The Electrolysis of Chemistry

Well, we’re finally here. The last blog post. The last of the unconquered waters of AP Chem. Right before sailing through AP Exam Reef and docking into the harbor one last time.

Before Spring Break, we had finished off the topic of electrochemistry – the Nernst equation to determine the E of the cell in nonstandard conditions and the second type of power cell, the electrolytic cell. First of all, the Nernst equation is actually a manipulation of the equation used to determine the ∆G of a reaction in nonstandard equations. Basically, it is said…

-nFE = ∆G ; n= moles of electrons, F is Faraday’s Constant

So replacing ∆G with -nFE and cancelling out the -nF yields the Nernst equation:

E = E^o -RT/nF ln Q

The key all lies within the natural log of Q. Basically, Q is the K of the system, which accounts how the different concentrations wouldaffect the spontaneity and E of the cell. 

On the opposite spectrum of spontaneity is the electrolytic cell, which is not driven by free energy whatsoever. Rather, the cell works when a current is run through (a.k.a. energy is put into the system). Because of this unique property, there is no need for a separation of chambers, but only a controllable power source.  Typically with electrolytic cells, electrolysis is performed, where it’s possible for the water that holds the substances to take part in the reactions. For example, around the anode of the cell, either the electrode substance or the water can be oxidized – it all depends on which half-reaction has the more positive E^o, and the same holds true for reduction at the cathode end.

     One of the biggest applications of electrolysis is the process of electroplating – basically running a current through a vat of water in order to oxidize the plating metal into solution and reducing it around the desired object. Typically, this occurs with gold and silver, but it’s possible for many substances to do this – the week before Spring Break’s lab focused on this aspect with copper.

As depicted above, we were electroplating copper from one strip to the other, depending on which one was the anode and which one was the cathode. This lab was particularly fun because 1) it was essentially the opposite of the first lab we ever did in AP Chem at the beginning of the year, and 2) we were allowed to copper plate some of our possessions, like dull pennies (returning the sheen) or non-copper plated coins. In the end, we were basically transferring copper from one plate to the other.

      As for any hardships on the Chem journey, I still often more than not get confused about flipping signs and whatnot. Do I flip the E and add, or leave it and subtract? Should I do both? Honestly, just thinking about confuses me albeit less intensely. Some usage of the Nernst equation also gave me trouble too, but that was mostly from forgetting to change some units here or there, or just plain old forgetting to include the temperature in the equation. Being a chemist sure has its moments, huh?

Well, it’s sad to see this blog reach its last post for the known future. Looking through the posts is like going down memory lane – I can’t believe it’s already been an entire school year. College is on the horizon, and I’m just speeding closer and closer with broken brakes. All things considered though, it’s been a hell of a ride. Now, all that there’s left for me to do is to ace the AP Chem test and graduate. I might as well increase the entropy of the universe while I’m at it. So, until next time!

Ensign Hung, signing off!

What Shocking News!

Believe it or not, we are in the final stretch for our Chemistry Odyssey, and the final frontier we are tackling is the concept of voltaic/galvanic cells and the reduction oxidation that allows for such cells to exist. The basic idea of a galvanic cell is that the cell separates the oxidizing half reaction from the reducing half reaction, isolating them in different chambers. Between the two reactions is a wire that transfers the necessary electrons from the oxidation to the reduction. There is also a “salt bridge” between the two chambers that serves to neutralize the built up electrical charge through salts, replenishing the electrons for further energy production. The entire process looks like this:

The side responsible for oxidation is called the anode – this is where the solid metal loses its electrons, dissolves from the electrode, and becomes an ion. The other side, the cathode, does the exact opposite – the ions gain electrons and are able to coagulate around the electrode. The actual energy from this process comes from the electrons that pass through the wire during the transfer.

