Chemical Processing II: Dyes, Dyeing, Machines, Dye-Fibre Thermodynamics and Kinetics, Colour Measurement and Fastness

This chapter answers the second paragraph of Section XE9.5 “Chemical Processing” of XE9 Textile Engineering and Fibre Science in the Engineering Sciences (XE) paper: “Classification of dyes; Dyeing of cotton, wool, silk, polyester, nylon and acrylic with appropriate classes of dyes; Dyeing of polyester/cotton and polyester/wool blends; Dyeing machines; Dyeing processes and machines for cotton knitted fabrics; Dye-fibre interaction; Introduction to thermodynamics and kinetics of dyeing; Brief idea about the relation between colour and chemical constitution; Beer-Lambert’s law; Kubelka-Munk theory and its application in colour measurement; Methods for determination of wash, light and rubbing fastness.” It teaches dye classes and the fibre each suits, the sequence for each fibre and blend, the machines, the forces that hold a dye, affinity, isotherms and rate of dyeing, colour and constitution, Beer-Lambert and Kubelka-Munk, and the fastness tests. The numericals are recipes and salt quantities, exhaustion from absorbance, concentration from Beer-Lambert, Nernst partition, Kubelka-Munk reflectance and an Arrhenius rate ratio.

1. Classification of dyes

Dyes are classified by application (how they reach and stay in the fibre), which is what the exam asks, and by chemical constitution (azo, anthraquinone, indigoid, triphenylmethane, phthalocyanine and so on). A dye is chosen from the fibre: the match between its ionic or polar character and the fibre’s functional groups decides the interaction, the fastness and the process. Cellulose has only OH groups and is weakly anionic in water; wool, silk and nylon have –NH₂ and –COOH groups and become cationic in acid; polyester is a hydrophobic ester with no ionic groups; acrylic carries anionic sulphonate or carboxylate end groups from its initiator or comonomer.

Application classes of dyes
ClassFibreHow it is held
DirectCelluloseHydrogen bonds and van der Waals forces; salt aids uptake; modest wash fastness
ReactiveCellulose (also wool, silk, nylon)Covalent bond with cellulose under alkali; hydrolysis competes
Vat, indigoCelluloseInsoluble pigment reduced to soluble leuco form, absorbed, re-oxidised inside
Sulphur, azoicCelluloseSulphur: reduced then re-oxidised; azoic: naphthol coupled with a diazo component in the fibre
DispersePolyester, acetate, nylonSparingly soluble non-ionic dye dissolves in the fibre as a solid solution
Acid, metal-complexWool, silk, nylonAnionic dye ionic-bonded to protonated amino groups; 1:2 complexes also use van der Waals forces
Basic (cationic)AcrylicDye cation pairs with anionic end groups
⚠️ Match the class to the fibre
The common mismatches are: disperse dyes for cotton (no affinity), reactive dyes for polyester (no reactive site), and acid dyes for acrylic (the fibre is anionic, so a cationic dye is needed). Wool can also be dyed with reactive and chrome dyes, and nylon with disperse dyes in pale shades.

2. Dyeing of cotton, wool, silk, polyester, nylon and acrylic

Cotton, reactive: the dye is first exhausted with common salt or Glauber’s salt (it screens the negative charge of wet cellulose, which would otherwise repel the anionic dye), then alkali (soda ash, and caustic for the less reactive types) forms cellulosate ions Cell–O⁻ that attack the dye and form a covalent bond, then the unfixed and hydrolysed dye is soaped off. Fixation is below 100% because water also attacks the dye. Direct dyes use salt and no alkali. Vat dyeing has four steps: vatting (reduce with sodium dithionite and NaOH to the soluble leuco anion), dyeing, oxidation (air or peroxide) back to the insoluble pigment and soaping. Indigo for denim is a vat dye applied in repeated dips and oxidised in air between them.

