OCR化学 方程式配平 放热吸热 沉淀 电解
OCR 的 GCSE Chemistry A (Gateway Science) 编号是 J248，C3 Chemical reactions 分成四块：C3.1 引入化学反应、C3.2 能量学、C3.3 反应类型、C3.4 电解。Foundation 考 Paper 1 和 Paper 2，Higher 考 Paper 3 和 Paper 4，每份 90 分、1 小时 45 分钟、占总成绩 50%，Paper 1 和 Paper 3 考 C1 到 C3 加 C7 实验技能。这一篇按 OCR 官网 J248 大纲的条目编号把 C3 的常考内容串一遍，方程式、能量变化、酸的反应和电解各占几段。
OCR GCSE Chemistry A (Gateway Science) is specification J248, and its C3 topic splits into C3.1 introducing chemical reactions, C3.2 energetics, C3.3 types of chemical reactions and C3.4 electrolysis. Papers 1 and 3 cover topics C1 to C3 plus C7 practical skills, each paper lasting 1 hour 45 minutes and worth 90 marks, or 50% of the qualification. Each section below is tied to the wording of the OCR specification so your notes can be checked against the board's own checklist.
化学式：符号、下标与括号
C3.1a 要求用化学符号写出元素与简单共价、离子化合物的化学式。元素符号第一个字母大写，第二个字母小写，Co 是钴，CO 是一氧化碳，两个写法指向完全不同的物质。离子化合物的下标由电荷配平决定，钠离子和氯离子各带一个电荷，写成 NaCl，镁离子带两个正电荷，需要两个氯离子，写成 MgCl2。多原子离子必须加括号，硝酸钙写成 Ca(NO3)2，括号外的 2 作用在整组 NO3 上。C3.1d 要求背熟常见离子：硝酸根 NO3-、硫酸根 SO4 2-、碳酸根 CO3 2-、氢氧根 OH-、铵根 NH4+。很多学生以为 Ca(OH)2 里的 2 只属于氧原子，其实它表示两个氢氧根。阅卷时把 CaNO32 这种漏括号的写法当成答案，一道化学式题就直接零分。
Specification point C3.1a asks you to build formulae from symbols for elements and for simple covalent and ionic compounds, and capitalisation carries meaning, so Co is cobalt while CO is carbon monoxide. Ionic formulae come from charge balance, which is why Mg2+ pulls in two chloride ions and why polyatomic ions such as nitrate take brackets in Ca(NO3)2. Learn the common ions cold, because a missing bracket is the most common way candidates lose the whole formula mark.
配平方程式与状态符号
配平只改系数，不改下标，这一点在 C3.1b 里写得很明确。甲烷燃烧写成 CH4 + 2O2 -> CO2 + 2H2O，碳、氢、氧三种原子的数目两边相等，检查顺序按金属、非金属、氢、氧走一遍，最后再数一次总原子数。C3.1f 要求用 s、l、g、aq 标出状态，室温下的水写 H2O(l)，水蒸气写 H2O(g)，稀盐酸写 HCl(aq) 而不是 HCl(l)。常见的错误是把 H2O 前面的系数搬到下标上，把 O2 改成 O4 来凑氧原子。改下标等于换了一种物质，方程式再平衡也失去意义。
Balancing changes coefficients only, never subscripts, and the safest routine is to work through metals, then non-metals, then hydrogen, then oxygen before a final atom count. State symbols follow the conditions of the reaction, so water is H2O(l) at room temperature, steam is H2O(g), and dilute hydrochloric acid is HCl(aq) rather than HCl(l). Turning O2 into O4 to make the oxygen tally look right silently changes the substance, and the equation no longer describes the reaction on the page.