From galvanic cells, an equation can be derived:

Eºcell = Ecathode – Eanode

We also know that if the resulting voltage of the cell is negative, that means it repels electrons and is NOT spontaneous. If the cell’s voltage is positive, it attracts electrons and is quite spontaneous. The reason for the spontaneity is that E can be related to G (Free Energy) through this equation:

ΔG = -nFE

n – number of electrons transferred
F – Faraday Constant:
E – Eºcell

     Voltaic cells aren’t some top-secret technology that the government is desperate to keep prying eyes off of – in fact, they’re everywhere (the cells, not the government). Voltaic/Galvanic cells are simply batteries, whether they be AAA’s, AA’s, or 9V’s. The everyday batteries are specifically known as dry cell batteries. Any self-respecting chemist can see the correlation between the galvanic cell and the common battery:

There’s a cathode, an anode, something resembling a salt bridge (the paste), and even a small separation between the two chambers via a manipulation of the anode’s shape (clever!).

     This particular topic, considering how specific it is, has a few difficulties in concept that I have trouble keeping straight. One of said difficulties is the confusion of cathode and anode – oftentimes, the diagram has the anode on the left and the cathode on the right, but when no such diagram exists, it can get a bit tricky. A key point in keeping the two identified is to find out what is going on REDOX-wise: check to see if any part of the half reaction is being oxidized or reduced. Another issue I encountered in my studies is whether or not I should flip the voltage sign when I flip the reaction. Through experience, I found out that the answer is NO. Like the stochiometric coefficients, the flipping of the reactants and products has no effect on the Ecell. It confused me a lot at the beginning (and costed me some points), but I understand now, and it’s better than not understanding at all.

Anyways, that’s all for the progress report on the Odyssey. Tune in next week for more adventures and possible advancements in Chemistry!

Disorderly Conduct is the Universe’s Specialty

Wow, it’s been a while sine the previous blog post -mostly thanks to exams – but it’s good to be back. During this (4-day) week at St. Xavier High School, we’ve ventured into the rough waters of Complex Ion Equilibrium and the Sea of Entropy. With complex ions, the major things we needed to know were that they typically contain a transition metal (such as Fe) and they have Keq >> 1.  An example would be the reaction between Iron(III) cations and Thiocyanate, which looks like this:

Fe^3+ (aq) + SCN^- (aq)  ⇌ Fe(SCN)^2+

This is known as the formation reaction of the Iron Thiocyanate. But remember that the Keq of this reaction is much greater than one, which makes the reaction basically one-way. But what about equilibrium? We can determine this by finding the Ksp of a similar reaction, preferable one that involves the Iron(III), and multiply it with the Kf (Keq of the formation reaction.). This also results in the addition of the two reactions together for the final equilibrium reaction, where spectator ions are eliminated:

     Fe^3+ (aq) + SCN^- (aq)  ⇌ Fe(SCN)^2+
+Fe(OH)3 (s) ⇌ Fe^3+ (aq) + 3OH^- (aq)
———————————-
Fe(OH)3 (s) + SCN^- (aq) ⇌ Fe(SCN)^2+ (aq) + 3OH^- (aq)

Keq = Kf x Ksp

As for the Sea of Entropy, it honestly is more like an ocean – this idea of the probability of the universe being a in a certain state certainly is on a larger scale than anything we’ve encountered before. Entropy largely depends on the 2nd Law of Thermodynamics, which states:

ΔSuniv = ΔSsys + ΔSsurr

Entropy can be calculated using this formula:

S = K ln

Where K = Boltzmann’s constant (1.38 x 10^-23 J/K) and W is the number of possible microstates. Entropy can be applied chemically too – this is A.P. Chemistry after all. From the 2nd Law of Thermodynamics, we can rearrange it into a measure of chemical potential, called Gibbs Free Energy.

ΔG = ΔH – TΔS

For understanding the value of ΔG, these basic rules need to be known.

