How To Use Red Mud As A Catalyst

Can chemists come up with better uses of mineral resources to make catalysts that are more sustainable? For a growing number of researchers, the answer is yes, and the key is taking advantage of materials that are already out of the ground. Red mud, the noxious by-product of the Bayer process for extracting aluminum from bauxite ore, makes a good case study.

The majority of material processed in mining operations ultimately goes to waste. For every ton of alumina extracted from bauxite, more than a ton of red mud is produced; aluminum mining leaves behind some 120 million metric tons per year of the salty, highly alkaline, heavy-metal-laden material, according to the International Aluminum Institute. Some 4 billion metric tons of the  material is lying about globally, much of it held in retention ponds.

Mining companies have long tried to find ways to recycle the environmentally problematic red mud. It is a classic problem in search of a solution. One approach is neutralizing red mud with seawater or treating it with CO2 or sulfur compounds. The modified materials have been tried as fill for mining and construction, as pigment and filler for bricks and cement, and as a sorbent for water treatment. Others have looked at extracting more aluminum from red mud, or obtaining other useful metals such as sodium, copper, and nickel. But so far there have been few safe and economical large-scale applications.

On a new front, some chemists are trying to go catalytic, focusing on iron oxide, the chief component of red mud. But given the purity and properties of red mud, researchers have found it typically is not an active enough catalyst to compete against existing commercial catalysts. That’s because the mineral composition, particle size, and surface properties are important in developing heterogeneous catalysts. With red mud, finding the right combination is a work in progress.

One early sign of success comes from Foster A. Agblevor of Utah State University’s USTA Bioenergy Center and coworkers in conjunction with Pacific Northwest National Laboratory researchers. They have been testing red mud as a bulk catalyst to replace zeolites in a fluidized-bed reactor to pyrolyze biomass to make crude oil.

The team processes the biocrude oil using a traditional catalytic hydrotreating process to make a gasoline- type fuel and has tested it on a lawn mower or lawn trimmer. “We are able to run an engine on the fuel without difficulty,” Agblevor says. The Utah State researchers have applied for a patent for their process. They are working with catalyst company Nexceris to scale up catalyst production and with Wildland Forestry & Environmental to harvest wood from pinyon-juniper range lands in the western U.S. to scale up biofuel production.

The team is also expanding the scope of using red mud beyond biomass pyrolysis, Agblevor says. The researchers have applied the catalyst to coal gasification, he notes, as well as to a process for catalytic pyrolysis of waste tires for fuel production. Despite raw red mud’s ultimate utility as a catalyst, its story points to other possibilities for recovering metals that have already been extracted and used. For example, industrial processing, the use of consumer goods and medicines, and even the wearing away of jewelry leads to measurable amounts of catalyst metals such as gold, silver, and platinum accumulating at wastewater treatment plants.

Weak Base – Strong Acid Titration Curves

The titration of a weak base (NH3) with a strong acid (HCl) is shown below. Note that the curve has the same shape as the weak acid-strong base curve, but it is inverted. Thus, the regions of the curve have the same features, but the pH decreases throughout the process:

Curve for a weak base-strong acid titration. Titrating 40.00 mL of 0.1000 M NH3 with a solution of 0.1000 M HCl leads to a curve whose shape is the same as that of the weak acid-strong base curve,

but inverted. The midpoint of the buffer region occurs when [NH3] = [NH4+].

Methyl red is a suitable indicator here.

1.  The initial solution is that of a weak base, so the pH starts out above 7.00.

2.  The pH decreases gradually in the buffer region, where significant amounts of base (NH3) and conjugate acid (NH4+) are present. At the midpoint of the buffer region, the pH equals the pKa of the ammonium ion.

3.  After the buffer region, the curve drops vertically to the equivalence point, at which all the NH3 has reacted and the solution contains only NH4+ and Cl-. Note that the pH at the equivalence point is below 7.00 because Cl- does not react with water and NH4+ is acidic:

                                 NH4+(aq) + H2O(l) ⇌  NH3(aq) + H3O+(aq)

4. Beyond the equivalence point, the pH decreases slowly as excess H3O+ is added.

For this titration also, we must be more careful in choosing the indicator than for a strong acid-strong base titration. Phenolphthalein changes colour too soon and too slowly to indicate the equivalence point; but methyl red lies on the steep portion of the curve and straddles the equivalence point, so it is a perfect choice.

Why Warm Water Freezes Faster Than Cold

Nearly 50 years ago, Erasto B. Mpemba and Denis G. Osborne reported that if samples of water at 90 °C and 25 °C are cooled, the one starting at 90 °C begins freezing first. Many explanations for the “Mpemba effect” have been proposed, including ones based on evaporation, temperature gradients, impurities, and dissolved gases.

In warm water, weak hydrogen bonds break (top, red squiggles), leaving fragments

that easily reorganize into an ice lattice (bottom), a new study says.

A new computational study suggests that the effect arises from the liquid’s hydrogen bond network (J. Chem. Theory Comput. 2016, DOI: 10.1021/acs.jctc.6b00735). Southern Methodist University’s Dieter Cremer and colleagues investigated clusters of 50 and 1,000 water molecules, characterizing the types and strengths of the clusters’ 350 and more than 1 million hydrogen bonds, respectively. In (H2O)1,000 , raising the temperature from 10 °C to 90 °C led to fewer hydrogen bonds, as weaker, predominately electrostatic bonds broke.

That left behind cluster fragments with strong hydrogen bonds with more covalent character and proportionately more “dangling” or terminal hydrogen bonds. That hydrogen bond combination enables the fragments to easily reorganize and form the hexagonal lattice of ice.

