יחידה 2: אנרגטיקה, קבוצות ותגובות אורגניות
Pearson Edexcel · International A-Level · Chemistry נושא 2 33:01 קריאת קול באנגלית · תרגום אנגלי + סינית שרוף בתוך הסרטון
פרקים
תרגום טקסט
Energy, attractions, electron transfer and collision models explain different observations.
能量、吸引力、电子转移和碰撞模型可以解释不同的实验现象。
Use Unit 1 mole ratios, polarity and structural formulae; state conditions before predicting products.
请运用第一单元中的摩尔比、极性和结构式知识。 在预测产物之前，先说明反应条件。
At constant pressure, delta H is the reaction heat change.
在恒压条件下，ΔH 表示反应热变化。
Exothermic: surroundings warm and reaction delta H is negative.
放热反应中，环境温度升高，且反应的 ΔH 为负值。
Standard data specify 100 kPa, temperature and physical states.
标准数据规定了 100 kPa、温度和物理状态。
Formation: one mole of compound from standard-state elements.
生成：由标准状态下的元素生成一摩尔化合物。
Combustion: complete burning of one mole of substance.
燃烧：一摩尔物质完全燃烧。
Neutralisation: one mole of water formed; atomisation: one mole of gaseous atoms.
中和：生成一摩尔水。 原子化：生成一摩尔气态原子。
q equals m c delta T, which is 100.0 grams times 4.18 joules per gram per degree Celsius, times 6.8 degrees Celsius, equaling 2.842 kilojoules.
q = mcΔT = 100.0 g × 4.18 J g⁻¹ °C⁻¹ × 6.8 °C = 2.842 kJ。
The number of moles of H2O equals concentration times volume, which is 1.00 mole per cubic decimeter times 0.0500 cubic decimeter, equaling 0.0500 mole.
H₂O的物质的量 n(H₂O) = cV = 1.00 mol dm⁻³ × 0.0500 dm³ = 0.0500 mol。
Delta H equals negative q divided by n, which is negative 56.8 kilojoules per mole. Use total heated mass and limiting amount.
ΔH = -q/n = -56.8 kJ mol⁻¹；使用总加热质量和限量反应物。
The solution gains 1.20 kJ while 0.0200 mol reacts.
溶液吸收了1.20 kJ的热量，同时有0.0200 mol发生反应。
What is the molar reaction enthalpy?
摩尔反应焓是多少？
Pause and justify your choice.
请先暂停，论证你的选择。
Delta H equals negative q divided by n, which is negative 1.20 kilojoules divided by 0.0200 moles, resulting in negative 60.0 kilojoules per mole.
ΔH = −q/n = −1.20 kJ / 0.0200 mol = −60.0 kJ·mol⁻¹。
B is correct.
B 正确。
A uses the solution sign. C multiplies by amount. D omits the mole basis.
A 使用了溶液符号；C 乘以了物质的量；D 遗漏了摩尔基准。
C plus O sub 2 yields C O sub 2, negative three hundred and ninety-four kilojoules per mole.
C 加 O₂ 生成 CO₂，负三百九十四千焦每摩尔。
Reverse C O plus one half O sub 2 yields C O sub 2, positive two hundred and eighty-three kilojoules per mole.
反转 CO 加二分之一 O₂ 生成 CO₂，正二百八十三千焦每摩尔。
Sum: C plus one half O sub 2 yields C O; delta H equals negative one hundred and eleven kilojoules per mole.
相加：C 加二分之一 O₂ 生成 CO；ΔH 等于负一百一十一千焦每摩尔。
Count bonds broken minus bonds formed, including coefficients.
计算断键数减去成键数，需包含化学计量系数。
For H2 plus Cl2 yields 2HCl: 436 plus 243 minus 2 times 431 equals negative 183 kilojoules per mole.
对于 H2 + Cl2 → 2HCl：436 + 243 − 2(431) = −183 kJ mol⁻¹。
Mean values describe gas-phase bonds in averaged environments; delta H does not determine rate.
平均值描述的是气相中平均环境下的键能；ΔH 不能决定反应速率。
Draw exothermic and endothermic enthalpy-level diagrams.
绘制放热和吸热的焓级图。
Label both levels and the signed ΔH arrow; the horizontal axis is reaction progress.
标注两个能级以及带符号的ΔH箭头，横轴为反应进程。
Exothermic products lie lower: the reaction ΔH arrow points down.
放热反应的产物位置较低，反应焓变箭头向下。
Endothermic products lie higher: the arrow points up.
吸热反应的产物位置较高，箭头向上。
Find C graphite plus 2 H sub 2 gas yields CH sub 4 gas using combustion data.
利用燃烧数据求C石墨加2H₂(g)生成CH₄(g)。
For carbon, the value is minus 394 kilojoules per mole.
碳的数值为负394千焦每摩尔。
For hydrogen gas, the value is minus 286 kilojoules per mole.
氢气的数值为负286千焦每摩尔。
For methane, the value is minus 891 kilojoules per mole.
甲烷的数值为负891千焦每摩尔。
Draw a common final state with the correct oxygen amount.
画出具有正确氧气量的共同终态。
Carbon plus two hydrogen moles: minus 394, plus 2 times minus 286, equals minus 966 kilojoules per mole.
碳加两摩尔氢气：-394 + 2 × (-286) = -966 kJ mol⁻¹。
Minus 966 equals delta H, plus minus 891, so delta H equals minus 75 kilojoules per mole.
-966 = ΔH + (-891)，所以 ΔH = -75 kJ mol⁻¹。
Before mixing at 120 s: 20.0 °C.
混合前，在120秒时温度为20.0摄氏度。
After: 25.2 °C at 150 s, then 24.9, 24.6, 24.3 °C at 180, 210, 240 s.
混合后，在150秒时温度为25.2摄氏度，随后在180、210和240秒时分别为24.9、24.6和24.3摄氏度。
Plot and extrapolate to mixing.
绘制数据图并外推至混合时刻。
Then use 80.0 g, c = 4.18 J g⁻¹ °C⁻¹ and 0.0400 mol water to find ΔH.
然后使用80.0克水、比热容为4.18焦耳每克每摄氏度以及0.0400摩尔水来计算焓变ΔH。
Extrapolate to 25.5 degrees Celsius at 120 seconds.
外推至120秒时温度为25.5摄氏度。
The corrected change in temperature is 5.5 degrees Celsius.
校正后的温度变化ΔT为5.5摄氏度。
q equals 80.0 times 4.18 times 5.5, which equals 1839.2 joules.
q等于80.0乘以4.18乘以5.5，即1839.2焦耳。
Delta H equals negative 1.8392 divided by 0.0400, which equals negative 46.0 kilojoules per mole.
ΔH等于负1.8392除以0.0400，即负46.0千焦每摩尔。
H bonded to N, O or F can attract a suitable lone pair on another molecule.
与氮、氧或氟形成氢键的原子，可吸引另一分子上合适的孤对电子。
The dashed attraction is between molecules; the O–H covalent bond stays within water.
虚线表示的是分子间的吸引力；而O–H共价键则保留在水分子内部。
More electrons usually strengthen London forces; alkane boiling temperatures rise with chain length.
更多电子通常会增强伦敦力。 烷烃的沸点随碳链增长而升高。
Branching reduces effective contact between isomers.
支链会降低异构体之间的有效接触面积。
Alcohols also hydrogen-bond; HF is unusually high compared with HCl/HBr/HI.
醇类也能形成氢键。 HF的沸点异常高，高于HCl、HBr和HI。
An open hydrogen-bonded ice structure gives lower density than liquid water.
开放的氢键冰结构密度低于液态水。
Water hydrates ions and hydrogen-bonds with small alcohols.
水可水合离子，并与小分子醇形成氢键。
A polar C–X bond alone does not make a halogenoalkane water-soluble.
仅凭极性的碳-卤键并不能使卤代烃溶于水。
Pause and draw one pair for each of the three liquids.
请先暂停，为这三种液体各画出一对分子。
Distinguish the covalent bonds inside a molecule from the hydrogen bond between molecules.
区分分子内部的共价键与分子间的氢键。
Show the acceptor lone pair and partial charges.
标出受体上的孤对电子和部分电荷。
Water keeps two covalent O–H bonds and two oxygen lone pairs in each molecule.
每个水分子包含两个共价O–H键和两对氧孤对电子。
The donor O–H hydrogen is partly positive; a neighbouring oxygen lone pair accepts the hydrogen bond.
供体的O–H氢带部分正电荷；相邻的氧孤对电子接受氢键。
The dashed line is an intermolecular attraction.
虚线表示分子间作用力。
Each ammonia molecule retains three N–H bonds and one nitrogen lone pair.
每个氨分子保留三个N-H键和一个氮孤对电子。
A nitrogen-bonded hydrogen is the donor; the neighbouring nitrogen lone pair is the acceptor.
与氮相连的氢是供体，相邻的氮孤对电子是受体。
This is a pair-interaction schematic rather than a three-dimensional shape model.
这是成对相互作用的示意图，而非三维形状模型。
Each HF molecule retains its covalent H–F bond and three fluorine lone pairs.
