Chemistry 103A; Sections 5, 6, 7, 8; Lecture 15, 27 Sep 00

Heats of Chemical Reactions - Summary

We have agreed that a chemical reaction equation will always be written in the form:

reactants ® products.
We have further agreed that the reactants are isolated, pure, at temperature, T, and pressure, p, and that the products are also isolated, pure, and at the same temperature and pressure.

Any heat involved in the reaction is measured and removed (or added) so that the products will be at the same temperature as the reactants.

One thing we didn't tell you. All components are assumed to be in their standard state. The standard state of a component is the most stable state of that component at the given temperature and pressure.

For example. At 25oC and 1 atmosphere pressure the most stable state of water is H2O(l). The most stable state of carbon is graphite and not diamond or buckminsterfullerene. The most stable state of oxygen is O2(g), and so on.

We usually symbolize quantities that refer to the standard with a superscript, o, so that heat of reaction for reactants and products in their standard states is usually written, DHo.
 
 

How To Measure Heats of Chemical Reactions

A common way to measure heats of chemical reactions is to use a "bomb calorimeter." A bomb calorimeter is a sealed metal container in which the reaction is carried out and the container (with its reaction mixture) are immersed in a large insulated water bath.

The reaction is allowed to run in the sealed container and any heat involved goes into (our out of) the water in the water bath.

Since we know the specific heat of the water in the water bath and we know how much water is in the water bath. We can find out how much heat the reaction generated or used by measuring the temperature change in the water bath.

(We would like the water bath to be large so that the temperature change is small. Recall that ideally we want the final temperature of the products to be the same as the initial temperature of the reactants.)

Here's an example of such an experiment.

We have a bomb calorimeter apparatus with heat capacity 3576 J/K. (This means that it takes 3576 J to raise the temperature of the apparatus by 1.00 K.)

We place 0.4539 g of glucose in the apparatus with an excess of O2 in order to run the reaction:

C6H12O6(s) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l).
We find that the temperature of the apparatus increases by 1.978 K. What is the DH for the above reaction in kJ/mol? (FW of glucose is 180.16 g/mol.)
 

State Functions

Several of the variables which we use to describe materials belong to a class of variables (functions) which we call state functions. Some of the state functions we have used or defined so far are: T, p, E, H, V, etc.

These variables or functions are characterized by the property that changes in these properties do not depend on how the change was made.

That is, for example, DT for a process does not depend on how the process was carried out. It only depends on the initial and final temperatures. Thus we wrote:

DT = Tfinal-Tinitial.
In other words, the change in temperature does not depend on how fast or how slow we heated our sample. It does not depend on any of the details of how we got from Tinitial to Tfinal, it just depends on the difference in the two quantities.

The same thing is true of any and all state functions. DT, Dp, DE, DH, DV, etc. are all "independent of path."

There are some quantities which DO depend on path. q and w depend on how we get from the initial state to the final state. It is beyond the scope of this course to prove this statement, but it is true.

Interestingly, the sum, q + w, is independent of path even though the individual q and w are not.

The fact that H is a state function is of great value to us. It allows us to run a reaction which gets hot, then remove the heat and separate the products to end up at T and p which are the same as the initial T and p and get a DH for the reaction.

This is because DH doesn't care how we got from reactants to products, it just cares what the reactants and products are.
 

Hess's Law

There are probably hundreds of thousands of possible chemical reactions. It is not feasible to try to measure the heats of reaction for all possible reactions that we might want to know.

We know that we can always add two or more (balanced) reactions to get an overall reaction. For example,

Reaction 1: a A ® c C

Reaction 2: b B ® d D

can be added to give an overall reaction,
Reactions 1 + 2: a A + b B ® c C + d D.
This doesn't look like it gained us very much, but consider the following practical example:
Reaction 3:          2 C + O2 ® 2 CO

Reaction 4:          2 CO + O2 ® 2 CO2.

