# Trends and Effective Nuclear charge

13/02/19 15:39

# Effective Nuclear Charge

As we have noted, elements in a column (group) on the periodic table tend to have similar chemistry. We posited earlier in the year that this has to do with similarities in how their electrons are arranged. We derived this just by thinking about the fact that chemistry happens at the level of electrons interacting or trading places, always to find a lower potential energy (PE).

Now, we are starting to use quantum to define the energies of these electrons. We have also learned (or, at least are beginning to learn) that there is periodicity in the pattern of the outermost electrons…all fits together.

So, we will be talking about "Valence" electrons, which are the ones on the outermost shell that is occupied (corresponds to the period on the table), and "inner" electrons, which we will call "shielding" electrons. They are in between the nucleus and the valence electrons, shielding the attraction between the valence electrons and the nucleus. In each group, the atoms have similar valence structure

## Most "trends" across a period can be explained easily by invoking the concept of "effective nuclear charge."

This idea seems harder than it is. It is a simple bookkeeping step to determine how much the outermost electrons of an element are attracted to that atom's nucleus: more nuclear charge, more attraction. Nuclear charge (z) is the same as the atomic number: the number of protons. However, the attraction will be reduced (shielded) by the presence of inner (non-valence) electrons. To restate, a valence electron is any electron in the outermost shell, or principle quantum number. By the process of elimination, a non-valence electron is one that is in one of the inner shells (has a principle quantum number less than that of the valence. So, take any element in period 3: any electron in either the S orbital or the P orbitals of level 3 is valence. For all period 3 elements, there are 10 non-valence, or shielding electrons. These are the electrons in the 1S, 2S, and 2P orbitals and are in the same configuration as Neon's ten electrons.

The equation for effective nuclear charge is:

z

_{eff}=z-core electrons

Core electrons are those of the previous noble gas plus any d and f electrons added since that noble gas. More on that later.

These are also called “shielding electrons” or the “shielding constant” because they interpose between the valence electrons and the nucleus, effectively shielding the attraction.

For sodium, there are 11 protons, but 10 shielding electrons. So, Zeff is 11-10 or 1. For potassium it is 19-18 or 1. For all elements in group 1, the effective nuclear charge will be 1.

For Mg, Z

_{eff}is 12 (its atomic number) minus 10, the shielding constant for all elements in period 3. So, for period 3, the number of core electrons does not change, but the number of protons increases, the effective nuclear charge increases. That must mean that the valence electrons are more attracted to the nucleus as you move across the period, and thus many properties of the atoms will follow a trend due to that increasing effect. Below are several important properties of atoms and ions and how they are affected by z

_{eff}.

- Atomic Radius: outermost electrons are mores strongly attracted the more the z
_{eff}. ∴ radius gets smaller moving left to right. Note also that a smaller radius gives rise to still greater attraction (the closer the electron is to the protons, the more attracted it is). - 1st ionization energy (the energy required to remove an electron to make the atom a 1+ ion): electrons more strongly attracted ∴ harder to remove as you move left to right.
- Electron affinity (how easy it is to add an electron to make the atom a 1- atom): more attraction ∴ easier to add another electron as you move left to right
- Electronegativity (how tightly they horde electrons when in a bond with another atom): more attraction as you proceed left to right ∴ more likely to steal or hog electrons in bond.

## Transition elements

These follow the rule. However, electrons are being added to inner levels and thus the shielding constant is increasing at the same rate as the number of protons. Thus, Zeff is essentially the same for all the transition metals (Z

_{eff}=2).

## For trends down a column (group) invoke the atomic radius.

As noted, the closer two charges are, the more they are attracted. As you move down a group Z

_{eff}is unchanged, but the principle quantum number (n) for the valence electrons goes up by 1 each step. "n" is related to the average distance from the nucleus. So, with each step down a group, the valence electrons are farther out, and less attracted:

- Atomic radius goes up as n goes up.
- 1st ionization energy: larger radius, less attraction, easier to remove ∴ loer 1st ionization energy
- Electron affinity: less attraction ∴ harder to add another electron
- Electronegativity: less attraction ∴ less likely to steal or hog electrons.