每个HF分子保留其共价H-F键和三个氟原子孤对电子。
The partly positive donor hydrogen is attracted to a neighbouring fluorine lone pair.
带部分正电荷的供体氢原子被邻近的氟原子孤对电子吸引。
Do not replace that dashed attraction with another covalent H–F bond.
请勿用另一个共价H-F键替代该虚线吸引力。
Oxidation loses electrons and raises oxidation number; reduction does the reverse.
氧化失去电子并升高氧化数，还原则相反。
The oxidising agent gains electrons and is reduced.
氧化剂获得电子并被还原。
O is −1 in peroxides; H is −1 in metal hydrides.
过氧化物中氧为负一，金属氢化物中氢为负一。
Sum values to the species charge.
各元素数值之和等于物种电荷。
Al to Al three plus plus three e minus, and Cu two plus plus two e minus to Cu.
铝转化为铝三价离子加三个电子，以及铜二价离子加两个电子转化为铜。
Multiply by 2 and 3 so six electrons cancel.
分别乘以2和3，使六个电子抵消。
2 Al plus 3 Cu two plus to 2 Al three plus plus 3 Cu.
2个铝加3个铜二价离子生成2个铝三价离子加3个铜。
Cu two plus is the oxidising agent.
铜二价离子是氧化剂。
Down Groups 1/2, radius and shielding increase; outer electrons are lost more easily.
在第1和第2基团中，半径和屏蔽效应增加；外层电子更容易失去。
Small high-charge-density cations polarise anions strongly, lowering carbonate/nitrate thermal stability.
小尺寸高电荷密度阳离子强烈极化阴离子，降低碳酸盐和硝酸盐的热稳定性。
Group 2 hydroxides become more soluble down the group; sulfates less soluble.
第2基团氢氧化物随基团向下溶解度增加；硫酸盐溶解度降低。
Li₂CO₃ and Group 2 carbonates give oxide + CO₂ on heating.
碳酸锂和第二族碳酸盐加热时生成氧化物与二氧化碳。
Most Group 1 nitrates give nitrite + O₂; lithium and Group 2 nitrates also give NO₂.
大多数第一族硝酸盐生成亚硝酸盐与氧气。 锂和第二族硝酸盐还会生成二氧化氮。
Flame-test colours arise from electron transitions, not the ordinary colour of the solid.
焰色反应的颜色源于电子跃迁，而非固体本身的颜色。
Equal volumes of saturated Mg(OH)₂ and Ba(OH)₂ are titrated with the same acid.
等体积的饱和氢氧化镁和氢氧化钡用同一种酸滴定。
Which needs more?
哪一个需要更多？
Equal Mg²⁺/Ba²⁺ concentrations receive excess sulfate.
等浓度的镁离子和钡离子加入过量硫酸盐。
Which gives more precipitate?
哪一个产生更多沉淀？
Explain both.
请解释两者。
Ba(OH)₂ is more soluble: its equal saturated volume contains more OH⁻ and needs more acid.
氢氧化钡的溶解度更高。 其等体积饱和溶液中含有更多的氢氧根离子，需要消耗更多的酸。
BaSO₄ is less soluble: equal initial cation amounts give more solid with excess sulfate.
硫酸钡的溶解度较低。 在初始阳离子量相等的情况下，加入过量硫酸根会生成更多沉淀。
Solubility controls dissolved amount; it is different from dissociation strength.
溶解度决定了溶解的量，这与解离强度不同。
Match measured amounts, particle size, apparatus and heating rate; compare sample temperature at a consistent decomposition criterion.
匹配测量量、颗粒大小、仪器和加热速率。 在一致的分解标准下比较样品温度。
CO₂: milky limewater; O₂: glowing splint relights; NO₂: brown and acidic.
二氧化碳：使澄清石灰水变浑浊；氧气：使带火星的木条复燃；二氧化氮：呈棕色且显酸性。
Clean flame-test wire and keep the flame region/loading consistent.
清洁火焰测试用铂丝，并保持火焰区域与载样量一致。
Na yellow, K lilac, Ca brick-red, Sr crimson-red; sodium carry-over can mask another colour.
钠呈黄色，钾呈淡紫色，钙呈砖红色，锶呈深红色；注意钠的残留可能掩盖其他颜色。
Carbonate + acid gives CO₂; confirm with limewater.
碳酸盐加酸产生二氧化碳，用澄清石灰水验证。
Sulfate: dilute HCl then BaCl₂ gives white BaSO₄.
硫酸盐：先加稀盐酸，再加氯化钡，生成白色硫酸钡沉淀。
Ammonium + warm NaOH gives NH₃; damp red litmus turns blue.