Both of these reactions have commercial and environmental importance.

Adding reactions 3 and 4 we get

2 C + 2 CO + 2 O2 ® 2 CO + 2 CO2,
but the CO's on either side cancel out to give us,
2 C + 2 O2 ® 2 CO2,
which also has commercial importance.

Hess's law says that when you add (or subtract) chemical reactions you can also add (or subtract) the corresponding DH's (or DHo's) to get a DH for the overall reaction.

In the above case this would look like:

Reaction 3:          2 C + O2 ® 2 CO     DH3o = - 221.050 kJ

Reaction 4:          2 CO + O2 ® 2 CO2.     DH4o = - 565.968 kJ

Adding reactions 3 and 4 (and canceling the 2 CO) we get
Reaction 5:          2 C + 2 O2 ® 2 CO2,
with DH5o = DH3o + DH4o = - 787.018 kJ.

Caution: Sometimes we must multiply one or more of the reactions we are adding (or subtracting) to allow some of the components to cancel.

For example:

Reaction 6:         S8 + 8 O2 ® 8 SO2     DH6o = - 2374.64 kJ

Reaction 7:         2 SO2 + O2 ® 2 SO3     DH7o = - 197.780 kJ.

In order for the SO2 to cancel out when we add these reactions we must multiply Reaction 7 by 4. If we do this:
Reaction 6 + 4 Reaction 7 = Reaction 8,

Reaction 8:          S8 + 12 O2 ® 8 SO3.

And DH8o = DH6o + 4 DH7o = - 3165.76 kJ.
 

Heats of formation

Hess's law is useful for doing the kind of calculations above, but it is most useful in providing a mechanism for calculating DHo for any reaction we desire using a relatively small data base (table) of selected DHo's.

The selected DHo's are called heats of formation.
The heat of formation of a compound is defined as DHo for the reaction which forms one mole of the compound from pure isolated elements in their standard states.

(It stands to reason that the heat of formation on an element in its standard state is zero. For example, the heat of formation for oxygen gas would be defined as DHo for the reaction

O2(g) ® O2(g).
Since this reaction is not a reaction, that is, nothing happens, it must be true that its DHo = 0.)

A few sample formation reactions are,

For liquid water:
H2(g) + 1/2 O2(g) ® H2O(l)      DHof = - 241.8 kJ/mol
(The values of these heats of formation are taken from your text. There is a short table of heats of formation on p270 and much longer one in Appendix L, starting on page A-31.

Note that the units are kJ/mol, because we are talking about the heat of formation of one mole of the compound.

Note also that this is the only place in this course where you can get away with coefficients in your balanced reactions which are not integers.)

For methane gas:
C(graphite) + 2 H2(g) ® CH4(g)   DHof = - 74.81 kJ/mol.
Here's an interesting one:
N2(g) + 2 H2(g) ® N2H4(l)      DHof = + 50.63 kJ/mol
(The formation of hydrazine is an endothermic process.)

So, how do we use these heats of formation?

Here's the general formula. Given a hypothetical reaction:

a A + b B ® c C + d D,
(where a, b, c, and d are small integers which balance the equation and A, B, C, and D are compounds),
DHorxn = c DHof,C + d DHof,D- a DHof,A- b DHof,B .


Example calculations:

2 NH3(g) ® N2H4(l) + H2(g)
NH3(g)     - 46.11 kJ/mol

N2H4(l)     + 50.63 kJ/mol


2 Fe2O3(s) + C(graphite) ® 4 Fe(s) + 3 CO2(g)

 
Fe2O3(s)   - 824.2 kJ/mol

CO2(g)     - 393.506 kJ/mol


Here's the reaction that keeps us alive:

C6H12O6(s,a-D-glucose) + 6 O2(g) ® 6 CO2(g) + 6 H2O(l)
C6H12O6(s,a-D-glucose)    - 1274 kJ/mol

H2O(l)     - 285.830 kJ/mol