Wool and silk, acid and metal-complex: the fibre protonates, –NH₃⁺ ··· ⁻O₃S–Dye. Levelling acid dyes go on at pH about 2 to 4 with sulphuric acid and Glauber’s salt, which slows uptake for even results, at or near the boil for wool; milling dyes and 1:2 metal-complex dyes need weaker acid or near-neutral baths and give better wet fastness. Wool is damaged by a very long boil, so time and temperature are controlled; silk is dyed below the boil (about 80 to 85 °C) with weaker acid. Nylon is dyed with acid dyes at pH about 4 to 6 near the boil: its dye sites are limited (the amino end groups), so it saturates, and differences in end-group content or heat history give streaky (barré) dyeing; disperse dyes serve pale shades.

Polyester (no ionic groups, compact, glassy) is dyed with disperse dyes at about 130 °C under pressure at pH 4.5 to 5.5, or with a carrier at the boil, or by the thermosol continuous process (pad, dry, then dry heat at about 200 °C for around a minute). Heat takes the fibre well above its glass transition so that chains move and dye can diffuse in. Surface dye is removed by reduction clearing (dithionite and caustic) for good wash and rubbing fastness. Acrylic is dyed with cationic dyes at pH 4 to 5.5 with a cationic retarder; the fibre takes dye rapidly only above its wet glass transition, so the temperature is raised slowly through the critical range (about 1 °C/min) to avoid unlevelness, and cooled slowly to avoid creasing.

3. Blends, dyeing machines and knitted cotton

Polyester/cotton needs two dye classes. The usual route is disperse dye for the polyester first (acid pH, 130 °C), a reduction clearing that also cleans disperse dye stained on cotton, then reactive, direct or vat dye for the cotton (alkaline). The order matters because reactive-dye alkali would attack many disperse dyes and because the high-temperature acid stage would damage a reactive dye already fixed. Polyester/wool is dyed with disperse dye at a lower temperature (about 100 to 105 °C) at pH 4.5 to 5.5 to spare the wool, then wool with acid or metal-complex dye; because wool takes up disperse dye and stains, the disperse dye must be chosen and cleared carefully (reduction clearing is limited, as it would harm wool).

Dyeing machines
MachinePrincipleNotes
WinchRope of fabric lifted and dropped through a bathSimple, gentle, high liquor ratio
JigOpen-width fabric passes between two rolls through a small bathNeeds tension: unsuitable for stretchy knits
Jet, overflow, soft-flowRope circulated by liquor flowLow tension, low liquor ratio; HT versions for polyester
Beam, package, hankLiquor forced through wound materialYarn and some fabrics; pressure-capable
Pad machines (continuous)Pad, then batch, steam or thermosolPad-batch for reactive dyes; pad-dry-thermosol for polyester

Cotton knits are dyed in rope form in jet, overflow or soft-flow machines (or winches) because a knit is stretchy and needs low tension and gentle handling to avoid creases and distortion, and the lower liquor ratio saves water, salt and chemicals. Reactive dyes are the main class: salt, then dye, then alkali dosed in stages for evenness, then wash-off, then dimension control. The liquor ratio is the mass-to-volume ratio of bath to fabric, and all “% owf” and “g/L” chemical quantities follow from it.

4. Dye-fibre interaction, thermodynamics and kinetics of dyeing

A dye is held by one or more of: ionic bonds (acid on wool and nylon, cationic on acrylic), covalent bonds (reactive on cellulose), hydrogen bonds and van der Waals/hydrophobic forces (direct on cellulose, metal-complex on wool), and solid solution (disperse in polyester, with no specific site). Dyeing is the transfer of dye from bath to fibre until equilibrium, so two questions arise: how much (thermodynamics: affinity and isotherm) and how fast (kinetics: diffusion).