铵盐加温热氢氧化钠产生氨气，湿润的红色石蕊试纸变蓝。
Rinse burette and pipette with their own solutions.
用各自的溶液润洗滴定管和移液管。
Fill the jet and remove the funnel.
充满尖嘴并移除漏斗。
Use concordant accurate titres, not the rough trial.
使用一致且准确的滴定体积，而非粗略试验值。
Two plus or minus zero point zero five centimetre cubed readings give plus or minus zero point one zero centimetre cubed maximum titre uncertainty.
两个正负零点零五立方厘米的读数，最大滴定体积不确定度为正负零点一零立方厘米。
Twenty-five point zero zero cubic centimetres of zero point zero five zero zero moles per cubic decimeter sodium carbonate uses twenty-four point six zero cubic centimetres of hydrochloric acid.
25.00 cm³ 浓度为 0.0500 mol dm⁻³ 的碳酸钠消耗了 24.60 cm³ 盐酸；比例为 1:2。
The ratio is one to two. The amount of carbonate equals concentration times volume, which is zero point zero zero one two five zero moles. The amount of hydrochloric acid is zero point zero zero two five zero zero moles.
n(碳酸根) = cV = 0.001250 mol；n(HCl) = 0.002500 mol。
The concentration of hydrochloric acid equals the amount divided by the volume, which is zero point zero zero two five zero zero moles divided by zero point zero two four six zero cubic decimeters, giving zero point one zero one six moles per cubic decimeter.
c(HCl) = n/V = 0.002500 mol / 0.02460 dm³ = 0.1016 mol dm⁻³。
A salt releases gas turning limewater milky with acid.
某盐遇酸释放气体，使澄清石灰水变浑浊。
A fresh aliquot gives ammonia with warm NaOH.
取新鲜试样与温热氢氧化钠反应生成氨气。
Identify the supported ions and write both NH₃ production and HCl-fume equations.
请鉴定所含离子，并写出产生氨气和氯化氢白雾的化学方程式。
Can you distinguish carbonate from hydrogencarbonate?
你能区分碳酸根和碳酸氢根吗？
NH4 plus OH minus yields NH3 plus H2O supports ammonium.
NH4+加OH-生成NH3和H2O，可证明铵离子的存在。
NH3 plus HCl yields NH4Cl explains the white fumes.
NH3与HCl反应生成NH4Cl，解释了白烟现象。
Both carbonate and hydrogencarbonate release CO2 with acid; the evidence does not select just one.
碳酸盐和碳酸氢盐遇酸均会释放CO2，仅凭此证据无法区分二者。
H two A has M equals one hundred twenty-six point zero grams per mole.
H₂A的摩尔质量M为126.0 g mol⁻¹。
Dissolve one point five seven five grams and make to two hundred fifty point zero cubic centimetres.
溶解1.575 g并定容至250.0 cm³。
Twenty-five point zero zero cubic centimetres of acid requires twenty-three point eight zero cubic centimetres of NaOH.
25.00 cm³该酸溶液消耗23.80 cm³ NaOH。
Use H two A plus two NaOH yields Na two A plus two H two O to find NaOH concentration.
利用反应方程式H₂A + 2NaOH → Na₂A + 2H₂O，求算NaOH的浓度。
The total moles of acid equals 1.575 grams divided by 126.0 grams per mole, which is 0.01250 moles.
酸的总物质的量等于1.575克除以126.0克每摩尔，即0.01250摩尔。
One tenth is pipetted.
移取其中十分之一。
The moles of sodium hydroxide equals two times 0.001250 moles, which is 0.002500 moles.
氢氧化钠的物质的量等于2乘以0.001250摩尔，即0.002500摩尔。
The concentration of sodium hydroxide equals 0.002500 moles divided by 0.02380 cubic decimeters, which is 0.1050 moles per cubic decimeter.
氢氧化钠的浓度等于0.002500摩尔除以0.02380立方分米，即0.1050摩尔每立方分米。
25.00 cm³ HCl uses 24.60 cm³ of 0.1000 mol dm⁻³ NaOH.
25.00 cm³ 的 HCl 消耗了 24.60 cm³ 浓度为 0.1000 mol dm⁻³ 的 NaOH。
Calculate concentration, then explain rinsing, meniscus readings and the phenolphthalein endpoint.
请计算浓度，然后解释润洗、弯液面读数以及酚酞终点。
The amount of sodium hydroxide is 0.1000 times 0.02460, which is 0.002460 moles; the concentration of hydrochloric acid is 0.002460 divided by 0.02500, which is 0.09840 moles per cubic decimetre.
氢氧化钠的物质的量等于零点一零零零乘以零点零二四六零，即零点零零二四六零摩尔。 盐酸的浓度等于零点零零二四六零除以零点零二五零零，即零点零九八四零摩尔每立方分米。
Rinse the pipette with the acid, the burette with the sodium hydroxide and the flask with deionised water; fill the jet and read at eye level.
用盐酸润洗移液管，用氢氧化钠润洗滴定管，用去离子水润洗锥形瓶。 充满尖嘴部分并在平视下读数。
Swirl and add dropwise near the first persistent pale-pink endpoint; repeat to concordance and exclude the rough titre.
摇动并逐滴加入，直至出现第一个持久的淡粉色终点。 重复实验至结果一致，并剔除粗测值。
Cl₂ displaces Br⁻ and I⁻; Br₂ displaces I⁻; I₂ displaces neither.
氯气可置换溴离子和碘离子；溴可置换碘离子；碘则不能置换任何离子。
Increasing radius/shielding weakens attraction for an incoming electron.
半径增大或屏蔽效应增强会削弱对进入电子的吸引力。
Iodine is brown in water but violet in a suitable non-polar organic solvent.
碘在水溶液中呈棕色，但在合适的非极性有机溶剂中呈紫色。
Hot concentrated alkali reacts with chlorine.
热浓碱与氯气反应。
Which oxyanion forms?
会生成哪种含氧酸根离子？
Pause and justify your choice.
请先暂停，论证你的选择。
Three Cl two plus six OH minus yields five Cl minus plus Cl O three minus plus three H two O.
3Cl₂ + 6OH⁻ → 5Cl⁻ + ClO₃⁻ + 3H₂O。
Answer B.
答案B。
Chlorine goes from zero to negative one and from zero to positive five.
氯从0价变为-1价，同时从0价变为+5价。
Cold dilute alkali instead forms Cl O minus.
冷稀碱液则生成ClO⁻。
Conditions belong in the answer.
条件应包含在答案中。
Acidify with HNO₃ before AgNO₃: AgCl white, AgBr cream, AgI yellow.
在加入硝酸银之前，先用硝酸酸化。 氯化银呈白色，溴化银呈淡黄色，碘化银呈黄色。
AgCl dissolves in dilute NH₃; AgBr needs concentrated NH₃; AgI remains.
氯化银溶于稀氨水；溴化银需浓氨水才能溶解；碘化银不溶。
With concentrated H₂SO₄, bromide and especially iodide can reduce sulfur; chloride normally only gives HCl.
与浓硫酸反应时，溴离子和尤其是碘离子可还原硫；氯离子通常只生成氯化氢。
Electronegativity concerns a shared bonding pair; oxidising power concerns electron gain in redox.
电负性涉及共享的成键电子对。 氧化能力涉及氧化还原反应中的电子获得。
Both decrease from Cl to Br to I as radius and shielding weaken attraction despite greater nuclear charge.
尽管核电荷增加，但从氯到溴再到碘，随着原子半径增大和屏蔽效应增强，吸引力减弱，两者均呈下降趋势。
Add a small amount of Br₂ to excess KI or KCl, then extract into cyclohexane.
向过量的KI或KCl中加入少量Br2，然后萃取到环己烷中。
Bromine is orange and iodine purple in this solvent.
溴在此溶剂中呈橙色，碘呈紫色。
Predict both layers and write the reaction.
预测两层液体的颜色并写出反应方程式。
Potassium iodide gives purple iodine, I2.
碘化钾生成紫色的碘，I₂。
The reaction is Br2 plus 2I minus yields 2Br minus plus I2.
反应式为：Br₂ + 2I⁻ → 2Br⁻ + I₂。
Potassium chloride retains orange bromine because bromine cannot oxidize chloride under these conditions.
氯化钾保持橙色的溴，因为在此条件下溴无法氧化氯离子。
HI reduces H₂SO₄ to SO₂, S or H₂S while forming I₂.
HI将H2SO4还原为SO2、S或H2S，同时生成I2。
Balance each route and state all three final sulfur oxidation numbers.
请配平每条反应路径，并写出三种最终产物的硫氧化数。
Two HI plus H two SO four yields I two plus SO two plus two H two O. Sulfur goes from positive six to positive four.
2HI + H₂SO₄ → I₂ + SO₂ + 2H₂O；硫：+6 → +4。
Six HI plus H two SO four yields three I two plus S plus four H two O. Sulfur goes from positive six to zero.
6HI + H₂SO₄ → 3I₂ + S + 4H₂O；硫：+6 → 0。
Eight HI plus H two SO four yields four I two plus H two S plus four H two O. Sulfur goes from positive six to negative two.
8HI + H₂SO₄ → 4I₂ + H₂S + 4H₂O；硫：+6 → −2。
Iodide is oxidized from negative one to zero. Oxidizing sulfuric acid consumes the desired HI.