Apart from learning what the name of this effect is and why it occurs, you can now answer two unit 1 past paper questions with the knowledge that:

1. Hydrogen bonds are largely electrostatic in nature.

2. Each water molecule forms four hydrogen bonds (from diagram).

Relative Acidities Of Alcohols - In Aqueous Solution

Alcohols have acidic character as the react with active metals like sodium or potassium liberating hydrogen. For example,

C2H5OH + Na → C2H5ONa + ½H2

However, alcohols are weak acids. This is because they have an electron-releasing alkyl group (+I effect) which increases electron density around oxygen so that the release of a proton is rendered difficult.

Acidic character of alcohols shows the following order:

primary alcohol > secondary alcohol > tertiary alcohol

The acidic character of alcohols depends on the release of H+ from O–H. The +I effect increases  from primary alcohols (having one alkyl group) to secondary alcohols (having two alkyl groups) to tertiary alcohols (having three alkyl groups).

As a result, in tertiary alcohols, the release of a proton is most hindered making them the weakest acids of the three classes of alcohols.

In the gas phase, order of acidity is the exact opposite of that given above. Currently, this ought to be beyond the scope of the CAPE syllabus.

Phenyl Radical - Structure And Reason For Reactivity

The group derived by loss of an H from benzene is a phenyl group abbreviated Ph.

The phenyl radical (C6H5·) is the prototypical σ-type aryl radical and one of the most common aromatic building blocks for larger ring molecules. Using a combination of rotational spectroscopy of singly substituted isotopic species and vibrational corrections calculated theoretically, an extremely accurate molecular structure has been determined.

Figure  1. Side-by-side comparison of the structures of benzene and the phenyl radical.

The phenyl radical (C6H5·) is a highly reactive species formed by the homolytic cleavage of a C–H bond in benzene (Figure 1), the prototypical aromatic  compound. It is one of the most common aromatic radicals, and plays a central role in many reactions, ranging from astronomy to combustion and biochemistry.

As the simplest aryl radical, it also serves as the benchmark for computational investigations of larger, open-shell ring molecules.

The reactivity of aryl radicals is due to the localization of the unpaired electron in a σ-type orbital at the C–H cleavage site, as indicated by the very high C–H bond dissociation energy of benzene (465 ± 3 kJ mol-1).

Purpose Of The Salt Bridge

This explanation makes reference the Zn/Cu2+ cell shown above.

The cell cannot operate unless the circuit is complete. The oxidation half-cell originally contains a neutral solution of Zn2+ and SO42- ions, but as Zn atoms in the bar lose electrons, the solution would develop a net positive charge from the Zn2+ ions entering. Similarly, in the reduction half-cell, the neutral solution of Cu2+ and SO42- ions would develop a net negative charge as Cu2+ ions leave the solution to form Cu atoms. A charge imbalance would arise and stop cell operation if the half-cells were not neutral. To avoid this situation and enable the cell to operate, the two half-cells are joined by a salt bridge, which acts as a "liquid wire," allowing ions to flow through both compartments and complete the circuit. The salt bridge shown in the diagram is an inverted U tube containing a solution of the nonreacting ions Na+ and SO42 - in a gel. The solution cannot pour out, but ions can diffuse through it into and out of the half-cells.

To maintain neutrality in the reduction half-cell (right; cathode compartment) as Cu2+ ions change to Cu atoms, Na+ ions move from the salt bridge into the solution (and some SO42- ions move from the solution into the salt bridge). Similarly, to maintain neutrality in the oxidation half-cell (left; anode compartment) as Zn atoms change to Zn2+ ions, SO42- ions move from the salt bridge into that solution (and some Zn2+ ions move from the solution into the salt bridge). Thus, as the diagram shows, the circuit is completed as electrons move left to right through the wire, while anions move right to left and cations move left to right through the salt bridge.

Partition Coefficient - What's The Numerator?

The process of a solute dissolved in one solvent being pulled out, or “extracted” into a new solvent actually involves an equilibrium process. At the time of initial contact, the solute will move from the original solvent to the extracting solvent at a particular rate, but, after a time, it will begin to move back to the original solvent at a particular rate. When the two rates are equal, we have equilibrium. We can thus write the following:

Aorig ⇌ Aext

in which A refers to analyte and orig and ext refer to original solvent and extracting solvent, respectively. If the analyte is more soluble in the extracting solvent than in the original solvent, then, at equilibrium, a greater percentage will be found in the extracting solvent and less in the original solvent. If the analyte is more soluble in the original solvent, then the greater percentage of analyte will be found in the original solvent. Thus, the amount that gets extracted depends on the relative distribution between the two layers, which, in turn, depends on the solubilities in the two layers. A distribution coefficient analogous to an equilibrium constant (also called the partition coefficient) can be defined as follows:

Often, the value of K is approximately equal to the ratio of the solubilities of A in the two solvents. If the value of K is very large, the transfer of solute to the extracting solvent is considered to be quantitative. A value around 1.0 would indicate equal distribution and a small value would indicate very little transfer. Uses of the distribution coefficient include:

1.the calculation of the amount of a solute that is extracted in a single extraction step,

2.the determination of the weight of the solute in the original solute (important if you are quantitating the solute in this solvent),

3.the calculation of the optimum volumes of both the extracting solvent and the original solution to be used,

4.the number of extractions needed to obtain a particular quantity or concentration in the extracting solvent, and

5.the percent extracted.

The following expansion of the previous equation is useful for these: