哈伯法制造及氮磷钾（NPK）肥料：比例、条件与盐类
引入
词汇
| English | 释义 | 拼音 |
|---|---|---|
| NPK fertiliser | 氮磷钾化肥 | dàn lín jiǎ huà féi |
| Haber process/ˈheɪbə ˈprəʊses/ | 哈伯法 | hā bó fǎ |
English
Does recycling gases make fertiliser resource-free?
- Returning unreacted gases improves feedstock use. It does not eliminate the energy needed to heat and compress gases or the resources needed to make salts.
- This Separate Chemistry lesson covers Haber manufacture and NPK fertilisers 氮磷钾化肥. The process loop is required at both tiers; equilibrium/rate evaluation is explicitly Higher tier (HT).
English
Follow the Haber loop
- The Haber process 哈伯法 manufactures ammonia. Nitrogen comes from air; hydrogen can come from natural gas reacted with steam. Purified gases pass over iron at about 450 °C and about 200 atmospheres: $N_2+3H_2\rightleftharpoons2NH_3$.
- Cooling liquefies ammonia for removal. Unreacted nitrogen and hydrogen return to the reactor. The reaction is reversible, so per-pass conversion is different from overall conversion after repeated recycling.
练习
经冷却后返回哈伯反应器的物料流是哪一股？
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
冷却使液氨液化并分离；未反应的氮气和氢气则被回收。该循环工艺可提高原料利用率，但加热和压缩仍需消耗能量。
English
Use the coefficient ratio on one pass
- Feed contains 2 mol nitrogen and 6 mol hydrogen. If only 1 mol nitrogen reacts, it consumes 3 mol hydrogen and forms 2 mol ammonia. Remaining gases are 1 mol nitrogen and 3 mol hydrogen.
- If all the original feed converted, the theoretical maximum would be 4 mol ammonia. Do not use this maximum as the amount actually made on the stated pass. The balanced equation conserves nitrogen and hydrogen atoms.
English
HT: balance rate, yield and cost
- The forward reaction is exothermic. Lower temperature favours equilibrium ammonia yield but slows reaction; higher temperature increases rate but lowers equilibrium yield. Interpret rate and equilibrium graphs as different quantities.
- Higher pressure favours ammonia because four reactant gas molecules become two product molecules, but compression and strong equipment cost energy and money. Iron speeds both directions and approach to equilibrium without changing its position. The industrial conditions are a compromise, not maximum possible equilibrium yield.
练习
在温度和压力不变的情况下，铁能提高氨的平衡产率。
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
铁催化剂同时加速正逆反应速率，使系统更快达到平衡，但不改变平衡位置。这一高温（HT）特性区分了反应速率与平衡产率。
English
Supply nitrogen, phosphorus and potassium
- NPK fertiliser is a formulation of salts supplying three nutrient elements. Ammonia can make ammonium salts and nitric acid. Neutralisation gives ammonium nitrate: $NH_3+HNO_3\rightarrow NH_4NO_3$.
- Potassium chloride, potassium sulfate and phosphate rock are mined. Phosphate rock is not directly suitable as fertiliser; treatment makes usable soluble salts. Phosphoric acid supplies no potassium by itself.
练习
氨气中和硝酸生成何种铵盐？
难度 1/5 · 今日首次完成可获 100 金币及少量额外奖励
NH₃ +HNO₃ →NH₄NO₃，即硝酸铵。它提供氮元素，而完整的NPK配方还同时提供磷和钾。
English
Distinguish three phosphate-rock treatments
- Nitric acid treatment produces calcium nitrate and phosphoric acid. Sulfuric acid makes single superphosphate containing calcium dihydrogen phosphate with calcium sulfate.
- Phosphoric acid makes triple superphosphate containing calcium dihydrogen phosphate without that sulfate component. Do not call every acid treatment simply “superphosphate”, and do not confuse a phosphorus source with a potassium source.
练习
将术语与其精确含义匹配。
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
利用化学结构、工艺或比较界限来区分这些术语。
English
Compare the same salt in lab and industry
- Carefully controlled laboratory neutralisation/titration followed by evaporation and crystallisation can produce a pure solid. A laboratory titration does not inherently give an impure product.
- Industry can integrate ammonia/acid production, recover heat and sustain large output, but needs equipment and emissions control. Compare supplied energy per unit, purity, scale, equipment, waste and costs; “industry is always better” is not an evaluation.
练习
哪两项主张正确限定了制造比较？
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
实验室滴定本身并不必然含杂质。工业具备规模与集成优势，但仍需能源与设备；请评估实际标准与数据。
English
Apply the method to a new feed
- Independent feed: 3 mol nitrogen and 9 mol hydrogen. If 2 mol nitrogen reacts on a pass, 6 mol hydrogen reacts, 4 mol ammonia forms and 1 mol nitrogen plus 3 mol hydrogen remain.
- A catalyst can reach equilibrium sooner but cannot increase the equilibrium amount at unchanged temperature and pressure. Give a reason based on rate or equilibrium, then a cost/condition that limits the industrial claim.
Key idea
Follow the process, use a balanced ratio, distinguish rate from equilibrium yield, and identify the actual nutrient salts and manufacturing comparison.
练习
若进料含有3 mol 氮气和9 mol 氢气。若单次通过反应器时2 mol 氮气发生反应，则生成多少摩尔的氨？
难度 2/5 · 今日首次完成可获 140 金币及少量额外奖励
N₂ + 3H₂ ⇌ 2NH₃。2 mol 氮气消耗6 mol 氢气并生成4 mol 氨，剩余1 mol 氮气和3 mol 氢气。按全部进料计算的理论最大值6 mol 并非题目所指的单次通过量。