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
(graphite)+O2(g)CO2(g)
Reaction 2 2H
2(g)+O2(g)2H2O(l)
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.

C
(graphite)+O2(g)CO2(g)
2H
2(g)+O2(g)2H2O(l)
CO
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:

C
(graphite)+2H2CH4

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:
C
(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:
CH
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).
C
graphite + O2CO2
(Formation reaction of CO2) and
H
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
of
, 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.