Question

1) An open tubular GC column has an inner diameter of 300 micrometer and it is coated inside with a layer of stationary phase that is 2.0 micrometer in thickness. On this column, a derivatized form of adenosine has a retention time of 300 sec, and un-retained ethane has a sharp spike at 32 sec. Please calculate the partition coefficient for the derivatized adenosine.

Answer: 3.1*10²

2)  For the same separation as described in the last question, please calculate the number of theoretical plates for the separation of the derivatized adenosine if the column length is 50 m and the peak width for the derivatized adenosine is 2.5 sec

Answer: 2.3 x 10⁵

3) . For the same separation as described in the last question, please calculate the plate height for the separation of the derivatized adenosine.

Answer: 2.2 x 10^-4 M

I wonder how they got. Can you show me step by step?

Answer 1

1)

we know that

rentention factor is given by

k = ( tr- tm / tm)

so

k = ( 300-32) / 32

k = 8.375

now

b = r / 2d

where

r = column radius

d = thickness

given

r = 150 x 10-6 m

d = 2 x 10-6 m

so

b = 150 / 2 x 2

b = 37.5

now

partition coefficient = k x b = 8.375 x 37.5 = 314

so

the partition coefficient is 3.1 x 10^2

2)

we know that

N = 16 (tr/W)^2

where

N = number of theoretical plates

tr = retention time

W = peak width

given

tr = 300 sec

W = 2.5 sec

so

using given values

we get

N = 16 ( 300 / 2.5)^2

N = 230400

N = 2.304 x 10^5

so

number of theoretical plates are 2.304 x 10^5

3)

we know that

H= L / N

where

L = length of column

given

L = 50 m

so

H = 50 / 2.304 x 10^5

H = 2.17 x 10-4 m

so

the plate height is 2.2 x 10-4 m