碘离子从−1被氧化至0；起氧化作用的硫酸会消耗目标产物HI。
The catalysed route has an intermediate and lower activation barriers.
催化途径具有中间产物和较低的活化能垒。
Reactant/product enthalpy levels and ΔH stay the same.
反应物和生成物的焓值以及ΔH保持不变。
For a fixed endpoint, comparative rate is one over t in s to the minus one.
对于固定终点，比较反应速率是1除以t，单位为s⁻¹。
Instantaneous gas rate is a tangent gradient, not final volume divided by total time.
瞬时气体反应速率是切线斜率，而非最终体积除以总时间。
Higher temperature increases the fraction beyond E sub a. A catalyst lowers E sub a without changing the distribution.
升高温度会增加超过活化能Ea的分子分数；催化剂降低Ea，但不改变分子分布。
Forward and reverse rates are equal in a closed system; concentrations need not be equal.
在封闭系统中，正逆反应速率相等，但浓度不必相等。
Higher pressure favours fewer gas moles; higher temperature favours the endothermic direction.
高压有利于气体摩尔数较少的方向；高温有利于吸热方向。
A catalyst reaches the same equilibrium sooner.
催化剂能更快达到同一平衡状态。
Industrial operation balances rate, yield and cost.
工业操作需权衡反应速率、产率与成本。
Draw the tangent at P, then choose two separated points on that straight line.
在点P处画出切线，然后在该直线上选取两个相距较远的点。
Find the instantaneous gas-production rate with units.
求出瞬时气体生成速率并带上单位。
Use the tangent points 20 seconds, 30 cubic centimetres and 60 seconds, 50 cubic centimetres.
使用切点20秒、30立方厘米，以及60秒、50立方厘米。
The rate is 50 minus 30, over 60 minus 20, which is 0.50 cubic centimetres per second.
速率等于50减30除以60减20，即0.50立方厘米每秒。
The line from the origin to P gives the earlier average, 1.0 cubic centimetre per second.
从原点到P给出的是之前的平均速率，1.0立方厘米每秒。
At fixed temperature, draw the energy distribution and two activation thresholds.
在固定温度下，画出能量分布曲线和两个活化能阈值。
Shade the sufficiently energetic fraction for each pathway.
对每条路径中能量足够的部分进行阴影填充。
State what remains unchanged.
说明哪些特征保持不变。
Catalysed Eₐ lies to the left; more collisions have enough energy.
催化反应的活化能E_a位于左侧。 更多碰撞具有足够的能量。
The same curve, total population and mean energy remain.
曲线、总粒子数和平均能量保持不变。
Equilibrium yield at this temperature is unchanged.
该温度下的平衡产率不变。
At 450 kelvin, the yield is 92 percent and the rate is 0.040 moles per minute.
在450 K时，产率为92%，速率为0.040 mol min⁻¹。
At 550 kelvin, the yield is 76 percent and the rate is 0.60 moles per minute.
在550 K时，产率为76%，速率为0.60 mol min⁻¹。
At 650 kelvin, the yield is 46 percent and the rate is 2.5 moles per minute.
在650 K时，产率为46%，速率为2.5 mol min⁻¹。
The requirements are a yield greater than or equal to 70 percent and a rate greater than or equal to 0.50 moles per minute.
要求产率 ≥ 70%且速率 ≥ 0.50 mol min⁻¹。
Choose the temperature that meets these criteria and reject the others using the data.
根据数据选择符合条件的温度并排除其他选项。
550 Kelvin passes: 76 percent is greater than or equal to 70 percent, and 0.60 moles per minute is greater than or equal to 0.50 moles per minute.
550 K 通过：76% ≥ 70%，且 0.60 mol min⁻¹ ≥ 0.50 mol min⁻¹。
450 Kelvin is too slow; 650 Kelvin gives too little equilibrium product.
450 K 反应太慢；650 K 平衡产物太少。
For an exothermic forward reaction, heating favours the reverse reaction but raises the energetic collision fraction.
对于放热正反应，升温有利于逆反应，但能提高有效碰撞分数。
Warm aqueous potassium hydroxide reacts with bromoethane to produce ethanol by substitution.
温热的氢氧化钾水溶液与溴乙烷发生取代反应生成乙醇。
Hot ethanolic potassium hydroxide reacts with bromoethane to produce ethene by elimination.
热的氢氧化钾乙醇溶液与溴乙烷发生消除反应生成乙烯。
Alcoholic potassium cyanide gives propanenitrile.
醇性氰化钾生成丙腈。
Alcoholic excess ammonia under pressure gives ethylamine.