  • Nernst isotherm (partition, no saturation): [D]f = K [D]s, with [D]f in the fibre and [D]s in the bath. It fits disperse dye in polyester. If the dye is present in total amount m₀ with fibre mass w and bath volume V, the dye in the fibre is x, then x/w = K (m₀ − x)/V.
  • Langmuir isotherm (limited sites S): [D]f = S K [D]s / (1 + K [D]s). It fits acid dye on wool or nylon and cationic dye on acrylic; at high [D]s the fibre saturates at [D]f = S, which sets the saturation value. Freundlich [D]f = k [D]sⁿ fits many direct dyes on cellulose.
  • Affinity is the standard chemical-potential difference: −Δμ° = RT ln K for a partition isotherm, where K is the ratio of the dye in the fibre to that in the bath with the stated standard states. A bigger K is a higher affinity. Heat of dyeing comes from the van’t Hoff relation d ln K / d(1/T) = −ΔH°/R: an exothermic dyeing (ΔH° negative, the usual case, for example acid dyes on wool) loses affinity as T rises, so temperature buys rate rather than exhaustion, while an endothermic one gains affinity.
  • Kinetics: diffusion inside the fibre is the slow step. Fick’s law gives a half-dyeing time t½ ∝ r²/D, so doubling the fibre radius r quadruples t½. D = D₀ exp(−Ea/RT) rises steeply with temperature, because diffusion in a glassy polymer has a high activation energy and plasticising helps. Agitation, fine fibre and higher temperature speed uptake; affinity governs the final amount, not the rate.

Recipe and exhaustion arithmetic. “2% owf” means 2 kg of dye per 100 kg of dry fabric. For 25 kg of fabric that is 0.5 kg; from a 10 g/L stock solution that is 50 litres. Bath volume = fabric mass × liquor ratio. Exhaustion E% = 100 (C₀ − C_f)/C₀ for a bath of unchanged volume, with C measured by absorbance (section 5). The fixation of a reactive dye is the fraction of the dye actually bonded, which is smaller than exhaustion since some exhausted dye is later hydrolysed and washed off.

5. Colour and constitution, Beer-Lambert and Kubelka-Munk

A dye looks coloured because it absorbs part of the visible spectrum (about 400 to 700 nm) and the eye sees the complementary light: absorption of blue appears yellow, of green appears magenta, of red appears cyan. Colour needs an extended conjugated system with a chromophore (azo –N=N–, nitro, carbonyl, quinoid, –C=C–). An auxochrome (–OH, –NH₂, –NR₂) intensifies and deepens the colour, and groups such as –SO₃H give water solubility and ionic character. Longer conjugation or stronger donor–acceptor push moves the absorption maximum to longer wavelength (bathochromic), the opposite is hypsochromic; higher absorption intensity is hyperchromic. Colour theory is credited to Witt’s chromophore–auxochrome picture.

Beer-Lambert law: A = log₁₀(I₀/I) = ε c l, with ε the molar absorptivity (L mol⁻¹ cm⁻¹), c the molar concentration and l the path length (cm). Transmittance T = I/I₀ and A = −log₁₀ T. It holds for dilute solutions of one absorbing species in monochromatic light; it fails by aggregation (dyes often dimerise), at high concentration, with fluorescence, or when the light is not monochromatic. Dye-bath concentration and thus exhaustion are found from the absorbance at λmax against a calibration, diluting to keep A below about 1.

Kubelka-Munk theory treats an opaque dyed textile as a light-absorbing (coefficient K) and light-scattering (coefficient S) layer. For a layer thick enough to be opaque, with R the reflectance (as a fraction) at the wavelength of maximum absorption: K/S = (1 − R)² / (2R). K/S rises as R falls, and for a given substrate K/S is roughly proportional to dye concentration at low to moderate depths, so K/S measures colour strength, helps to match shades by recipe prediction and to follow exhaustion from the fabric. The theory needs an opaque sample, uniform scattering and negligible surface reflection (a Saunderson correction handles it), and it departs from linearity in deep shades. Colour difference is reported in the CIELAB system as ΔE = √(ΔL² + Δa² + Δb²).

⚠️ Two different light laws
Beer-Lambert applies to a transparent solution (A proportional to c, with A = −log T). Kubelka-Munk applies to an opaque scattering solid and uses reflectance, with K/S proportional to c. Using A = εcl on a fabric, or using R in place of T in the solution law, is the standard error. Also, K/S increases when R decreases: darker means higher K/S.

6. Fastness: wash, light and rubbing

Fastness is resistance of a dyeing to a treatment, graded by comparing the change in colour of the specimen and the staining of an adjacent white cloth against the grey scales, where 5 means no change or no staining and 1 means severe. Wash fastness (the ISO 105 C series, AATCC counterparts) agitates the dyed specimen sewn to a multifibre strip in a standard detergent solution at a set temperature and time, with steel balls for abrasion, then rinses and dries it. It depends on dye–fibre bond strength and on how well unfixed dye was washed off (soaping, reduction clearing).

