FE Chemical Practice Question Solution included after investigation
CASE E-070 · FE

The 70% Conversion That Produced Only 50% Yield

Seventy percent of A reacted, but only fifty percent of the feed became the desired product B. Conversion was not the same as yield.

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Case E-070 engineering failure visual
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2

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3

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4

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5

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Case Brief

SAME PRINCIPLES. HIGHER STANDARDS.
PROBLEM STATEMENT

Use a 100 mol A feed basis. Parallel reactions convert A one-for-one to desired product B or undesired product C. At the outlet, 30 mol A remains, 50 mol B has formed, and 20 mol C has formed. Define conversion as moles A reacted per mole A fed, desired-product yield as moles B formed per mole A fed, and selectivity as moles B formed per mole C formed. Calculate all three quantities.

Chemical Reaction Engineering Chemical Design Verification Root Cause Analysis Safe Correction
E-070
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CASE CLOSED · ENGINEERING VERDICT

What actually failed — and what should change.

What failed

Reactant conversion was reported as desired-product yield.

What was missed

Twenty moles of A were consumed by the undesired parallel pathway.

Corrected model

Calculate conversion, yield, and selectivity from their stated bases separately.

QUICK ANSWER

On a 100 mol A feed basis, 30 mol A remaining means 70% conversion. If 50 mol B and 20 mol C form, the stated desired B yield is 50/100=50%, while B/C selectivity is 50/20=2.5.

ENGINEERING TAKEAWAY

Conversion tracks reactant consumption; yield and selectivity track how reacted material is distributed among products. Definitions must be stated explicitly.

CASE QUESTIONS

Questions this case should settle.

Is conversion the same as product yield?

No. Conversion measures how much reactant is consumed; yield measures desired product according to a stated basis.

What is B/C selectivity when 50 mol B and 20 mol C form?

2.5 mol B per mol C.

NEXT FE PRACTICE QUESTION

The Rate Constant That Increased Sixfold at 330 K

Chemical Reaction Engineering · 5–10 min

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