Statistical theory of multiple-site linear wall-adsorption capillary Chromatography

Statistical theory of multiple-site linear wall-adsorption capillary Chromatography

Journal of Chromatography A, 1216 (2009) 1132–1139 Contents lists available at ScienceDirect Journal of Chromatography A journal homepage: www.elsev...

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Journal of Chromatography A, 1216 (2009) 1132–1139

Contents lists available at ScienceDirect

Journal of Chromatography A journal homepage: www.elsevier.com/locate/chroma

Statistical theory of multiple-site linear wall-adsorption capillary Chromatography Yinliang Chen ∗ , Yuanming Tang Institute of Nuclear Physics and Chemistry, Chinese Academy of Engineering Physics, Mianyang, Sichuan 621900, China

a r t i c l e

i n f o

Article history: Received 9 October 2008 Received in revised form 29 November 2008 Accepted 17 December 2008 Available online 25 December 2008 Keywords: Chromatography Capillary Multiple site Statistical theory Elution curve Moment Retention

a b s t r a c t Based on the mass-balance principle, a particular diffusion equation to describe the movement of solute molecules in the stagnant layer of multiple-site solid surfaces is constructed. From the equation, the moments of residence time in a step on multiple-site surfaces are derived. Similarly, the moments in a step in the mobile phase are also derived from a diffusion-drift equation. According to the probability theory, there exists a general relationship between the moments of an elution curve and the moments in a step. Through this relationship, the expressions of the elution-curve moments are derived from the step moments. In this paper, the details related to multiple-site linear wall-adsorption capillary chromatography are described and added in the equations to determine the step moments. The resultant expressions of the elution-curve moments involve various factors, such as adsorption–desorption rate constants, equilibrium constants, axial and radial dispersions in the mobile phase. Afterwards, the moment expressions are used to analyze the peak tailing. The results show that a small quantity of sites with a slow desorption rate will lead to a large peak asymmetry. © 2008 Elsevier B.V. All rights reserved.

1. Introduction Multiple sites on solid surfaces will lead to the tailing phenomena of chromatographic peaks and lower the efficiency of columns. Giddings and Eyring first studied the problem [1–3]. Supposing that the residence times on different types of sites and in the mobile phase are all exponentially distributed, they derived an analytical expression of two-site elution curves based on the probability theory under the condition of constant time spent in the mobile phase and concluded that a small quantity of strong-adsorption sites would lead to peak tailing in linear chromatography. McQuarrie [4] extended Giddings’ consideration to this case that the time spent in the mobile phase is of Gaessian distribution, derived an analytical expression of two-site elution curves also and suggested that the expressions can be approached by Cram-Charlier series in general cases. He further proposed that the moments of elution curves can be determined by differentiating its Laplace transform. Dondi, Felinger and coworkers [5–9] utilized the characteristic function (CF) method of probability into chromatography, derived a Fourier transform of the probability density function of the total time in a column and found the moments by differentiating the Fourier transform. The CF method that transfers the study of elution

∗ Corresponding author. E-mail address: chen [email protected] (Y. Chen). 0021-9673/$ – see front matter © 2008 Elsevier B.V. All rights reserved. doi:10.1016/j.chroma.2008.12.047

curve from time domain to frequency domain largely simplified the kinetic problem of chromatography. The method earlier also used the assumption that total time spent in the mobile phase is of a Gaussian distribution [9], which partially solves the axial dispersion problem. Later, the method used a concept of first passage time distribution [10,11] and involved the axial dispersion completely. The CF method is readily extended to linear adsorption chromatography with multiple-site stationary phases and is used to find the skew and excess and to explain the influence of desorption rate constants on peak tailing [9,12]. The distribution of the residence time in a step depends on all the details of the surrounding surface and the containing mobile phase [2] and is very complex. The assumption that the residence time on sites is exponentially distributed holds accurately, but the same assumption about the residence time in mobile phases is only a rough approach. Ref. [13] presented a rigorous model to determine the probability density function (PDF) of the traveling distance and residence time in a step. Although it is difficult to give an analytical expression of PDF, the corresponding exact expressions of moments can be easily obtained from this model. Then the moments of an elution curve can be derived from the step moments by means of the probability theory. The purpose of this paper is to apply the step moment model [13] into multiple-site linear wall-adsorption capillary chromatography [14] and to derive the expressions of the retention time, the height equivalent to a theoretical plate and various moments. Two

Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

1133

concepts of steps, hitting steps and adsorption steps, are used and the results are compared. The results are also compared with those in the literature. 2. General relationship According to the random walk model [15], once a solute molecule enters a column, it will pass through the column by a series of steps. During a step, the molecule moves a random displacement along the column and spends a random time. Denoting the traveling distance and the residence time in the ith step by (i , the molecule to pass through the column  i ), then the total time for n is the sum of  i , i.e.  , where n is the total steps, which is a i=1 i random number. If the distribution of the residence time in a step in the mobile phase is exponential, n will follow a Poisson distribution [1,9]. But if not, how about the distribution of n? In spite of the different distribution of the residence time in a step in the mobile phase, it is obvious that the following inequality for a random n always holds: n 

i ≤ L <

n+1 

i=1

i

(1)

i=1

where L is the length of the column. This inequality states that a molecule still stays in the column After it moves n steps and leaves the column after a more step. Here n is defined as the steps for a molecule to pass through the column. Rigorously, the total time n n+1  and  . In spent in the column should be between i=1 i i=1 i practice, since n is very large and it is reasonable to ignore the influence of less than a step. So the total time  c for a molecule to pass through a column can be written as

 n+1

c =

i

(2)

i=1

Now our problem turns to find the probability density function and the moments of  c under the limitation of Eq. (1). This problem has been resolved and the moments of  c , i.e. the moments of elution curve are obtained as follows [13]: tR = Lc1 =

¯ L ¯

M2 = Lc2 =

M3 =

L 2 { − 2c1 M11 (, ) + c12 2 } ¯ 

(4)

L {m3 () − 3c1 M12 (, ) + 3c12 M21 (, ) − c13 M3 () ¯ − 3c2 (M11 (, ) − c1 2 )}

M4 =

(3)

(5)

L {[M4 () − 4c1 − M13 (, ) + 6c12 M22 (, ) − 4c13 M31 (, ) ¯ + c14 M4 ()] − 6c2 [M12 (, ) − 2c1 M21 (, ) + c12 M3 ()] − 4c3 [M11 (, ) − c1 2 ] + 3c22 (2 − ¯ 2 )}

Fig. 1. Section of a column.

two parts can be treated separately and then combined. Fig. 1 illustrates the cross section of a wall-adsorption capillary column, in which we draw a dash circle as an auxiliary line and define the space between the column wall and the auxiliary line as stagnant layer. The stagnant layer is considered as a part of stationary phase. Once a molecule enters the stagnant layer, it means that it enters the stationary phase. The stagnant layer is only an intermediate concept and tends to zero in the ultimate results. The concentration distribution of solute molecules in a stagnant layer follows the diffusion equation: ∂C1 (r, t) ∂2 C1 (r, t) = Dm ∂t ∂r 2

Since the stagnant layer is very thin, it can be considered as a lamella and 1-dimension diffusion equation is enough. The ordinary initial and boundary conditions for this equation are as follows:

⎧ C1 (r, 0) = 0 ⎪  ⎪ ⎪ ⎪ ∂C1 (r, t)  ⎪ ⎪ ⎨ Dm ∂r 

3. Step moments on multiple-site solid surfaces The traveling distance and residence time of a molecule in a step contain two parts, the stationary and the mobile phases. The



= r=R

 (kdj Csj (t) − kaj C1 (R, t))

j

⎪ dCsj (t) ⎪  ⎪ = −kdj Csj (t) + kaj C1 (R, t) ⎪ ⎪ ⎪ ⎩ dt

(8)

Csj (0) = 0

where Cl (r, t) denotes the concentration of solute molecules at r position and at t time in the stagnant layer, Csj (t) the concentration on sites of type j, Dm the diffusive coefficient in the mobile phase, R the inner radius of capillary columns, R1 the inner radius of the stagnant layer, kdj is the desorption rate constant on sites of  is the corresponding adsorption rate constant, cm/s. type j, s−1 , kaj Note that in Eq. (8), Csj (t) is area concentration and Cl (r,t) volume  must be in cm/s [18]. concentration, so kaj The problem of interest is to determine how long the molecules stay in the stagnant layer. In order to do so, lets imagine that a unit amount of solute molecules cross the auxiliary line in Fig. 1 and enter the stagnant layer at t = 0. This can be stated by the following equation:



(6)

where, tR is the retention time, M2 and M3 are the second and third central moments of elution curve, respectively, M4 = M4 − 3 M22 is the fourth cumulant (M4 is the fourth central moment), ¯ and ¯ are the mean values of step displacements  and step residence times , Mnm (, ) is the joint central moment of the nth  and the mth .

(7)

−Dm

∂Cl (r, t)   ∂r

= r=R1

ı(t) 2R1

(9)

where ı(t) is the Dirac ␦-function. After entering the stagnant layer, the molecules diffuse in it and gradually leave it later. Once a molecule leaves the stagnant layer, it is no longer considered, as if the outside of the stagnant layer is in vacuum. Therefore, at the interface of r = R1 , there is [13,16]



Dm

∂C1 (r, t)   ∂r

= r=R1

v 2

C1 (R1 , t)

(10)

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Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

where v is the mean speed of solute molecules. Eqs. (9) and (10) can be combined as

 ∂C1 (r, t)  Dm  ∂r

v

=

2

r=R1

C1 (R1 , t) −

ı(t) 2R1

(11)

