Edexcel化学 动态平衡 Kc Kp 勒夏特列原理
平衡这一章在 Edexcel 9CH0 里横跨 Topic 10 与 Topic 11，Paper 1 的题目不长，分值却不小。学生丢分大多落在同一批地方：Kc 表达式写错、单位漏写、把浓度变化当成了平衡常数的变化。下面按考点顺序把这两章拆开讲，每个知识点配一个可以照着算的例子，数据都能自己复算。
Equilibrium covers two topics in the Pearson Edexcel specification, and on Paper 1 it arrives as short, calculation-heavy questions that carry a fair share of marks. Marks tend to disappear in the same places: a badly built expression, missing units, and the assumption that changing a concentration changes the constant. What follows follows the specification order and uses numbers you can rework yourself.
可逆反应与动态平衡的判据
封闭体系里，正反应速率与逆反应速率相等时，体系处在动态平衡。动态两个字指的是两个方向的反应都还在进行，只是净变化为零。宏观上看，浓度、压强、颜色都保持不变，微观上分子仍在碰撞与转化。判断是否达到平衡，题目常给三条依据：浓度不再随时间改变、正逆反应速率相等、条件不变时颜色或总压稳定。很多学生以为反应停了才叫平衡，其实恰好相反，平衡时两个方向都没有停。Edexcel 9CH0 的 Topic 10 要求判断平衡状态，也要求把反应速率与平衡位置分开描述，这两件事在答案里混在一起就会失分。
At dynamic equilibrium the forward and reverse rates are equal, and that can only happen inside a closed vessel where nothing escapes. Nothing has stopped at that point: the two directions cancel out, so concentrations, pressure and colour stay steady while molecules keep reacting. Many students read equilibrium as the reaction being over, and that reading is the quickest way to lose marks on a rate graph, because the specification expects you to separate how fast a reaction goes from where it finally settles.
Kc 表达式的写法与纯固体处理
平衡常数 Kc 的表达式是生成物浓度之积除以反应物浓度之积，每种物质的浓度按方程式里的系数取幂。写表达式只用平衡时的浓度，不用初始浓度，这一步在计算题里最容易被跳过。系数是 2 就平方一次，系数是 3 就三次方，不要拿系数直接当乘数。纯固体与纯液体的浓度视为常数，不写进表达式，碳酸钙分解这类反应的分母里不会出现碳酸钙。同一温度下 Kc 只与温度有关，与投料比例、容器大小、是否加催化剂都无关。方程式反过来写，Kc 变成原来的倒数；系数整体乘以 2，Kc 变成原来的平方。这几条变换在选择题里出现频率很高，写表达式时先抄方程式再逐个取幂，比背结论稳。
The equilibrium constant is built from product concentrations over reactant concentrations, with every term raised to the power of its coefficient in the equation. Only equilibrium concentrations belong in the expression, which is why a calculation that quietly uses starting values collapses at the substitution stage. Pure solids and pure liquids are left out because their concentrations are effectively constant, so the decomposition of calcium carbonate carries no carbonate term at all; reversing an equation inverts the constant, and doubling every coefficient squares it.
Kc 的单位与有效数字
Kc 的单位由上下两侧浓度幂次之差决定，考试要求写单位，漏写通常直接扣一分。氢与碘生成碘化氢的反应两侧幂次相同，Kc 没有单位。氮与氢生成氨的反应里，生成物是二次方、反应物合计四次方，单位就是 dm6 mol-2。写单位时可以先把幂次相减，再把 mol dm-3 的指数换成分式，比死记结论可靠。有效数字按题目给的数据取，题目给的浓度都是三位，答案就写两到三位。数值特别大或特别小时用标准形式，写成小数点后一长串在阅卷时容易被判为格式问题。常见的错误是单位只写 mol dm-3，把二次方那一步整段漏掉。
The units of Kc come from subtracting the powers of concentration on the two sides, so the constant is unitless when those powers match. In ammonia synthesis the product term is squared against four reactant terms, leaving dm6 mol-2, and a missing unit costs a mark rather than being treated as a rounding slip. Working the powers out each time beats memorising cases, and answers should be quoted to two or three significant figures to match the precision of the data given.
一道 Kc 计算：731 K 下的碘化氢体系
在 731 K 下，1.00 dm3 容器里起始各放入 0.100 mol 的氢与碘，平衡时测得碘化氢为 0.160 mol。按方程式的 1 比 1 比 2 关系，生成 0.160 mol 碘化氢要消耗 0.080 mol 氢与 0.080 mol 碘。平衡时两种反应物各剩 0.020 mol，除以体积 1.00 dm3 得 0.020 mol dm-3。代入表达式，分子是 0.160 的平方，分母是 0.020 乘 0.020，结果是 64。这个反应两侧幂次相同，所以 Kc 没有单位。算完把平衡浓度代回表达式复算一次，能发现抄错数字这类低级失误。若下一问改问转化率，用消耗量除以起始量，0.080 除以 0.100 得 80%。
A 1.00 dm3 vessel held 0.100 mol of hydrogen and 0.100 mol of iodine at 731 K, and the equilibrium mixture contained 0.160 mol of hydrogen iodide. The equation consumes one mole of each reactant for every two moles of product, so forming 0.160 mol of HI costs 0.080 mol of each starting material; the remaining 0.020 mol in 1.00 dm3 gives 0.020 mol dm-3 for both, and the constant works out at 64 with no units because the powers cancel. Dividing the amount consumed by the starting amount gives an 80 per cent conversion, which is usually what the next part of the question wants.
