Exercise set

Chemical Reactor Conversion and Sizing Exercises

Worked chemical reactor exercises for conversion, selectivity, yield, CSTR and PFR sizing, kinetics, recycle, heat removal, equilibrium and release checks.

Branch: Chemical Engineering
Content: Exercise set
Updated: Jul 03, 2026
Revision: v1.0.0 · reviewed

These exercises focus on chemical reactor conversion, selectivity, yield, kinetic sizing, ideal CSTR and PFR models, recycle effects, reaction heat removal and release checks. The purpose is not to make an ideal reactor model look exact. The purpose is to decide what the model can support before scale-up, commissioning, production release or safety review.

Assume simplified nominal values unless an exercise states otherwise. Real reactor decisions require verified kinetics, phase behaviour, mixing evidence, heat-transfer data, catalyst condition, residence-time distribution, pressure drop, relief basis, control authority and operating procedures.

Release Evidence Notes

Reactor calculations must state whether conversion is single-pass, overall, equilibrium-limited, selectivity-limited or heat-removal-limited. A reactor can meet conversion and still fail product quality, impurity formation, hot-spot control, catalyst life, relief basis or downstream separation. Tie every calculation to measured feed composition, temperature, residence time, catalyst state, cooling duty and sampling evidence.

Engineering Boundary Notes

Ideal CSTR and PFR equations are screening models. They are useful only when their assumptions are stated. A CSTR assumes well-mixed outlet composition. A PFR assumes axial concentration change without back-mixing. Batch data need time, temperature and mixing context. Recycle changes overall conversion but can hide single-pass limitations.

Common Release Mistakes

quoting conversion without stating feed basis and limiting reactant;
confusing selectivity to desired product with yield on fresh feed;
using CSTR equations for a poorly mixed vessel;
using PFR equations for a reactor with bypassing or channeling;
ignoring heat removal for an exothermic conversion target;
accepting equilibrium conversion as if more residence time could overcome it;
omitting uncertainty when conversion is close to a release limit.

Scenario Map

Scenario	Main calculation	Engineering decision
Conversion and yield	$X_A$ , selectivity and product rate	Check chemistry and product target.
CSTR sizing	$V=F_{A0}X/(-r_A)$	Decide mixed-tank volume or staging.
PFR sizing	first-order plug-flow relation	Estimate tubular or packed-bed volume.
Batch kinetics	time-conversion relation	Set hold time or retest action.
Heat release	$	\Delta H_R
Equilibrium and guard band	conversion limit and uncertainty	Decide release, hold or redesign.

Validation Package Checklist

feed composition, limiting reactant and conversion basis;
reactor type, volume, active catalyst or liquid holdup;
temperature, pressure, phase and mixing evidence;
rate law source and fitted temperature range;
heat-removal duty and utility margin;
product assay, side-product or impurity evidence;
guarded conversion rule with uncertainty and action limits.

Exercise 1: Reactant Conversion

A reactor feed contains reactant $A$ at:

F_{A0}=120\ \text{mol/min}

The outlet contains:

F_A=18\ \text{mol/min}

Compute conversion.

Solution

\displaystyle X_A=\frac{F_{A0}-F_A}{F_{A0}}=\frac{120-18}{120}=0.850=85.0\%

Engineering Comment

Conversion is high, but it does not prove selectivity, heat-removal adequacy or product quality.

Plausibility Check

The outlet has much less reactant than the feed, so conversion should be high.

Exercise 2: Desired Product Formation

Use Exercise 1. Of the consumed $A$ , $92\%$ forms product $P$ on a one-to-one molar basis. Compute product rate.

Solution

Reactant consumed:

F_{A,cons}=120-18=102\ \text{mol/min}

Desired product:

F_P=0.92(102)=93.8\ \text{mol/min}

Engineering Comment

The remaining $8\%$ may become side products, heavies, emissions or fouling precursors. Selectivity is a release variable, not a footnote.

Plausibility Check

Product rate is lower than consumed reactant because selectivity is below $100\%$ .

Exercise 3: Yield on Fresh Feed

A reactor receives $150\ \text{mol/min}$ of reactant $A$ and produces $105\ \text{mol/min}$ of desired product on a one-to-one basis. Compute yield on fresh feed.

Solution

\displaystyle Y_P=\frac{F_P}{F_{A0}}=\frac{105}{150}=0.700=70.0\%

Engineering Comment

Yield on feed is different from selectivity on consumed reactant. A high selectivity claim can still hide low feed utilization if conversion is poor.

Plausibility Check

The yield is below $100\%$ because product flow is lower than reactant feed.

Exercise 4: CSTR First-Order Conversion

A liquid CSTR has residence time:

\tau=4.17\ \text{h}

The first-order rate constant is:

k=0.45\ \text{h}^{-1}

For a first-order CSTR:

\displaystyle X=\frac{k\tau}{1+k\tau}

Compute conversion.

