Calcylator
Reaction conversion

Reaction conversion:
how much reactant was actually used up

Conversion measures how much reactant disappeared, not how much useful product appeared. Learn to separate the two and the plant numbers start to make sense.

Calcylator Editorial Team

Updated · 4 min read

Conversion as a fraction of what went in

A reactor is fed a stream of raw material and delivers a stream that is part product, part leftover reactant and sometimes by-products. Conversion asks a narrow question about that process: of the key reactant that entered, what share is no longer there when the stream leaves?

It is always defined for one named reactant, usually the limiting or the most valuable one. A reaction can show 95% conversion of one reagent and 40% of another, and both statements are correct, so reports must name the species. Conversion is expressed as a fraction or a percentage and ranges from 0, nothing reacted, to 1 or 100%, everything reacted.

The equation and its two forms

Conversion of reactant A =amount of A consumedamount of A fed
amount consumed:
feed in minus amount leaving, in moles or mol/h
amount fed:
what entered the reactor in the same units
Multiply by 100 for a percentage.

Since the amount consumed is the difference between what went in and what came out, the equation is equally written as (fed − out) ÷ fed. That second form is the one used in practice, because plants measure the composition of the outlet stream, and consumption is deduced from it rather than measured directly.

Batch reactors use the same equation with the amounts at start and at the end of the run. Continuous reactors use flow rates in mol per hour, and the flow units cancel in the ratio. If the volume or total flow changes between inlet and outlet, as in many gas reactions, convert to moles first; concentrations alone can mislead.

Worked example: a continuous reactor

A reactor receives 150 mol/h of reactant A. The outlet analysis shows 33 mol/h of A still present. The consumption is therefore 150 − 33 = 117 mol/h.

  • A fed

    150 mol/h

  • A leaving

    33 mol/h

  • A consumed

    150 − 33 = 117 mol/h

Conversion of A

78%

117 ÷ 150 = 0.78.

Now suppose the desired product P forms in a one-to-one ratio with A and the outlet stream contains 105 mol/h of P. Not all of the consumed A became P, since 117 − 105 = 12 mol/h went to by-products. Two further numbers describe this.

Same reactor, three different percentages
MeasureDefinitionValue here
Conversion of Aconsumed ÷ fed117 ÷ 150 = 78%
Yield of PP formed ÷ A fed (1:1)105 ÷ 150 = 70%
Selectivity to PP formed ÷ A consumed (1:1)105 ÷ 117 ≈ 89.7%

Yield equals conversion multiplied by selectivity: 0.78 × 0.897 is 0.70. That chain is the reason a high-conversion reactor is not automatically a good one.

Single-pass and overall conversion

Many plants recycle unreacted feed. Single-pass conversion looks at one trip through the reactor and might be as low as 20 to 30% for equilibrium-limited reactions such as ammonia synthesis. Overall conversion looks at the whole plant boundary, comparing fresh feed in with material leaving, and can be above 95% because the recycle gives the leftover reactant another chance.

The two numbers answer different questions. Single-pass conversion drives reactor size and the recycle load. Overall conversion drives raw-material cost. When a figure is quoted without a label, ask which one it is.

  • Low single-pass conversion with high recycle needs a bigger separator and more pumping.
  • High single-pass conversion may need a larger reactor, more catalyst or more heat management.
  • An overall conversion near 100% needs an efficient purge for impurities that would otherwise build up in the recycle loop.

How conversion builds up over time in a batch vessel

In a batch reactor, conversion is not a fixed property of the chemistry; it grows as the clock runs. Take a first-order reaction that uses up 30% of the remaining reactant every ten minutes, starting from 2.00 mol/L.

Conversion creeps up with diminishing returns
TimeReactant left (mol/L)Conversion
0 min2.000%
10 min1.4030.0%
20 min0.9851.0%
30 min0.68665.7%

Each ten-minute block removes the same fraction of what is left, not the same amount, so the last block adds less than the first. That curve explains why pushing for the final few percent of conversion is expensive: the reaction time, and so the reactor volume, climbs steeply for a small gain.

Gas-phase reactors add a trap. In N₂ + 3 H₂ → 2 NH₃, four moles of gas become two, so total flow shrinks along the reactor. Computing conversion from outlet mole fractions without allowing for that shrinkage will give the wrong figure; use absolute molar flow rates, or an inert tracer, as the basis.

What stops conversion reaching 100%

  • Equilibrium, where the reverse reaction balances the forward one and no further net change occurs under those conditions.
  • Residence time, because a reactant that passes through quickly has less opportunity to react.
  • Temperature and catalyst activity, which govern how fast the reaction runs.
  • Mixing and contact, particularly in heterogeneous systems with two phases.

A student result above 100% almost always signals an error, such as mismatched units in feed and outlet or a different basis for the two streams. Negative conversion means more reactant left than entered, which could come from a measurement error or from the reactant being formed in a side reaction.

Reporting it so nobody is misled

State the species, whether it is single-pass or overall, the basis in moles or mass, and the time over which flows were averaged. Rounding to the nearest whole percent is normally enough, since the analytical error of an outlet measurement is rarely better than half a percent.

Where conversion is quoted for an exam or a process specification, the denominator matters most. Some plants base it on the design feed, others on the actual feed that day, and the two differ whenever the plant runs off-rate. Writing the basis in the report prevents a lot of avoidable argument.

A general conversion tool can help turn units in a worksheet, for instance kg per hour into mol per hour, before the percentages are worked out; the conversion ratio itself is a single division you can check by hand.

Common questions

What is the formula for percentage conversion?

Conversion equals reactant consumed divided by reactant fed, times 100. If 150 mol/h enters and 33 mol/h leaves unreacted, 117 mol/h was consumed, and the conversion is 117 ÷ 150 × 100 = 78%. Always name the reactant it refers to.

What is the difference between conversion and yield?

Conversion tracks how much reactant is used up, while yield tracks how much desired product forms relative to the feed. A reactor can convert 78% of the feed but yield only 70% product if some of the consumed reactant forms by-products.

What is selectivity in a reaction?

Selectivity is the desired product formed divided by the reactant consumed, accounting for stoichiometry. With 105 mol/h of product from 117 mol/h consumed in a one-to-one reaction, selectivity is about 89.7%. Yield equals conversion times selectivity.

Can conversion be higher than 100%?

No, not in a correct calculation. A value above 100% means the amount consumed exceeds the feed, usually due to unit mismatch, different bases for inlet and outlet, or measurement error. Recheck flow rates and compositions.

What is single-pass conversion?

It is the conversion of reactant during one pass through the reactor, before any recycle. It can be modest, such as 25%, while overall plant conversion is far higher because unreacted material is separated and sent back to the reactor.

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