离子方程式与旁观离子
C3.1e 要求写出配平的离子方程式。硝酸银溶液与氯化钠溶液混合，真正发生变化的是银离子和氯离子，写成 Ag+ + Cl- -> AgCl(s)，钠离子和硝酸根留在溶液里，属于旁观离子，直接从方程式里删掉。写之前先把可溶的强电解质拆成离子，难溶的沉淀、气体和水保留分子形式，这一步做错，后面配平再准也会被判错。电荷要配平，左边一个正电荷加一个负电荷等于零，右边是中性的沉淀。学生最常把 AgCl 拆成 Ag+ 加 Cl-，忘了沉淀在溶液里并不以自由离子形式存在。
C3.1e asks for balanced ionic equations, and the method is to split every soluble strong electrolyte into ions and then delete the spectators, which is how silver nitrate plus sodium chloride collapses to Ag+ + Cl- -> AgCl(s). Insoluble precipitates, gases and water stay written as molecules, and charge has to balance across the arrow as well as atoms. Writing AgCl as separate ions is the classic error, because a precipitate is not present as free ions in the solution.
质量守恒与表观质量变化
C3.1b 用质量守恒解释方程式的合理性，反应前后原子的种类和数目都不变。敞口容器里镁条燃烧，固体质量变大，因为镁结合了空气中的氧；碳酸钙与稀盐酸在烧杯里反应，总质量变小，因为二氧化碳逸散到空气里。把反应放进带气球的锥形瓶或者密封瓶，天平读数就不动了。很多学生以为质量守恒只在密闭体系成立，其实守恒一直成立，只是没称到跑掉的气体。答题时要把逸出或结合的气体名字写出来，只写质量变了不给分。
Conservation of mass underpins every equation in C3.1b, and apparent mass changes only reflect gas leaving or joining an open system. Burning magnesium gains mass as oxygen from the air is incorporated, while calcium carbonate plus dilute hydrochloric acid loses mass because carbon dioxide escapes into the room. Seal the reaction in a conical flask with a balloon and the balance stops moving, so name the gas in your answer instead of saying that the mass changed.
摩尔与阿伏加德罗常数
C3.1g 要求用标准形式写出阿伏加德罗常数的值，6.02 x 10^23 每摩尔，并且知道一摩尔的微粒数就是这个数。一个原子的质量等于相对原子质量除以阿伏加德罗常数，碳原子大约是 12 除以 6.02 x 10^23，结果落在 10 的负 23 次方克这个量级。C3.1h 用 n 等于 m 除以 Mr 把质量换成摩尔，二氧化碳的 Mr 是 44，所以 4.4 克对应 0.1 摩尔，分子数是 6.02 x 10^22。常见的错误是拿 Mr 直接当质量用，或者把 10^23 写成 10^22 少了一个数量级。Higher 卷会把限量试剂和方程式的系数比结合起来考。
C3.1g requires the Avogadro constant written in standard form as 6.02 x 10^23 per mole, together with the idea that one mole contains exactly that many particles. Dividing relative atomic mass by the Avogadro constant gives the mass of a single atom, which lands around 10 to the minus 23 grams for carbon. For C3.1h, n equals m divided by Mr, so 4.4 g of carbon dioxide with Mr 44 is 0.1 mol and therefore 6.02 x 10^22 molecules.
放热与吸热：温度变化判断
C3.2a 靠周围温度的变化判断反应方向。温度升高说明反应放热，能量以热的形式传给环境，燃烧、中和、金属与酸的反应都属于这一类。温度降低说明反应吸热，从环境吸走热量，碳酸钙受热分解、柠檬酸与碳酸氢钠的反应是常见例子。做实验时把温度计插在反应混合物里读最高值，容器要尽量隔热，敞口烧杯散热太快，读数偏低。常见的错误是把吸热反应写成温度不变，其实温度会掉，只是幅度可能只有几度。题目问的是周围温度，不是反应物本身的温度。
C3.2a separates the two directions using the temperature change of the surroundings, so a rise means energy is released as heat and a fall means energy is taken in. Combustion, neutralisation and metal plus acid reactions warm the mixture, whereas thermal decomposition of calcium carbonate and citric acid with sodium hydrogencarbonate cool it. Insulate the container and record the largest temperature change, because an open beaker loses heat too quickly for a reliable reading.