## Exceptions:

These require you to consider the exact electron under consideration. For this, we will look in more detail at the quantum description.

# Quantum 1

01/02/19 14:11

# Quantum 1

Everything you know is wrong. Well, not everything. Just most of it. If there is one take-home lesson from this is that the assumption we all make that two opposite conditions cannot both be true at the same time is not valid, or at least not useful, when examining the world of subatomic particles and EM radiation. Think of it this way: our interaction with the BIG world averages out all the little events and we don’t notice them. Think about atmospheric pressure. The collisions of air molecules with your body average out and you don’t feel like you are being pummeled. But, if you were a spec of dust...you might have a different experience. Matter and energy, as we experience them, seem to be two different things with different properties. Actually, they are manifestations of the same thing.

## Nature of light:

- All electromagnetic (EM) radiation travels a the same speed in a vacuum (2.9979x10
^{8}m/sec. 3.00x10^{8}m/sec will do ) - We characterize EM radiation in terms of two parameters other than speed, wavelength, which is given the Greek symbol λ, (lower case lambda) and frequency, which is given the Greek symbol ν (lower case nu)
- Wavelength is exactly that, the distance from the peak of one wave to the peak of the next and is given in meters (visible light has wavelengths in the 4-7x10
^{-7}range, while UV light is shorter and X-rays much shorter. FM radio has values in the 3-10 meter range. - Frequency is “cycles per second;” just how many waves will pass a given point per second. It has units of 1/seconds or sec
^{-1}. - Logically, if I multiplied λ in meters by ν in 1/s I would get m/s, which is speed. Therefore νλ=c, the speed of light. Also λ∝1/ν That is, as wavelength goes up, frequency decreases.
- Light has a dual nature. In some ways, it behaves like a wave (shows diffraction patterns and refraction), in others it behaves like a particle (can travel in a vacuum, and has a measurable momentum).

## Some More annoying things about light:

Max Planck did an experiment in which he heated objects until they glowed. He then increased the energy he put in (by electricity) and expected the total energy he got out to go up as a straight line. Instead, it went up as a step function (that's not quite true…I'm presenting a clarifying view). More properly, the total energy was always some multiple of the same value for any frequency. That is, the height of the step was always the same for any wavelength.

He found that the value was equal to hν where “h” is Planck’s constant and ν is the frequency of light emitted. The value of Planck’s constant is 6.63e-34 J*s. The total energy emitted is given by nhν where “n” is some integer. Note, since λν=c, hν=hc/λ.

## Energy is "quantized"

Energy, therefore, was quantized. There seemed to be “packets” of energy of discreet amounts give by E=hν. These packets are known as “quanta,” which just means “amount.” Energy is said to be “quantized.” Notice that if I multiply frequency (1/s) times Planck’s constant (J*s), seconds cancel and the answer is in Joules. The interesting thing here is that our whole definition of atoms from Dalton and the discovery of an electron all depended on the idea that if there is some fundamental indivisible thing...that’s the particle. So...these quanta of energy come really close to our definition of “particle.”## The Photo-electric effect:

