Using Hess's law

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 +1/2O2 H2O(l). Note that I use 1/2O2 because I want the equation to be for one mole of the compound, water. The formation of CO2 is just the burning of Carbon (as graphite) with O2 to make CO2 The ΔHfo 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’ Law

Hess’ 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)+2H2CH4 (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
Reaction 2 2H
Reaction3 CH
4(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.

2(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:


Conceptually, it is as if I am doing the target reaction in steps: Graphite burns in oxygen to produce CO
2 (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 ΔH
o 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:
(graphite) + O2(gas) CO2(gas)
You can measure that easily. But, the formation of CH
4 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 ΔH
o f to predict it.

In this case, you take the path from the reactants to the products through their elements. So consider the following:
3OH + 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: CH
3OH2H2 +Cgraphite + 1/2O2
(the reverse of the formation reaction for CH3OH, the ΔH of which can be found in a table).
graphite + O2CO2
(Formation reaction of CO2) and
2 + 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 ΔH
, 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.


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 CO2 and H2O, or the reaction of Fe2O3 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 CO2 and H2O). 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:
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
T=q-ΔPV or q-PΔV if pressure is constant.

GasLaw Smartboard

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

Gas Laws Intro

Gas Laws

We went through a lot of derivations, but the take-home lessons are really not that complicated. There are three basic rules relating to the volume of a gas, all of which are intuitive.
  1. If you add more molecules (or moles, so that they are easier to count) to a balloon, the volume increases. If you remove molecules, the volume decreases.
  2. If you raise the temperature of a gas in a balloon, the volume will increase. If you decrease the temperature, the volume will decrease.
  3. If you increase the pressure on a balloon, the volume will decrease. If you decrease the pressure on the outside of a balloon, its volume will increase.
Okay, these things are obvious, as I said. Now we need to come up with ways to quantify these relationships. We will derive equations.

Volume versus moles

Increasing the number of moles of gas will increase the volume. In fact, if you double the number of moles, the volume will double. You can plot this on a graph of Volume (on Y) versus “n” (number of moles, on X), you get a straight line. This is a simple process. Put in some air, measure the volume. Then put in more, measure the new volume. You then plot the volume (in liters) at each number of moles. The slope of that line is in units of L/mol and is a proportionality constant relating volume to moles. The equation is V = k(n), where k is the slope of the line. A plot is included below. The number (22.4L/mol) is what the slope of the line would be for a gas at 1.00 atmosphere of pressure and 273K. You may note from that statement that we would have a different slope if we did the same experiment at a different temperature or pressure. This equation works only if temperature and pressure are constant. At other temperatures or pressures, you would have different equations.

Volume versus Temperature

If you increase the temperature, the volume also should increase. If we use the absolute scale, we get a graph that looks just like the one we got for volume versus “n.” In this case, the line would be V = k
2*T where k2 is the slope of the line in Liters/Kelvin (it’s critical that we use Kelvin for this application). If you use Celsius, you would get the following graph

You may notice that there is a point at which the volume is predicted to go to zero. We really don’t expect that to happen, of course. Sooner or later, we would expect that the gas would condense and become a liquid, or even a solid, at which point the volume would not decrease much more. However, that point at which the volume is predicted to go to zero is -273.15 C, which we define as “absolute zero, or 0.0 Kelvin (0.0 K). If we re-plot the same data using the Kelvin scale (just add 273 to degrees celsius), we get the following plot:

Using Kelvin, you have no negative temperature values. Also, we get a straightforward relationship that if we double the temperature (in K), we double the volume. Once again, we get a straight line with a slope in units of Liters/Kelvin (liters per Kelvin). That slope is the proportionality constant. Again, this line is derived for a particular number of moles and particular pressure.

Volume versus Pressure

If you increase the pressure, in this case, the volume will decrease, while volume will increase if the pressure is decreased. This is known as an “inverse relationship.” So, to get a straight line for this plot, you have to plot Volume versus 1/P.

You can the equation V=k3*1/P, which can also be written as P*V =k3, where k3 is the slope of the line and has units of Liter * atmospheres (or liters x whatever unit we use for pressure). All three of these “special” gas laws work only if the other two quantities are held constant.
In other words, the line for volume being proportional 1/P assumes that the temperature and number of moles are constant.

Ideal Gas Law

It would be great if we could derive a single equation that relates all of the parameters. That turns out to be pretty easy: you multiply the three equations and simplify the result. In the end, you get PV=nRT The ideal gas law. P is “pressure (usually in atmosphere in this class), V is volume (always in Liters in this class), ”n“ is number of moles and
T is temperature ALWAYS in Kelvin). R is the Ideal Gas constant. The units of R, for us, will usually be liter*atmospheres/mol*Kelvin. In those units, the value is 0.08206 L*atm/mol*K. Using this equation is really quite simple. If I give you any 3 of the unknowns, you can solve for the fourth. For example, if I tell you that pressure is 1.2 atm, temp is 278K and volume is 22.0L, you can calculate how many moles of gas there are in the balloon. As you would for any equation, you just rearrange the equation to get the unknown thing on one side, and leave all the known parameters on the other. Then plug in the values.

Using the equations:

Using this equation is really quite simple. If I give you any 3 of the unknowns, you can solve for the fourth. For example, if I tell you that pressure is 1.2 atm, temp is 278K and volume is 22.0L, you can calculate how many moles of gas there are in the balloon.
As you would for any equation, you just rearrange the equation to get the unknown thing on one side, and leave all the known parameters on the other. Then plug in the values. so, in this case, you are solving for “n” and the equation is which would be.

Notice that the units cancel properly.

Special Laws

There are also times in which I will not give you enough of the unknowns for you to solve for the last one. Instead I will ask you to tell me the change in one of the parameters given a change in some others. An example would be if I had a 0.57L balloon that was at 285K and I increased the temperature to 305K, what would be the new volume? Once again, the approach would be to rearrange the equation to get all of the parameters that are changing (in this case volume and temp) on one side and the ones that are not changing on the other. So, PV=nRT becomes:

At the first condition, the value is:

But, the values of n, R and P are all constant, so the value of nR/P has to be constant also. That means, that for any temperature and volume, the quotient of V/T has to be constant unless . At the new temperature, 305K, V/305=0.002. You just solve for V.
If you want to skip a step or two, you just set the two conditions equal to each other. That is, you don't even have to know that V/T=.002. Even if you didn't know the pressure and couldn't solve for the value, you would still know that V/T would be constant, as long as P or n do not change. So, as long as that is true:

Just solve for V2.

You can do the same for other "special" cases. Suppose I am holding temperature and number of moles constant, but changing pressure. Let's say, I put a sealed balloon in an oven (code phrases: sealed means "n" is constant…"rigid" means "V" is constant..which doesn't apply here).
You have 0.45L of volume in the balloon at 1.0 atm. Assuming temperature is constant, what is the new volume if I increase the pressure in the room to 1.8atm?
Here, we know n, R and T are constant. PV is changing. So, even though I don't know the temperature and therefore cannot solve explicitly for "n," I do know that P times V will remain constant, just as I showed in the demo today.
So, you can write P
1V1=P2V2 and solve for the unknown.
In this case, PV=1.0x0.45=1.8xV
2. V2=0.25L

RedOx II

Added Balancing Redox in Base
I'm being pretty lazy here and using the Washington University, St. Luis website

Read More…