CHM 101 GENERAL CHEMISTRY
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FALL
QUARTER 2008 |
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Section
2 |
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Lecture
Notes |
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(last revised: |
6.3
Hess Law: Hess Law takes advantage of the fact that enthalpy is a state
function. Hess Law applies to the change in enthalpy that takes place when
some set of reactants is converted to some set of products. Simply stated,
Hess Law says:
The change in enthalpy (ΔH) is the
same whether the process takes place in a single step or in a series of steps.
- An Example of Hess Law (pp. 242-3): Let us consider the oxidation of
N2 to produce nitrogen dioxide (NO2). We can write
it as a single-step reaction in which 2 oxygen molecules combine with 1
nitrogen molecule to form 2 nitrogen dioxide molecules:
N2 (g)
+ 2O2 (g) > 2NO2
(g) ΔH1 = 68 kJ
The alternative is to regard
the reaction as taking place in two steps. In the first step, 1 molecule of
oxygen combines with 1 molecule of nitrogen to form 2 nitrogen monoxide
molecules. Then in the second step, a second oxygen molecule combines with the
2 nitrogen monoxide molecules to form the final products, 2 nitrogen dioxide
molecules. We write separate equations (and enthalpy changes) for the two
steps, then we add them together to obtain the final,
equation and enthalpy change for the overall reaction:
|
N2 (g) + O2 (g) |
> |
2NO
(g) |
ΔH2 = |
180 kJ |
|
2NO (g)
+ O2 (g) |
> |
2NO2
(g) |
ΔH3 = |
112 kJ |
|
N2 (g) + 2O2 (g)
|
> |
2NO2
(g) |
ΔH2 + ΔH3 = |
(180 112) kJ |
|
N2 (g) + 2O2 (g) |
> |
2NO2
(g) |
ΔH1 = |
68 kJ |
Note that the net reaction, and its enthalpy change are identical, whether we
run the reaction in a single step or in two steps. A comparison of the one step
process versus the two step process is shown in Figure 6.7 from your text:
While the foregoing example
demonstrates Hess Law, it does not prove it. Any proof lies in the consistency
between Hess Law and the First Law.
- Characteristics of Enthalpy Changes: Suppose we know the stoichiometry and the
change in enthalpy for some particular reaction. Take, for example:
Xe (g) +
2F2 (g) > XeF4
(s) ΔH = 251 kJ
Suppose we needed to use
this reaction in a Hess Law problem, but at double the quantities of reactants
and products. What happens to the change in enthalpy? Because enthalpy change is
an extensive property, it is proportional to the amounts of material. In this
case, we know the enthalpy change for the production of 1 mole of XeF4
(s). So, if we make 2 moles instead
of 1, we double the change in enthalpy. The equation for that case is:
2Xe (g) +
4F2 (g) > 2XeF4
(s) ΔH = 502 kJ
Suppose instead, we
needed to use the (first) reaction in another Hess Law problem, but in
reverse? What then, would be the change in enthalpy? In the (first) reaction
the system transfers 251 kJ of heat to the surroundings, so one would expect
that the reverse reaction would require that the same amount of heat would be
transferred from the surroundings into the system. Thus ΔH would have the same magnitude, but the opposite
sign:
XeF4 (s) > Xe (g) + 2F2
(g) ΔH = +251 kJ
These characteristics
apply in general, and can be expressed as rules:
o 1) The magnitude of ΔH is directly proportional to the quantities of reagents (reactants and
products) involved in the reaction. If the coefficients in the reaction are
multiplied by some factor, the enthalpy change must be multiplied by the same
factor.
o 2) If the direction of the reaction is reversed,
the sign of ΔH is
also reversed. (Note that you could derive this rule from the first rule by
using 1 as the factor by which the coefficients and ΔH are multiplied.)
- Sample Exercise 6.7 (p. 244): Graphite
and diamond are two allotropic forms of carbon shown in
the following illustration:
Their properties are
strikingly different as are their molecular structures. Graphite is a soft,
black, slippery substance used for lubrication and as pencil lead; Diamond is
a brilliant, crystal-clear substance used for gemstones and as an industrial
abrasive. The enthalpies of combustion for graphite and for diamond are,
respectively, 394 kJ/mol and 396 kJ/mol. Calculate ΔH for the conversion of graphite to diamond. (This
problem illustrates the use of Hess Law to compute an enthalpy change that is
most impractical to measure directly.)