Einstein interpreted an experiment in which others shined light of different wavelengths onto a metal surface and found that electrons could be ejected from the surface. If one turned up the intensity of the light, more electrons were ejected, but they all had the same kinetic energy. If one increased the frequency of the light (lowered the wavelength), the electrons came off at higher velocity, therefore higher kinetic energy (since the mass of the electron is constant, higher KE means higher velocity). Einstein found that the kinetic energy of the electrons was related to frequency by the equation hν=BE+KE where ν is the frequency of light used and BE is the ionization energy (Binding Energy). It may seem obvious now, but it was not always clear how a ray of light could impart energy to a particle. In order to make sense of these results, and those of Planck, Einstein proposed that energy was transferred to the electron by a particle of light called a “photon” and that the energy of a photon was given by the equation Ephoton=hν. Therefore, the reason light was quantized is that each quantum of energy corresponds to the energy of a photon. One electron was ejected by a collision with one photon. The kinetic energy of the electron after the collision was equal to the kinetic energy of the photon after the collision minus that amount needed to remove the electron. Based on this, it was possible to calculate what the momentum of the photon had to be in order to impart energy to the electron at the measured level. Using the equations one could show that the momentum (p) of the photon was given by p=hν/c=h/λ. This suggests that a photon has mass, which it really doesn’t. It does have momentum.

## Your Lab

I don’t want to lose track of the big picture in the details of some fairly simple calculations. The actual calculations are not hard, they just involve some numbers that seem non-sensically small.

Here is the big picture: we treated light like a wave, using standard equations that describe any wave and diffraction and got a number for the wavelength. Then, we looked at light as a particle (photon) with momentum, a completely different way of looking at it, and we got the same answer.

How can two completely different ways of viewing light both be right. This is one of those “Jimmy Moments” where you want to look at me and say: “No, the blue dog is up, damn it!” It underscores the point that a theory’s job is not to make you feel better about understanding things. Its job is to make predictions that can be tested. Bohr’s theory is clearly inadequate for the reasons I stated earlier. It does, however, make some astonishing predictions that turn out to be correct.

The equations:

You use trigonometry to figure out theta and from that, you figure out the wavelength from the diffraction angle and width of the slit (5.15E-6 in this case). First you have to figure out what theta is. For each bright line, you measured the length of the opposite in the right triangle (let’s say it was 37 cm for the red). You measured the adjacent , which was the same for each (100 cm). The ratio of opposite over adjacent is therefore 0.37. That’s related to the angle though the “tangent.” So you use the inverse of tangent (which is usually called “atan” for arctan or tan-1). arctan 0.37 is 20.3 degrees. So that’s theta. Then you take sin of that times 5.15e-6, and you get finally the wavelength at 6.59e-7m (659nm or rounded to 660nm).

The other numbers should be about 490nm for green and 430nm for Blue/violet.

## The Quantum Side

This is not too hard. We'll cover it later in the week.

In fact, the math is easier.

Step 1 is to figure out the energy of each level in the Bohr atom. The amount of energy on level 1 is The answer is in joules because the ionization energy is (2.18x10

^{-18}J ) constant is in Joules.

For Level 2 it is and for level 3 it’s 1/32 times 2.18 E

^{-18}etc. Figure out the value in joules of levels 2 through 5.

Now, find the difference between the value for 2 and the value for 3 (it should be about 3.03x10

^{-19}). Do the same for the difference between 4 and 2, then 5 and 2. In each case, this is the energy of the photon that must be emitted.

Now that you have the Energy of each photon, use that to solve for the wavelength using “h” is Planck’s constant (6.63x10-34 J*s) and c is the speed of light (3.00x108m/s). Multiply hxc (you should get 1.99x10-25) and divide that by each of the energies you got for the differences between 3 and 2 etc. You should get wavelengths that are similar to the red, teal and blue lines.

## That last little bit you thought was still right…

The Bohr model has a lot of truth in it:

- Bohr’s idea that the electron exist only in discrete energy states is correct.
- He is also correct that the emission or absorption of a photon occurs when an electron transitions from one energy state to another.
- Where he was wrong was in that his mathematical approach did not explain the emission spectra of any other atom, and he had no real explanation as to why the electron should only be in discreet states.

