Collision rates, catalysts and equilibrium choices
Diperkenalkan| English | Makna |
|---|---|
| equilibrium/ˌiːkwɪˈlɪbrɪəm/ | kesetimbangan |
| catalyst/ˈkætəlɪst/ | Katalisator |
| activation energy/ˌæktɪˈveɪʃn ˈenədʒi/ | energi aktivasi |
| tangent/ˈtændʒənt/ | garis singgung |
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.
A reacting gas is compressed at fixed temperature. Which collision-theory statement is supported?
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Particles are closer and collide more frequently. Fixed temperature preserves the energy distribution, so compression alone does not increase the energetic fraction.
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.
Two matched experiments reach the same endpoint in 15 s and 45 s. Calculate the relative rate of the 15 s experiment divided by that of the 45 s experiment.
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Relative rate is proportional to 1/time for the same endpoint amount. Ratio = (1/15)/(1/45) = 45/15 = 3.
A tangent to a gas-volume curve passes through (10 s, 25 cm³) and (50 s, 49 cm³). Calculate its gradient in cm³ per s.
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Gradient = volume change/time change = (49 − 25)/(50 − 10) = 24/40 = 0.60 cm³ per s. Use points on the tangent, not an origin-to-curve secant.
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.
At fixed temperature a catalyst increases the equilibrium concentration of product by raising the mean molecular energy.
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A catalyst lowers route barriers without changing the temperature distribution. It speeds both directions; equilibrium position and mean energy stay unchanged at fixed conditions.
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.
At 450/550/650 K, yields are 92/76/46% and relative rates 0.040/0.60/2.5. Which meets yield ≥70% AND rate ≥0.50?
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450 K fails the rate limit; 650 K fails the yield limit. Only 550 K satisfies both. The supplied limits do not establish a universal economic optimum.