活化能与反应能量曲线
C3.2c 把活化能定义为反应发生所需要的最低能量，分子碰撞时的动能要超过这个门槛，反应才能进行。C3.2b 要求画出并标注反应曲线，横轴是反应进程，纵轴是能量，曲线从反应物出发爬到一个峰值，再降到生成物。放热反应的生成物能量低于反应物，峰值与反应物之间的高度差就是活化能；吸热反应反过来，生成物能量高于反应物。加入催化剂会降低峰值，活化能变小，但反应物和生成物的能量差不变，所以放热的仍然放热。很多学生以为催化剂能让吸热反应变成放热反应，这是把活化能和总能量变化搞混了。
Activation energy, defined in C3.2c, is the minimum energy colliding particles need before a reaction can proceed, and it appears on the profile as the height from the reactants up to the peak. In an exothermic profile the products sit lower than the reactants, while an endothermic one ends higher, and the vertical gap between the two levels is the overall energy change. A catalyst lowers the peak without shifting either level, so it cannot turn an endothermic reaction into an exothermic one.
键能计算能量变化
C3.2d 用键能算能量变化，断键要吸收能量，成键要放出能量。氢气与氯气生成氯化氢，H-H 键能 436 千焦每摩尔，Cl-Cl 键能 243 千焦每摩尔，断裂这两个键一共吸收 679 千焦。生成两摩尔氯化氢要形成两个 H-Cl 键，每个 432 千焦，一共放出 864 千焦。吸收减放出，679 减 864 等于负 185，说明每摩尔反应放热 185 千焦。算的时候先数键的个数再乘键能，漏掉系数是最常见的错误，氢气里只有一个 H-H 键，生成物里却有两个 H-Cl 键。题目给的键能单位是千焦每摩尔，答案也要带单位。
Bond energies give the energy change in C3.2d, because breaking bonds absorbs energy and making bonds releases it. For H2 + Cl2 -> 2HCl, breaking 436 and 243 kJ per mole costs 679 kJ, while forming two H-Cl bonds at 432 kJ each releases 864 kJ. The net value is 679 minus 864, so the reaction releases 185 kJ per mole, and forgetting that two H-Cl bonds form is where most of the marks disappear.
金属与酸、碳酸盐与酸
C3.3 把酸的反应分成几类。金属与稀酸反应生成盐和氢气，镁、锌、铁都能反应，铜排在氢后面，加进稀盐酸里没有气泡。氢气用点燃的木条检验，发出爆鸣声说明是氢气。碳酸盐与酸反应生成盐、水和二氧化碳，石灰水变浑浊是二氧化碳的检验方法。写方程式时盐的名字由酸决定，盐酸给氯化物，硫酸给硫酸盐，硝酸给硝酸盐。镁与盐酸写成 Mg + 2HCl -> MgCl2 + H2，注意氢气是双原子分子。学生最常把气泡直接写成氧气，金属与酸放出的是氢气，这一点在实验题里反复考。
C3.3 groups acid reactions into families, and the first is metal plus dilute acid giving a salt and hydrogen, which is why magnesium, zinc and iron fizz while copper sits there doing nothing. A lighted splint gives the squeaky pop that identifies hydrogen, and carbonate plus acid gives a salt, water and carbon dioxide that turns limewater milky. Name the salt from the acid used, hydrochloric for chlorides and sulfuric for sulfates, and remember hydrogen is diatomic when you write Mg + 2HCl -> MgCl2 + H2.