o We start by writing equations for the conversion
reaction and for the two combustion reactions:
|
C (graphite) |
> |
C (diamond) |
ΔH = |
? |
|
C (graphite) + O2 (g) |
> |
CO2 (g) |
ΔH = |
394 kJ |
|
C (diamond) + O2 (g) |
> |
CO2 (g) |
ΔH = |
396 kJ |
o Now we solve the problem by reversing the reaction
for the combustion of diamond and adding it to the reaction for the combustion
of graphite:
|
C (graphite) + O2 (g) |
> |
CO2 (g) |
ΔH = |
394 kJ |
|
CO2 (g) |
> |
C (diamond) +
O2 (g) |
ΔH = |
+396 kJ |
|
C (graphite) |
> |
C (diamond) |
ΔH = |
(396 394) kJ |
|
C (graphite) |
> |
C (diamond) |
ΔH = |
2 kJ |
- Sample Exercise 6.8 (p. 245): (This exercise illustrates the kind
of systematic procedure often necessary for the solution of a complex
Hess Law problem.) Calculate ΔH for
the synthesis of diborane (B2H6 (g)) from its elements, boron (B (s)) and hydrogen (H2 (g)):
2B (s) +
3H2 (g) > B2H6 (g)
Use the following data for
the calculation:
|
(a) |
2B (s) + 3/2O2
(g) |
> |
B2O3 (s) |
ΔH = |
1273 kJ |
|
(b) |
B2H6 (g) + 3O2
(g) |
> |
B2O3 (s) + 3H2O(g) |
ΔH = |
2035 kJ |
|
(c) |
H2 (g) + ½O2 (g) |
> |
H2O (l) |
ΔH = |
286 kJ |
|
(d) |
H2O (l) |
> |
H2O(g) |
ΔH = |
+44 kJ |
o Our objective is to combine the four reactions so
that we assemble our reactants, B (s)
and H2 (g), on the left
hand side and our product, B2H6 (g), on the right, while we make all the intermediate reagents cancel
out. The solution in the text begins by combining (a) with the reverse of (b),
and that is perfectly valid. But it is also perfectly valid to begin by
combining (a) with 3 x (c), and that is what we will do.
|
(a) |
2B (s) + 3/2O2
(g) |
> |
B2O3 (s) |
ΔH = |
1273 kJ |
|
3 x (c) |
3H2 (g) + 3/2O2 (g) |
> |
3H2O (l) |
ΔH = |
3(286 kJ) |
|
(1) |
2B (s) + 3H2 (g) |
> |
B2O3 (s) |
ΔH = |
1273 858 kJ |
This accomplishes the objective
of assembling the right amounts of both reactants on the left hand side.
o Now we can continue by combining our first
intermediate result (1) with the reverse of (b).
|
(1) |
2B (s) + 3H2 (g) |
> |
B2O3 (s) |
ΔH = |
2131 kJ |
|
(b) |
B2O3 (s) + 3H2O(g) |
> |
B2H6
(g) + 3O2 (g) |
ΔH = |
+ 2035 kJ |
|
|
2B (s) + 3H2 (g) |
> |
B2H6
(g) |
ΔH = |
96 kJ |
|
(2) |
2B (s) + 3H2 (g) |
> |
B2H6
(g) |
ΔH = |
96 kJ |
This accomplishes the
objective of assembling the right amount of product on the right hand side. It
also gets rid of most of the intermediate reagents. But we still need to
eliminate 3H2O(g) from the left hand side and 3H2O
(l) from the right.
o Now we can finish by combining our second
intermediate result (2) with 3 x reaction (d).
|
(2) |
2B (s) + 3H2 (g) |
> |
B2H6
(g) |
ΔH = |
96 kJ |
|
3 x(d) |
3H2O (l) |
> |
3H2O(g) |
ΔH = |
+3x44 kJ |
|
|
2B (s) + 3H2 (g) |
> |
B2H6
(g) |
ΔH = |
96 + 132 kJ |
|
|
2B (s) + 3H2 (g) |
> |
B2H6
(g) |
ΔH = |
+36 kJ |
o We are done! The synthesis of one mole of diborane
from its elements is slightly endothermic. Assuming that all the measured
enthalpy changes for reactions (a) (d) were made under standard conditions (1
atm pressure and 25 °C), we have thus determined that the standard enthalpy of formation of
diborane is +36 kJ/mol. We will discuss standard enthalpies of formation in
detail in the next section.
6.4
Standard Enthalpies of Formation: We saw in Section 6.2 that constant pressure
calorimetry is a method we can apply to make measurements of enthalpy changes (i.
e., heats of reaction) for reactions that take place in solution under ambient
pressure. Additionally, constant volume calorimetry (into which we did not
delve to any great extent) can be used to determine heats of combustion for a
wide variety of substances. But what do we do about reactions that do not lend themselves to calorimetry? How do we obtain enthalpy changes
for such reactions? (A good example is the reaction for the conversion of
graphite into diamond, where we obtained the enthalpy change from the heats of
combustion of graphite and of diamond.) Would it help if someone has already
tabulated the enthalpy changes for a list of reactions that encompass all the
compounds we might ever be interested in? Actually, there is such a list in the
back of your text.
- Standard Enthalpy of Formation, a Definition: The standard
enthalpy of formation (ΔHf°) of a compound is defined as the change in enthalpy that accompanies the
formation of one mole of the compound from its elements with all
substances in their standard states. The degree sign in the
symbol, ΔHf°, indicates that this is the case.
