UNIT 2 CHEMISTRY · PEARSON AAQ · D1.5
Application of Chemical Kinetics, Energetics and Equilibrium to Industry
FREE CHEMISTRY REVISION · TOPIC 48
Application of Chemical Kinetics, Energetics and Equilibrium to Industry
Pearson BTEC Level 3 AAQ Applied Science
Unit 2: Principles and Applications of Chemistry · D1.5
What you will learn
Combine collision theory, energy changes and equilibrium to justify industrial operating conditions. A useful process must make enough saleable product in an acceptable time, with manageable energy use, cost and risk. Pearson does not require memorising particular chemical industries; every process in this lesson is an illustrative teaching case.
1. Three ideas answer different questions
| Question | Relevant idea | Consequence |
|---|---|---|
| How quickly is product made? | Kinetics: collision frequency, activation energy and catalysts | Measure a rate, conversion after a stated time, or production per hour. |
| What composition is possible at equilibrium? | Equilibrium constant and Le Chatelier’s principle | State temperature, pressure and the reaction direction. |
| What heat must be supplied or removed? | Energetics: sign and size of ΔH | Plan heating, cooling and heat recovery. |
Equilibrium yield is not reaction rate. A large equilibrium yield can take too long to approach. Conversely, rapid reaction does not guarantee a favourable equilibrium composition. A yield measured after a short time is not automatically the equilibrium yield.
2. Choose and control temperature
Increasing temperature usually increases reaction rates because a greater fraction of particles can overcome the activation-energy barrier. For an exothermic product-forming direction, heating also favours the endothermic reverse direction, reducing equilibrium product yield. These effects can oppose one another. For an endothermic product-forming reaction, higher temperature can favour both rate and equilibrium yield, but heating cost, side reactions and material stability can still limit the temperature.
| Temperature / K | Equilibrium yield / % | Yield after 10 min / % |
|---|---|---|
| 350 | 95.0 | 9.0 |
| 450 | 83.0 | 43.4 |
| 550 | 71.0 | 70.7 |
| 650 | 59.0 | 59.0 |
Worked evaluation 1 — identify a compromise
At 350 K, the equilibrium yield is 95.0%, but only 9.0% is present after ten minutes. At 550 K, the corresponding values are 71.0% and 70.7%. At 650 K, the ten-minute yield falls to 59.0% despite faster approach to equilibrium.
For a ten-minute residence-time limit, 550 K is the best of these four tested settings. It combines fast approach with a higher product yield than 650 K. It is not proven to be the exact optimum: test temperatures around 550 K and include energy use, safety and uncertainty. Choosing 350 K solely because its equilibrium yield is highest ignores the time constraint.
Control temperature with monitored heating and cooling, heat exchangers and appropriate control systems. Exothermic processes may need continuous heat removal to prevent overheating; recovered heat can preheat feed. A nominal temperature is useful only if it is maintained throughout the reactor.
3. Choose and control pressure
Compression of a gaseous equilibrium at fixed temperature favours the side with fewer gas moles. Count gaseous coefficients in the balanced equation; pressure does not always improve product yield. Equal gas-mole totals give no ideal-gas equilibrium shift. Higher reactant partial pressures can increase rate when the rate depends on those reactants, but the kinetic effect must be supported by the rate equation or data.
Higher pressure requires compression energy and suitable pressure-rated plant, with associated capital, maintenance and containment demands. Compare the extra production with those costs. Control reacting partial pressures and flow; adding an inert gas at constant volume does not reproduce compression of the reacting mixture.
Worked evaluation 2 — diminishing pressure benefits
An illustrative exothermic reaction is A(g) + 2B(g) ⇌ C(g), so product formation reduces gas moles from three to one. At the same temperature, feed and observation method:
| Pressure / atm | Equilibrium C yield / % | Compression energy / kWh per kmol feed |
|---|---|---|
| 10 | 50 | 2 |
| 30 | 75 | 7 |
| 60 | 82 | 16 |
From 10 to 30 atm the gain is 25 percentage points for 5 extra kWh per kmol feed. From 30 to 60 atm it is only 7 points for 9 extra kWh. If at least 70% equilibrium yield is required and the compression budget is at most 10 kWh per kmol feed, 30 atm meets both constraints; 60 atm exceeds the budget. The table does not provide a reaction rate, so it cannot alone establish whether production time is acceptable.
4. Select and maintain a catalyst
A suitable catalyst provides a pathway with lower activation energy and speeds both directions of a reversible reaction. At unchanged temperature and pressure it changes neither K nor equilibrium yield. It may allow an acceptable rate at a lower temperature or shorter residence time. A selective catalyst can reduce unwanted competing reactions; it must also remain active, withstand operating conditions and be recoverable or replaceable.