中和反应与盐的命名
酸与碱反应生成盐和水，本质是氢离子与氢氧根离子结合成水，离子方程式写成 H+ + OH- -> H2O。氢氧化钠与盐酸生成氯化钠，氢氧化钠与硫酸生成硫酸钠，盐的名字跟着酸走。指示剂用来判断终点，酚酞在碱性溶液里是粉红色，加到恰好褪色说明酸和碱正好反应完。用滴定管可以控制加入量，用滴管做课堂实验只能粗略判断。常见的错误是以为中和以后溶液一定中性，强酸与弱碱生成的盐会水解，pH 不一定等于 7。题目问现象时写颜色变化，问本质时写离子方程式。
Neutralisation produces a salt and water, and its ionic core is H+ + OH- -> H2O, which is the line examiners look for when the question asks what happens at particle level. Sodium hydroxide with hydrochloric acid gives sodium chloride, and the same alkali with sulfuric acid gives sodium sulfate, so the salt name follows the acid rather than the metal. Phenolphthalein is pink in alkali and colourless at the end point, and assuming the final solution must be pH 7 is an overreach.
沉淀反应与溶解性规则
两种溶液混合生成不溶的固体，就是沉淀反应。溶解性规则要先背下来：钠盐、钾盐、铵盐和硝酸盐都可溶，大多数氯化物可溶，氯化银和氯化铅除外，大多数硫酸盐可溶，硫酸钡和硫酸铅除外，碳酸盐大多不溶，只有钠、钾、铵的碳酸盐可溶。检验氯离子用硝酸银溶液，出现白色沉淀说明有氯离子；检验硫酸根用氯化钡溶液，白色沉淀说明有硫酸根。写方程式时沉淀标 (s)，溶液里的离子标 (aq)。很多学生以为所有白色沉淀都一样，其实要配合试剂才能判断是哪种离子，单看颜色分不出来。
A precipitation reaction forms an insoluble solid when two solutions are mixed, and the solubility rules do most of the work: sodium, potassium, ammonium and nitrate salts dissolve, most chlorides dissolve except silver and lead, and most sulfates dissolve except barium and lead. Silver nitrate gives a white precipitate with chloride ions, and barium chloride does the same for sulfate ions. Label the precipitate (s) and the remaining ions (aq), because a white solid on its own does not identify the ion without the reagent that produced it.
电解：电极与产物判断
C3.4 讲电解，用电流把化合物分解。熔融氯化钠里只有钠离子和氯离子，阴极带负电，吸引钠离子，得到金属钠，阳极带正电，吸引氯离子，得到氯气。氯化钠水溶液里还有氢离子和氢氧根离子，阴极上氢离子得电子变成氢气，因为钠比氢活泼，会留在溶液里。阳极上氯离子失去电子生成氯气，溶液很稀时氢氧根放电生成氧气。阴极接电源负极，阳极接正极，这个对应关系不要记反。判断产物时先看电解质是熔融还是水溶液，再看金属在活动性顺序里的位置。学生最容易把阴阳极的电荷搞混，画装置图时在电极旁边标上正负号。
Electrolysis in C3.4 splits a compound with a direct current, and molten sodium chloride is the simplest case because only Na+ and Cl- are present, giving sodium at the negative cathode and chlorine at the positive anode. In aqueous sodium chloride the water contributes H+ and OH-, so hydrogen is released at the cathode while the more reactive sodium stays in solution. Check whether the electrolyte is molten or aqueous, and label the polarity of each electrode on your diagram, because reversed charges wreck the product prediction.
C3 答题顺序：先配平再算能量
卷面上 C3 的题按顺序做未必省时间。先把方程式配平，后面算摩尔、算键能、判断电解产物都靠这一步，系数错了后面全错。写现象题先写看到什么，再写结论，气泡、沉淀、颜色变化各占一分。算能量变化的题把断键和成键分开列，单位写在等号右边。文字题要出现题目里的物质名和条件，只写反应发生了不给分。做完回头检查三个地方：下标有没有被改、状态符号有没有漏、沉淀有没有标 s。
Work through C3 in an order that protects the marks, balancing every equation first because mole calculations, bond energy sums and electrolysis predictions all depend on those coefficients. In observation questions describe what you see and then state the conclusion, since fizzing, precipitates and colour changes each carry a mark. Before you finish, check subscripts, state symbols and whether every precipitate is marked (s).
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