Stoichiometric Imbalance and Chain Stoppers

Introduction

The Carothers equation shows that step-growth chains only become long when conversion is extremely close to complete. That result assumed a perfect 1:1 balance of the two reacting groups. Real production never achieves perfect balance by accident: a small weighing error, a volatile monomer lost to the vapour, or an impurity with a single reactive group all upset the ratio. This page shows how even a tiny imbalance caps the chain length — and how chemists exploit exactly that effect to control molar mass deliberately.

Core explanation

Defining r. Consider a polymerisation of A–A with B–B, for example a diacid with a diol. Let N A be the number of A groups and N B the number of B groups present at the start, with B in excess. The stoichiometric ratio is

r = N A ÷ N B (so r ≤ 1)

A perfectly balanced system has r = 1. Using 1.0% more B–B than A–A gives r = 1/1.01 ≈ 0.990.

Counting molecules. Let p be the extent of reaction of the A groups (the minority group). The total number of monomer molecules at the start is (N A + N B)/2 = N A(1 + 1/r)/2. Each reaction between an A and a B group joins two molecules and so reduces the number of molecules by one. After N A·p reactions, the number of molecules is

N = N A(1 + 1/r)/2 − N A·p

Dividing the starting number by the number remaining gives the modified Carothers equation :

Xn = (1 + r) ÷ (1 + r − 2rp)

When r = 1 this collapses back to the ordinary form, Xn = 1/(1 − p).

The ceiling on chain length. The most important consequence appears when the minority group is used up completely (p = 1):

Xn(max) = (1 + r) ÷ (1 − r)

With r = 0.99 the ceiling is 1.99/0.01 = 199 repeat units, however long the reaction continues. With r = 0.95 it falls to 39. Every chain end eventually carries a B group, and B cannot react with B, so growth stops.

Monofunctional reagents. A molecule with a single reactive group — for example ethanoic acid in a polyamide synthesis — reacts with a chain end and leaves an unreactive cap. Such a molecule is a chain stopper . Its effect is included by redefining r. If N B′ molecules of a monofunctional B reagent are added to a balanced A–A/B–B mixture,

r = N A ÷ (N B + 2N B′)

The factor of 2 appears because each monofunctional molecule removes one growing end but, unlike a bifunctional molecule, contributes no new end: it has the same limiting effect as a whole extra B–B molecule. The same equation also covers A–B monomers such as amino acids, where r = N A/(N A + 2N B′).

Why industry wants a ceiling. Very long chains are not always desirable. Melt viscosity rises steeply with molar mass, so a polymer that is too long cannot be spun into fibre or moulded. A controlled amount of chain stopper sets a target Xn and also makes the product stable: capped ends cannot keep reacting when the polymer is later remelted, so its molar mass does not drift during processing.

Formulae

r = N A ÷ N B (r ≤ 1). Xn = (1 + r) ÷ (1 + r − 2rp). At p = 1: Xn = (1 + r) ÷ (1 − r). With a monofunctional reagent B′: r = N A ÷ (N B + 2N B′). Mn ≈ Xn × M₀, where M₀ is the mean mass of the structural units.

Step-by-step reasoning

To predict the chain length of an imbalanced step-growth polymerisation:

1. Count the moles of each functional group, not the moles of each monomer. 2. Identify the minority group and set r = minority ÷ majority, adding 2 × moles of any monofunctional reagent to the majority count. 3. Decide the extent of reaction p of the minority group. 4. Substitute into Xn = (1 + r)/(1 + r − 2rp). 5. Convert to Mn by multiplying by the mean mass of a structural unit.

Visual explanation

Picture a growing chain as a line of beads with one hook at each end. With balanced hooks, lines keep meeting and joining. With excess B, eventually every line ends in two B hooks that cannot join each other. A chain stopper is a bead with only one hook: once it clips on, that end is sealed.

Real-world analogy

Imagine a dance where partners must be one leader and one follower. If there are 100 leaders and 101 followers, nearly everyone pairs, but one follower is always left over. In a polymer, the "left-over" groups sit at every chain end, and once only those remain, no further joining is possible.

Real-world example

Nylon-6,6 is made from hexane-1,6-diamine and hexanedioic acid. To guarantee a 1:1 ratio, manufacturers first crystallise the nylon salt , the ammonium carboxylate formed from the two monomers in exact proportion. A small, measured amount of ethanoic acid is then added as a chain stopper so that fibre-grade polymer has a consistent, spinnable melt viscosity.

Why?

Why does a tiny excess matter so much? Because Xn(max) = (1 + r)/(1 − r) depends on 1 − r, a small number. Changing r from 0.999 to 0.99 multiplies 1 − r by ten, so the maximum chain length falls roughly tenfold, from about 2000 to about 200.

Common misconception

"Adding more of one monomer pushes the reaction further and makes longer chains." For ordinary equilibria an excess of one reagent can drive conversion, but in step-growth polymerisation an excess of one bifunctional monomer caps the chain ends and shortens the chains.

Worked example

Question: A polyester is made with r = 0.990. Calculate Xn when p = 0.990 and when p = 1.

Reasoning: At p = 0.990: 1 + r − 2rp = 1.990 − 2(0.990)(0.990) = 1.990 − 1.9602 = 0.0298. Xn = 1.990/0.0298 = 66.8. At p = 1: Xn = 1.990/(1 − 0.990) = 199.

Answer: About 67 at 99.0% conversion, rising to a limit of 199 at complete conversion.

Quick check

1. What is the maximum Xn for a step-growth polymerisation with r = 0.98? Answer: Xn(max) = (1 + 0.98)/(1 − 0.98) = 1.98/0.02 = 99 repeat units.

Exam focus

Always count functional groups, not molecules. Remember the factor of 2 for monofunctional reagents, and check that r is less than or equal to 1. Examiners often ask you to explain, not just calculate, why an excess of one monomer lowers the molar mass.

Advanced insight

The same imbalance argument explains why step-growth polymers are so sensitive to purity. A monomer that is 99.5% pure, with the impurity carrying one reactive group, already limits Xn to a few hundred. It also explains a useful synthetic trick: deliberately using an excess of a diol gives oligomers with hydroxyl groups at both ends, called telechelic prepolymers, which are the building blocks of polyurethanes.

Summary

In an A–A + B–B polymerisation the ratio r = N A/N B (r ≤ 1) limits chain length through Xn = (1 + r)/(1 + r − 2rp). Even at complete conversion, Xn cannot exceed (1 + r)/(1 − r). Monofunctional chain stoppers are included by adding twice their amount to the majority group. Industry uses nylon salts for exact balance and measured chain stoppers to fix a target molar mass.

Practice questions

1. State the modified Carothers equation and show that it reduces to Xn = 1/(1 − p) when r = 1. Answer: Xn = (1 + r)/(1 + r − 2rp). With r = 1 it becomes 2/(2 − 2p) = 1/(1 − p). 2. A mixture contains 1.000 mol of a diacid and 1.000 mol of a diol plus 0.010 mol of ethanoic acid. Calculate r. Answer: Acid groups are the majority: N B + 2N B′ = 2.000 + 0.020 = 2.020 mol; OH groups N A = 2.000 mol; r = 2.000/2.020 = 0.990. 3. Explain why a nylon salt is prepared before nylon-6,6 is polymerised. Answer: Crystallising the salt fixes an exact 1:1 ratio of amine and acid groups, so r is very close to 1 and high molar mass can be reached. 4. Give two reasons why manufacturers add chain stoppers deliberately. Answer: To set a target molar mass and melt viscosity suitable for processing, and to cap the chain ends so the molar mass does not change when the polymer is remelted.