ΔG < 0, the reaction associated is spontaneous
ΔG > 0, the reaction is non-spontaneous
ΔG = 0, the reaction is at equilibrium

When thinking of Gibbs Free Energy, the best way to visualize it as the slopes of a valley. As you fall down one end, when ΔG is negative, it takes no energy to move. But when you try to climb the other end, it takes energy to do so. At the very bottom of the valley, ΔG is zero, indicating that the reaction can be nudged forwards or backwards, as demonstrated below.

In demonstrating the concepts we’ve learned this past week, our current lab is to determine the solubility product of an ionic compound – this compound is Ca(NO3)2, which forms a complex ion with the presence of OH^- and forms a precipitate. The entire point of the lab is to perform serial dilution of both the Ca(NO3)2 and the OH in separate trials to see how the formation of the precipitate changes (more precisely, where the precipitate stops forming in the series.) From this lab, we have to use our knowledge of complex ions to determine the concentration of whatever is dissolved in the final slot (#12) using the Ksp. It was quite interesting the see the series of solutions, starting from the cloudiest and working its way to transparency.

I am having a few difficulties conceptually this time around, mostly concerning with the nuances found in Gibbs Free Energy. For example, when determining if a reaction is spontaneous, there are many levels of spontaneity – @ high temps, @ low temps, @ all temps, @ no temps, and even backwards! Keep track of all the variables needed to make this prediction can get me call cross-eyed if I’m not careful of what I’m doing. I suppose a good way to counter this is to simply write out all of the factors of the Gibbs Free Energy equation out to externalize the info and not worry about it clouding my mind. I’ll give it a try.

Anyways, thanks for reading this journal log. Lucky for you, I can’t spill any ink from the ship’s rocking. Well, I mean, I could, but it wouldn’t show up on the blog. Or would it?

Signing off!

Calling my Buffs

Another 4-day week has passed in our AP Chemistry Odyssey as we continue diving into the depths of acids and bases. This time around, we discussed the concept of chemical buffers. A buffer is a system of a weak acid/base and its conjugate in salt form. In solution, this is the typical acid/base equilibrium that one would see in dissolving a weak acid/base.  However, the true power behind the buffer is the ability to maintain the solution’s pH around a constant range.

The weak acid component neutralizes any base:

HA(aq) + OH-(aq) → A-(aq) + H2O(l)

And the conjugate salt eliminates any acid:

A-(aq) + H+(aq) → HA

Buffers are an application of the Common Ion Effect. This is a shift in its equilibrium position that occurs when an ion involved with the Keq is added to the solution. For a buffer, the common ion is the part that isn’t the H+ cation or the metallic component of the salt. (e.g. C2H3COO- is a common ion. Its acid is C2H3COOH, and its salt can be NaC2H3COO)

Visually, this is what a buffer looks like:

The free HCO3- anions in the solution combine with the H+ cations that would manifest upon being dissolved, forming carbonic acid and keeping the H+ from freely floating around and making the solution acidic.

Calculation-wise, this only makes minor changes. The only noteworthy one is that when using a RICE table, you have to account for the fact that there is already some base/acid present for the Initial values that you normally set as zero.

Although there is no lab to speak of this week, it is important to mention how incredibly important buffers are to solution-based experiments as well as all living organisms. In a solution based experiment, perhaps it is necessary to keep the pH of the solution mostly constant to ensure that no unwanted change will occur to the data. For example, say that an unknown compound is extremely sensitive to pH changes outside of its natural range of 6.3 – using a buffer to hold the solution there allows researchers to work freely with running tests, such as introducing iodine, benedict’s solution, alcohol, etc.

Buffers are fundamental to many biological systems throughout the world: take the human circulatory system, for example. When we breathe, there is a constant exchange of O2 and CO2. However, the presence of CO2 within our bloodstream holds a great risk, as CO2 has the capability to lethally change the blood’s pH to make it either too acidic or too basic. This is why our blood has a buffer called the bicarbonate buffer system, in which carbonic acid (H2CO3) and the bicarbonate ion (HCO3-) work together in order to maintain the blood’s pH slightly above 7. Excess waste CO2 reacts with water to form carbonic acid, and too much stress on carbonic acid forces it to form more bicarbonate in order to maintain pH. The entire process is illustrated below.

