Equilibrium Yield with a Limiting Feed

Introduction

Stoichiometry tells you the most product a reaction could possibly give: the limiting reactant runs out and the reaction stops. Equilibrium tells you something different: a reversible reaction stops short, because the reverse reaction catches up with the forward one. Many problems blend the two ideas, and students often confuse them. This page shows how to calculate both numbers separately and then compare them, so that "percentage yield" has a clear meaning in a reversible system.

Core explanation

Two different ceilings. For a reversible reaction there are two limits on the amount of product. The first is the stoichiometric maximum , fixed purely by the feed amounts and the balanced equation. The second is the equilibrium amount , fixed by the equilibrium constant K. The equilibrium amount is always less than the stoichiometric maximum, because K is finite; it approaches the maximum only when K is very large.

Worked illustration. Consider H₂(g) + I₂(g) ⇌ 2HI(g) in a sealed 1.00 dm³ vessel, with a feed of 1.00 mol H₂ and 0.500 mol I₂, and suppose Kc = 64 at the operating temperature (a round value of the right order for this reaction at a few hundred degrees Celsius).

- Limiting reactant: I₂, since the ratio is 1 : 1 and there is less I₂. Stoichiometric maximum of HI = 2 × 0.500 = 1.00 mol. - Equilibrium table, with extent x: H₂ = 1.00 − x, I₂ = 0.500 − x, HI = 2x. - Because Δn(gas) = 0, the volume cancels and Kc = (2x)² / ((1.00 − x)(0.500 − x)) = 64. - Expanding: 4x² = 64(0.500 − 1.50x + x²), so 60x² − 96x + 32 = 0, which simplifies to 15x² − 24x + 8 = 0. - Roots: x = (24 ± √96) / 30, giving x = 0.473 or x = 1.13.

Choosing the root. The extent cannot exceed the amount of the limiting reactant, so x must lie between 0 and 0.500. The root 1.13 would make both reactant amounts negative and is rejected. So x = 0.473 mol and HI at equilibrium = 0.947 mol.

Comparing. The equilibrium yield is 0.947 / 1.00 × 100 = 94.7% of the stoichiometric maximum. Equivalently, the equilibrium conversion of I₂ is 0.473 / 0.500 = 94.7%. The conversion of H₂ is only 47.3%, because H₂ was in excess. Always state which reactant a conversion refers to.

Pushing conversion higher. Adding more of the excess reactant shifts the position of equilibrium and raises the conversion of the limiting reactant, but it can never raise the product above the stoichiometric maximum set by that limiting reactant.

Step-by-step reasoning

1. Balance the equation and identify the limiting reactant from the feed. 2. Calculate the stoichiometric maximum amount of product. 3. Set up an equilibrium table in terms of the extent x. 4. Substitute into the K expression and solve for x. 5. Keep only the root that gives non-negative amounts for every species. 6. Express the product amount as a percentage of the stoichiometric maximum.

Visual explanation

Draw a bar for the stoichiometric maximum and a shorter bar beneath it for the equilibrium amount. The gap between the two bars is the product "lost" to reversibility, not to side reactions or poor technique. Increasing the excess feed shortens the gap, but the upper bar stays exactly where it is.

Real-world analogy

Imagine a car park with 50 spaces. The number of spaces is the stoichiometric maximum. Even on a busy day, some cars are always leaving as others arrive, so perhaps 47 spaces are full at any moment. The steady number of parked cars is like the equilibrium amount: it depends on the arrival and departure rates, and it can never exceed the capacity.

Real-world example

In ammonia synthesis, the hydrogen and nitrogen feed could in principle give far more ammonia than a single pass through the converter actually produces. Industrial plants accept a modest equilibrium conversion per pass, remove the ammonia by condensation, and recycle the unreacted gases, so the overall conversion approaches the stoichiometric figure.

Why?

Why can the equilibrium amount never reach the stoichiometric maximum? Reaching the maximum would require the concentration of the limiting reactant to fall to zero. The K expression would then have a zero in its denominator, which no finite K can match, so some limiting reactant always remains.

Common misconception

"A 95% yield means the chemist lost 5% of the product." In a reversible system the shortfall may be entirely due to equilibrium. The unreacted material is still present in the vessel; it has not been spilled or wasted.

Worked example

Question: For ethanoic acid + ethanol ⇌ ethyl ethanoate + water, take Kc = 4.0. Starting from 1.00 mol acid and 3.00 mol ethanol (no ester or water), find the equilibrium amount of ester and the percentage yield.

Reasoning: The acid is limiting, so the stoichiometric maximum is 1.00 mol ester. Table: acid 1.00 − x, ethanol 3.00 − x, ester x, water x. Volume cancels: x² / ((1.00 − x)(3.00 − x)) = 4.0. Expanding gives 3x² − 16x + 12 = 0, so x = (16 ± √112) / 6 = 0.903 or 4.43. The second root exceeds 1.00 and is rejected.

Answer: 0.903 mol ester, a yield of 90.3% of the stoichiometric maximum (compared with 66.7% for a 1 : 1 feed).

Quick check

1. A feed gives a stoichiometric maximum of 2.0 mol product, and 1.5 mol is present at equilibrium. What is the equilibrium yield? Answer: 1.5 / 2.0 × 100 = 75% of the stoichiometric maximum.

Exam focus

Examiners reward a clear statement of the limiting reactant, a labelled equilibrium table, and an explicit reason for rejecting a root. Say which reactant any "conversion" refers to, and do not quote a percentage yield above 100%.

Advanced insight

When Δn(gas) is not zero, the volume or total pressure does not cancel and the conversion depends on the operating pressure as well as on K. Equilibrium yield can also be limited further by competing side reactions, in which case selectivity and equilibrium must be treated together, often by solving simultaneous equations for two extents.

Summary

The stoichiometric maximum comes from the limiting reactant alone; the equilibrium amount comes from solving the K expression. The physically valid root keeps every amount non-negative. Percentage equilibrium yield compares the two. Excess of one reactant raises conversion of the other but never lifts the product above the stoichiometric ceiling.

Practice questions

1. For H₂ + I₂ ⇌ 2HI with 1.00 mol of each reactant and Kc = 64, find the equilibrium amount of HI. Answer: (2x)² / (1.00 − x)² = 64, so 2x / (1.00 − x) = 8, x = 0.800 and HI = 1.60 mol (80% of the 2.00 mol maximum). 2. Why is a root of x = 1.13 mol rejected when the feed contains 0.500 mol of the limiting reactant? Answer: It would leave a negative amount of the limiting reactant, which is physically impossible. 3. For the esterification with Kc = 4.0 and a 1 : 1 feed of 1.00 mol each, what is the percentage yield? Answer: x / (1.00 − x) = 2, so x = 0.667 mol, a yield of 66.7%. 4. Can adding a large excess of ethanol give more than 1.00 mol ester from 1.00 mol ethanoic acid? Explain. Answer: No; the acid is limiting, so 1.00 mol ester is the absolute ceiling. Excess ethanol only moves the equilibrium amount closer to it.