Part 2: Industrial yield, rate and process decisions

Use equilibrium and kinetics together to evaluate process data, including temperature, pressure, catalyst, separation, recycling and cost.

High equilibrium yield and fast production are different objectives

For an exothermic product-forming equilibrium, a lower temperature can give a larger equilibrium product fraction. But lower temperature also reduces the fraction of collisions able to cross activation barriers. A process that gives a high yield after an impractically long time can make less product per hour than a warmer process with a lower single-pass equilibrium yield.

A working temperature is therefore often a compromise between rate, equilibrium composition, energy cost and catalyst performance. Explain both directions: why low temperature helps equilibrium yield and why it harms rate. Calling a temperature 'a compromise' without those causes does not evaluate the choice.

Pressure decisions are also conditional. For an equilibrium with fewer gaseous moles on the product side, high pressure can improve yield and rate, but compression consumes energy and high-pressure equipment is costly. For equal gas coefficients, there is no equilibrium-yield benefit from compression in the ideal-gas model. Use the given equation before applying an industrial slogan.

Original data comparison: use more than one column

The declining equilibrium amount as temperature rises is consistent with an exothermic forward reaction. The much shorter times show improved kinetics. Judging only yield would select 350 °C; judging only the simple product/time proxy would select 450 °C. Neither column alone proves the most profitable operating condition.

The proxy ignores heating/cooling time, loading, separation, recycling, product degradation, energy costs and catalyst lifetime. If a hotter process required expensive separation of a very dilute product, its larger reaction-stage throughput might not translate into the cheapest final product. A defensible answer states what the data support and what additional information is needed.

The word 'near-equilibrium' matters: ideal kinetic models approach equilibrium asymptotically, so a practical time requires a specified tolerance. Compare times measured by the same criterion, rather than treating arbitrary stopping times as directly comparable.

Apply the same reasoning to different equations

State symbols must match the reactor conditions. In the hot gas-phase hydration of ethene, ethanol is a gas in the equilibrium equation; it can then be condensed during downstream cooling. Writing it as liquid while discussing a homogeneous hot gas mixture obscures the actual system.

Industrial operating conditions vary with technology and plant design. Topic 10 expects application and evaluation of the supplied data rather than a claim that one remembered temperature or pressure is optimal for every process. A catalyst's identity or cost should be connected to the particular process rather than imported from an unrelated example.

A catalyst can improve the process without moving equilibrium

At a fixed temperature a catalyst does not change the equilibrium constant or final equilibrium composition. It can reach an economically useful conversion sooner, allowing more throughput, or allow a lower temperature while maintaining an acceptable rate. In an exothermic synthesis, that chosen lower operating temperature can increase equilibrium yield; the temperature change is responsible for that shift.

A catalyst therefore creates flexibility in selecting conditions. Include catalyst manufacture, replacement, poisoning and recovery in a cost evaluation. A high purchase price is not necessarily a disadvantage if the catalyst is reused for many batches and saves more energy or reduces unwanted products.

Separate product and recycle unreacted material

A reactor need not convert all starting material in one pass. Cooling can condense a product from a gas mixture, while unreacted gases return to the reactor. Removing product from contact with reactants prevents its immediate reverse reaction; feeding the unreacted material back provides another opportunity for conversion. This can give a high overall recovery despite a modest single-pass yield.

Original model: a pass converts 20.0% of the limiting feed and product is completely removed. From 100 mol feed-equivalent, pass 1 makes 20.0 mol product-equivalent and leaves 80.0; pass 2 converts 20.0% of 80.0 = 16.0, for a cumulative 36.0; pass 3 converts 12.8, giving 48.8. It is wrong to add 20 percentage points each pass because each pass has less unreacted material.

The model assumes perfect separation, no side reactions and no losses. Real plants may need a purge to prevent inert impurities accumulating, which also loses some reactants; recycling consumes energy. A flowing plant is not one uniformly closed vessel at equilibrium, even though equilibrium limits the composition attainable in its reactor stages.

Build an evaluation around evidence and trade-offs

Start with the equation and sign of ΔH, identify the gas-mole change, then extract the relevant trend from the data. Connect the trend to equilibrium or collision theory. Finally weigh the benefit against a specific cost or practical limit and make a conclusion tied to the objective, such as production per hour, raw-material use or energy cost.

For example: a table shows raising pressure increases yield from 90% to 92% but doubles compression cost. It supports a small yield benefit with a large stated energy penalty; a lower pressure could be preferable if the value of the extra product is smaller than the extra cost. Without product price, flow rate and other expenses, a precise profitability claim is not justified.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official Edexcel mark allocations.

Sources