There are a few things that plague me in terms of calculations: sometimes, I’m just not sure what value to put in, and if these values account for all the circumstances in the problem. For example, there was a RIC table problem where I needed to find the moles of each component of the buffer. I ran into difficulty, however, when I went looking for the Liters of solution in order to multiply it with the Molarity. Thankfully, it’s been mostly small, almost non-issues that trouble me. Of course, I can improve simply with experience in doing more problems, and having each problem become progressively harder and harder. Looking for hints and tips in the book would help, too, but we’ll see if I become that desperate for a rescue.

Credits for the Featured Image go to Miss Marisa Sanders (possibly related to good ‘ole Bernie?)

What even…

Is it Time to Drop the Base?

Oh man, another crazy four-day week has flown by for us AP Chemists -between working with the concept of weak acids/bases and finishing up a lab that had us determine the Ka (acid ionization constant) of 4 unknown weak acids, there hasn’t been a single moment to take a breath and enjoy the chemical landscape. But still, we manage the trudge on.

We started off the weak [sic] by expanding our calculation knowledge of bases – similar to calculating acids, but there are some pitfalls that we must avoid. First of all, we have to remember that pH is the measure of [H3O+], not [OH-]. So after using the RICE table and finding the concentration of hydroxide, putting it in the -log[OH-] would yield pOH, not pH. We have to go a step further and subtract the pOH from 14 to get the pH.

Another important concept we learned was the idea of an acid/base having a conjugate, and the characteristics of this conjugate based on the original substance. Basically, the stronger the original acid/base, the weaker its conjugate. Strong acids produce a conjugate acid so weak, that it’s considered neutral, and vice versa. The reason why is because the K value of the conjugate is extremely smaller, much smaller than the Kw of H2O (1.0 x 10^-14). Knowing all of these characteristics and rules, we are able to informally determine the pH of a solution by splitting up the substance into ions, and finding its conjugate. This process is demonstrated below.

     A week in AP Chemistry would be pretty uneventful without one of its trademark labs. Lucky for us, we did a lab that had us determine the Ka of 4 unknown weak acids. They were all in the solid state, so we would have to dissolve the acid ourselves in 50 mL of distilled water. From there, we would split the solution into two 25 mL samples, and titrate one of the samples with NaOH base until the indicator inside of the solution turns a faint fuchsia to indicate the solution’s neutralized status. Then, we reintroduce the two samples in a beaker and find the pH.

Because of that little procedure, we have successfully made the concentration of the weak acid and the concentration of its conjugate base equal to each other. So when calculating the Ka, which is basically equilibrium, the concentration of the conjugate base on top cancels out the weak acid on the bottom, leaving the Ka as the concentration of the hydronium particles, which can be found via the pH we measured.

Although the calculations involved in this week’s lesson isn’t too difficult, there are times where I become completely entrenched and lost within the math, like the Minotaur in the labyrinth – like forgetting that I actually calculated pOH, not setting up the Ka/Kb equilibrium equation properly, or even forgetting some crucial exponents. I also find it difficult to estimate the pH/[H3O+/OH-] of a solution just by looking at its Ka or Kb. Being able to recognize what number means what in the world of acids and bases is really important. So for once, it’s the conceptual side of Chemistry that’s tripping me up. Hopefully, I’ll understand the concept behind this once I read the material more thoroughly and discuss the meaning of the concepts with other Chem classmates.