反应商 Q 与平衡移动方向
把任意时刻的浓度代进与 Kc 相同的表达式，得到的是反应商 Q，不是平衡常数。Q 小于 Kc 时生成物相对不足，净反应向正方向进行，直到 Q 追上 Kc。Q 大于 Kc 时生成物偏多，净反应向逆方向进行。Q 等于 Kc 时体系已经在平衡，宏观浓度不再改变。很多学生以为 Q 与 Kc 是两个不同的公式，其实表达式完全一样，差别只在用哪一刻的数据。题目给出起始浓度让判断方向，先算 Q 再与 Kc 比大小，比凭感觉猜方向可靠得多。这一步也解释了加反应物为什么让平衡右移：投料瞬间 Q 变小，体系要靠多生成产物把 Q 拉回 Kc。
Feeding whatever concentrations happen to be present into the same expression gives the reaction quotient, which is the constant read at a different moment rather than a different formula. When Q sits below Kc the mixture is short of products and the net change runs forward; when Q exceeds Kc the reverse direction takes over, and equality means the system has already settled. Using Q removes the guesswork from questions about which way a mixture shifts, and it explains why adding a reactant pushes the position right, since the quotient drops the instant the addition is made.
浓度与压强改变时平衡怎么动
增大某一种反应物的浓度，平衡向消耗它的方向移动，Kc 的数值不变。减小生成物浓度效果相同，平衡也会右移。压强的影响只对气体分子数不同的反应成立：加压时平衡向气体分子数少的一侧移动，减压时反过来。两侧气体分子数相等时，改变压强只改变浓度，平衡位置不动，Kc 也不动。恒容容器里充入惰性气体，各组分浓度没变，平衡不移动。恒压条件下充入惰性气体，总体积变大，各组分分压下降，平衡向气体分子数多的一侧移动。常见的错误是把压强当成直接改变 Kc 的因素，实际改的只是浓度商，Kc 仍由温度决定。
Adding more of one reactant shifts the position towards the side that consumes it and leaves the value of Kc untouched. Pressure only matters when the two sides hold different numbers of gas molecules, since compression favours the side with fewer; an inert gas added at constant volume changes nothing, while the same gas added at constant pressure dilutes the mixture and pushes the position towards the side with more molecules. Treating pressure as something that alters the constant is a common slip, because pressure only moves the quotient.
温度是唯一改变 Kc 数值的因素
温度是唯一能改变 Kc 数值的因素，判断方向要看反应的焓变符号。正反应放热时，升温使平衡左移，Kc 变小；降温使平衡右移，Kc 变大。正反应吸热时方向相反，升温右移，Kc 变大。Edexcel 的 specification 在 Topic 11 里明确要求用 Kc 或 Kp 数值的变化来解释温度对平衡位置的影响，只写左移或右移拿不到全分。作答时先写焓变符号，再说升温有利于吸热方向，最后落到 Kc 数值的变化，这条链条把三个得分点串在一起。很多学生以为升温总是加快反应所以产率更高，对放热反应恰恰相反，升温提高速率却压低平衡产率。
Temperature is the one variable that changes the value of the constant, and the direction depends on the sign of the enthalpy change. For an exothermic forward reaction, heating shifts the position left and lowers Kc, while cooling raises it; an endothermic forward reaction behaves the other way round. The Pearson Edexcel specification asks for the change in the constant itself as the explanation, so an answer that only reports a shift, without quoting what happens to Kc, leaves marks on the table.
催化剂不改变平衡位置与产率
催化剂降低正逆两个方向的活化能，降低的幅度相同，所以两个方向的速率同步加快。结果是体系更快到达平衡，平衡位置与 Kc 都不变，产率也不会提高。工业上使用催化剂是为了在给定温度下缩短达到平衡所需的时间，让单位时间的产量上升。常见的错误是写催化剂使平衡右移或提高转化率，这类表述在阅卷里通常得零分。题目若问加催化剂后平衡浓度如何变化，答案是到达平衡更快，平衡浓度与原来相同。把速率问题与平衡位置问题分开写，这一章的表述题基本不会混。接触法与哈伯法都用到这一条：催化剂只买时间，不买产率。
A catalyst lowers the activation energy of both directions by the same amount, so the two rates rise together and the mixture simply arrives at equilibrium sooner. The position and the value of Kc stay exactly where they were, which means yield does not improve; industry uses catalysts to shorten the time needed at a given temperature and so raise output per hour. Writing that a catalyst pushes the equilibrium to the right is a standard zero-mark answer, because it confuses how quickly a system gets there with where it ends up.
