0.60，总压为 2.0 atm，则分压分别是 0.40 atm、0.40 atm 与 1.20 atm。Kp 等于 p(SO3) 的平方除以 p(SO2) 的平方乘 p(O2)，代入得 1.44 除以 0.16 乘 0.40，结果是 22.5 atm^-1，单位由三次方相减得到。非均相的气体平衡同样把固体项去掉，例如碳与二氧化碳反应生成一氧化碳时，Kp 只含两种气体的分压。算分压之前先核对总物质的量，这一步最容易出错，因为有液体或固体生成时它们不计入气体总量。
Gas phase equilibria are expressed through partial pressures, where the partial pressure of one gas is its mole fraction multiplied by the total pressure, and the mole fraction is that gas's amount in moles divided by the total amount of gas present. Topic 11 of the Edexcel specification requires Kp to be written from partial pressures in atm with a numerical value and a unit, so for the oxidation of sulfur dioxide, 2SO2 reacting with O2 to give 2SO3, mole fractions of 0.20, 0.20 and 0.60 at a total pressure of 2.0 atm give partial pressures of 0.40, 0.40 and 1.20 atm. Substituting into Kp, the square of the sulfur trioxide pressure over the square of the sulfur dioxide pressure times the oxygen pressure, returns 1.44 divided by 0.064 and a value of 22.5 atm^-1, with the unit falling out of the powers. Heterogeneous gas equilibria drop the solid term, as when carbon reacts with carbon dioxide to give carbon monoxide, and the total amount of gas has to be checked before any partial pressure is worked out.
工业条件取舍：哈伯法与接触法 / Industrial Compromises: Haber and Contact
工业合成要在速率、产率与成本之间取舍，Edexcel 的 Topic 10 与 Topic 11 都用工业流程作为平衡移动的应用背景。哈伯法用铁催化剂、约 450 摄氏度与 200 atm 的条件合成氨，高温提高反应速率但降低放热反应的产率，高压提高产率却增加设备与能耗成本，实际路线是折中值。氨在流程中被不断冷凝分出，移走产物这一步让平衡持续右移，这一点与降温提高产率的理由常常被写在同一道题里。接触法用五氧化二钒催化二氧化硫氧化，约 400 到 450 摄氏度、接近常压的条件同样来自权衡：该反应放热，降温有利于三氧化硫生成，但温度太低催化剂活性不足。判断这类题的答案是否完整，看有没有把速率与产率两条线分开写，再写明平衡移动的方向。错得最多的是把催化剂写成提高产率的手段，实际上它只影响达到平衡需要的时间。
Industrial processes are a study in trade-offs between rate, yield and cost, and the Edexcel specification uses them as the applied setting for shifting an equilibrium. The Haber process runs nitrogen and hydrogen over an iron catalyst at roughly 450 degrees Celsius and 200 atm, where the high temperature buys speed at the expense of yield in an exothermic reaction and the high pressure buys yield at the expense of equipment and energy, so the operating point is a compromise. Ammonia is continuously condensed out of the loop, and that removal of product keeps driving the position right, which is often asked alongside the yield argument for lowering the temperature. The Contact process oxidises sulfur dioxide over vanadium(V) oxide at about 400 to 450 degrees Celsius and close to atmospheric pressure, since cooling favours sulfur trioxide formation but leaves the catalyst too sluggish below that range. Answers lose marks when the rate argument and the yield argument are merged into one sentence instead of being separated.
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