Solution

k\tau=0.45(4.17)=1.88

\displaystyle X=\frac{1.88}{1+1.88}=0.653=65.3\%

Engineering Comment

The result assumes well-mixed behaviour and no heat or mass-transfer limitation. Poor mixing can reduce effective conversion.

Plausibility Check

For a CSTR, conversion approaches one as $k\tau$ becomes large; $k\tau=1.88$ gives a moderate conversion.

Exercise 5: CSTR Volume for Target Conversion

A first-order liquid reaction has $k=0.60\ \text{h}^{-1}$ and feed flow $Q=2.0\ \text{m}^3/\text{h}$ . Estimate CSTR volume for $X=75\%$ .

Solution

For a first-order CSTR:

\displaystyle X=\frac{kV/Q}{1+kV/Q}

Solve for residence time:

\displaystyle \tau=\frac{X}{k(1-X)}=\frac{0.75}{0.60(0.25)}=5.0\ \text{h}

V=Q\tau=2.0(5.0)=10.0\ \text{m}^3

Engineering Comment

The volume is large because CSTR conversion becomes inefficient near high conversion for first-order kinetics.

Plausibility Check

Targeting higher conversion requires disproportionally more residence time.

Exercise 6: Two Equal CSTRs in Series

Two equal CSTRs each have $k\tau=1.0$ . For first-order reaction, each stage conversion on its inlet is:

\displaystyle X_s=\frac{1}{1+1}=0.5

Compute overall conversion.

Solution

Unreacted fraction after one stage:

1-X_s=0.5

After two stages:

1-X_{overall}=0.5^2=0.25

X_{overall}=0.75=75\%

Engineering Comment

Staging improves conversion compared with one CSTR of the same stage residence time because concentration is higher in the first vessel.

Plausibility Check

Overall conversion is greater than one-stage conversion and below $100\%$ .

Exercise 7: PFR First-Order Conversion

A plug-flow reactor has first-order rate constant $k=0.45\ \text{h}^{-1}$ and residence time $\tau=4.17\ \text{h}$ . For a first-order PFR:

X=1-e^{-k\tau}

Compute conversion.

Solution

X=1-e^{-0.45(4.17)}=1-e^{-1.88}=0.847=84.7\%

Engineering Comment

The PFR gives higher conversion than a CSTR at the same $k\tau$ for first-order kinetics, but only if plug-flow assumptions are credible.

Plausibility Check

The result is higher than the CSTR conversion from Exercise 4, as expected.

Exercise 8: PFR Volume From Target Conversion

A first-order liquid PFR treats $1.5\ \text{m}^3/\text{h}$ with $k=0.50\ \text{h}^{-1}$ . Estimate volume for $X=80\%$ .

Solution

For first-order PFR:

\displaystyle \tau=\frac{-\ln(1-X)}{k}

\displaystyle \tau=\frac{-\ln(0.20)}{0.50}=3.22\ \text{h}

V=Q\tau=1.5(3.22)=4.83\ \text{m}^3

Engineering Comment

The calculation is useful for screening a tubular reactor, but heat transfer and pressure drop must be checked before design release.

Plausibility Check

The PFR volume is lower than an equivalent high-conversion CSTR would require.

Exercise 9: Batch First-Order Hold Time

A batch reaction follows:

X=1-e^{-kt}

with $k=0.35\ \text{h}^{-1}$ . Compute time for $90\%$ conversion.

Solution

\displaystyle t=\frac{-\ln(1-X)}{k}=\frac{-\ln(0.10)}{0.35}=6.58\ \text{h}

Engineering Comment

The hold time is only valid at the tested temperature and mixing condition. Heat-up and sampling delay may add to batch cycle time.

Plausibility Check

Reaching $90\%$ conversion takes several time constants, which is expected.

Exercise 10: Rate Constant From Batch Data

A batch reactor reaches $X=60\%$ after $3.0\ \text{h}$ for a first-order reaction. Estimate $k$ .

Solution

1-X=e^{-kt}

\displaystyle k=\frac{-\ln(1-X)}{t}=\frac{-\ln(0.40)}{3.0}=0.305\ \text{h}^{-1}

Engineering Comment

This fitted value should not be extrapolated to another temperature without activation-energy evidence.

Plausibility Check

$k$ has units of inverse time and is of order $0.3\ \text{h}^{-1}$ for conversion over hours.

Exercise 11: Overall Conversion With Recycle

A reactor has single-pass conversion $40\%$ . Unreacted reactant is separated and recycled perfectly. Fresh feed is $100\ \text{mol/min}$ . Purge is neglected. What is overall conversion of fresh reactant at steady state?