Light fastness (ISO 105 B02) exposes the specimen with a set of blue wool references numbered 1 (lowest) to 8 (highest) to a xenon-arc lamp that imitates daylight; the rating is the reference that fades to the same extent. Each higher blue-wool number lasts about twice as long, so rating 6 is about twice as light-fast as rating 5. Rubbing (crocking) fastness (ISO 105 X12) rubs the specimen with a white cotton cloth on a crockmeter, dry and then wet, under a standard load and number of strokes; the staining of the white cloth is graded on the grey scale. Rubbing fastness is poor for deep shades with much unfixed surface dye, and wet rubbing is usually worse than dry.

ℹ️ Reading the ratings
Wash and rubbing results are two ratings each (colour change and staining) on the 1 to 5 grey scales; light fastness is a single number on the 1 to 8 blue wool scale. A higher number is better in every case.

Key takeaways

  • Match dye to fibre: direct, reactive, vat, sulphur and azoic for cellulose; acid and metal-complex for wool, silk and nylon; disperse for polyester; cationic for acrylic. Reactive dye needs salt then alkali, and hydrolysis lowers fixation.
  • Polyester is dyed at about 130 °C (or with carrier, or thermosol) because heat takes it above its glass transition; acrylic is heated slowly through its critical range. In polyester/cotton, disperse dye goes on first, then the cotton dye.
  • Knits go in jet, overflow or soft-flow machines at low tension; bath volume is fabric mass × liquor ratio, and % owf quantities follow from fabric mass, not from bath volume.
  • Nernst: [D]f = K[D]s (no saturation); Langmuir: [D]f = SK[D]s/(1 + K[D]s) with saturation S; −Δμ° = RT ln K; van’t Hoff gives ΔH°; t½ ∝ r²/D and D = D₀exp(−Ea/RT).
  • Colour needs conjugation with a chromophore and auxochrome; longer conjugation shifts absorption to longer wavelength. Beer-Lambert A = εcl is for transparent solutions; Kubelka-Munk K/S = (1 − R)²/(2R) is for opaque fabrics and measures colour strength.
  • Fastness is graded on grey scales (1 to 5) for change and staining in wash and rubbing tests, and on blue wool (1 to 8, each step about twice as durable) for light; unfixed surface dye ruins wash and rubbing results.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. Which class of dye forms a covalent bond with cellulose under alkaline conditions?

    1. Direct dyes, through hydrogen bonds aided by salt and heat
    2. Reactive dyes, through reaction with the cellulosate ion formed in alkali
    3. Disperse dyes, through dissolving in the amorphous region as a solid solution
    4. Basic dyes, through ionic pairing with the sulphonate groups of the cotton
    Show answer

    Answer: B — Reactive dyes, through reaction with the cellulosate ion formed in alkali

    Reactive dyes carry a group (for example a chlorotriazine or a vinyl sulphone) that reacts with Cell–O⁻ formed by alkali, giving a covalent bond. Direct dyes are held by physical forces. Disperse dyes dissolve in hydrophobic fibres, and cotton has no sulphonate groups for basic dyes.

  2. In the exhaust dyeing of cotton with a reactive dye, salt is added before the alkali. What is the main role of the salt?

    1. It reacts with the dye to make it insoluble so that it deposits on the fibre surface, where it is then fixed by the alkali
    2. It raises the pH of the bath so that the cellulose ionises and reacts with the dye
    3. It acts as a catalyst for the covalent bond between the dye and the cellulose
    4. It screens the negative charge of wet cellulose, which would repel the anionic dye, and so promotes exhaustion
    Show answer

    Answer: D — It screens the negative charge of wet cellulose, which would repel the anionic dye, and so promotes exhaustion

    Cellulose develops a negative surface charge in water and the dye is anionic, so salt screens the repulsion and drives the dye into the fibre before fixation. Raising pH is the job of the alkali, which is added afterwards. Salt does not insolubilise or catalyse the dye.

