How Does Langmuir Kinetics Apply to Adsorption Processes?

[Zealduke]

Homework Statement

Adsorption of many solutes is governed by Langmuir kinetics given below.

Equation 1
dq/dt=k_a (q_m−q )c−k_dq

If a batch of adsorbent pellets of mass M is added to a solution of adsorbate, at an initial concentration ofc0 and volume V, the mass balance yields:

Equation 2
c= c_o−(M/V)q

Substituting (2) into (1) gives:

Equation 3
dq/dt = aq² + bq + f

a = k_a (M/V)

b = -(k_a(M/V)q_m + k_d + k_ac_o)

f = k_aq_m

Subject to the initial condition that q (t=0 )=0 Solve Equation (3) by simple quadrature (integration). To do this, first separate the variables q and c and integrate analytically.

The long time solution of Equation (1) should give the Langmuir isotherm.
q*= (qm(k_a/k_d)c) / ( 1 + ( k_a/k_d)c )

The Attempt at a Solution

So after plugging in all the data, to isolate q and c I subtracted k_aq_m and divided by q yielding:

(-k_aq_m / q) = (k_a(M/V)q) - k_a(M/V)q_m - k_d - k_ac_o

Then I added k_d to both sides in the hopes to simplify some of the data into the variable c. Not really sure what to do next or if I'm even headed in the right direction. Any thoughts or help would be really appreciated

 

[Goddar]

Hi.
Seems confusing indeed, since apparently c and q are functions of each other... are you sure you're not asked to separate q and t? because they appear to be the only variables and then equation 3 is integrated easily:
dq/(aq^2 + bq + f) = dt,
then complete the square on the left and change variable, integrate both sides. You can plug back the constants at the end if you like...

 

[Zealduke]

I appreciate the help, I'm going to run with what you've presented since my professor really doesn't have it together

So as far as completing the square I've got it to (q+(bq/4a))^2 = -f (bq/4a)^2 and it looks like it's about to get ugly, so I feel I'm on the wrong track. Does it look right so far? It's been ages since I've done any calculus, so my next question would be with what do I change the variable with?

 

[Goddar]

If you factor out the "a" and take it to the right, you get:
dq/[(q + b/2a)^2 – (b/2a)^2 + f/a] = a•dt
So with x = q + b/2a, dx = dq and g^2 = f/a – (b/2a)^2
your equation has the form: dx/(x^2 + g^2) = a•dt,
which you might find familiar. if you don't, just look up the integral (the right one is trivial of course), it's a very common identity. You may even look up the initial one, i don't know if you're supposed to carry the whole integration...

 

[paranormal]

.
Hi there,

Thank you for sharing your attempt at solving this Langmuir kinetics problem. It seems like you have made some progress in isolating the variables and trying to simplify the equation. However, there are a few errors in your steps that I would like to point out.

Firstly, when you subtracted kaqm and divided by q, you should also divide the other terms by q as well. This will give you:

(-kaqm / q) = (ka(M/V)q / q) - (ka(M/V)qm / q) - (kd /q) - (kaco / q)

Simplifying this further, we get:

(-kaqm / q) = ka(M/V) - ka(M/V)qm - (kd /q) - (kaco / q)

Next, you added kd to both sides, which is not necessary. Instead, you can rearrange the terms to get:

ka(M/V)qm = (-kaqm / q) - (kd /q) - (kaco / q)

Now, we can substitute this into the original equation (3) to get:

dq/dt = (-kaqm / q) - (kd /q) - (kaco / q) + kaq + f

Or, in a simplified form:

dq/dt = (-kaqm / q) + (ka - kd)q + f

From here, you can integrate both sides to get:

∫ dq/q = ∫ (-kaqm / q)dt + ∫ (ka - kd)qdt + ∫ fdt

Solving these integrals will give you the long time solution for q, which is the Langmuir isotherm. I hope this helps guide you in the right direction. Good luck!