 

A Return to the Basics

After 3 quarters of exploring the unknowns of chemistry, we finally go back to familiar waters to re-explore the depths of acids and bases – namely, the what actually makes acids and bases act the way they do. The answer? Both acids and bases cause the presence of H3O+ and OH- ions, respectively. The definitions used to describe acids and bases vary, as there are at least 2 of them: the Arrhenius definition, and the Br⌀nsted-Lowry definition. Arrhenius’s is simple – acids have an extra hydrogen that form the H+ ions, and bases have an OH that separates from the entire complex. Both are always in solution. Br⌀nsted-Lowry, instead, uses a broader definition of a proton acceptor (base) or proton donor (acid). Donors give away their electron to what is usually water, making hydronium (H3O+) and increasing the [H+ ].

HCl(aq) + H2O(l) → Cl-(aq) + H3O+(aq)

Meanwhile, acceptors take a proton/H+ away from what is usually water, and that results in OH- forming from the water.

NH3(aq) + H2O(l) → NH4+(aq) + OH-(aq)

Both the original acid/ base becomes a conjugate base/acid, since they now are able to do the opposite of their original proton ability. As for the water, this can be considered amphoteric, as it has the ability to be both an acid and a base under the Br⌀nsted-Lowry definition. We also learned that H3O+ and OH- occur naturally in water via a reversible reaction, thus having an equilibrium constant (Kw) of 1.0 x 10^-14. So:

10 x 10^-14 = [H3O+][OH-]

Now, we have all the necessary knowledge to calculate the pH and pOH from the concentrations of either hydronium or hydroxide. The resultant calculation system can be put in a square, like so:

Where the antilog is simply 10^-pH or pOH

     We explored this idea of acids and bases as in equilibrium in a lab we performed this past week. Our objective was to determine the Ka of several unknown acids by making the concentrations of the acid and the nonacid part of the original equal to each other, which would cancel both out in the Ka equation, leaving only the concentration of the H3O+. From there, we would collect the pH and calculate [H3O+], which would be the Ka of the acid.

      As far as the concept and the calculations go, I haven’t really run into any trouble. They’re all pretty straightforward, with nothing complex or difficult to remember. However, some problems that incorporate other chemistry concepts do give me trouble, since I’m not exactly sure how to apply the acids/bases concept to the other concept. But, with all of the problems and labs that I’m doing, it shouldn’t take too long at all before I can nail acids and bases down for good.

Chemical Nirvana

Hey there fellow chemists (or at least, I hope you are fellow chemists)!

This week in AP Chemistry, we continued with the theme of chemical equilibrium. Namely, we explored some more of its aspects. For one, we dove deeper into the idea of equilibrium in terms of partial pressures: We already knew that the Kp = Pressure of reactants/Pressure of products. But now, we can also discover the Kp through this equation:

Kp = Kc (RT)^∆n

where ∆n = (total moles of product – total moles of reactant), R= ideal gas constant, T= temperature in K, and Kc= Keq.

We also explored the realm of RICE: with the values of the stoichiometric coefficients, some initial concentrations, and a final equilibrium concentration, we can find the change of concentration and use it as a basis for the other changes, modified by the stoichiometric coefficients.

This week, we also engaged in a lab to determine the equilibrium constant of FeSCN2+. To do this, we first concocted several solutions of Fe(NO3)3 of varying molarity and added it to KSCN with varying molarity as well. Once these substances are completely reacted, each solution was placed into a colorimeter to test for absorbance, which will be used to determine the molar extinction coefficient via a absorbance vs. concentration graph. Another trial was also done, but with some water to dilute both substances as well.

There are a few problems that I’ve encountered so far this week. For one, the RICE table can become a bit confusing if I’m not properly tracking all of my numbers. This holds especially true when I need to multiply/divide the change in molarity by stochiometric coefficients constantly. Then there’s also the problem of finding the equilibrium concentrations when given only the Kc/Keq. The reason this is giving me trouble is because using the SOLVER on my calculator is slightly tricky business, thanks to all of the parentheses needed to yield the right equation for the desired result. Luckily, both of these problems are pretty minor in terms of scope – nothing to do with not grasping the conceptual ideas. So I think I’ll be fine as long as I watch where I tread. Because honestly, this is how I feel like kinetics is to me…