Kp 表达式与分压的换算
气相反应的平衡常数可以写成 Kp，用分压代替浓度。Edexcel 的 specification 规定分压以大气压为单位，表达式里每一项写成该气体的分压。某气体的分压等于它的摩尔分数乘以总压，摩尔分数是该气体的物质的量除以平衡时气体总物质的量。写 Kp 时同样按系数取幂，纯固体不写进去。分压的单位是 atm，Kp 的单位由幂次差决定，可能是 atm-2，也可能没有单位。题目给摩尔数与总压时，先算总摩尔数再算摩尔分数，顺序颠倒容易把分母弄错。若题目给的是浓度与温度，可以先用理想气体的关系换算成分压再代入表达式。
For gas phase systems the constant can be written as Kp, with partial pressures in atmospheres replacing concentrations, as the specification requires. Each partial pressure is the mole fraction of that gas multiplied by the total pressure, and the mole fraction comes from the amount of that gas divided by the total amount present at equilibrium. Powers follow the stoichiometry exactly as they do for Kc and solids are omitted, so the units fall out of the power difference and can be atm-2 for a reaction that halves its gas molecules.
一道 Kp 计算：合成氨的平衡混合物
合成氨的反应由氮与氢生成氨，某温度下平衡混合物含 0.200 mol 氮、0.600 mol 氢与 0.400 mol 氨，总压为 100 atm。气体总量为 1.200 mol，氮的摩尔分数是 0.200 除以 1.200，得 0.1667，分压为 16.67 atm。氢的摩尔分数是 0.5，分压 50 atm，氨的摩尔分数是 0.3333，分压 33.33 atm。代入 Kp，分子是 33.33 的平方，分母是 16.67 乘以 50 的三次方，结果约为 5.3 乘 10 的负 4 次方。反应两侧气体分子数从 4 变成 2，所以单位是 atm-2。计算过程保留足够位数再四舍五入，中途只留两位小数会把末位算偏。
An ammonia equilibrium mixture held 0.200 mol of nitrogen, 0.600 mol of hydrogen and 0.400 mol of ammonia under a total pressure of 100 atm. Dividing each amount by the 1.200 mol total gives mole fractions of 0.1667, 0.5 and 0.3333, which become partial pressures of 16.67, 50 and 33.33 atm. Substituting those, with the ammonia term squared and the hydrogen term cubed, gives about 5.3 x 10-4, and the unit is atm-2 because four moles of gas on the left become two on the right.
哈伯法与接触法的条件取舍
哈伯法合成氨用铁作催化剂，条件约为 450 摄氏度与 200 个大气压。正反应放热且气体分子数减少，低温高压有利于产率，温度太低速率上不去，工业上取一个折中温度。接触法氧化二氧化硫用五氧化二钒催化，温度同样在 450 摄氏度附近，压力只需略高于常压，因为常压下转化率已经很高，再加压的设备成本换不来多少收益。两个例子考的是同一个思路：把平衡与速率放在一起考虑，而不是只看产率。作答时先写反应放热与分子数变化，再说条件怎样影响平衡与速率，最后说工业为什么这样选。很多学生只背数字，题目一换成别的反应就不会推。
The Haber process runs at roughly 450 degrees Celsius and 200 atmospheres over an iron catalyst, a compromise between a yield that prefers low temperature and a rate that needs heat. The Contact process oxidises sulfur dioxide over vanadium pentoxide at a similar temperature but only slightly above atmospheric pressure, since conversion is already high there and extra compression would cost more than it returns. Both cases test one habit of mind: read the enthalpy change and the mole change first, then explain the operating window instead of reciting a remembered set of numbers.
常见扣分点与图像题的读法
平衡题的扣分位置很集中：浓度代错、单位漏写、有效数字过多、把纯固体写进表达式、把速率变化说成平衡移动。图像题先看横纵坐标，浓度对时间趋于水平说明达到平衡，速率对时间两条线趋于相等说明两个方向速率相同，这两类图问的不是同一件事。题目要求解释曲线斜率的变化时，答案要落到速率，不要落到产率。计算题把每一步代入的数值写清楚，阅卷按步骤给分，最终数值错但过程对仍能拿部分分。常见的错误是写出浓度之比却不写幂次，或者把 Kc 与 Kp 的单位混用。拿这几处对着自己的错题本过一遍，比再刷十道同类型题有用。
Marks disappear in the same few places: substituting starting concentrations, dropping units, quoting too many significant figures, leaving a solid in the expression, or describing a rate change as a shift in position. On a concentration against time graph the curve flattening out signals equilibrium, while on a rate against time graph the two lines converging means the rates have become equal, and those two graphs want different answers. Writing each substitution on its own line protects the method marks, since Pearson examiners award them step by step even when the final value goes wrong.
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