Worked evaluation 3 — catalyst performance at a fixed time
At one fixed temperature and pressure, the equilibrium product yield is 80%. Without catalyst, the measured yield after five minutes is 30%; catalyst X gives 68%, and catalyst Y gives 72%.
Both catalysts improve the approach to the same 80% equilibrium yield. Y gives 4 percentage points more product after five minutes than X. This does not prove Y is preferable overall: compare purchase cost, lifetime, poisoning, recovery and selectivity. A result above 80% would require checking the stated conditions, product definition or measurements before drawing a conclusion.
5. Compare production per unit time
State what the data measure. Batch production must include filling, heating, cooling, separation and turnaround where supplied. A continuous process may need conversion per pass, flow rate, selectivity, recycle and separation energy. Do not compare a batch percentage directly with a flow in kg h⁻¹.
Worked evaluation 4 — the highest batch yield is not the highest output
Each batch has a theoretical maximum of 1.00 tonne of desired product, and the reported yields are isolated product yields.
| Setting | Yield / % | Reaction time / h | Other time / h |
|---|---|---|---|
| Low temperature | 90 | 6.0 | 1.0 |
| Moderate temperature | 75 | 1.5 | 1.0 |
Low-temperature output = 1.00 × 0.90/(6.0 + 1.0) = 0.129 tonne h⁻¹. Moderate-temperature output = 1.00 × 0.75/(1.5 + 1.0) = 0.300 tonne h⁻¹. The moderate setting produces about 2.33 times as much product per hour under these assumptions, despite lower yield per batch. Check capacity, energy use, product quality and safe operation before selecting it.
Worked evaluation 5 — recycle changes overall conversion
A separator removes all product after each pass. A reactor converts 40.0% of A per pass, and 90.0% of the unreacted A is recovered and returned. At steady state, fresh A feed is 100 mol h⁻¹. Assume no side reactions and that A consumption and desired product are in a 1:1 mole ratio.
If total reactor A feed is F, unreacted A = 0.600F; recycled A = 0.900 × 0.600F = 0.540F. Hence F = 100 + 0.540F, so F = 100/0.460 = 217 mol h⁻¹ (3 s.f.). Desired product = 0.400F = 87.0 mol h⁻¹; unrecovered A = 0.100 × 0.600F = 13.0 mol h⁻¹.
Overall conversion relative to fresh feed is 87.0%, even though per-pass conversion is 40.0%. Recycling has not changed K or per-pass conversion under the stated assumptions. It adds circulation, separation and energy demands. Check the balance: 87.0 + 13.0 ≈ 100 mol h⁻¹.
6. Evaluate evidence rather than a single number
Worked evaluation 6 — a defensible recommendation
Process trials give the following illustrative results, all for the same feed rate and product purity:
| Trial | Product / kg h⁻¹ | Energy / kWh per kg product | Catalyst life / h |
|---|---|---|---|
| P | 120 ± 3 | 4.0 | 500 |
| Q | 125 ± 4 | 5.5 | 250 |
| R | 110 ± 3 | 3.0 | 700 |
If output must exceed 115 kg h⁻¹ and energy use must be at most 4.5 kWh kg⁻¹, P meets both nominal requirements; Q exceeds the energy limit and R misses the output requirement. The output ranges for P (117–123) and Q (121–129) overlap, so these trials do not establish a clear output advantage for Q. P also has twice Q’s stated catalyst life. Recommend P within the supplied constraints, then confirm repeatability, long-term performance and risk controls.
The ± ranges are supplied measurement ranges, not enough information to perform a formal significance test. Do not claim certainty from them.
- Compare data collected with controlled feed, temperature, pressure, catalyst amount and time.
- Separate equilibrium measurements from fixed-time measurements.
- Quote numerical differences with units and identify uncertainty.
- Explain both the benefit and the cost of a change.
- State missing evidence: energy, purity, catalyst lifetime, recycle losses or production rate.
- End with a conditional judgement linked to the stated process objective.
Common misconceptions
- Highest equilibrium yield means best process: acceptable time and output matter too.
- Higher temperature always increases product yield: equilibrium and rate can respond differently.
- Higher pressure always shifts equilibrium: gaseous mole ratios determine the direction.
- A catalyst raises equilibrium yield: it changes the approach to equilibrium at fixed conditions.
- Recycle raises K: it can improve overall resource use without changing K.
- One best data point proves an optimum: compare uncertainty and neighbouring conditions.
Examination tips
- Use a claim → numerical evidence → chemical reason → limitation structure.
- Name the reaction direction and sign of ΔH before discussing temperature.
- Count gas coefficients before discussing pressure.
- Show the denominator when calculating output per hour.
- Use supplied process information; memorising named industries is not required.
Check your understanding
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