在压力下使用过量醇性氨气生成乙胺。
OH minus lone pair to carbon bearing Br.
氢氧根离子的孤对电子指向连接溴的碳原子。 C–Br键指向溴原子。
C minus Br bond to Br. With NH3, an alkylammonium ion forms first. A second NH3 removes a proton.
与氨反应时，首先生成烷基铵离子；第二个氨分子移除一个质子。
For matched hydrolysis, weaker C minus I reacts faster than C minus Br then C minus Cl.
在对比水解反应中，较弱的C–I键反应速率快于C–Br，再快于C–Cl。
Control temperature and concentrations.
需控制温度和浓度。
Name CH₃CHBrCH₃ and CH₃CH(OH)CH₂CH₃.
命名CH3CHBrCH3和CH3CH(OH)CH2CH3。
Draw each displayed and skeletal formula.
画出它们的结构式和键线式。
Check every carbon valence and the O–H hydrogen.
检查每个碳的价键以及O-H氢原子。
The central carbon has two carbon neighbours, one Br and one H.
中心碳原子有两个碳邻位，一个溴和一个氢。
The skeletal corner and both ends represent three carbons.
骨架拐角和两端共代表三个碳原子。
Four carbons form the main chain; OH is on carbon 2.
四个碳原子构成主链。 羟基位于2号碳上。
The OH-bearing carbon has two carbon neighbours, so the alcohol is secondary.
连有羟基的碳原子有两个碳原子相邻，因此该醇为仲醇。
Use 1-bromopropane and aqueous OH⁻.
使用1-溴丙烷和氢氧根水溶液。
Show the donating pair, two curly arrows, C–Br polarity and products.
标出供体电子对、两个弯曲箭头、碳与溴之间的极性，以及产物。
A lone pair moves to the carbon bearing bromine.
孤对电子移向连接溴的碳原子。
The C–Br bond pair moves to bromine.
C–Br 键的电子对移向溴原子。
Propan-1-ol and Br minus form.
生成丙-1-醇和 Br⁻。
The total charge remains negative one.
总电荷保持为 −1。
Use excess alcoholic NH₃ under pressure with 1-bromopropane.
在加压条件下，使用过量的醇溶液氨与一溴丙烷反应。
Draw attack and proton transfer; retain every charge and show the second ammonia.
画出进攻和质子转移过程。 保留所有电荷，并画出第二个氨分子。
NH₃ donates its lone pair to carbon while the C–Br pair goes to Br.
NH3将其孤对电子提供给碳原子，同时C-Br键转移至Br。
The nitrogen has four bonds in CH₃CH₂CH₂NH₃⁺; bromide balances its charge.
在CH3CH2CH2NH3+中，氮原子形成四个化学键；溴离子平衡其电荷。
A second ammonia lone pair attacks a nitrogen-bound hydrogen.
第二个氨的孤对电子进攻与氮结合的氢。
The N-H bond returns to the nitrogen atom.
N-H键返回到氮原子上。
Propan-1-amine and ammonium ion form, while bromide remains.
生成丙-1-胺和铵离子，溴离子保留。
The amine lone pair allows further substitution.
胺的孤对电子允许进一步取代。
Distillation removes volatile aldehyde before further oxidation.
蒸馏在进一步氧化前除去挥发性醛。
Reflux returns condensed vapour so excess oxidant can make the acid.
回流使冷凝蒸气返回，以便过量的氧化剂生成酸。
Primary alcohol, arrow aldehyde or acid. Secondary, arrow ketone. Tertiary resists the stated oxidant. Aldehyde reduces warm Fehling's solution. Acid gives carbon dioxide with carbonate. Use molecular ion mass, IR absorptions and chemical tests together. A carbonyl peak alone is ambiguous.
伯醇生成醛或酸；仲醇生成酮；叔醇耐该氧化剂。 醛可还原温热的斐林试剂；酸与碳酸盐反应生成二氧化碳。 需结合分子离子质量、红外吸收峰和化学测试综合判断；仅凭羰基峰无法确定结构。
The organic product has density 0.87 g cm⁻³; the aqueous phase has density 1.00 g cm⁻³.
有机产物的密度为0.87 g cm⁻³；水相的密度为1.00 g cm⁻³。
Identify the layer, then explain separation, drying, salt removal and boiling-temperature evidence.
请识别分层，然后解释分离、干燥、除盐及沸点证据。
The stated product is the upper layer; separation removes the bulk aqueous phase.
目标产物位于上层。 分离操作可除去大量水相。
A suitable anhydrous salt removes residual water; filter or decant before distillation.