Eq. (11) is a special condition in the step-moment theory. If there is no such a condition, it is impossible to derive the probability density function and the moments of residence time in a step from the general diffusion equation. It is known from Eq. (10) that in one unit of interval  at time t the amount of leaving molecules is 2R1 Dm

 

∂Cl (r,t) ∂r

r=R1

= RvC1 (R1 , t). Because the total amount of

molecules into the stagnant layer is one, the amount of RvC1 (R1 , t) is equal to the probability density function fs (t) of the residence time in a step in the stagnant layer. When R1 → R, there is fs (t) = RvCl (R, t)

(12)

From this, it can be seen that once we solve Eq. (7) under the condition of Eqs. (8) and (11), we can find the probability density function of residence time in a step and find the moments of arbitrary order. Generally, it is more difficult to directly calculate the probability density function than to find the moments. For ordinary purpose, we only need to find the moments. If the probability density function is required, it can be approached by Edgeworth-Cramer series after finding the moments. Making Laplace transform of Eqs. (8) and (11) with respect to t and eliminating Csj , we have

 ⎧  p ∂C˜ 1 (r, p)  ⎪ ⎪ D =− j k C˜ 1 (R, p) m ⎨  p + kdj aj ∂r r=R ∂C˜ 1 (r, p)  1 v ⎪ ⎪ D = C˜ 1 (R1 , p) − m ⎩  2 2R1 ∂r

(13)

r=R1

where C˜ l (r, p) is the Laplace transform of Cl (r,t). Let R1 → R, from Eq. (13) it follows that C˜ 1 (R, p) =

1 Rv



1+

2

v

p  kaj j p + kdj

−1

(14)

f˜s (p) =

1+

2

v

p k v p + kdj aj

−1

f˜s (p) = 1 −

p j p + kdj

v

 kaj

(16)

According to the characteristic function theory, which can be referred to any text about the probability theory, from Eq. (16) we have



¯ s = −

df˜s (p)  dp 

2  kaj 2K kR = = v v v kdj 

= p=0

(17)

j

where  s is the residence time in a step in stationary phase, ¯ s is the  /k = K is the distribution coefficient on sites mean value of  s , kaj dj j of type j, K =



Kj is the total distribution coefficient, k = 2K/R is the

j

capacity factor, kj = 2kj /R is the partial capacity factor contributed by sites of type j. Similarly, we have M20 (s )¯ s = f˜s (0) =

2kR

vk d

(18)

(20)

vk d3

where Mn0 (s ) denotes the nth moment of  s about origin, k dn denotes the reciprocal of the averaged 1/kdn with weight of kj , i.e. 1 k dn

=

1  Kj 1  kj = n n K kdj kdj k j

(21)

j

4. Step moments in the mobile phase Corresponding to the definition of stationary phase, the region of r < R1 in Fig. 1 is defined as mobile phase. The concentration distribution of solute molecules in mobile phase obeys the diffusion-drift equation [17]: ∂C(r, z, t) = Dm ∂t



1 ∂C(r, z, t) ∂2 C(r, z, t) ∂2 C(r, z, t) + + 2 r ∂r ∂r ∂z 2

− w(r)

∂C(r, z, t) ∂z



(22)

The initial condition is C(r, z, 0) = 0

(23)

where C(r, z, t) denotes the concentration of solute molecules at time t and at position (r, z) in the mobile phase, w(r) represents the linear velocity of mobile phase at r, r is the radial coordinate, z the axial coordinate. Eqs. (22) and (23) are the ordinary diffusion equation and its initial conditions, respectively. Our purpose is to use these ordinary equations to describe the problem of the probability density function of traveling distance and residence time of a molecule in the mobile phase, here we need a boundary condition like Eq. (11). Suppose that at t = 0 and z = 0 at the interface R1 there is a unit amount of solute molecules coming from the stationary phase to the mobile phase, this means ∂C(r, z, t)   ∂r

(15)

are very small and can be neglected completely in the expressions of elution-curve moments. So when expanding Eq. (15) in series of 1/v, we can retain the first two terms only, 2

24kR

(4)

M40 (s ) = f˜s (0) =

Dm

From Ref. [13] it is seen that the terms containing 1/v, 1/v2 , and so on

(19)

vk d2



Therefore the Laplace transform of fs (t) is



6kR

(3)

M30 (s ) = −f˜s (0) =

= r=R1

1 ı(z)ı(t) 2R1

(24)

And when t > 0, there is a continuous molecular current diffusing out of the mobile phase from interface R1 , which can be described by



−Dm

∂C(r, z, t)   ∂r

= r=R1

v 2

C(R1 , z, t)

(25)

Eqs. (24) and (25) can be combined to



Dm

∂C(r, z, t)   ∂r

+ r=R1

v 2

C(R1 , z, t) =

1 ı(z)ı(t) 2R1

(26)

Because the total amount of molecules into the mobile phase is set as 1, the amount of RvC(R1 ,z,t) is equal to the probability density function fm m (z, t) in number, where fm m (z, t) denotes the PDF of the traveling distance and residence time of a solute molecule in a step in mobile phases. When R1 → R, there is fm ,m (z, t) = RvC(R, z, t)

