Yield and Selectivity in Linked Reactions

Introduction

A reaction can consume most of its feed yet make little desired product. This occurs when side reactions use the same starting material. Three percentages then answer three different questions: how much feed reacted, how much of the reacted feed followed the desired path, and how much desired product was obtained relative to the feed. A numerical problem is solvable only when the basis of each percentage is stated or can be inferred from clearly specified equations.

Core explanation

Consider competing one-to-one pathways A → B, the desired product, and A → C, an unwanted side product. Start with nA,0 moles and end with nA,f, nB,f and nC,f, assuming no products initially and no other pathways. Moles of A consumed are nA,0 − nA,f = nB,f + nC,f. Conversion X is (nA,0 − nA,f)/nA,0. Desired-path selectivity as a fraction of converted A is S = nB,f/(nB,f + nC,f). Desired B yield on a fed-A basis is Y = nB,f/nA,0. These definitions give Y = X S for this simple pair of one-to-one reactions.

Some sources define selectivity instead as the ratio of desired to undesired product, nB/nC. Both usages occur in chemical engineering. A ratio of 3:1 corresponds to a desired fraction 3/(3 + 1) = 0.75, not 3 or 300%. State the definition before computing. If reactions have coefficients other than one-to-one, compare reaction extents or reactant-equivalent amounts, not unadjusted product moles. For example, if 2A → B, one mole B represents two moles A consumed. A naive B/(B+C) could misstate the fraction of A following that path.

Suppose 10.0 mol A enters, 2.0 mol remains, 6.0 mol B forms and 2.0 mol C forms. Conversion is (10.0 − 2.0)/10.0 = 0.800. Desired-path selectivity is 6.0/(6.0 + 2.0) = 0.750, and yield is 6.0/10.0 = 0.600. The identity 0.800 × 0.750 = 0.600 checks the bookkeeping. A high conversion of 80% therefore does not mean an 80% yield of B. If B is partly lost during isolation, an isolated B yield is lower still; distinguish reaction formation from recovered product.

When a problem gives percent yield but no side-product amounts, it does not by itself reveal conversion or selectivity. A 60% isolated yield might mean complete reaction with 60% selectivity, 75% conversion with 80% selectivity, or high chemical formation with collection loss. Additional measurements are required. Purity is another separate input property; correct an impure A feed before using nA,0 as the denominator for chemical conversion or yield.

Step-by-step reasoning

1. Write every stated reaction, with coefficients and desired/undesired products marked. 2. Convert feed and final measurements to consistent mole amounts of active chemicals. 3. Calculate reactant consumed, then conversion on the specified feed basis. 4. Allocate consumed reactant among reaction pathways using product moles and coefficients. 5. Compute the stated selectivity and yield definitions; check all pathway amounts sum to consumption.

Visual explanation

Imagine a flow bar of 10 mol A splitting into three branches. Two moles leave unchanged, six mole-equivalents lead to B, and two lead to C. The converted branch is eight of ten, while the desired branch is six of those eight. Label the three ratios 8/10, 6/8 and 6/10 beside the branches to show conversion, desired fraction and yield.

Real-world analogy

A bakery uses ten trays of batter. Two trays stay unused, six become saleable cakes, and two burn. Using batter on eight trays is 80% conversion; six of the eight used trays produce saleable cakes, a 75% success fraction; six of ten original trays produce saleable cakes, a 60% overall yield. These percentages have distinct denominators.

Real-world example

Industrial oxidation can produce a desired oxygenated molecule alongside more oxidised byproducts. Raising temperature may consume more starting material while also favouring unwanted paths. The plant therefore tracks both conversion and desired-product selectivity. Maximising conversion alone can worsen raw-material efficiency and separation load.

Why?

Why does Y = XS hold in the simple model? Multiply (consumed A/fed A) by (A converted to B/consumed A). The consumed-A amount cancels, leaving A converted to B/fed A. The identity is a unit and basis check, not a universal law for every choice of yield or selectivity definition.

Common misconception

“Conversion is the same as percent yield.” Conversion counts all consumption of A, including unwanted pathways. Yield counts desired B on its specified theoretical basis. Another trap is treating a 3:1 B-to-C selectivity ratio as 75% without showing the conversion to a fraction.

Worked example

Feed 50.0 mol A to reactions A → B and A → C. Analysis finds 10.0 mol A, 30.0 mol B and 10.0 mol C at the reactor outlet. A consumed = 50.0 − 10.0 = 40.0 mol, equal to 30.0 + 10.0. Conversion X = 40.0/50.0 = 80.0%. Desired fraction of converted A S = 30.0/40.0 = 75.0%; the B:C product ratio is 30.0/10.0 = 3:1. B yield on fed A Y = 30.0/50.0 = 60.0%. If 27.0 mol B is isolated, isolated yield is 27.0/50.0 = 54.0%, while reactor selectivity remains 75.0% if no further reaction occurred during isolation.

Quick check

1. If conversion is 0.90 and desired-path selectivity is 0.80 in the one-to-one two-path model, what is desired-product yield on feed? Answer: Y = XS = 0.90 × 0.80 = 0.72, or 72% of fed A equivalents.

Exam focus

Write a named denominator for every percentage. Check coefficient-based reactant equivalents for non-one-to-one reactions and separate reactor product formation from isolated product recovery. An identity such as Y = XS is valid only for compatible definitions.

Advanced insight

In a reaction network, a selectivity vector can allocate consumed feed across several independent extents. Product measurements alone may fail to identify those extents when multiple pathways produce the same product. Material balances then require additional measurements or assumptions, such as byproduct analysis or isotope tracing.

Summary

Conversion measures feed disappearance, selectivity measures pathway preference, and yield measures desired product on a chosen input basis. In a simple one-to-one pair of competing reactions, yield equals conversion times desired-path selectivity. Always state the definitions and account for side products and collection losses separately.

Practice questions

1. Starting from 20 mol A, 5 mol remains and 12 mol B forms in A → B and A → C. How much C formed? Answer: A consumed is 15 mol; 12 mol makes B, so 3 mol C forms. 2. Find conversion and desired-path selectivity for that mixture. Answer: Conversion is 15/20 = 75%; desired fraction is 12/15 = 80%. 3. What is the B yield on fed A, and what B:C ratio corresponds to those amounts? Answer: Yield is 12/20 = 60%; B:C = 12:3 = 4:1.