### So, how do we calculate the energy of an electron?

- de Broglie and Schröedinger decided that they could take the equations developed to describe the energy of waves of fluid, or sound, and apply them to the behavior of the electron.
- The result was what is known as a “wave function,” which is abbreviated Ψ (psi) and is also known as an orbital. For each electron of an atom, there is a unique form of wave equation, which describes the energy of that electron. While this in not an “orbit,” this serves the same role as the “orbit” in the Bohr model, and so is called an orbital.
- An orbital does not tell you much about where an electron is, or where it is going. However, there is a form called Ψ
^{2}, which predicts the probability of an electron being in any particular place. - Since we have had to rely on the wave mechanical model to describe the electron, it is difficult to then solve for the position of the electron (position implies it is a particle).

Werner Heisenberg came up with the following equation

Δx X Δmv≥h/4π

where Δ

*x*is the uncertainty of the location of the electron and Δ

*mv*is the uncertainty in the momentum (momentum=mass x velocity) of the electron. h/4π (Planck's Constant times 4 pi) is a small number (5.28x10-35). But, when we are talking about an electron’s mass and velocity, the number is big enough to make a difference. The result of this is that as you know more about an electrons position, you know less about its momentum and vice versa.

SO…we now can calculate the energy of an electron of other elements at any energy state (orbital)

What is the test of this?

We look to see whether it accurately tells us the wavelength of the bright-line spectrum of other elements…

**it does.**

We also now know why an electron only exists in discreet states: this is a characteristic of waves. Once you define certain parameters (like, where the nodes are, like in a guitar string), the distribution of the electron is constrained.

### What is an orbital

?- The wave function describes the energy of an electron. What are the types of energy we need to take into account? a. Kinetic Energy of the electron (it’s moving and it has mass)
- Potential Energy attracting the electron towards the nucleus (opposites attract).
- Potential Energy resulting from the
*repulsion*from other electrons. - Bohr dealt pretty well with the first two. However, he never was able to deal with the third. The reason his equations work for hydrogen is that there
*are*no other electrons to repulse. - The wave function can be thought of as a “ticket” for a seat in a theater.

Level 1 is closest to the stage, and has only section “s” which has only 1 box. Those are the best seats, but there are only two of them.

When you go to level 2, you have section “s” which has only one box (2 seats in it) and section “p” which has three boxes, total of six seats. The next best seats are on level 3, again there is a section s (one box), a section p (three boxes) and now even a section d with 5 boxes in it. This goes on. Oh, one last thing: the ticket printer has a perverse sense of humor in terms of how he names each box and seat. The single box in any ‘s’ section is called box “0,” the three boxes in the ‘p’ section are called -1, 0 and +1; the 5 boxes in section ‘d’ are called -2, -1, 0, 1, 2 and the 7 sections in section f are -3, -2, -1, 0, 1, 2, 3. And the seats (remember, only two per box) are called either +1/2 or -1/2. So, if you get a great seat, you might be in level 1, section s, box 0 and seat +1/2. Maybe I’m in level 2, section p, box -1, seat +1/2. And, just confuse things more, section “s” is also known as section “0,” p is known as 1, d as 2 and f as 3....yeah, we needed that complication. But, remember, these are really part of equations, so they have to have some numerical value.

### Quantum Numbers (without the theater metaphor)

The first or principle quantum number is known as “n,” and can have positive integer values: 1; 2; 3; 4 etc. These play the same exact role as “n” in the Bohr equation. What Bohr missed is that at each primary level, there were multiple possible states for the electron to be in, and each differs slightly in energy. 2. Second quantum number is the angular momentum number and is symbolized as

*l*. It loosely can be called the “shape” of the orbital. It depends somewhat on the principle quantum number. The values are whole numbers from 0 to (n –1). So, if principle quantum number is 1, the only value for the second is 0 (n-1). However, on the second quantum level, n=2, so

*l*can be 0 or 1. If n=3,

*l*can be 0; 1or 2. Oh, and just to be confusing, there are names given to these values.

*l=0*is known as an “s” (spherical) orbital.