(27)

Making Fourier-Laplace transform of Eqs. (22) and (26), we have

⎧ ˜ ω, p) ˜ ω, p) 2 w(r)iω − p d2 C(r, 1 dC(r, ⎪ ⎪ ˜ ω, p) = ω − + C(r, ⎨ Dm dr dr 2 r (28) ˜ w, p)  dC(r, 1 ⎪  + v C(R, ˜ ⎪ ω, p) = ⎩ Dm  2 2R dr r=R

Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

where the Fourier-Laplace transform is







5. Moments of elution curve



C(r, z, t)eiωz−pt dzdt

˜ ω, p) = C(r, 0

(29)

−∞

It is difficult to directly solve Eq. (28), but possible to expand it into series of ω and p: ˜ ω, p) = C(r,

∞ ∞   1 n=0 m=0

n

n!m!

yn,m (r)(iω) (−p)m

(30)

Substituting Eqs. (30) into (28) and rearranging it, by comparing the coefficient of ωn pm we have

⎧ 2 ⎨ d yn,m (r) + 1 dyn,m (r) = − r

dr 2

dr

1 (myn,m−1 (r) + nw(r)yn−1,m (r)) − n(n − 1)yn−2,m (r) Dm

⎩ Dm y (R) + yn,m (R) = ın,m n,m 2 2R

The traveling distance  and residence time  per step in formulas (1) and (2) consist of two parts, related to that of the stationary and mobile phases, respectively:  = m + s

(40)

 = m + s

(41)

where s ≡ 0 in this paper. From Eqs. (40) and (41), the origin moments of (, ␶) are obtained as 0 (, ) = Mn,m

m  j=0

v

(31) where ın,m = 1 for n = 0 and m = 0, otherwise, ın,m = 0. W(r) is set to be a parabola:



w(r) = 2u

1−

r2 R2

dr

r

dr

m! 0 (m , m ) M 0 (s )Mn,j j!(m − j)! m−j

(42)

The central moments can be derived from the origin moments by Mn,m (, ) =

m n   i=0 j=0



n! m! ¯ m−j M 0 (, )(−) ¯ n−i (−) i!(n − i)! j!(m − j)! i,j (43)

(32)

where u is the average of linear velocity of the mobile phase. If representing the right-hand side of the first equation in Eq. (31) by f(r), it becomes the equation of the following type:

dy(r) d

1135

= rf (r)

(33)

Eq. (33) is easily solved. In order to solve Eq. (31), firstly let n = m = 0, this corresponds to f(r) = 0 in Eq. (33). So y(r) in Eq. (33), i.e. y0,0 (r) in this case, can be solved. Next let n = 0 and m = 1, in this case f(r) = −y0,0 (r)/Dm , in a similar way to solving y0,0 (r) we can find y0,1 (r), and so on. In this way, we can find yn,m (r) with any n and m. From Eq. (27), the Fourier-Laplace transform of fm ,m (x, t) is obtained as

 in (−1)m

˜ ω, p) = Rv f˜m m (ω, p) = RvC(R,

n

n

n!m!

where, ¯ = ¯ m , ¯ = ¯ s + ¯ m . Substituting Mn,m (, ) from Eq. (43) into Eqs. (3)–(6), it is obtained that tR = (1 + k)tM



M2 = tM

2 Dm (1 + k) u2

 M3 = tM

+

2

+

2 (1 + k) 12Dm u4



1 + 2 u

yn,m (r)ωn pm

(44)

4Dm k d

(45)

3

12Dm k(1 + k) (1 + 6k + 12k2 + 7k3 )R2 + 2 k d

k(3 + 11k)R2

(34)

(1 + 6 k + 11 k2 ) R2 2k + 24Dm k d



+

6k k d2

+



(3 + 37k + 165k2 + 251k3 )R4



2 960Dm

(46)

According to the characteristic function theory, it follows that



n

0 Mn,m (m , m ) = (−i) (−1)m

   

∂n+m ˜ f , (ω, p) ∂ωn ∂pm m m

= Rvyn,m (R)

 M4

ω=0,p=0

(35)

where (m ,  m ) is the traveling distance and residence time in a step 0 ( ,  ) of in mobile phase. In this way we find the moments Mn,m m m any orders from the diffusion-drift equation with a special condition, the results are listed in appendix A. Below we list the first 4 moments: ¯ m = ¯ m =

Ru

+

2 k(1 + k) 144Dm

 +

1 u2

2



R

(37)

v

0 M0,2 (m , m ) =

=

R3 4Dm v 2Dm R

v

+

k d

(55 + 604k + 2504k2 + 4372k3 + 2599k4 )R4 480Dm

6k(2 + 8k + 7k2 )R2

 (38)

1+

11R2 u2 2 48Dm



1 + Dm

(39)

where ¯ m and ¯ m are the mean values of m and  m , respectively.

