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Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.
- a)0.67
- b)1.67
- c)2.67
- d)None of the mentioned
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
Find the change in concentration for a steady state equimolar counter ...
Explanation: N flux of A = D(AB)/z * (Concentration difference)
Concentration difference = 5/3 =1.67.
Concentration difference = 5/3 =1.67.
Most Upvoted Answer
Find the change in concentration for a steady state equimolar counter ...
Change in Concentration for Steady State Equimolar Counter Diffusion
Given:
Diffusion coefficient, D(AB) = 6 sq.m/sec
Change in distance, Δx = 2 m
Flux of A, NA = 5 mol/sq.m sec
To find the change in concentration for steady-state equimolar counter diffusion, we can use Fick's first law of diffusion, which states that the molar flux of a component is proportional to the concentration gradient of that component.
The equation for Fick's first law of diffusion is:
NA = -D(AB) * (dCA/dx)
Where:
NA is the molar flux of component A
D(AB) is the diffusion coefficient of component A in component B
dCA/dx is the concentration gradient of component A
We can rearrange the equation to solve for the concentration gradient:
(dCA/dx) = -NA / D(AB)
Given that the molar flux of A, NA, is 5 mol/sq.m sec and the diffusion coefficient D(AB) is 6 sq.m/sec, we can substitute these values into the equation to find the concentration gradient:
(dCA/dx) = -(5 mol/sq.m sec) / (6 sq.m/sec)
Simplifying the equation gives:
(dCA/dx) = -0.83 mol/m^3
The negative sign indicates that the concentration gradient is in the opposite direction of the flux.
To find the change in concentration, we can multiply the concentration gradient by the change in distance:
ΔCA = (dCA/dx) * Δx
Substituting the values:
ΔCA = (-0.83 mol/m^3) * (2 m)
Simplifying the equation gives:
ΔCA = -1.67 mol
The negative sign indicates that the concentration of component A has decreased by 1.67 mol.
Therefore, the change in concentration for a steady state equimolar counter diffusion is 1.67 mol.
Hence, option B (1.67) is the correct answer.
Given:
Diffusion coefficient, D(AB) = 6 sq.m/sec
Change in distance, Δx = 2 m
Flux of A, NA = 5 mol/sq.m sec
To find the change in concentration for steady-state equimolar counter diffusion, we can use Fick's first law of diffusion, which states that the molar flux of a component is proportional to the concentration gradient of that component.
The equation for Fick's first law of diffusion is:
NA = -D(AB) * (dCA/dx)
Where:
NA is the molar flux of component A
D(AB) is the diffusion coefficient of component A in component B
dCA/dx is the concentration gradient of component A
We can rearrange the equation to solve for the concentration gradient:
(dCA/dx) = -NA / D(AB)
Given that the molar flux of A, NA, is 5 mol/sq.m sec and the diffusion coefficient D(AB) is 6 sq.m/sec, we can substitute these values into the equation to find the concentration gradient:
(dCA/dx) = -(5 mol/sq.m sec) / (6 sq.m/sec)
Simplifying the equation gives:
(dCA/dx) = -0.83 mol/m^3
The negative sign indicates that the concentration gradient is in the opposite direction of the flux.
To find the change in concentration, we can multiply the concentration gradient by the change in distance:
ΔCA = (dCA/dx) * Δx
Substituting the values:
ΔCA = (-0.83 mol/m^3) * (2 m)
Simplifying the equation gives:
ΔCA = -1.67 mol
The negative sign indicates that the concentration of component A has decreased by 1.67 mol.
Therefore, the change in concentration for a steady state equimolar counter diffusion is 1.67 mol.
Hence, option B (1.67) is the correct answer.
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Find the change in concentration for a steady state equimolar counter diffusion if D(AB)= 6 sq.m/sec, the change in distance is 2 m and the N flux of A is 5 mol/sq.m sec.a)0.67b)1.67c)2.67d)None of the mentionedCorrect answer is option 'B'. Can you explain this answer?
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