Solution

With perfect separation and no purge or loss, unreacted reactant returns until it reacts. Overall conversion of fresh feed approaches:

X_{overall}=100\%

Engineering Comment

This ideal result hides real constraints: purge, side reactions, catalyst deactivation, recycle compressor capacity, impurity buildup and residence-time limits.

Plausibility Check

Perfect recycle can make overall conversion much higher than single-pass conversion, but only in an ideal material balance.

Exercise 12: Recycle With Purge Loss

Fresh reactant feed is $100\ \text{mol/min}$ . Product formation is $88\ \text{mol/min}$ and reactant lost in purge is $7\ \text{mol/min}$ . Compute overall conversion to useful product.

Solution

Useful conversion on fresh feed:

\displaystyle X_{useful}=\frac{88}{100}=88\%

Reactant not converted to product:

100-88=12\ \text{mol/min}

Of this, $7\ \text{mol/min}$ is purge loss.

Engineering Comment

Purge protects the loop but reduces yield. The purge composition and treatment system must be part of the reactor performance review.

Plausibility Check

Useful conversion plus unconverted/lost reactant accounts for the fresh feed.

Exercise 13: Exothermic Reaction Heat Removal

Reactant consumption is $80\ \text{mol/min}$ . Heat of reaction is:

-\Delta H_R=65\ \text{kJ/mol}

Compute heat-release rate.

Solution

\dot{Q}_{rxn}=80(65)=5200\ \text{kJ/min}

Convert to kW:

\displaystyle \dot{Q}_{rxn}=\frac{5200}{60}=86.7\ \text{kW}

Engineering Comment

Cooling capacity must exceed this heat release with margin. If conversion increases during startup, heat release can exceed nominal design.

Plausibility Check

Tens of mol/min times tens of kJ/mol gives thousands of kJ/min.

Exercise 14: Adiabatic Temperature Rise

An exothermic batch releases $4200\ \text{kJ}$ into $1800\ \text{kg}$ of liquid with:

c_p=3.5\ \text{kJ/(kg K)}

Estimate adiabatic temperature rise.

Solution

\displaystyle \Delta T_{ad}=\frac{Q}{mc_p}=\frac{4200}{1800(3.5)}=0.667\ \text{K}

Engineering Comment

This heat release is small relative to batch heat capacity. A larger heat of reaction, concentrated feed or poor mixing could change the safety conclusion.

Plausibility Check

Large liquid mass and heat capacity make the temperature rise small.

Exercise 15: Equilibrium Conversion Limit

A reversible reaction has equilibrium conversion $X_{eq}=72\%$ at the operating temperature. The production target is $80\%$ conversion without separation or temperature change. Does added reactor volume solve the problem?

Solution

The target exceeds equilibrium:

80\%>72\%

Additional residence time cannot push conversion beyond equilibrium under the same conditions.

Engineering Comment

The process needs a different temperature, pressure, product removal, reactant excess or separation strategy, not merely a larger reactor.

Plausibility Check

Equilibrium is a thermodynamic limit, not a kinetic time delay.

Exercise 16: Guarded Conversion Acceptance

Measured conversion is $84.0\%$ . Expanded uncertainty is $1.8\%$ . The release limit is $82.0\%$ . Use:

X_{guarded}=X-U

Solution

X_{guarded}=84.0\%-1.8\%=82.2\%

Since $82.2\%>82.0\%$ , the conversion passes narrowly.

Engineering Comment

The result should trigger continued monitoring because the guarded margin is only $0.2\%$ .

Plausibility Check

The raw conversion has a $2.0\%$ margin, but uncertainty consumes almost all of it.

Exercise 17: Catalyst Deactivation Screen

A catalyst bed loses $3.0\%$ activity per week. Initial conversion is $86\%$ . Estimate conversion after four weeks using a linear activity screen.

Solution

Activity after four weeks:

a=1-4(0.030)=0.88

Approximate conversion:

X=0.88(86\%)=75.7\%

Engineering Comment

This simple screen indicates a likely run-length limit. Real deactivation may be nonlinear and affected by poisons, fouling or regeneration.

Plausibility Check

Conversion decreases because activity decreases.

Exercise 18: Reactor Release Gate

A reactor must meet $82\%$ guarded conversion, remove at least $90\ \text{kW}$ of heat and keep selectivity above $91\%$ . The review results are:

X_{guarded}=82.2\%,\quad Q_{cooling}=96\ \text{kW},\quad S=93\%

Decide release status.

Solution

Conversion margin:

82.2\%-82.0\%=0.2\%

Cooling margin:

96-90=6\ \text{kW}

Selectivity margin:

93\%-91\%=2\%

Engineering Comment

All gates pass, but conversion margin is very narrow. Release should be conditional on feed assay, temperature control, catalyst condition and repeat sampling.

Plausibility Check

The smallest positive margin controls the confidence level, so the result is conditional release rather than broad approval.

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