CSIR-NET Polymer Chemistry — Molar Mass Averages, Carothers and Kinetics
A polymer sample is not one compound. It is a mixture of chains of different lengths, and almost everything CSIR-NET asks about polymers follows from that single fact: why there are several molar-mass averages, why dispersity is never below 1, why step-growth polymerisation needs extraordinary conversion to give a useful chain, and why a living polymerisation behaves so differently from a free-radical one. This article works those numbers out in full.
Why one polymer has several molar masses
Mn counts every chain equally — it is an ordinary mean over molecules, which is why any method that counts particles (osmometry, end-group analysis, freezing-point depression) returns Mn. Mw weights each chain by its own mass, so heavy chains dominate, and any method that responds to mass (light scattering, sedimentation) returns Mw. Since the heavy chains are always over-weighted in Mw, Mw ≥ Mn always, and dispersity Đ ≥ 1 always. Đ = 1 exactly means every chain has the same length — a monodisperse sample, which in practice only proteins and a few carefully made living polymers approach.
Worked example 1 — a deliberately crude two-component blend. Take 100 chains of M = 10 000 g mol−1 and 100 chains of M = 100 000 g mol−1.
Mn:
Σ NiMi = (100 × 10 000) + (100 × 100 000)
= 1 000 000 + 10 000 000 = 1.10 × 107
Σ Ni = 200
Mn = 1.10 × 107 ÷ 200 =
55 000 g mol−1
Mw:
Σ NiMi² = 100 × (10 000)² + 100 × (100 000)²
= 100 × 108 + 100 × 1010
= 1.00 × 1010 + 1.00 × 1012 = 1.01 × 1012
Mw = 1.01 × 1012 ÷ 1.10 × 107 =
91 818 g mol−1
Dispersity: Đ = 91 818 ÷ 55 000 = 1.67
Half the chains are ten times heavier than the other half, and the two averages differ by 67 %. That gap is the information — it tells you the sample is broad, and no single number could have.
Degree of polymerisation and the repeat unit
Worked example 2 — polystyrene. The repeat unit is
−CH2−CH(C6H5)− = C8H8.
C: 8 × 12.011 = 96.088
H: 8 × 1.008 = 8.064
M0 = 96.088 + 8.064 = 104.152 g mol−1
For a sample with Mn = 208 000 g mol−1:
DPn = 208 000 ÷ 104.152 = 1997, so roughly two thousand styrene
units per chain on average. Here the monomer (styrene, C8H8) and the
repeat unit have the same formula, because addition polymerisation loses nothing.
Step-growth is different. For nylon-6,6 the repeat unit is what remains
after two water molecules are eliminated from one diamine and one diacid:
hexamethylenediamine C6H16N2:
(6 × 12.011) + (16 × 1.008) + (2 × 14.007) = 72.066 + 16.128 + 28.014 = 116.208
adipic acid C6H10O4:
72.066 + 10.080 + 63.996 = 146.142
two water molecules: 2 × 18.015 = 36.030
M0 = 116.208 + 146.142 − 36.030 =
226.320 g mol−1
Forgetting to subtract the eliminated water is one of the most reliable ways to lose a mark in this topic.
The Carothers equation — why step-growth needs 99 % conversion
| Conversion p | DPn = 1/(1 − p) | Comment |
|---|---|---|
| 0.50 | 2 | Still a dimer on average |
| 0.90 | 10 | An oligomer, mechanically useless |
| 0.95 | 20 | Still not a plastic |
| 0.99 | 100 | Barely a usable material |
| 0.999 | 1000 | A genuine engineering polymer |
This table is the whole reason step-growth polymerisations are run with rigorously pure, exactly balanced monomers and driven hard to remove the condensation by-product. Chain-growth polymerisation has no such requirement — a high polymer forms from the very first moments, because each chain grows to full length in a fraction of a second and conversion only controls how many chains exist, not how long they are.
Worked example 3 — a 1 % excess of one monomer. Take r = 0.99.
At complete conversion (p = 1):
DPn = (1 + 0.99) ÷ (1 + 0.99 − 2 × 0.99 × 1)
= 1.99 ÷ (1.99 − 1.98) = 1.99 ÷ 0.01 = 199
Even with perfect conversion, a 1 % imbalance caps the chain at about 200 units. The excess monomer ends up on both ends of every chain and stops growth.
At p = 0.99 as well:
2rp = 2 × 0.99 × 0.99 = 1.9602
DPn = 1.99 ÷ (1.99 − 1.9602) = 1.99 ÷ 0.0298 = 66.8
The two limitations multiply. This is exactly the calculation a NET numerical wants, and the trap is using the simple 1/(1 − p) form and answering 100.
Chain-growth kinetics in one line
Assuming a steady state in radicals, the rate of a free-radical polymerisation is first order in monomer and half order in initiator. The half order is the fingerprint: it comes from bimolecular termination, so if a question reports Rp ∝ [I]1/2 you are looking at conventional free-radical chemistry. The kinetic chain length ν = Rp/Ri then varies as [M]/[I]1/2 — raising initiator concentration speeds the reaction up but makes the chains shorter, a trade-off worth stating in an answer.
