碰撞频率、催化剂与平衡选择
引入| English | 释义 | 拼音 |
|---|---|---|
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | 平衡 | píng héng |
| catalyst/ˈkætəlɪst/ | 催化剂 | cuī huà jì |
| activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ | 活化能 | huó huà néng |
| tangent/ˈtændʒənt/ | 切线 | qiè xiàn |
Why can the fastest conditions give a poorer yield?
- An industrial reaction must make product quickly and give enough product at equilibrium 平衡. A hotter reactor can speed an exothermic reaction while reducing its equilibrium yield.
- Separate collision frequency, molecular energy and equilibrium position before deciding which change helps. A catalyst 催化剂 changes the route and the time to equilibrium, not the equilibrium yield at fixed conditions.
Explain successful collisions
- Reaction requires collisions with suitable orientation and enough energy to overcome activation energy 活化能. At fixed temperature, higher concentration or gas pressure makes collisions more frequent; a larger exposed solid surface provides more accessible reaction sites.
- Increasing temperature gives more frequent collisions and, especially, a larger fraction with enough energy. On equal-particle-number energy distributions, the hotter curve is broader with a lower peak and more area beyond the same activation-energy threshold; total area stays the same.
某反应气体在恒温下被压缩。下列哪项关于碰撞理论的陈述是正确的？
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粒子间距减小且碰撞频率增加。恒温保持能量分布不变，因此仅靠压缩不会提高高能粒子的比例。
Measure a rate at the stated time
- For the same fixed endpoint, relative rate is proportional to $1/t$: 20 s versus 50 s gives rate ratio $(1/20)/(1/50)=2.5$. This comparison assumes the same endpoint amount and matched conditions.
- For gas volume against time, draw a tangent 切线 at the requested point and choose two well-separated points on that tangent. Points (20 s, 30 cm³) and (60 s, 50 cm³) give gradient $\dfrac{50-30}{60-20}=0.50\,\dfrac{\mathrm{cm}^3}{\mathrm s}$. The line from the origin to P gives an average rate, not the instantaneous tangent rate.
With equal endpoint amounts, a halogenoalkane taking 5 s has 36 times the relative rate of one taking 180 s: (1/5)/(1/180) = 36. An endpoint comparison does not prove a complete rate law or isolate bond strength if other conditions differ.
两组匹配实验分别在15 s和45 s达到同一终点。计算15 s实验的相对速率除以45 s实验的相对速率。
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对于相同的终点量，相对速率与1/时间成正比。比值 = (1/15)/(1/45) = 45/15 = 3。
气体体积曲线的一条切线经过(10 s, 25 cm³)和(50 s, 49 cm³)。以cm³每s为单位计算其斜率。
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斜率 = 体积变化/时间变化 = (49 − 25)/(50 − 10) = 24/40 = 0.60 cm³每s。使用切线上的点，而非原点至曲线的割线。
Change the route without changing the energy distribution
- At fixed temperature, a catalyst gives an alternative route with lower activation barriers; the molecular-energy distribution itself stays unchanged. More particles can react, and both forward and reverse reactions become faster.
- Draw catalysed and uncatalysed reaction profiles with the same start/end levels and ΔH. A two-step catalysed route has two peaks and an intermediate valley. A selective route can improve atom economy; faster operation may use less energy, but neither benefit is guaranteed for every catalyst.
在固定温度下，催化剂通过提高平均分子能来增加产物的平衡浓度。
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催化剂在不改变温度分布的情况下降低反应途径势垒。它加速正逆两个方向；在固定条件下，平衡位置和平均能量均保持不变。
Justify equilibrium shifts and a numerical compromise
- In a closed system at dynamic equilibrium, forward and reverse rates are equal and concentrations stay constant; concentrations need not be equal. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), increasing pressure favours fewer gas moles, and adding N₂ favours its consumption. For an exothermic forward reaction, cooling favours products but slows reaction.
- Supplied plant data at 450, 550 and 650 K give yields 92%, 76%, 46% and relative rates 0.040, 0.60, 2.5. If yield must be at least 70% and rate at least 0.50, only 550 K meets both. These requirements justify that choice; the table alone does not establish a global economic optimum.
Pressure does not shift equilibrium when the two gas sides have equal mole totals. A catalyst does not favour products at equilibrium. Constant concentration means continuing reactions balance, not that molecular reactions stop.
Use collision theory for rate, tangent or fixed-endpoint data for measurement, and reversible-reaction reasoning for yield. Test industrial choices against the actual numerical constraints.
在450/550/650 K时，产率分别为92/76/46%，相对速率分别为0.040/0.60/2.5。哪个条件满足产率≥70%且速率≥0.50？
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450 K不满足速率限制；650 K不满足产率限制。仅550 K同时满足两者。所给限制并未确立普遍的经济最优值。