合适的无水盐用于去除残留水分。 蒸馏前需过滤或倾析。
Measure vapour temperature at the condenser entrance and compare at the same pressure.
在冷凝器入口处测量蒸气温度，并在相同压力下进行比较。
A narrow range supports purity, not unique identity.
窄沸程表明纯度高，但不能证明唯一身份。
C₃H₆O has molecular ion 58, fragment 29, carbonyl absorption and positive warm Fehling’s test.
C3H6O的分子离子为58，碎片峰为29，有羰基吸收峰，且热费林试剂测试呈阳性。
Check the mass, propose two ion formulae of mass 29 and choose propanal or propanone from all evidence.
请核对质量数，提出两个质量为29的离子式，并根据所有证据在丙醛和丙酮之间做出选择。
Both isomers have relative molecular mass 3 times 12, plus 6, plus 16, which is 58.
两种异构体的相对分子质量均为 3 乘以 12 加 6 加 16，等于 58。
The C 2 H 5 plus ion and the C H O plus ion both have mass 29.
C2H5 正离子和 CHO 正离子的质量数均为 29。
Positive Fehling's evidence supports propanal.
费林试剂呈阳性反应支持丙醛的结论。
One fragment or carbonyl alone cannot uniquely establish the isomer.
仅凭单一碎片或羰基无法唯一确定该异构体。
Use the supplied approximate absorption regions.
使用提供的近似吸收区域。
Match each bond environment and keep overlapping regions distinct.
匹配每种键环境，并保持重叠区域区分开。
Use the supplied approximate absorption regions.
使用提供的近似吸收区域。
Match each bond environment and keep overlapping regions distinct.
匹配每种键环境，并保持重叠区域相互区分。
Use the supplied tables for CH₂=CHCH₂NH₂ and HOCH₂CHO.
使用提供的表格，针对 CH₂ = CHCH₂NH₂ 和 HOCH₂CHO。
List all relevant C–H environments and each additional functional-group absorption.
列出所有相关的 C–H 环境以及每个额外的官能团吸收峰。
For prop-2-en-1-amine, look for alkene and alkane C–H, the carbon–carbon double bond and N–H.
对于丙-2-烯-1-胺，寻找烯烃和烷烃的 C–H、碳碳双键以及 N–H。
For hydroxyethanal, look for alkane and aldehyde C–H, a broad alcohol O–H and the carbon–oxygen double bond.
对于羟基乙醛，寻找烷烃和醛的 C–H、宽的醇 O–H 以及碳氧双键。
Use the given ranges; an alcohol O–H differs from the very broad acid O–H of ethanoic acid.
使用给定的范围；醇的 O–H 与乙酸极宽的酸 O–H 不同。
A/B: white BaSO₄.
A与B：白色BaSO4。
Only A releases NH₃ with warm NaOH.
只有A在加热NaOH时释放NH3。
C/D: M⁺ 58 and carbonyl; only C gives positive Fehling’s.
C与D：M+为58且含羰基；只有C的斐林试剂测试呈阳性。
Candidates: sodium sulfate, ammonium sulfate, propanal and propanone.
候选物质：硫酸钠、硫酸铵、丙醛和丙酮。
Assign each and write the two ionic equations.
请分别鉴定每种物质，并写出两个离子方程式。
A is ammonium sulfate.
A是硫酸铵。
B is sodium sulfate.
B是硫酸钠。
C is propanal.
C是丙醛。
D is propanone.
D是丙酮。
Ammonium ion plus hydroxide ion yields ammonia and water.
铵根离子加氢氧根离子生成氨气和水。
Barium ion plus sulfate ion yields solid barium sulfate.
钡离子加硫酸根离子生成硫酸钡沉淀。
Use fresh aliquots.
使用新鲜分取液。
Shared mass and carbonyl evidence do not distinguish between C and D.
共享的质量和羰基证据无法区分C和D。
The aldehyde reducing test does.
醛的还原性测试可以区分。
C₃H₆O has a carbonyl IR absorption and gives a positive Fehling’s test.
分子式 C3H6O 具有羰基红外吸收峰，且斐林试剂测试呈阳性。
Which structure fits all evidence?
哪种结构符合所有证据？
Pause and justify your choice.
请先暂停，完成这道原创练习再继续。
B: propanal has formula C₃H₆O, a carbonyl and a reducing aldehyde group.
B：丙醛的分子式为C3H6O，含有羰基和还原性醛基。
A is propanone: formula and carbonyl fit, but Fehling’s result does not.
A是丙酮：其分子式和羰基符合，但斐林试剂结果不符。
C and D have different formulae.
C和D的分子式不同。
No one test is a whole structural proof.
没有任何单一测试能作为完整的结构证明。