  3. The correct order of the steps in applying a vat dye to cotton is

    1. oxidation to the leuco form, dyeing, reduction to the insoluble pigment and soaping, in a bath kept strongly acidic
    2. soaping, dyeing with the insoluble pigment, then neutralisation with acid
    3. reduction to the soluble leuco form in alkali, dyeing, oxidation to the insoluble pigment and soaping
    4. diazotisation, coupling in the fibre and then reduction clearing
    Show answer

    Answer: C — reduction to the soluble leuco form in alkali, dyeing, oxidation to the insoluble pigment and soaping

    Vat dyes are insoluble pigments, so they are reduced (sodium dithionite and NaOH) to the soluble leuco anion that has affinity for cellulose, which is oxidised back inside the fibre and soaped. The reverse order cannot dye, and diazotisation-coupling is the azoic route.

  4. Why can polyester be dyed with disperse dyes at 130 °C under pressure but dyes only very slowly at 60 °C?

    1. Above the glass transition the chain segments become mobile, which opens free volume for dye diffusion into the compact fibre
    2. At 130 °C the ester groups hydrolyse and become ionic, so the fibre can bond to the dye covalently, which fixes the dye permanently in the fibre structure
    3. At 130 °C the disperse dye becomes fully ionised and is attracted to the fibre by charge
    4. At 60 °C the disperse dye is chemically destroyed by the acetic acid in the bath
    Show answer

    Answer: A — Above the glass transition the chain segments become mobile, which opens free volume for dye diffusion into the compact fibre

    Polyester is compact and glassy; heat well above its glass transition frees chain motion and also raises dye solubility and diffusion, so uptake is fast. Disperse dyes are non-ionic and the fibre has no ionic sites, so charge attraction and bonding explanations are wrong; the dye is stable in the weakly acidic bath.

  5. When acrylic is dyed with a cationic dye, the temperature is raised very slowly (about 1 °C/min) through the range just above the fibre’s wet glass transition. The reason is that

    1. the fibre melts rapidly in that range and must be kept intact by slow heating, so that the filaments do not fuse together in the dyebath
    2. the dye uptake rises sharply there and a fast rise gives rapid, unlevel dyeing that is hard to correct
    3. the cationic dye decomposes quickly above that temperature in an acidic bath
    4. the anionic end groups of the fibre are lost if heated rapidly
    Show answer

    Answer: B — the dye uptake rises sharply there and a fast rise gives rapid, unlevel dyeing that is hard to correct

    Cationic dye binds strongly to the anionic sites and, once the chains become mobile, is taken up fast and does not migrate afterwards, so unlevelness is permanent unless heating is slow in the critical range. The fibre does not melt at the dyeing temperature, and the end groups are not lost.

  6. For dyeing polyester/cotton with a disperse dye for the polyester and a reactive dye for the cotton, the polyester is normally dyed first because

    1. the acidic 130 °C disperse stage would damage reactive dye already on the cotton, and clearing then removes disperse stain
    2. reactive dye can only react with cellulose after the polyester has been fully dissolved
    3. disperse dye must be applied in strong alkali, which is available only at the start of the process
    4. polyester loses its affinity for disperse dye once cotton is dyed, because the cotton coats it with salt and so blocks the entry of the dye
    Show answer

    Answer: A — the acidic 130 °C disperse stage would damage reactive dye already on the cotton, and clearing then removes disperse stain

    The high-temperature acid disperse stage would damage a reactive dye bonded earlier, and reduction clearing after it also cleans disperse dye stained on cotton. The other statements are false: disperse dyeing is acidic, not strongly alkaline, and polyester is never dissolved.

  7. A dye shows an uptake isotherm that rises linearly with bath concentration with no saturation over the whole range studied, while a second dye on the same fibre levels off at a fixed limit. Which interpretation is consistent?