*l=*1 is a “p” (petal) orbital,

*l*=2 is a “d”(I don’t know why) orbital and “3” is “f”. If n=1, then we can only have an s orbital. If n=2, we can have an “s” and a “p.” On n=3, we have s, p and d. Level 4 has s, p, d, and f. 3. The next quantum number is known as m

*l*, or the “magnetic” quantum number. The values for m

*l*, go from –

*l*, 0, +

*l.*That is, an “s” orbital (

*l*=0), there is only one magnetic state. But, if we have a “p” orbital, (

*l*=1) we can have –1, 0, and +1 as possible values. This number essentially tells you the orientation of the orbital in space. A sphere can only have one orientation. But the “petals” can be either in the x, y or z plane. That gives you three orientations, called -1, 0 and 1. For d and f, the orientations are a bit harder to explain, but there are 5 and 7 orientations respectively. 4. Finally, each m

*l*can have two electrons in it, in what is known as a “spin pair.” This is called the “spin” quantum number and for

**any**m

*l*there are 2 and only two possible states, called either + and or +1/2 and –1/2. 5. So, if n=1, there is only one possible orbital, the S (

*l*=0) state, and it can hold two electrons (or has two sites that could be occupied). If n=2, I have an “s” (

*l*=0) and a “p” (

*l*=1). The p orbitals comprise 3 three possible m

*l*values, each of which has 2 possible “slots” for electrons(spin quantum number). Thus, the 3 p orbitals combined can have 6 electrons in them. Here’s the fun part: if you do the math, it turns out that this set of numbers explains the shape of the periodic table.??? There is only room for two electrons on level 1 (in the single “s” orbital). On period 1, you have only two elements with 1 and 2 electrons (H and He). On level two (period 2), you can have an s (2 electrons) and 3 p orbitals (6 electrons total). That’s 8 elements, which is how many you have. They even come in a pair ( s orbital) on the left and a group of 6 on the right (p orbitals). For a reason I’ll explain soon, you have “d” orbitals on level three...but they don’t fill until after the level 4 s electrons, in period 4. So, level 3 has the same pattern as level 2 (2 on the left, 6 on the right for 8 total). Level 4 starts out with its own “s” orbital, then goes to fill the level 3 d orbitals. That means that these electrons are added

*inside*the level 4s electrons. There are 5 “boxes” for a total of 10 electrons...one for each of the 10 elements in the transition metals on a given period. The same pattern repeats with level 5 being the first time you get electrons in the f orbitals. These are actually the 4 f orbitals. There are 7 orbitals (boxes) for a total of 14 electrons as found in those odd rows always at the bottom (they actually belong up in the regular table). Cool stuff...well to me anyway.

# Using Hess's law

18/01/19 13:03

## Enthalpies of formation:

Rather than determine the ΔH for every reaction, we have tables of standards involving all sorts of chemicals. These are called “Standard Enthalpies of Formation” abbreviated ΔH

^{o}

_{f}. The superscript "o" stands for "standard conditions. For the rest of the page, I've had trouble formatting it correctly. The role these play is rather like that of “altitude above sea level” when we are talking about altitude. It is not as though there is no altitude when you are at sea level. Imagine this: you are climbing down into Death Valley and you reach sea level. The Valley floor is still 283 feet below you. Would you say “gee, I cannot fall any farther because I am already at sea level” ? Of course not. You know you still can fall. Sea Level is just some arbitrary (if useful) value we set to zero.

