Lesson 3 · 30 min

Column efficiency: plates, band broadening and resolution

Bands spread as they travel. Plate height, the van Deemter curve and the resolution equation show why doubling the column improves resolution by only 41 % — and why changing selectivity is usually the better lever.

In this lesson

  • Compute the plate number N and plate height H from peak widths, and explain why widths grow with the square root of column length
  • Use the van Deemter equation to explain how flow rate affects efficiency and find the optimum velocity
  • Compute the resolution Rs and use the Purnell equation to decide whether efficiency, selectivity or retention is worth improving
  • Estimate the number of plates needed for baseline separation

Two compounds have a selectivity of only α = 1.05 — say, isomers that differ in the position of one methyl group. Separating them to baseline at k = 3 takes about 28 000 theoretical plates. An old-fashioned 2 m packed GC column with a few thousand plates is helpless; a 30 m capillary with more than a hundred thousand does it easily. Liquid chromatography shrank its packing from 10 µm to 1.7 µm chasing the same quantity: narrower bands.

Predict first

Everything else stays the same, but the column is made twice as long. By what factor does the resolution Rs of two peaks change?

Pick an answer and lock it in before you look. A wrong guess teaches more than no guess.

Plate number and plate height

As a band travels, each molecule meets its own string of random events: a wider channel here, a longer stay in the stationary phase there. The sum of random steps is Gaussian, and its variance grows in proportion to the distance travelled: σ2=HL\sigma^2 = H L. The constant, the plate height HH, is smaller for narrower bands and a more efficient column. Column length divided by plate height is the number of theoretical plates N=L/HN = L/H — the counterpart of the last lesson’s tubes.

N can be read straight from a chromatogram: for a Gaussian peak N=(tR/σ)2N = (t_\mathrm{R}/\sigma)^2. In practice one uses the base width wbw_\mathrm{b} (between the tangents’ intersections with the baseline, 4σ) or the half-height width w1/2w_{1/2} (2.355σ).

N=(tRσ)2=16(tRwb)2=5.545(tRw1/2)2H=LNN = \left(\frac{t_\mathrm{R}}{\sigma}\right)^2 = 16\left(\frac{t_\mathrm{R}}{w_\mathrm{b}}\right)^2 = 5.545\left(\frac{t_\mathrm{R}}{w_{1/2}}\right)^2\qquad H = \frac{L}{N}
The three forms agree for a Gaussian peak; 5.545 = 8 ln 2.
σ\sigma
standard deviation of the peak (in time)
wb,w1/2w_\mathrm{b}, w_{1/2}
base width and width at half height
LL
column length

The van Deemter equation: three kinds of spreading

In 1956 the Dutch chemical engineer van Deemter split the plate height into three terms:

  • A (eddy diffusion): in a packed bed, molecules take channels of different lengths between particles; the term does not depend on flow and scales with particle diameter. Open capillaries have no A term.
  • B/u (longitudinal diffusion): the band diffuses back and forth along the column; the longer it stays, the wider it gets, so it matters most at low velocity u. Diffusion in gases is about 10 000 times faster than in liquids, so this term matters most in GC.
  • C·u (mass transfer): getting in and out of the stationary phase and through the pores takes time; if the flow is too fast, equilibrium lags and the band is smeared. Smaller particles and thinner films shrink C.

The sum falls and then rises, and its minimum marks the optimum velocity.

H=A+Bu+C uuopt=BCHmin=A+2BCH = A + \frac{B}{u} + C\,u\qquad u_\text{opt} = \sqrt{\frac{B}{C}}\qquad H_\text{min} = A + 2\sqrt{BC}
Setting dH/du = 0 gives the optimum velocity. In packed columns A ∝ d_p, B ∝ D_m and C ∝ d_p²/D_m.
uu
mobile-phase linear velocity (length/dead time)
AA
eddy (multipath) diffusion term
BB
longitudinal diffusion, ∝ diffusion coefficient in the mobile phase
CC
mass-transfer term
H = A + B/u + C·u · computed from particle size and diffusion

The van Deemter curve: flow rate and plate height

Particle size
12345678Linear velocity u / (mm/s)010203040Plate height H / µm
  • H (sum)
  • A: eddy (multipath) diffusion
  • B/u: longitudinal diffusion
  • C·u: mass transfer
  • Other particles / gases
  • Optimum

Near the optimum: B/u equals C·u and H is smallest.

u · H
1.00 mm/s · 9.5 µm
A · B/u · C·u
5.0 · 2.0 · 2.5 µm
Plates N (15 cm)
15800
Dead time t_M
2.5 min
Optimum velocity
0.89 mm/s
Minimum plate height
9.5 µm
LC: the reduced van Deemter equation h = a + b/ν + c·ν with a = 1, b = 2, c = 0.1 (typical of a well-packed column), ν = u·d_p/D_m, D_m = 1.0 × 10⁻⁵ cm²/s; column length 15 cm. GC: Golay’s equation for an open tube (no A term), 0.25 mm i.d., 0.25 µm film, n-octane at k = 3 and 100 °C, D_m from the Fuller equation at a mean pressure of 1.5 atm; length 30 m. Real columns need measured coefficients; the pressure gradient and extra-column effects are ignored.
Drag the handle to change the velocity and see which term dominates. Switch from 5 µm to 1.7 µm particles: the minimum H falls to about a third, the optimum velocity rises about threefold and the right-hand side flattens — small particles allow faster analyses. Then switch to GC and compare nitrogen, helium and hydrogen.

Resolution: efficiency, selectivity and retention

Whether two peaks separate is measured by the resolution Rs: the difference in retention time divided by the average base width. At Rs = 1.0 the peaks overlap by about 2 %; at Rs = 1.5 they are essentially baseline-separated. Substituting tR=tM(1+k)t_\mathrm{R} = t_\mathrm{M}(1+k) and σ=tR/N\sigma = t_\mathrm{R}/\sqrt{N} gives the Purnell equation, which splits resolution into three independent factors: efficiency N/4\sqrt{N}/4, selectivity (α−1)/α(\alpha-1)/\alpha and retention k/(1+k)k/(1+k).

The levers are very unequal. Doubling N buys only 41 %; beyond k ≈ 5 the factor k/(1+k)k/(1+k) hardly grows and the run just gets longer; but raising α from 1.05 to 1.10 nearly doubles Rs. That is why a chromatographer’s first move is usually to change the stationary or mobile phase to change the selectivity.

Rs=tR,2−tR,112(wb,1+wb,2)=N4⋅α−1α⋅k21+k2R_\mathrm{s} = \frac{t_{\mathrm{R},2} - t_{\mathrm{R},1}}{\tfrac12(w_{\mathrm{b},1} + w_{\mathrm{b},2})} = \frac{\sqrt{N}}{4}\cdot\frac{\alpha - 1}{\alpha}\cdot\frac{k_2}{1 + k_2}
The Purnell equation (N and k of the later peak). Solving for N estimates the plates needed for a target: N = 16Rs²(α/(α−1))²((1+k₂)/k₂)².
RsR_\mathrm{s}
resolution; 1.5 means baseline separation
N/4\sqrt{N}/4
efficiency factor
(α−1)/α(\alpha-1)/\alpha
selectivity factor
k2/(1+k2)k_2/(1+k_2)
retention factor term
Rs = (√N/4)·((α − 1)/α)·(k₂/(1 + k₂))

The three knobs of resolution

t_M 01234Time / t_MDetector signal
  • What the detector sees
  • Peak 1 (k₁ = 2.86)
  • Peak 2 (k₂ = 3.00)

Rs = 0.91 · Badly overlapped (Rs < 1.0)

Efficiency √N/4 25.0
Selectivity (α − 1)/α 0.048
Retention k₂/(1 + k₂) 0.750
Rs from peak positions and widths
0.909
Purnell equation
0.893
N needed for Rs = 1.5
28200
Two Gaussian peaks of equal area: t = t_M(1 + k), σ = t/√N (the same N for both). Rs is computed exactly as (t₂ − t₁)/(2(σ₁ + σ₂)); the Purnell equation is its form for nearly equal widths, and the two agree closely when α is near 1.
Start from N = 10 000, α = 1.05, k = 3 (Rs ≈ 0.89). Push Rs to 1.5 by moving one knob at a time: how large must N be? α? Can k do it at all? Watch the three bars.

Worked example

Efficiency and resolution from peak widths

On a 15.0 cm LC column a peak has tR=8.00t_\mathrm{R} = 8.00 min and a base width of 0.20 min. Find N and H. If a pair of compounds on this column has α = 1.10 and k = 4.0 for the later peak, what is the resolution?

Concept check

A difficult pair has α = 1.02, k = 5 and Rs = 0.6. Which change helps most?

Try the calculation

How many theoretical plates are needed for Rs = 1.5 when α = 1.05 and k₂ = 3.0?

Key takeaways

  • Peak variance grows linearly with length: σ2=HL\sigma^2 = HL and N=L/H=16(tR/wb)2N = L/H = 16(t_\mathrm{R}/w_\mathrm{b})^2.
  • The van Deemter equation H=A+B/u+CuH = A + B/u + Cu has an optimum velocity B/C\sqrt{B/C}; small particles lower both A and C.
  • Rs=(N/4) ((α−1)/α) (k/(1+k))R_\mathrm{s} = (\sqrt{N}/4)\,((\alpha-1)/\alpha)\,(k/(1+k)); Rs = 1.5 means baseline separation.
  • Rs grows only as √N; changing the selectivity α is usually the most effective improvement.
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Data and references

  1. J. J. van Deemter, F. J. Zuiderweg, A. Klinkenberg, “Longitudinal diffusion and resistance to mass transfer as causes of nonideality in chromatography”, Chem. Eng. Sci. 5, 271–289 (1956).
  2. M. J. E. Golay, in Gas Chromatography 1958, D. H. Desty (ed.), Butterworths (1958), pp. 36–55.
  3. J. H. Purnell, “The correlation of separating power and efficiency of gas-chromatographic columns”, J. Chem. Soc. 1268–1274 (1960).
  4. D. C. Harris, Quantitative Chemical Analysis, 9th ed. (2016), Ch. 23 (efficiency of separation, plate height, resolution).
  5. L. R. Snyder, J. J. Kirkland, J. W. Dolan, Introduction to Modern Liquid Chromatography, 3rd ed. (Wiley, 2010), Ch. 2.