    1. The first is a Nernst partition (no specific sites); the second is a Langmuir isotherm (limited sites)
    2. The first fits a Langmuir isotherm with a limited number of sites and the second a Nernst partition without sites
    3. Both must be Nernst, because every dye partitions between bath and fibre without a limit
    4. Both isotherms show ionic bonding, because only an ionic bond can give a linear or a levelling-off curve
    Show answer

    Answer: A — The first is a Nernst partition (no specific sites); the second is a Langmuir isotherm (limited sites)

    A straight line through the origin is [D]f = K[D]s, a partition into the fibre without specific sites (for example disperse dye in polyester); a plateau at the number of sites is the Langmuir form (for example acid dye on nylon). The curve shape reports partition or site-limited uptake, not the type of bond.

  8. A dyeing is assessed at blue wool rating 4 and another dyeing of a different dye at rating 6 in the same light-fastness test. Which statement is correct?

    1. The rating 6 dyeing is more light-fast, needing about four times the exposure, as each step about doubles it
    2. The rating 4 dyeing is more light-fast, since a lower blue wool number means a greater resistance
    3. The two dyeings are equal, since blue wool ratings only record the colour of the dye and not the fading
    4. The rating 6 dyeing is more light-fast, needing exactly twice the exposure to fade equally, since the scale is linear
    Show answer

    Answer: A — The rating 6 dyeing is more light-fast, needing about four times the exposure, as each step about doubles it

    Blue wool numbers run from 1 (lowest) to 8 (highest) and each step lasts about twice as long, so two steps is about four times. A lower number is a worse fastness, and the scale is geometric, not linear.

  9. Which statements about colour and chemical constitution are correct?

    1. A dye that absorbs blue light appears yellow because the eye sees the transmitted or reflected complementary light
    2. Lengthening the conjugated system generally shifts the absorption maximum to longer wavelength (bathochromic shift)
    3. A sulphonic acid group is the chromophore that produces the colour of an azo dye
    4. Auxochromes such as –OH and –NH₂ cannot alter the absorption of a dye because they are saturated groups
    Show answer

    Answer: A — A dye that absorbs blue light appears yellow because the eye sees the transmitted or reflected complementary light; B — Lengthening the conjugated system generally shifts the absorption maximum to longer wavelength (bathochromic shift)

    Complementary-colour perception and the bathochromic effect of longer conjugation are standard. The azo group –N=N– is the chromophore, while –SO₃H mainly gives solubility; auxochromes such as –OH and –NH₂ have lone pairs that interact with the conjugated system and do shift and intensify the absorption.

  10. Which of the following statements about the Beer-Lambert law and Kubelka-Munk theory are correct?

    1. In the Beer-Lambert law the absorbance is proportional to the concentration of the dye for a fixed path length and wavelength, in dilute solution
    2. In the Kubelka-Munk equation K/S = (1 − R)²/(2R), a lower reflectance corresponds to a higher K/S
    3. Kubelka-Munk theory is applied to a clear dye solution in a cuvette
    4. Dye aggregation in a concentrated bath has no effect on Beer-Lambert linearity
    Show answer

    Answer: A — In the Beer-Lambert law the absorbance is proportional to the concentration of the dye for a fixed path length and wavelength, in dilute solution; B — In the Kubelka-Munk equation K/S = (1 − R)²/(2R), a lower reflectance corresponds to a higher K/S

    Linear A against c holds in dilute solution, and K/S rises as R falls. Kubelka-Munk applies to opaque scattering textiles, not to clear solutions, and aggregation is a classic cause of deviation from Beer-Lambert linearity.

  11. Which statements about the testing of fastness are correct?

    1. Colour change and staining are graded on grey scales in which 5 is the best rating
    2. A multifibre strip sewn to the specimen in a wash test shows which fibres are stained by the dye released
    3. Light fastness is graded on the grey scale, where 1 is the best
    4. Wet rubbing is carried out with a dry white cloth so that no moisture reaches the specimen
    Show answer

    Answer: A — Colour change and staining are graded on grey scales in which 5 is the best rating; B — A multifibre strip sewn to the specimen in a wash test shows which fibres are stained by the dye released

    Grey scales run 1 to 5 with 5 best, and the multifibre strip records staining on several fibre types. Light fastness uses the 1 to 8 blue wool scale in which 8 is best, and wet rubbing uses a wet cloth, hence the name.

