# Tag Info

12

There are two common arguments presented as for why $\Delta H < 0$: Argument (1): Well, indeed I see nothing wrong with the argument presented by the textbook. If adsorption takes place spontaneously, then one can conclude that the change in Gibbs free energy of the process is indeed negative. Since, the entropy change associated with process is ...

11

Activated charcoal is a great adsorbent because of it's huge surface area. While it doesn't bind very many ions/atoms/molecules per surface area (which is the characteristic of a 'good' adsorbent), due to very big surface area per unit of mass it can adsorb a lot of particles. Actually, process of 'activating' charcoal is designed to maximize the surface ...

11

The second law of thermodynamics states that the entropy of the universe always increases. $$\mathrm{d}S > 0$$ In the case of adsorption the entropy of the system; the gas being adsorbed; decreases but the entropy of the surroundings;the rest of the gas and the surface (and everything else in the universe); increases and this outweighs the decrease in ...

11

Critical temperature is kind of measure of the strength of intermolecular van der Waals force. Comparison of actual and critical temperature is kind of comparison of average kinetic energy of molecules with energy needed to break the intermolecular bonds. So temperature itself does not say anything about the bonding strength. It only indirectly determine ...

9

OK, let's step through your questions one at a time. First off, what do they mean by steady-state number here? Intermediates formed in reactions usually exist in low concentration. The steady-state assumption assumes that the concentration of this (low-concentration) intermediate will not change appreciably over the course of the reaction once steady ...

8

First, a note on the Bell-Evans-Polanyi (BEP) equation: The form you've written it in disguises what the constant $\alpha$ tells you. From the wikipedia article, it's said that $\alpha$ is a number between 0 and 1 which tells you how close the transition state is to the reactant state. Thus, this constant actually tells you quite a lot about the shape of ...

8

The IUPAC defintion isn't more limited than the ones you've read. Instead, it is more general. Your confusion results from not being familiar with the term "condensed phase". You think a condensed phase is a 'more condensed liquid', and consequently think IUPAC is referring to a surface formed between a liquid and a 'more condensed liquid': "...

7

Charcoal isn't a particularly good adsorbent even though it is chemically very similar to activated carbon. Activated carbon is usually made by more specialised processes that guarantee the final product will have a very large surface area (often >1,000m2/g). Manufacture usually involves pyrolysation with hot gasses, but many forms are also further ...

6

Without knowing exactly what you were told its hard to be exact, but here is a description of the situation. The Langmuir model makes assumptions (a) adsorption is complete when the surface is filled with one gas molecule per site, (b) all adsorption sites are equivalent (i.e. the same) and (c) adsorption and desorption are separate processes, i.e. one ...

6

At hydrous equilibrium will the "stronger" desiccant contain virtually all of the water? Or is the equilibrium distribution of water a function of the desiccants' relative "hygroscopy," rates of sorption, or some other factor(s)? "Rates" aren't relevant for questions about equilibrium. I am trying to imagine a series of tests in which two different ...

6

When burning a blunt, joint, doobie or whatever, THC evaporates (bp of tetrahydrocannabinoles is around 155 °C). Depending on the length of the "object" , related to the temperature gradient between the front and the mouth piece, and the type of the "filling" and its adsorptive properties a part of the vaporized material might condense and/or get absorbed ...

6

I see three reasons: Activated carbon is commonly used to adsorb organics, that should bind well on coal. For many purposes, activated carbon is treated with potassium or iodine to provide ions for charged centers. Charring organic stuff is a cheap and easy way to create large surfaces for adsorption.

6

As acknowledged in your question, there are a large number of factors at play here. The most basic of these is that there are three fundamental mechanisms by which freezing of a droplet can initiate when in contact with a superhydrophobic surface: 1) heterogeneous nucleation where an ice freezing nucleus (IFN) is present in the water droplet, 2) ...

5

Cyclodextrin hydroxyls are the external surface (hydrophilic and hydrogen-bonding). They circle the larger open end. The carbon and hindered ether skeleton is internal (hydrophobic). A cyclodextrin looks like a bonded hollow detergent micelle in water. Before you believe advertising hype, remember that underivatized cyclodextrins are nephrotoxic, ...

5

Brunauer–Emmett–Teller (BET) explains the physical adsorption of gas molecules on a solid surface, and doesn't address chemisorption. Even the wikipedia link that you provided, makes that quite explicit. However, you are correct to think that it is in a extension of the Langmuir adsorption model, which only accounts for monolayers. The assumption that ...

