Keep total dead volume below 5 % of the irradiated reactor volume — for a 100 mL gas-phase reactor, 5 mL, about 6 m of 3.2 mm ID tubing. Counting it as reactor volume makes an apparent first-order rate constant 10 % low at 10 % dead volume and 20 % low at 20 %.
Almost nobody budgets dead volume, because it is distributed across fittings that each look trivial. A 1 m length of 3.2 mm ID tube holds 8 mL — more than the headspace of many lab cells. The table below collects measured values for the parts that appear in a typical gas-phase rig.
| Component | Typical internal volume | Notes |
|---|---|---|
| 1/8 in. tubing, 3.2 mm ID | 80 µL per cm | 1 m = 8 mL; usually 3–5 lengths per rig |
| 1/16 in. tubing, 1.6 mm ID | 20 µL per cm | Same length holds 4× less |
| 6-port sampling valve | 100–300 µL | Rotor grooves plus connecting bores |
| GC sample loop | 0.25–1 mL | Sits downstream but lengthens response |
| Pressure transmitter leg, 3 mm × 100 mm | 0.7 mL | Dead leg, not swept by flow |
| Inline particulate filter | 0.3–1 mL | Housing dominates, not the frit |
| Moisture or O2 trap | 5–20 mL | Never place between reactor and detector |
| Gas bag or buffer volume | 100–1000 mL | Converts the rig into a batch system |
A rig built without attention lands at 15–30 mL: two metres of 3.2 mm line, one valve, a filter, a transducer leg. Against a 100 mL cell that is 15–30 % dead volume; against a 20 mL microreactor it is more dead volume than reactor.
Steady-state error. With volumetric flow Q and total wetted volume Vtot, the transit time is τtot = Vtot/Q. Only the illuminated fraction reacts, so the true illuminated residence time is τill = τtot(1 − f), where f is the dead-volume fraction. Reporting k from τtot gives kapp = k(1 − f): at f = 0.10 the reported constant is 10 % low and the correction factor is 1/(1−f) = 1.11; at f = 0.20 the correction is 1.25. The same trap applies to apparent quantum yield, since it is normalised to photons absorbed per unit time in the illuminated volume.
Transient error. Dead volume behaves as a well-mixed buffer with time constant τd = Vd/Q. A step change reaches 95 % of its final value only after 3τd. With 20 mL of dead volume at 10 mL/min, τd = 2 min and 95 % arrives at 6 min. A light-on/light-off transient recorded over 10 min is then mostly purging, not kinetics — and the "induction period" that appears at the start of every run is often just the rig equilibrating.
Wall adsorption adds a tail on top of this. CO2 and water adsorb on stainless steel and PTFE, so a step in CO2 concentration can take 10–20 min to settle even with negligible dead volume. Use argon or helium as the tracer when you want the geometric answer.
Two methods, both worth running once per rig configuration.
Run the tracer test at the flow rate you will actually use. Dead volume is a fixed volume but its damage scales as 1/Q: the same 20 mL that costs 6 min at 10 mL/min costs 60 min at 1 mL/min, which is why low-flow gas-phase photocatalysis is so sensitive to plumbing.
Fit the sampling valve directly on the reactor outlet with a minimum-length stub, not in a rack 40 cm away. Replace 3.2 mm lines with 1.6 mm between reactor and valve; at equal length that is a 4× reduction. Delete the inline filter — if fines are a problem, put the frit inside the reactor body where it is swept. Mount the pressure transmitter on a short 1 mm bore leg or use a flush-diaphragm sensor. Keep traps and back-pressure regulators downstream of the sampling point.
After that, target 5 mL or 5 % of reactor volume, whichever is smaller. Verify with a second tracer run rather than trusting the drawing: the number that counts is the one at the detector.
No. At steady state the same molar flow enters and leaves, so measured conversion is unaffected as long as the flow has stabilised. What changes is the residence time you divide by — and therefore the rate constant, turnover frequency and apparent quantum yield you calculate from that conversion.
Not for the reactor, because it sits downstream of the sampling point and is isolated between injections. It does matter for transient measurements: a 1 mL loop at 5 mL/min adds about 12 s of smearing, and if the loop is continuously swept the effect doubles. For step-response work, sample from a tee as close to the reactor as possible.
Only when the dead volume is a swept plug-flow segment in series, where subtraction is exact. If it is a well-mixed buffer — a bag, a large filter housing, a wide tee — subtraction underestimates the smearing and you must treat it as a CSTR in series with the reactor instead.
Under 1 mL, which in practice means 1.6 mm tubing only, no filter between reactor and valve, and a valve mounted within 10 cm. At 2 mL/min that 1 mL is already 30 s of τd and 90 s to 95 % — comparable to the residence time itself.
