Lecture 4 Flashcards
(31 cards)
Agenda for this Lecture
This lecture will serve as the end of chapter 4. Today we will talk about 3 different systems: Liquid-Solid, Liquid-Gas, and Solid-Gas.
We will discuss each VERY briefly, and then we will show examples for them
What are the 3 different separation techniques for Solid-Liquid systems?
- Leaching
2 Crystallization
- Adsorption (main focus)
What is leaching?
In leaching (solid–liquid extraction), a multicomponent solid mixture is separated by contacting the solid with a solvent that selectively dissolves some solid species. Although this operation is similar to liquid–liquid extraction, leaching is a more difficult operation to simulate in that diffusion in
solids is very slow compared to diffusion in liquids, making it difficult to achieve equilibrium. Also, it is impossible to completely separate a solid phase from a liquid phase. A solids-free liquid phase can be obtained, but the solids will always be
accompanied by some liquid. In comparison, complete separation of two liquid phases is easy to achieve by settling vessels with draw-offs or by continuous centrifugation
What is crystallization?
Crystallization or precipitation of a component from a liquid mixture is an operation in which equilibrium can be
achieved, but a sharp phase separation is again impossible. A drying step is needed because crystals occlude liquid.
What is adsorption (again main focus)?
Adsorption, a third application of liquid-solid systems, employs porous solid particles that do not undergo phase or
composition change. Instead, they selectively adsorb liquid species on their exterior and interior surfaces. After a contact time sufficient to approach equilibrium, adsorbed species are desorbed and the solid adsorbent particles are regenerated. Ion exchange and chromatography are variations of adsorption.
Exactly what Micheal means about a porous solid particle is shown in written notes!
Before diving deeper into adsorption let us make some clear definitions of what is what because the terms confuse me
- The solid particle is called the adsorbENT
-The adsorbed components are called
solutes when in the liquid and adsorbATEs when adsorbed.
What are some features (3yani usual conditions) in a solid-liquid adsorption?
- Solid adsorbents are essentially
insoluble in the liquid. - Other component(s) of the solution (The liquid phase) are the solvent and carrier, which are assumed not to adsorb.
What happens when a liquid contacts porous solid particles?
Adsorption takes place on the external and internal solid surfaces until equilibrium is reached.
What are the adsorption isotherms
When it comes to adsorption equilibrium curves there is no theory based on the molecular properties of the solute and solid that is universally embraced. Laboratory measurements are necessary to provide data for plotting isothermal equilibrium curves, called adsorption isotherms. These play a very important role in understanding adsorption systems! We will now further talk about them
What is the general x-axis and y-axis of an adsorption isotherms
weird question yes but the units can be tricky so better to make it clear.
The y-axis of the adsorption isotherm is the adsorbate q*. The units are mmole of the species (liquid) adsorped per gram os the adsorbent (the solid)
The x-axis of the adsorption isotherm is the equilibrium concentration c. The unites are millimoles of the SOLUTE per liter of aqueous solution.
What is an example of an adsorption isotherm?
Look at digital notes. As seen (IMP) Higher concentrations of solute in the solution result in higher adsorbate concentrations (duh). The Adsorbate increases rapidly
at first and increases slowly as saturation is approached.
What does the extent of adsorption depend on?
depends on the adsorption process
used and the absorptivity of the solid. As the microporous structure of activated Solid provides a high internal surface area per unit mass of the solid, and therefore a high capacity for adsorption
What information can we get out of isotherms?
Adsorption isotherms are used to determine the amount of adsorbent required to selectively remove solute from a liquid.
Derivation:
Consider the ideal, single-stage batch adsorption process shown in digital notes, where solid adsorbent C and a liquid
mixture of carrier A and solute B are charged to a vessel and brought to equilibrium. Let: c(F)B = concentration of solute in the feed; cB = concentration of solute in the product liquid;q∗B = equilibrium concentration of adsorbate on adsorbent; Q = volume of liquid (assumed to remain constant during adsorption); and S = mass of adsorbent particles on a solute free basis. The rest of the derivation are in written notes.
An example of how we can determine the equilibrium point using the intersection of the material balance?
Look at written notes
What happens in liquid adsorption when more adsorbent is used (More S) (EXAM Q)?
From the material balance, we see a Lower intercept with the y-axis, and the slope becomes less negative. This leads to increased Percent adsorption (shown in digital notes)
What would happen in liquid adsorption when the initial solute concentration in the
liquid mixture would be higher (More c(F)B) (EXAM Q)?
Higher intercept with y-axis, slope would remain the same. Due to the curvature of the isotherm Percent adsorption decreases, again shown in the digital notes
Break moment
Now we will make the shift to gas-liquid systems
What are the different gas-liquid systems?
Only one was mentioned which is aBsorption (not aDsorption as discussed before)
How is a gas-liquid system any different from the pre-discussed vapour-liquid systems?
In the vapour–liquid systems described previously, components in the vapour phase were all condensable at the system
temperature. At near ambient pressure, their K-values are determined by Raoult’s law. The K-value equation requires a value of the vapour pressure of the component. If a component in the vapour has a critical temperature below the system temperature (meaning that the vapour has surpassed its critical point), it is non-condensable, its vapour pressure does not exist, and Raoults law does not apply.
If a vapour includes one or more non-condensable components, it is commonly called a gas.
Note that a vapour past its critical point can still be changed into a solid as we know from the very similar diagram shown in digital notes.
Now you might ask yourself, ‘hmmmm the gas is non-condensable, then how tf is the only gas-liquid system an absorption one? How can we even do anything if Raoutl’s law doesn’t apply? hmmmmm’
Non-condensable components can dissolve, to some extent, into a liquid phase
containing other components.
Equilibrium K-values of noncondensable components, at near ambient pressure (0.1-10 atm), the following version of
Henry’s law is applicable:
Ki = yi/ xi = Hi / P
where Hi = Henry’s law constant for non-condensable component i, which depends mainly on the temperature and
liquid-phase composition. Hi has the same units as P.
Using an example, how can we determine the composition of a species in the gas and liquid phase, WHEN Henry’s law is applicable?
Look at written notes
When is Henry’s law not applicable? And how can we determine the composition of liquid gas systems under this constraint?
Henry’s law is not applicable to
gases at high pressure or for non-condensable components with a high solubility in the liquid phase, e.g., ammonia
in water. Then, experimental data from systems T and Pm are needed. Y3ani via experimental data like the mole faction of the species in the gas plotted against the mole fraction of the species in liquid. This is shown in written notes.
Break moment
Now we move into solid-gas systems
What are the different gas-liquid systems?
- Desublimation
- Gas aDsprbtion