+

2

Dm (1 + k) (15 + 70k + 91k2 )R2 2





(36)

v

0 M2,0 (m , m )

= tM

4

3 (1 + k) 120Dm 1 + 4 u6 u

k d

11k2 R2 2 2k d

+

+ Dm

24k2 k d

k(3 + 11k)R2 k d2

2

+

48k(1 + k)

 +



k d2

24k k d3

(−53 + 204k + 6866k2 + 33516k3 + 50427k4 )R6 3 322560Dm

 (47)

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Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

The height equivalent to a theoretical plate H is as follows according to its definition H=

2 Dm u + 2 u (1 + k)





(1 + 6k + 11k2 )R2 24Dm

+

2k k d

M3 =



(48)

M4 =

In Eqs. (45)–(48) the terms containing v have been omitted [13].

 M3 = tM

2 Dm (1 + k) u2

M4 = tM

2



24k2 k d

2

+



(49)

k d +

u2 k d

+

6k k d2

(50)

M3 = tM

48k(1 + k)



k d2

12Dm k(1 + k) u2 k d

 M4

= tm

2k k d

j

(56)

dj

+



24k

(51)

k d3

 Dm u2

24k2 k d2

+



+

6k k d2

48k(1 + k) k d2

(52)



 +

24k k d3

tM = 2n

1  kaj ka k2 j

dj

ka

(53)

(54)

(57)

Eqs. (54)–(56) become identical to those of Ref. [12]. But here the probability pj is proportional to the adsorption rate constant on sites of type j, not the abundance of the sites, as stated in the reference. The reaction of solute molecules and sites of type j can be described by the following chemical reaction: k

aj

Aj + S Aj S

(58)

kdj

where Aj represents the site of type j, S the solute molecules in the mobile phase, Aj S the compound of solute molecules and j-type sites. According to the kinetics of chemical reactions, the concentration of Aj S obeys the following equation: dCsj dt

 (CAj − Csj )Cs − kdj Csj = kaj

(59)

where Csj and Cs denote the concentrations of Aj S and S respectively,  and k CAj the initial concentration of Aj , kaj dj the rate constants of the positive and counter reactions. When the concentration of compound Aj S is very low, Eq. (59) becomes  CAj Cs − kdj Csj = kaj

(60)

By comparing Eq. (60) with Eq. (8), we have

pj =

After comparing Eqs. (52) and (53) with Eqs. (50) and (51), it is found that the assumption of the Gaussian distribution only partially involves the axial dispersion in it. If deleting the all terms related to diffusions, Eqs. (45)–(47) become M2 =

1  kaj ka k4

kaj

  = kaj CAj kaj

2 k(1 + k)2 144Dm

u4 k d

pj =

dt

Ref. [9] supposed that the total time spent in the mobile phase is of the Gaussian distribution and the resulting expression of 2nd moment is identical to Eq. (49), the 3rd and 4th moments are as follows:



k d3

tM = 24n

(55)

dj

j

k the total adsorption rate constant, n = ka tM the average ka = j aj steps for solute molecules to pass through a column. Let

dCsj



12Dm k(1 + k)

+

3 (1 + k)4 120Dm u6

Dm + 2 u

2k

+

2 (1 + k)3 12Dm u4



24k

1  kaj ka k3



For one-site adsorption chromatography, the k dn in Eqs. (45)–(48) become kdn and Eqs. (45)–(48) become identical to that of Ref. [13]. When we make the comparison, we must remove all the terms related to diffusion in stationary phase in the equations of Ref. [13] because the diffusion in stationary phase does not exist in the case of adsorption chromatography. In Eqs. (45) and (48), the terms containing Dm in numerators is caused by the axial diffusion in the mobile phase, so can be called axial-diffusion terms. Similarly, the terms containing Dm in denominators is caused by the radial diffusion and called radial-diffusion terms, and the terms containing k d are denoted as desorption-rate terms. In the same way, each term in the expressions of the third and fourth moments, M3 and M4 , has special meaning and can be attributed to one or several factors, such as the pure axial diffusion, the pure radial diffusion, the pure desorption rate and their interactions. In the characteristic function method, while the first passage time distribution is used, the axial dispersion and the adsorption rate constants are all involved [10,11] and the resulting moments for multiple-site are equivalent to that of Eqs. (45)–(47) after deleting the terms related to the radial dispersion in the mobile phase: M2 = tM

k d2

tM = 6n

 /R is the adsorption rate constant of j-type sites, where kaj = 2kaj

6. Discussion



6k

 C kaj Aj  C k j aj Aj



(61) (62)