 

 

 

 

 

Chemical Zen

This 4-day week, we explored the concept of chemical equilibrium. To understand equilibrium, we have to first understand that there many chemical reactions that can flow both ways, depending on the circumstances. These reactions are known as reversible reactions. Typically, reversible reactions have two rates: one for the reactants, and one for the products (since it’s reversible). Initially, these two rates will probably be completely different from one another, but as time goes on, the reaction will eventually reach a state called Dynamic Equilibrium, where both rates stop changing and the amounts of both reactants and products stay constant. When graphed, it looks like this:

https://edublognss.wordpress.com/2013/04/16/chemical-equilibrium/

A constant for such a reversible reaction, called the Keq, can be derived from this equation:

For reaction aA + bB ⇔ cC + bB,

Keq = [C]^c [D]^d/[A]^a [B]^b

However, it is very important to note the state of matter that every participating substance is in. If a reactant/product is a liquid or a solid, then it would not be factored into the Keq, since the concentration of a solid in a liquid or a liquid in a liquid (water in water, for example) wouldn’t be able to change much.

Although not directly relating to the idea of a chemical equilibrium, this week’s crystal violet lab still uses many of the same concepts, since it is also related to the rate of reactions. Basically, a set amount of crystal violet was reacted to 3 different concentrations of NaOH. In order to record the rate of change, the reactions were placed in individual cuvettes and observed through a colorimeter, where absorbance and Beer’s law was used to determine the rate of the reaction. By determining the rate of all 3 reactions, the k constant for the overall rate law can be found.

Since this has been a conceptual based week, there hasn’t been a lot of mathematical errors that I could’ve committed, so it’s been a bit easier on me in terms of difficulty. I still do, however, experience difficulty in remembering some of the concepts of chemical equilibrium, like the idea that solids and gases aren’t included in the Keq, and partial pressures can be used to determine the Keq.I guess keeping all of these little details in line comes with time spent on working with equilibrium, so I’ll bear with it until it sticks. The phet activity we’re doing right now is actually really helping.

Featured image credit: http://uoitbiology12u2014.weebly.com/dynamic-equilibrium.html

 

When a Chemical Reaction Gets a Speeding Ticket

After a long hiatus for the holiday season, its time to jump right back into the Odyssey of A.P. Chemistry. This past week,  our crew has been navigating through the waters of chemical kinetics: the area of studying a reaction’s rate and mechanisms. We kicked things off by taking a look at the rate of reactions, where we observe the concentration v. time curve and find both average and instantaneous rates of reactions.

Source

For finding the average rate:  (-)∆[A]/∆t

For finding instantaneous rate: d[A]/ dt,                                                  where A is some reactant

A little bit more complex is the rate law, where the rate of a set of concentrations is known. Normally, the rate law of a reaction looks like this:

Rate = k[A]^m[B]^n

The exponents m and n are the order of the reaction, which shows how changing the concentration of the reactant will affect the rate. Zero-order reactions are those that changing the concentration of the reactant won’t change the rate. First-orders have a proportionally direct relationship between concentration and speed, and second order has a quadratic relationship between concentration and rate (double reactant, quadruple rate). There is such a thing called third-order, and it follows the same trend as first and second. The k is simply the rate constant that changes depending on reaction order and concentrations.

Finally, we covered the topic of factors that affect reaction rate, as well as the mechanisms of a reaction. Basically, chemists pull a chemical reaction apart into smaller, theoretical reactions. They do this to determine the step with the lowest rate, as that is the reaction step that defines the speed of the entire reaction. This reaction could be slow due to a number of reasons: temperature, concentration, presence of a catalyst, even the surface area of the reactants.