The enthalpy of formation is the enthalpy involved in forming one mole of a compound from its elements in their most stable state. So, for water, it would be the enthalpy of the reaction H

_{2}+1/2O

_{2}→ H

_{2}O

_{(l)}. Note that I use 1/2O

_{2}because I want the equation to be for one mole of the compound, water. The formation of CO

_{2}is just the burning of Carbon (as graphite) with O

_{2}to make CO

_{2}The ΔH

_{fo of any element in its most stable form is zero. So, Carbon as graphite, O2 H2 etc all have a ΔHfo of zero. Note: This does not mean that these elements have zero enthalpy, any more than being at sea level on a cliff in Death Valley means you cannot fall. Hess’ LawHess’ law is really very straightforward: it says that the change in enthalpy between reactants and products is the same no matter what path you take. So, whether the reaction from A→B proceeds in one step or ten, the difference in enthalpy is the same. The analogy I sometimes make was that when I drive from my house to the school in the morning, change in altitude is -10m (When I go home, my change in altitude is +10m). That change in altitude is the same no matter what route I take from home to school. That makes both altitude and enthalpy something called a State Function, which just means that a particular state is independent of the route taken to get there (when we are in the classroom, we are all at the same altitude, no matter what path we took to get there. The practical (if you can call it that) application of this fact is that we can use the ΔH of reactions that are easy to measure to calculate differences that are harder to measure directly. Here’s an example: Suppose you want to know the change in altitude associated with climbing up a sheer rock wall, but you cannot climb it. To fit with our enthalpy model, you don’t have a device to measure your altitude, you only have one that measures change. In order to determine the difference in altitude, you find a route that takes you from your starting point to point “B” 50 feet below (ΔHeight=-50). Then you climb up a gentle slope (”C”) that goes around the back of the mountain, all the way up to the top (”D”) and find your ΔH is +350 from point “B.” Finally, you slide down a little slope for a ΔH of -25 to the finish point “E.” If you add all these up together, -50+350-25=+275, which is the change in height from A to E. You just add up the changes and you get the difference between the start and finish points. Enthalpy works just like that. You could make variation on this in which you don’t slide from D to E. Instead, you have a friend that has climbed from E to D and you get his ΔH. He records a +25 feet. You just change the sign, since you are interested in the other direction. Then you add it up as before. That would work too. Ok, so how do you use Hess’ law exactly? You can find a path between two points in which you can measure the ΔH of each of the steps along the way, then add them up. How do you use this? The first way is, as I said, to determine the ΔH of several reactions (using a calorimeter) that can be added up to the one you want, as you have done already. Here is a good example:Suppose you want to know the change in enthalpy of the reaction: C(graphite)+2H2→CH4 (which I'll call the "target reaction") but cannot make the reaction happen in a calorimeter. Well, consider these reactions (Guide reactions): Reaction 1 C(graphite)+O2(g)→CO2(g) Reaction 2 2H2(g)+O2(g)→2H2O(l) Reaction3 CH4(g)+ 2O2(g)→ CO2(g)+2H2O(l) Each of these is easy to run in a calorimeter and get the ΔH. If need be, I can easily change the sign of the ΔH I get to determine the enthalpy of running the reaction in the opposite direction. So, I can re-write them so that the products of each are reactants of another. The goal is to line them up so that the product of the target reaction is on the product side of one of the guide reactions, and the reactants of the target are on the reactant side of their equation. Change the sign of the ΔH if you have to flip a reaction. Also, if you need to multiply through by some amount to get the right amount of product or reactant, make sure you multiply the ΔH as well.C(graphite)+O2(g)→CO2(g) 2H2(g)+O2(g)→2H2O(l) CO2(g)+2H2O(l) → CH4(g)+ 2O2(g) You can add these three, canceling out things that are on both sides, and you get the target reaction:C(graphite)+2H2→CH4 Conceptually, it is as if I am doing the target reaction in steps: Graphite burns in oxygen to produce CO2 (reaction 1) Hydrogen and Oxygen combine to make water (reaction 2) and the CO2 and water (products of reactions 1 and 2) combine to make CH4 (reversed reaction 3). Now, I’m going to do a little sleight of hand: That last step actually produced the oxygen used in reactions 1 and 2. That may not make sense sequentially, but mathematically, it works out. If you prefer, we just “regenerate” the oxygen, so there is no net change in the oxygen content. Mathematically, it’s easy, once you’ve followed the steps to generate the three reactions, you just add up the three reactions, crossing out anything that appears on both sides of the final equation and you get the target reaction. That means that you can add up the ΔH’s of the reactions (be careful with signs and stoichiometry) and get the ΔH of the target reaction.The second way is to use a table of ΔHo f assembled from lots of previous work. Using enthalpies of formation to determine the enthalpy of a reaction A standard formation reaction is the formation of a compound from its constituent elements (in their most stable form). Sometimes that reaction can be measured directly, and sometimes it must be calculated using other, more easily measured reactions. For example, the formation of CO2 from carbon and oxygen would be this:C(graphite) + O2(gas) → CO2(gas)You can measure that easily. But, the formation of CH4 from Carbon and Hydrogen does not happen (but, we know how to determine it, as shown above).For any reaction that you just don't feel like running, you can use the tables of ΔHo f to predict it. In this case, you take the path from the reactants to the products through their elements. So consider the following:CH3OH + 3/2O2→ CO2 + 2H2O This is the target reaction, if you will. Now, I could just burn the ethanol. But, I can also imagine the following path: CH3OH→2H2 +Cgraphite + 1/2O2 (the reverse of the formation reaction for CH3OH, the ΔH of which can be found in a table).Cgraphite + O2→CO2 (Formation reaction of CO2) andH2 + 1/2O2 → H2O (formation reaction of H2O) You get the numbers for each of the reactions from the table. The CH3OH is converted to 2H2 +Cgraphite + 1/2O2 which combines with the oxygen to form the CO2 and 2H2O. Mathematically, you can do it like this: Add up the for the products (CO2 and 2H2O) and subtract the sum of the for the reactants. That's really the same as we have been doing. You are really just adding up the ΔHof, but changing the sign of the ones that "deconstruct" the reactants.Make sure you keep track of the number of moles of each. That’s the concept. you really have to practice to get it down.There is a webassign due on Wednesday. }