In a living polymerisation (anionic, and controlled radical methods such as ATRP or RAFT) there is no termination and no chain transfer. Every chain starts at the same time and grows at the same rate, so the length distribution is Poisson and
For DPn = 100 that gives Đ = 1.01 — essentially monodisperse. Conventional free-radical polymers typically come out well above 1.5. Living systems also allow block copolymers, because a second monomer can be added when the first is exhausted and the chain ends are still active.
Copolymers and reactivity ratios
Each r compares how fast a growing chain end adds its own monomer to how fast it adds the other. The product r1r2 classifies the copolymer immediately:
| Condition | Copolymer type |
|---|---|
| r1r2 = 1 (both ≈ 1) | Ideal / random — each end shows no preference |
| r1 = r2 = 0 | Perfectly alternating — each end only adds the other monomer |
| r1r2 → 0 (but not zero) | Strong alternating tendency |
| r1 > 1 and r2 > 1 | Blocky — each end prefers its own monomer |
| r1 ≫ 1, r2 ≪ 1 | Composition drifts strongly with conversion |
Viscometry and the Mark–Houwink relation
K and a depend on the polymer, the solvent and the temperature, and are supplied in the question — never quote them from memory. For a flexible chain in a good solvent a lies between about 0.5 and 0.8, and in that range Mn ≤ Mv ≤ Mw. When a = 1 exactly, Mv equals Mw.
Worked example 4 — viscosity-average molar mass. A question gives [η] = 0.85 dL g−1, K = 1.0 × 10−4 dL g−1 and a = 0.75.
[η]/K = 0.85 ÷ 1.0 × 10−4 = 8500
1/a = 1 ÷ 0.75 = 1.3333
ln 8500 = 9.0478; 9.0478 × 1.3333 = 12.0638
Mv = e12.0638 = 1.73 × 105 g mol−1
Note the exponent 1/a = 1.33 magnifies errors: a 10 % error in [η] becomes a 13 % error in Mv. Keep full precision until the final line.
Physical properties you should be able to reason about
- Tg and Tm. Every polymer has a glass transition Tg; only those with enough regularity to crystallise also have a melting point Tm. An atactic polymer normally shows Tg alone.
- Flory–Fox: Tg = Tg,∞ − K/Mn. Short chains have more free-volume-rich ends, so Tg rises with molar mass and then levels off.
- Tacticity. Isotactic (all substituents on one side), syndiotactic (alternating) and atactic (random). Ziegler–Natta and metallocene catalysts give stereoregular, crystallisable chains; free-radical polymerisation of a vinyl monomer normally gives atactic material.
- Crosslinking. Thermoplastics are linear or branched and re-melt; thermosets are network solids and do not. A gel point is reached at a conversion predicted by the Carothers or Flory–Stockmayer treatment when the average functionality exceeds two.
Mistakes that cost marks
- Reporting Đ < 1. Mathematically impossible. If you get it, you have divided Mn by Mw instead of the other way round.
- Using the monomer mass as the repeat-unit mass in a condensation polymer. Subtract the small molecule lost — water for polyamides and polyesters, HCl for some polycarbonate routes.
- Applying the Carothers equation to chain growth. It is a step-growth result. A free-radical polymerisation gives high polymer at 5 % conversion.
- Assigning the wrong average to a technique. Osmometry and end-group analysis → Mn; light scattering → Mw; dilute-solution viscometry → Mv. Size-exclusion chromatography gives the whole distribution, but only against the calibration standard used, which must be stated.
- Quoting Mark–Houwink constants from memory. They are specific to a polymer–solvent–temperature combination. Use only what the question supplies.
- Confusing kinetic chain length with degree of polymerisation. They are equal only when termination is entirely by disproportionation; with combination, DPn is about twice ν.
Where this appears in the paper
| Exam | Typical polymer task |
|---|---|
| CSIR-NET Chemical Sciences | Mn/Mw/Đ numericals, Carothers with imbalance, reactivity-ratio interpretation, living vs radical comparison |
| GATE Chemistry | Molar-mass averages, polymerisation kinetics orders, tacticity and Tg reasoning |
| IIT-JAM / CUET-PG | Types of polymerisation, named polymers and their monomers, simple average calculations |
| CBSE Class 12 | Addition vs condensation polymers, common commercial polymers and uses |
The CSIR-NET paper has Part A, Part B and Part C sections; for the current question count, marks and negative-marking rules, check the official notification for your session rather than any summary, this one included.
Every calculation above starts from a repeat-unit molar mass. Get C8H8 or C6H16N2 wrong and the degree of polymerisation, the averages and the dispersity are all wrong together. The Molar Mass & Composition tool takes a formula and returns the element-wise breakdown, so you can confirm M0 in seconds before starting the polymer arithmetic — and for a condensation repeat unit, compute both monomers and subtract the eliminated water yourself.
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