Question

Equimolar counter-diffusion is occurring at steady state in a tube 0.11 m long containing N2 and CO gases at a total pressure of 1.0 atm absolute. The partial pressure of N2 is 80 mm Hg at one end and 10 mm Hg at the other end. D_AB = 2.05 times 10^5 m^2/s.

a. Calculate the flux in kg mol/s.m^2 at 298 K for N2.
b. Repeat part a, at 473 K. Does the flux increase?
c. Repeat part a, at 298 K, but for a total pressure of 3.0 atm abs. The partial pressures of N2 remain 80 and 10 mm Hg. Does the flux change?
d. Calculate the CO flux for part c.

Answer 1

a. Therefore, the flux in kg mol/s.m² at N₂ =
`35.398 *10^(-29)kgmol/sm^2`

b. Therefore; when temperature = 473k , the flux (J) decreases.

c. Hence, when T = 298K, but the total pressure = 3.0 atm , the flux increases.

d. The CO flux for part C =
`212.6*10^(-29)kgmol/sm^2`

Step-by-step explanation:

Given that :

Equimolar counter-diffusion
`(d_x)` = 0.11 m

`D_(AB) =`
`2.05 *10^(-5)m^2s^(-1)`

T = 298 K

For ideal gas equation:

PV = nRT

Making V the subject of the formula:

`V = (nRT)/(P)`

where the number of moles (n) =
`(mass)/(molarmass)`

∴ the V =
`(mass of N_2)/(Molar mass of N_2) *(RT)/(P)`

V =
`(14 gmol^(-1))/(28 gmol^(-1)) *(8.314KPadm^3K^(-1)mol^(-1))/(101325Pa)`

V =
`(0.5*2477.572KPadm^3mol^(-1))/(101325Pa)`

V =
`0.01223Kdm^(3)mol^(-1)`

V =
`12.23 dm^3mol^(-1)`

V =
`(12.23dm^3mol^(-1))/(6.023*10^(23)mol^(-1))`

V =
`2.031*10^(23)dm^3`

V =
`2.031*10^(-23)*10^(-3)m^3`

V =
`2.031*10^(-26)m^3`

The volume of N₂ =
`2.031*10^(-26)m^3`

Density of
`P_(N_2) = (mass of N_2)/(Volume)`

=
`((14gmol^(-1))/(6.023*10^(23)mol^(-1)) )/(2.031*10^(-26)m^3)`

`P_(N_2) = (2.324*10^(-23))/(2.031*10^(-26)m^3)`

=
`1.44*10^3gm^(-3)`

`P_(N_2) = 1144 gm^(-3)`

Now, The flux (J) =
`D_(AB)*(P_(N_2))/(d_x)`

J =
`(2.05*10^(-5)m^2s^(-1)*1144gm^(-3))/(0.11m)`

J =
`21320*10^(-5)g/sm^2`

J =
`21320*10^(-8)kg/sm^2`

J =
`3539.8 *10^(-31)kgmol/sm^2`

J =
`35.398 *10^(-29)kgmol/sm^2`

Therefore, the flux in kg mol/s.m² at N₂ =
`35.398 *10^(-29)kgmol/sm^2`

b. At T = 473 K

`V_(N_2)= (0.5*8.314kPaK^(-1)mol^(-1)*473)/(101325Pa)`

`V_(N_2) =19.41 dm^3mol^(-1)`

`V_(N_2) = 19.41 *10^(-3)m^3mol^(-1)`

`P_(N_2)= (14gmol^(-1))/(19.41*10^(-3)m^3mol^(-1))`

`P_(N_2)= 0.7123*10^3gm^(-3)`

J =
`(2.05*10^(-5)m^2s^(-1)*0.7213*10^3gm^(-3))/(0.11m)`

J =
`13.44 *10^(-2) g/sm^2`

J =
`2.232*10^(-25)gmol/sm^2`

J =
`2.232*10^(-28)kgmol/sm^2`

J =
`22.32 *10^(-29) kgmol/sm^2`

Therefore; when temperature = 473k , the flux (J) decreases.

c. At P = 3 atm = 3×101325 Pa

T = 298 K

`V_(N_2) = (0.5 *8.314KPa K^(-1)mol^(-1)*298K)/(3*101325Pa)`

`V_(N_2) = 4.08dm^3mol^(-1)`

`V_(N_2) = 4.08*10^(-3)m^3mol^(-1)`

`P_(N_2) = (14g/mol)/(4.02*10^(-3)m^3mol^(-1))`

`P_(N_2) =3.43 *10^3gm^(-3)`

J =
`(2.05*10^(-5)m^2s^(-1)*3.43*10^3gm^(-3))/(0.11m)`

J =
`63.92*10^(-2)g/sm^2`

J =
`63.92*10^(-5)kg/sm^2`

To moles; we have:

J =
`10.61*10^(-28)kgmol/sm^2`

J =
`106.1 *10^(-29)kgmol/sm^2`

Hence, when T = 298K, but the total pressure = 3.0 atm , the flux increases.

d. We can determine the CO flux for part c as follows:

`P_(CO) = (mass of CO)/(Volume)`

`P_(CO) =`
`(28.01g/mol)/(4.08*10^(-3)m^3mol^(-1))`

`P_(CO) =`
`6.87*10^3gm^(-3)`

J =
`\frac{2.05*10^(-5)m^2s^-{1}*6.87*10^3gm^(-3)} {0.11m}`

J =
`128.03*10^(-2)g/sm^2`

J =
`128.03*10^(-5)kg/sm^2`

J =
`212.6*10^(-29)kgmol/sm^2`

The CO flux for part C =
`212.6*10^(-29)kgmol/sm^2`