5

Never forget to interpret in terms of physical/chemical forces, by order of energy: van der Waals, hydrophobic (pseudo forces), hydrogen bonds, disulfide bridges, covalent, electrostatic (for the most frequent). Indeed, in charcoal, there is a large number of $\pi$-electrons that can make bonds, in particular. And yes, the porosity is important. But ...

5

The ratio $$\frac{x}{m}= k P ^{1/n}$$ can be re-expressed in terms of the area $A$, surface-to-volume ratio $\lambda$, and density $\rho$ (note that $m=A\rho/\lambda)$: $$\frac{x}{A}= k' P ^{1/n}$$ where $$k'=k\rho/\lambda$$ The advantage of using $k$ as opposed to $k'$ is that you don't need to determine $\lambda$ to know how much (as in, what mass) ...

4

Charcoal (although activated carbon, a specially processed microporous charcoal product is better) is a material that is mostly carbon. The black allotrope of carbon is graphite, which is made of overlapping planar sheets of graphene. Graphite is very nonpolar (since it has only carbon atoms). Thus, nonpolar substances will adsorb. Most organic substances ...

4

Charcoal is a good absorbent because its quite selective and cheap. Its high surface area means nothing if it doesn't bond to the components that you want absorbed and not a lot else- witness a household sponge which has high surface area and mercury- the mercury just slides off. Charcoal is used in the gold mining/processing industry. The gold(and other ...

4

The author of the article has named the phenomenon incorrectly. Adsorption can indeed result in a lower volume than might be expected, for example when a gas interacts with a high surface area solid. In the case of hydrogen storage, for example, the adsorbed phase is dense enough at low temperatures that much more hydrogen can be stored in a container filled ...

4

Yes, it is meaningful, but often ignored for practical reasons. In the lab, e.g. in a desiccator, you would just use a large excess of the desiccant of choice which would always work. It’s often more meaningful to classify them as to whether they are acidic, basic or neutral. But let’s assume we wanted to create that scale. For simplicity reasons, let’s ...

4

For langmuir isotherm: Let say we are using linear langmuir isotherm. The formula is : $\frac{1}{q_e} = \frac{1}{q_m} + \frac{1}{q_mKL}$ find the value of $q_e$. Using the formula $q_e = \frac{(C_i-C_f)V}{\text{mass}}$ of adsorbent, and then plot the graph $\frac{1}{q_e}$ versus $\frac{1}{C_f}$. It would be easier by using excel. From the graph, we can ...

4

Why is this strange? Consider: $$\ce{2A <=>[k_1][k_{-1}] A_2}$$ Forward rate is second order: $k_1[\ce{A}]^{2}$. Reverse rate is first order $k_{-1}[\ce{A2}]$.

4

Short Answer The following equilibrium roughly represents what goes on during physisorption of a gases onto a a surface. $$\ce{Gas + Surface <=>[adsorption][desorption] Gas--Surface}$$ The effect of pressure can be understood on the basis of Le Chatelier's principle, which states that if a stress (such as a change in pressure) is applied to a ...

4

Well, it usually depends on the temperature: if temperature is too low, molecules cannot move so freely so they are localized on top of the solid without much movement. However, in general cases the adsorbed gas behaves like a liquid in two dimensions. Near the surface, the gas molecules are 'trapped' in the force field of the solid, so they don't escape so ...

4

Your problem is not the unit conversion, but the units you have used. The adsorption cross-section of $\ce{N2}$ ($s$) has been found using BET method is $\pu{0.162 nm^2}$ (Langmuir). Yet, you have used $\pu{0.162 nm^3}$, more than a factor of $1 \times 10^9~\mathrm{nm/m}$.

4

One version of Freundlich adsorption isotherm equation is: $$\frac xm = Kp^{\frac12},$$ which can also be written as: $$\log \frac xm = \log K + \frac12\log p$$ This is a straight line equation of type $( y = c + mx)$ as given in the question. However, I think the question has made a mistake saying the slope is $2$, but instead it should be $\frac12$. ...

3

On colliding with a surface the molecules do not necessarily bounce off because, depending on the collision energy, Van-der-Waals forces can act to hold the gas molecule close to the surface. These are typically London dispersion forces but depending on type of surface and molecule, any type of VDW force could occur. This process is called physisorption and ...

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