# Thermochemistry

16/01/19 16:51

## Intro and principles

All types of energy are interconvertible. The when a reaction takes place that releases energy, such as the burning of paper in air to produce CO

_{2}and H

_{2}O, or the reaction of Fe

_{2}O

_{3}with Aluminum (thermite), there is energy required to break the bonds initially, and energy released when the new bonds are formed. If the energy of the new bonds is lower on the potential energy hill, there will be a corresponding release of energy to the surroundings. Where does the energy to break the initial bonds come from? Usually it is kinetic energy. It can be kinetic energy from the movement of the molecules themselves. That increases with higher temperature. We can start paper burning by raising the temperature of the paper until the molecules collide with enough energy to break the bonds (which allows the carbon and hydrogen to combine with oxygen in the air to make CO

_{2}and H

_{2}O). I can also just use kinetic energy as in the energy I put into moving the two balls that smash into each other. In the second case, the molecules get enough kinetic energy by virtue of riding on the canon balls. It’s all the same.

## System and Surrounding

Energy comes in many forms, and can be converted among the forms, but not created or destroyed (this is known as the First Law of Thermodynamics). In order to keep track of energy, we need to define how to follow the changes and define a set of terms.

Here’s how: There is the

*system*, which is the reaction or change at the center of what we are following; there is the rest of the universe, or simply, the

*surrounding*corner of the universe. We are on the outside, watching the system. So, we are the universe, and we report what happens to the system. While there are many ways that we could follow changes in energy (in a car, we can talk about the work done in moving the car, which is the result of release of chemical potential energy in the bonds of gasoline and oxygen being made available when converted to carbon dioxide and water). But, as a practical matter, it’s much easier to follow heat. Heat is abbreviated "q." Under carefully controlled conditions heat (q) and change in enthalpy (ΔH) are just about the same.