  12. A lot of 40 kg of cotton knitted fabric is dyed with a reactive dye in a jet at a liquor ratio of 1:12 (1 kg fabric to 12 litres of liquor). The recipe calls for 50 g/L of common salt in the bath. The mass of salt needed is ____ kg.

    Numerical answer — type the value.

    Show answer

    Answer: 24

    Bath volume = 40 × 12 = 480 L. Salt = 480 L × 50 g/L = 24,000 g = 24 kg. The salt depends on the bath volume (g/L), unlike dye and alkali quoted as % owf.

  13. In a dyeing trial, 5.0 mL of the initial dyebath was made up to 100 mL with water and gave an absorbance of 0.80 at λmax. At the end of dyeing, 10.0 mL of the residual bath was made up to 100 mL and gave an absorbance of 0.30 under identical conditions. The bath volume did not change and the Beer-Lambert law holds. The percentage exhaustion, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 81

    Concentration in the bath is proportional to A × dilution factor. Initial: 0.80 × (100/5) = 16.0. Final: 0.30 × (100/10) = 3.0. E = (16 − 3)/16 × 100 = 81.25%, which is 81. Comparing the raw absorbances (0.80 and 0.30) would give 62.5% and ignores the different dilutions.

  14. A solution of a dye gives an absorbance of 0.46 in a cell of path length 1.00 cm at λmax. The molar absorptivity is 2.3 × 10⁴ L mol⁻¹ cm⁻¹ and the molar mass of the dye is 450 g/mol. Assuming the Beer-Lambert law, the concentration of the dye is ____ mg/L.

    Numerical answer — type the value.

    Show answer

    Answer: 9

    c = A/(ε l) = 0.46 / (2.3 × 10⁴ × 1.00) = 2.0 × 10⁻⁵ mol/L. Mass concentration = 2.0 × 10⁻⁵ × 450 = 9.0 × 10⁻³ g/L = 9.0 mg/L.

  15. Disperse dye partitions between polyester and the bath according to [D]f = K [D]s, with [D]f in g of dye per kg of fibre, [D]s in g of dye per litre of bath and K = 40 L/kg. A dyeing uses 2 kg of fibre, 20 L of bath and 5 g of dye in total, and reaches equilibrium. The percentage of the dye taken up by the fibre is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 80

    Let x grams be in the fibre. [D]f = x/2 g/kg and [D]s = (5 − x)/20 g/L. Then x/2 = 40 (5 − x)/20 = 2 (5 − x), so x = 4 (5 − x), 5x = 20 and x = 4 g. Exhaustion = 4/5 × 100 = 80%. Check: [D]f = 2, [D]s = 0.05, K = 2/0.05 = 40.

  16. A dyeing at 1% owf on cotton has a reflectance of 0.20 (fraction) at λmax. Assume the substrate’s own K/S is negligible and that K/S is directly proportional to the dye concentration. The reflectance, in per cent, of the 3% owf dyeing of the same dye is ____ (answer to two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 8.69

    K/S at 1% = (1 − 0.20)²/(2 × 0.20) = 0.64/0.40 = 1.60. At 3%, K/S = 4.80. Solve (1 − R)² = 9.6 R, that is R² − 11.6 R + 1 = 0, so R = [11.6 − √(11.6² − 4)]/2 = 0.08686, which is 8.69%. Tripling the concentration does not divide R by three, because K/S is not R.

  17. The diffusion coefficient of a disperse dye in polyester follows D = D₀ exp(−Ea/RT) with Ea = 90 kJ/mol and R = 8.314 J mol⁻¹ K⁻¹ (take T = °C + 273.15). The ratio D(130 °C)/D(100 °C) is ____ (answer to one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 8.7

    ln(D₂/D₁) = (Ea/R)(1/T₁ − 1/T₂) with T₁ = 373.15 K and T₂ = 403.15 K. 1/T₁ − 1/T₂ = 0.0026799 − 0.0024805 = 1.994 × 10⁻⁴ K⁻¹. Ea/R = 90,000/8.314 = 10,825 K. So ln ratio = 2.159 and the ratio = 8.66, which is 8.7. Using °C instead of kelvin would be the usual slip.