From Eq. (62) it is seen that the probability pj of a solute molecule being adsorbed by sites of type j is proportional to the product of the reaction rate constant and concentration of j-type sites. pj represents the adsorption-reaction fraction on sites of type j. When the concentration of j-type sites is low and the reaction rate constant is high, its adsorption–reaction fraction pj can not be low. Otherwise, when the concentration of j-type sites is high and the reaction rate constant is low, the adsorption–reaction fraction can not be high. Certainly, the discussion here must be limited to the field of linear chromatography. Table 1 gives several examples for one-site, from which it is seen that how much the effect of the radial diffusion is. In the table the column parameters are set to be: L = 1000 cm, R = 0.016 cm, u = 50 cm/s, k = 1 or 10, Dm = 0.1 cm2 /s or 0.01 cm2 /s, kd = 106 /s or 102 /s. In Ref. [10], the CF method has given the expressions of plate height, skew and the characteristic function of the band profiles. From this it is easy to find the 3rd and 4th moments. The expressions of H, M3 and M4 are then converted to the forms of this paper, which are equivalent to Eqs. (49)–(51). In the column of Ref [10] in the table the results are calculated by Eqs. (49)–(51), which don’t consider the radial diffusion only, and in the column of our work, the results are calculated by Eqs. (45)–(47), which consider all factors. It can be seen from Table 1 that the results from the our work and Ref. [10] are closer to each other when both of k and kd are smaller

Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

1137

Table 1 The moments from different conditions. Parameters k 1 1 1 10

Dm 0.1 0.1 0.01 0.1

Method 1 kd

M2 6

10 102 106 106

0.0064 0.41 0.00068 0.19

M3

Fig. 2. Skew vs p2 . (kd1 = 106 /s).

−6

3.1 × 10 0.012 3.3 × 10−8 0.00051

and Dm is lager, and the results from the two methods are largely different when k and kd increase and Dm decrease. It is obvious that there must be much larger differences between Eqs. (45)–(47) and Eqs. (54)–(56), in which the diffusions both in the axial direction and in the radial direction are all ignored completely, and between Eqs. (45)–(47) and Eqs. (52)–(53), in which the axial dispersion is partially involved, than that shown in Table 1. It can be derived from Appendix B that the radial dispersion can be ignored and the distribution of residence time per step in the mobile phase becomes exponential when R2 /2Dm « 1/ka . Since R2 /2Dm represents the time required to diffuse a distance of R by molecules and 1/ka is the mean residence time per step in the mobile phase, the above inference can be stated that the radial diffusion can be ignored and the distribution of residence time per step in the mobile phase becomes exponential only when the diffusion time is much less than the average residence time. Fig. 2 is the relationship of skew vs p2 . The skew reflects the symmetry and the tailing of a chromatographic peak, which is defined by M3 /(M2 )3/2 , where M2 and M3 are calculated by Eqs. (45) and (46). The larger the skew is, the more serious the tailing is. In the case of two sites, p2 = ka2 /ka , where ka2 represents the adsorption rate constant of the slow site and p2 represents the adsorption–reaction fraction of the component. From Fig. 2 it is seen that under the given kd1 and kd2 the skew increases with increasing p2 on the whole, and under the given p2 and kd1 the skew increases with decrease of kd2 . Fig. 2 shows that small slow component will lead to asymmetry of peaks. Roughly speaking, when kd2 < 10/s, a serious asymmetry will occur. Fig. 2 also shows that the curve of skew vs p2 has a maximum and the p2 at maximum decrease with decrease of kd2 . All of these agree with the deduction of Giddings [1–3] and Fornstedt [19] and the results of characteristic function method [12], but here more exact expressions of moments are used and the results will be more reliable. It is notable that the mean residence times per step in the stationary and mobile phases are kR/v and R/v by Eqs. (17) and (37), respectively. But according to the ordinary conception, they should be 1/kd (or k/ka for multiple-site) and 1/ka . There is a large difference between them. This is due to the different definitions of a step, see Fig. 3. In this paper and in Ref. [13] the process between two

Method 2 M4

M2 −9

2.5 × 10 0.00049 2.7 × 10−12 2.3 × 10−6

0.045 0.44 0.38 2.7

M4

M3 −5

9.2 × 10 0.014 0.0063 0.046

2.4 × 10−7 0.0006 9.6 × 10−5 0.00095

successive hittings of a molecule from the mobile phase onto the column wall is defined as a step. According to this definition, a step of a solute molecule in the mobile phase refers to the process from entering the mobile phase to hitting the column wall, corresponding to the processes from a to b, b to c, c to d and so on in Fig. 3. According to the ordinary concept, a step in the mobile phase refers to the process from desorbed to adsorbed by the column wall [1,9], corresponding to the process from a to g in Fig. 3. Generally it needs to take many times of hitting for a molecule to be adsorbed, since hit a wall does not mean to be adsorbed by the wall. It can be predicted that there are no differences about the results from the hitting-step concept and from the adsorption-step concept, because formulas (1) and (2) are universal and do not require any prior assumptions of steps.In Appendix B, if ignoring the influences of all the axial and radial diffusions in the mobile phase, the obtained moment expressions of the traveling distance and residence time in a step in the mobile phase are identical to the results from the assumption that the residence time in a step is of exponentially distributed and the velocity of the mobile phase is constant [1]. If only omitting the radial-diffusion terms, the obtained expressions are equivalent to the results from supposing that the residence time in a step is exponentially distributed and considering the axial dispersion [12]. The final results from the hitting-step concept and the adsorption-step concept are identical, but the method of the hitting-step has an advantage as follows. The derivation of the general relationship of elution curve moments and the step moments contains an assumption that the column length is long enough and the total steps are large enough [13]. For a weak-adsorption solute, the total steps based on adsorption-step concept may not be large enough to satisfy the assumption. But based on the hitting-step

Fig. 3. Illustration of hitting-step and adsorption-step. 䊉 represent a adsorption point.