**Heat**is the flow of thermal energy from one body to another. Generally, it is good enough to follow heat as a stand in for enthalpy change. We can never measure the enthalpy of a particular state of matter. But we can measure

*changes*in the enthalpy as matter changes by following how it affects the surroundings. Because of conservation of energy, we assume that whatever heat is gained by the surrounding must have been lost in the potential energy of the bonds in the system. Similarly, if energy is lost by the surroundings (it cools down), that energy has to have been gained somewhere, and it is the form of an increase in chemical potential energy in the system.

### The annoying bit

Okay, so here’s the somewhat odd part: we can only directly measure the changes in the surroundings; however, we like to report what has happened to the system. So, let’s say the surroundings (the water in a calorimeter, or your hand, for example) gain in heat. That must mean the system (the chemical reaction)

*lost*that same amount of energy. So we report the change in enthalpy as negative (the system lost energy...yes, the surroundings gained, but we report what happened to the system). So, the reactants must have higher potential energy than the products, in this case. So this is a

*negative change in enthalpy*from the standpoint of the system. We call this a ‒ΔH. Since energy is coming out of the reaction, we call it “exothermic.” On the other hand, if the surrounding loses energy, that energy must have gone into the system. The system must have gained energy (+ΔH) or

*endothermic*.

**Stoichiometry**You can use stoichiometry to calculate enthalpy in or out. For an endothermic reaction, it is as though the enthalpy is a reactant that must be put in, while for an exothermic reaction, it is like a product. If it takes 44 KJ to convert one mole of water from liquid to vapor, then converting two moles would be 88KJ. So, this is no harder than any other stoichiometry (no easier, unfortunately). You just figure out how many moles you have of whatever thing you are being asked to burn, or whatever, then multiply it by the number of KJ/mol.

**How do you know the enthalpy of a conversion**? For the most part, you can only do that with calorimetry. This basically runs the reaction or state change either in water or in some other material whose temperature you can measure, called the calorimeter. You run the reaction in a calorimeter and determine the heat flow to the water. If the temperature of the water in the calorimeter goes up, the reaction lost that heat (exothermic reaction). If the water temp goes down, that heat left the water and went into the system (endothermic). If the temperature of the water goes up, heat flowed into the water from the reaction, which must therefore have

*lost*that energy. The reaction is therefore exothermic. For reasons I would rather not get into in detail, if we measure heat under constant pressure, we know that the heat is the same as the enthalpy of the reaction. Just pay attention to the sign. Suppose you run a reaction in a calorimeter and find that 100.0g of water in it increased from 21.2 oC to 32.5 oC. Suppose I run a reaction and determine that reacting 0.5 mole of some chemical with another resulted in the change of temperature above (which will correspond to about 4723 Joules (I'll show you tomorrow how to do that) What would be the ΔH of the reaction per mol of reactant?

## Heat and Work

In a calorimeter, as I said, we aim to be totally

*inefficient*. That is, have all the energy come out as heat and do no work (or, a constrained type called PV work).

But, the total energy change of the system that is not in such a calorimeter has to be viewed as the total heat and work.

So, the explosion that goes on in the cylinder of your car's engine heats up the engine block, which heats up the radiator fluid, which passes the heat to the atmosphere. I can quantify that. But also, there is work done moving the car. Or, more directly, there is work done as PV work (remember, PV has units of energy).

Total energy change is heat plus work:

E

_{T}=

*q+w*

That's true no matter how you measure the work. However, when using PV work, there is a sign convention you must follow. Since work done by the system on the surroundings will

*increase*the pressure or volume (imagine an explosion in a cylinder), ΔPV (or, at constant pressure, PΔV) has a positive value. But, since it's work done by the system on the surroundings, we want that to have a negative. So, we change the sign. So w= -PΔV and

E

_{T}=

*q-*ΔPV or q-PΔV if pressure is constant.

# GasLaw Smartboard

11/12/18 07:46

Here is a link to the smart board entries, as requested. It has some other stuff on it, including some redox. Read More…