There is an issue that plagues my journey this week. The calculation of the instantaneous rate is giving me grief because any method of calculation that I use (besides the correct one) is too inaccurate, and therefore unreliable. Yet, I’m not entirely sure what everyone else is doing for instantaneous: are they using derivatives (even though I’m pretty sure we’re not supposed to)? I should probably ask around next time this class meets so I can pinpoint exactly what I need to do. But other then that, I believe I have been doing well. The first half of our current lab (the violet lab) is pretty interesting as well, since we have to use our knowledge of reaction rates, light absorbance, and general cleverness to find the rate of reaction (and maybe even the k value).

Anyways, go Chemistry!

 

 

Atomic Relationships

It’s been months sailing through the sea of chemistry, and the world that surrounds us no longer resembles the world when we first left this September. No longer are we memorizing polyatomic ions or doing percent mass. Now, we are analyzing the forces that molecules interact use to interact with each other and starting our dive into the Abyss of Alloys.

All molecules are affected by something called intermolecular forces: attraction. This can occur in 3 ways: it can be a dipole-dipole attraction, where the partially negative end of one polar molecule is attracted to the partially positive end of another polar molecule. There exists a stronger version of this dipole-dipole: hydrogen bonding, the second intermolecular force. The reason why H-bonding gets its own distinction is because of how the proton of the hydrogen atom is virtually naked when bonded with a very electronegative atom, such as oxygen or nitrogen. And finally, we have the dispersion forces, which are present in all atoms as it is based on the momentary dipole moment that results from all of the electrons on an atom concentrating in one point in the electron cloud, making a brief partial positive and negative end. This forces are critical in the physical behaviors of substances, as it determines the boiling point, melting point, and density of a substance. Higher intermolecular forces means higher boiling and melting point, as it is more difficult to make the vapor pressure of a substance equal to the atmospheric pressure if the molecules are extremely attracted to one another.

After dealing with polarity and its subsets, we found ourselves in the Abyss of Alloys, where we came to understand the true nature of metals. Basically, all metals loosely hold onto their valence electrons, so when a huge number of them come together, these very loosely held elections merge together into a sea of electrons, as shown below. This sea effectively masks the positive nuclei from each other, so they are able to come very close and slide past each other, which gives a metal its unique characteristics of malleability, ductility, conductivity, etc. Metals can also be put into solutions called alloys, where the certain characteristics, such as melting/boiling points and tensile strength, are enhanced due to the combination. There are two types: Substitutional Alloys are when metal atoms of similar size replace each other in the structure in a regular order, and Interstitial Alloys are when a metal and typically a small nonmetal combine in such a way that the nonmetal fills in the cracks between the metals in the structure of the solution, so a new shape is formed that is often much stronger than the original two elements.

In real life, we learned that interstitial alloys are used in the making of many things, such as the katana: ancient Japanese swordsmiths would combine iron sand with charcoal to form a very strong alloy called steel, which is perfect for the cutting edge of a sword, as it’s hard enough to cut well, but malleable enough to endure abuse of any kind. Basically, iron sand and charcoal are poured into an ancient furnace, and in this extreme heat, the structure of iron expands. The carbon in charcoal seeps into the expanded iron structure, and once the iron cools, it tries to revert to its original shape, but the carbon atoms are stuck inbetween the irons and it forms a new, much stronger shape. The process is hard, tedious, painstaking, and requires a lot of precision and detail, and it is a testament to the ancient Japanese swordsmiths for be able to forge a weapon of such technical excellence without needing the understand the scientific genius behind the process.

Swords and atoms aside, this week has had its fair share of difficulties: the intermolecular forces made enough sense, but the relationship between the strength of the forces and the boiling/melting points can get confused in my head pretty easily, but I know now that its generally the stronger the force, the harder it is melt/vaporize. But still, it doesn’t hurt to keep drilling myself on the concept, and it makes sense in my head. The reason why melting/boiling points are higher is because if the forces are stronger, it’s more difficult for the particles to energize enough to escape from the solution and establish the required vapor pressure to facilitate escape. And as for alloys, I’m pretty solid.