1138

Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

concept, the total steps are always large enough that the above assumption can always be satisfied and the results are reliable. The fact that the results based on above two step concepts are equal shows that the above assumption is not needed and the expressions can be used in a short column. This may lead to an important use. When the property of columns varies with positions, for example, in gas chromatography the pressure, the linear velocity and the diffusion coefficient may be all different at different positions, the column can be divided into many short sections so that the property becomes uniform and the expressions of elution-curve moments can be used in each section, and the expressions for the whole column can be obtained by integration over all the short sections. 7. Conclusion

The authors are grateful to Prof. Yibei Fu and Prof. Ying Sun for their support. We also thank Doctor Haile Lei for his serious correction of the English of the manuscript. Appendix A. The step moments in the mobile phase based on the hitting-step concept From Eqs. (31), the moments per step in the mobile phase are obtained as follows: Ru

(A1)

v R

(A2)

v

0 M0,2 (m , m ) =

0 M2,0 (m , m ) =

0 (m , m ) = M1,2

0 (m , m ) = M0,3

0 M3,0 (m , m )

=

R3 4Dm v 2Dm R

v

0 (m , m ) = M1,3

0 M2,2 (m , m ) =

0 (m , m ) = M0,4

0 M4,0 (m , m ) =

R3 u 3Dm v

(A5)



R3 2v

1+

61 R2 u2 2 120 Dm

(A6)

17R5 u

(A7)

2v 96Dm

R5 2v 8Dm

2R3 u

(A8)

1+

v 17R5 u 16Dm v



59R2 u2

(A9)

2 320Dm

1364R2 u2

1+

(A10)

2 5355Dm

R7 u 3v 8Dm

R5 4Dm v

(A11)

1+

3167R2 u2

(A12)

2 4320Dm

11R7

(A13)

3v 128Dm

3Dm R3

v

1+

61R2 u2 2 60Dm

+

129737R4 u4



4 967680Dm

(A14)

In above equations only the terms of 1/v are retained. The results are identical to those of Ref. [13]. But note that the parameter ˇm = R2 /(2Dm ¯ m ) in Ref [13] is miswritten and should be corrected to ˇm = R2 /(Dm ¯ m ). For liquid-solid adsorption chromatography, generally only the highest power of Ru/Dm needs to be retained. And for gas-solid adsorption chromatography, Ru/Dm may be either greater or less than 1, so all the terms of Ru/Dm need to be retained. Appendix B. Derivation based on the adsorption-step concept In spite of the different definition of a step, the concentrations of solute molecules in mobile phases all obey a diffusion-drift equation like Eq. (22). But for different definitions the boundary condition is different. For the definition based on adsorption, the whole equation can be written as

2

⎧ ∂C(r, z, t) 1 ∂C(r, z, t) ∂C(r, z, t) ∂ C(r, z, t) ∂2 C(r, z, t) ⎪ + − w(r) = Dm + ⎪ 2 2 r ∂r ∂z ∂r ∂z ⎪ ⎨ ∂t

Acknowledgements

¯ m =

0 M2,1 (m , m ) =

0 (m , m ) = M3,1

In this paper a set of mass balance equations describing elutioncurve moments of multiple-site linear wall-adsorption capillary chromatography are successfully constructed. By solving these equations the exact moments per step then the exact elution-curve moments are obtained. It is natural that the resulting moments by this way contain various factors. When the mean desorption rate constant is very small, the terms contributed by desorption rate constants will become the dominate ones in the expressions of elution-curve moments, and the contributions of the other factors can be ignored. In this case, the expressions of moments are simplified so that it only contains the desorption rate constant, and agrees with that from the characteristic function method completely [12]. The peak symmetry of linear adsorption-chromatography is related to the desorption rate constant on sites closely. Roughly, the rate constant less than 10/s may lead to a serious asymmetry. The symmetry is also related to the adsorption fraction of the slow site, generally the asymmetry increases with the fraction, but not monotonously. At a proper low adsorption fraction, the asymmetry will reach maximum. The step moment method divides a chromatographic problem into three parts consisting of the general relationship of step moments and elution-curve moments, the step moments in stationary phase and the step moments in mobile phase and much simplifies the problem. The results from applying this method into the uniform and multiple-site stationary phases are self-consistent. In comparison with the other methods, the step-moment method is more universal and readily involves various factors in it.

