Chemistry, 10.03.2020 03:11 devinmoore8686

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.

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Answer from: eweqwoewoji

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 = $212.6*10^{-29}kgmol/sm^2$

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:

$V = \frac{nRT}{P}$

where the number of moles (n) = $\frac{mass}{molarmass}$

∴ the V = $\frac{mass of N_2}{Molar mass of N_2} *\frac{RT}{P}$

V = $\frac{14 gmol^{-1}}{28 gmol^{-1}} *\frac{8.314KPadm^3K^{-1}mol^{-1}}{101325Pa}$

V = $\frac{0.5*2477.572KPadm^3mol^{-1}}{101325Pa}$

V = $0.01223Kdm^{3}mol^{-1}$

V = $12.23 dm^3mol^{-1}$

V = $\frac{12.23dm^3mol^{-1}}{6.023*10^{23}mol^{-1}}$

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 $P_{N_2} = \frac{mass of N_2}{Volume}$

= $\frac{\frac{14gmol^{-1}}{6.023*10^{23}mol^{-1}} }{2.031*10^{-26}m^3}$

$P_{N_2} = \frac{2.324*10^{-23}}{2.031*10^{-26}m^3}$

= $1.44*10^3gm^{-3}$

$P_{N_2} = 1144 gm^{-3}$

Now, The flux (J) = $D_{AB}*\frac{P_{N_2}}{d_x}$

J = $\frac{2.05*10^{-5}m^2s^{-1}*1144gm^{-3}}{0.11m}$

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

$V_{N_2}= \frac{0.5*8.314kPaK^{-1}mol^{-1}*473}{101325Pa}$

$V_{N_2} =19.41 dm^3mol^{-1}$

$V_{N_2} = 19.41 *10^{-3}m^3mol^{-1}$

$P_{N_2}= \frac{14gmol^{-1}}{19.41*10^{-3}m^3mol^{-1}}$

$P_{N_2}= 0.7123*10^3gm^{-3}$

J = $\frac{2.05*10^{-5}m^2s^{-1}*0.7213*10^3gm^{-3}}{0.11m}$

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

$V_{N_2} = \frac{0.5 *8.314KPa K^{-1}mol^{-1}*298K}{3*101325Pa}$

$V_{N_2} = 4.08dm^3mol^{-1}$

$V_{N_2} = 4.08*10^{-3}m^3mol^{-1}$

$P_{N_2} = \frac{14g/mol}{4.02*10^{-3}m^3mol^{-1}}$

$P_{N_2} =3.43 *10^3gm^{-3}$

J = $\frac{2.05*10^{-5}m^2s^{-1}*3.43*10^3gm^{-3}}{0.11m}$

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:

$P_{CO} = \frac{mass of CO}{Volume}$

$P_{CO} =$ $\frac{28.01g/mol}{4.08*10^{-3}m^3mol^{-1}}$

$P_{CO} =$ $6.87*10^3gm^{-3}$

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 = $212.6*10^{-29}kgmol/sm^2$

Answer from: Quest

6.4 cm (max) - 25 cm (min)

Answer from: Quest

definition

entropy is a measure of the randomness or the extent of disorder of a chemical process.

enthalpy is a measure of the heat change of a reaction occurring at a constant pressure.

measurement units:

entropy is measured in joule per kelvin (jk⁻¹) whereas enthalpy is measured in joule per mole (jmol⁻¹)

requirements:

entropy has no requirements or limits, and its change is measured by the division between the heat change of the chemical process and the temperature.

in contrast, enthalpy can only be related to the heat change of the reaction under standard conditions.

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Equimolar counter-diffusion is occurring at steady state in a tube 0.11 m long containing N2 and CO...

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