¯ m =

0 (m , m ) = M1,1

(A3)

1+

11R2 u2 2 48Dm

(A4)

C(r, z, 0) = 0

C(r, 0, t) = 0 ⎪  ⎪ ⎪ ⎩ Dm ∂C(r, z, t)  ∂r

(B1) + ka C(R, z, t) = r=R

1 ı(z)ı(t) 2R



where ka = k is the total adsorption–reaction rate constant. j aj The condition above shows that one unit of solute molecules comes into the mobile phase at t = 0 and z = 0, and a diffusing current of molecules equal to adsorbed amount by the wall goes out of mobile phase for t > 0. Under this condition the total outside diffusing amount at z and t is equal to the probability density function of the traveling distance and residence time in a step: fm ,m (x, t) = 2Rka C(R, t)

(B2)

The method to solve Eq. (B1) is similar to that used to Eq. (22), the results are as follows: u ¯ m = (B3) ka ¯ m =

1 ka

(B4)

Y. Chen, Y. Tang / J. Chromatogr. A 1216 (2009) 1132–1139

0 M0,2 (m , m ) =

0 M2,0 (m , m ) =

0 M1,1 (m , m ) =

0 M2,1 (m , m ) =

0 (m , m ) = M1,2

0 (m , m ) M0,3

=

0 (m , m ) = M3,0

0 M3,1 (m , m ) =

2

+

ka2

R2 4Dm ka

(B5)

2Dm 2u2 11R2 u2 + 2 + ka 24Dm ka ka 2u

+

ka2

4Dm 6u

+

ka3 6

+

ka3

R2 6u2 9R2 u2 61 R4 u2 + 3 + + 2 2k 240 2ka 4Dm ka Dm ka a

3R2

+

ka2

36Dm u ka3

+

ka

11R2 u

+

ka2

763R4 u3

+

 pj j kdj

(B10) M30 (s ) = +

24u3 ka4

11R2 u3 4Dm ka2

+

+

59R4 u3 2k 160Dm a

(B11)

17R4 u 57R2 u3 + 16Dm ka 4Dm ka3

341R6 u3

+

2 k2 240Dm a

ka3

pj kdj e−kdj t

=

(B17)

1  kaj 1 k = kj = ka ka ka j kdj j

(B20)M20 (s ) =

2k 8Dm a

6u3



where pj represents the probability of a solute molecule being adsorbed exactly by the site of type j when it is adsorbed. The step moments in the stationary phase are obtained from Eq. (B17) as

(B9)

R4

2R2 u

fs (t) =

¯ s =

2k 96Dm a

+

2Dm ka2

(B8)

17R4 u

+

6Dm ka2

12Dm u

+

(B7)

11R2 u

Correspondingly, the probability density function of residence time on solid surfaces of multiple-site based on adsorption steps is as follows:

j

R2 u 3Dm ka +

ka2

(B6)

1139

(B12)

3k 1260Dm a

M40 (s ) =

 6pj j

3 kdj



2pj j k2 dj

=

 24pj j

4 kdj

=

6k ka k d2

=

24k ka k d3

2 ka



kaj j k2

dj

=

2k 1 ka k

(B19)



kj j kdj

=

2k



ka kd

Similarly, (B21)

(B22)

Substituting Eqs. (B3)–(B16), (B19)–(B22) into Eqs. (42) and (43) gives the central moments of Mn,m (, ), and again substituting Mn,m (␩,␶) into Eqs. (3)–(6), the moments of the elution curve are obtained and the expressions are identical to Eqs. (44)–(48) completely. References

0 M2,2 (m , m ) =

12Dm ka3

2 k2 720Dm a

24u ka4

+

+

ka2

1727R4 u2

+

0 M1,3 (m , m ) =

+

3R2

+

21R2 u 2Dm ka3

24u2 ka4

R4 49R2 u2 + + 4Dm ka 4Dm ka3

3167R6 u2

(B13)

3k 17280Dm a

+

29R4 u 2 k2 16Dm a

+

R6 u 3k 8Dm a

(B14)

[1] [2] [3] [4] [5] [6]

[7] [8]

0 M0,4 (m , m ) =

0 M4,0 (m , m ) =

24 ka4

+

9R2 Dm ka3

2 24Dm

ka2 +

+

+

11R4 2 k2 8Dm a

+

11R6 3k 128Dm a

(B15)

3Dm R2 72Dm u2 27R2 u2 24u4 + + + 3 2 ka ka ka4 ka

61R4 u2 33R2 u4 2021R4 u4 129737R6 u4 + + + 2 k2 3k 20Dm ka 480Dm 2Dm ka3 322560Dm a a

[9] [10] [11] [12] [13] [14] [15] [16]

(B16)

[17]

But there is a difference that here all terms are retained and in the solution of Eq. (22) only the terms of the first order of 1/v are retained.

[18] [19]

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