- Standard States: The definition for standard enthalpy of
formation stipulates that all substances be in their standard states. So
we need to define the conditions that make a state standard:
o All
Substances: A temperature of
exactly 25 °C is required.
o Any
Gaseous Compound: The pressure
must be exactly 1 atmosphere. (This may soon change. The IUPAC has adopted a
new standard pressure of exactly 1 bar, or 100,000 Pa. By comparison, 1
atmosphere = 101,325
o Any Pure
Substance in a
o Any
Substance in Solution: The
standard state is a concentration of exactly one mole per liter (1 M).
o An Element: The standard state of an element is the form in
which the element exists under the conditions of 1 atmosphere pressure and 25 °C. (The
standard state for oxygen is O2 (g)
at 1 atm.; the standard state for carbon is graphite; the standard state for
mercury is Hg (l); the standard state
for sodium is Na (s), etc.)
- Examples of Standard Enthalpies of Formation: We follow the discussion on p. 247 of your
text:
o Nitrogen
Dioxide (NO2 (g)): We write the reaction for the
formation of one mole of product:
½ N2 (g) + O2 (g)
> NO2 (g) ΔHf° = 34 kJ/mol
The standard states are all
gases at 25 °C. The picture shows a sample of NO2 (g) in a
o Methanol (CH3OH (l)): We write the reaction for the formation of one mole of
product:
C (s) +
2H2 (g) + ½ O2
(g) > CH3OH (l) ΔHf° = 239 kJ/mol
The
standard state of carbon is graphite; the standard states of hydrogen and
oxygen are both gases; and the standard state of methanol is the liquid, with
all reagents at a temperature of 25 °C.
o Oxygen (O2 (g)): We write this reaction as:
O2 (g)
> O2 (g) ΔHf°
= 0 kJ/mol
Notice
that the enthalpy of formation of oxygen, or any other element, in its standard
state is necessarily zero, by virtue
of no real reaction having taken place.
o Lists of
Values of Standard Enthalpies of Formation: A short list is given in Table 6.2 in the text:
A
more complete list can be found in Appendix 4, on pp. A19-A22. The list
includes two other properties, ΔGf°,
and S°. We will discuss these properties later in
the year.
- Using Standard Enthalpies to Calculate
Standard Enthalpy Changes
(Heats of Reaction): We will follow the example from the text, the
combustion of methane. We wish to calculate the standard enthalpy of
combustion:
o We start by writing the reaction. (It should be an
old friend by now.)
CH4 (g)
+ 2O2 (g) > CO2
(g) + 2H2O (l)
Note that the water is in
its liquid state. This follows from our requirement that the reaction
begin and end at 25 °C. This does not restrict it from taking place (as
we know it does) at a much higher temperature. We simply invoke the fact that
enthalpy is a state function, and we can pick any path, as long as it starts
with reactants at standard conditions and ends with products at standard
conditions.
o We can set the path so as to allow us to break the
reaction up into a series of reactions whose enthalpy changes can be found in
our tables. A particularly useful series of reactions is the one where we
assume that our reactants un-form themselves back to their elements (in
standard states) then re-form as the products. We see this diagrammed in Figure
6.8 from your text
o This analysis gives us the following set of
reactions:
|
(a) |
CH4 (g) |
> |
C
(s) + 2H2 (g) |
ΔHa° = |
75 kJ |
|
(b) |
2O2 (g) |
> |
2O2
(g) |
ΔHb° = |
0 kJ |
|
(c) |
C (s) +
O2 (g) |
> |
CO2
(g) |
ΔHc° = |
394 kJ |
|
(d) |
2H2 (g) + O2 (g) |
> |
2H2O
(l) |
ΔHd° = |
572 kJ |
Reaction (a) is the
reverse of the reaction to form 1 mole of methane from its elements. According to
Appendix A4, p. A20, methane has a heat of formation of 75 kJ/mol, so we change
the sign to get ΔHa° = 75 kJ.
Reaction
(c) is the reaction to form 1 mole of CO2
(g) from its elements and we can find
ΔHc° = 394 kJ in Table 6.2.
Reaction (d) is the reaction
to form 2 moles of liquid water from its elements, so we double the value of
286 kJ/mol we find in Table 6.2 for the standard enthalpy of formation of
water and obtain ΔHd° = 572 kJ.
(We have already
discussed why ΔHb° = 0 kJ.)
o Now we can recombine the four reactions and add the
4 enthalpy changes to get the final result:
|
(a) |
CH4 (g) |
> |
|
ΔHa° = |
75 kJ |
|
(b) |
2O2 (g) |
> |
|
ΔHb° = |
0 kJ |
|
(c) |
|
> |
CO2
(g) |
ΔHc° = |
394 kJ |
|
(d) |
|
> |
2H2O
(l) |
ΔHd° = |
572 kJ |
|
|
CH4 (g) + 2O2 (g) |
> |
CO2
(g) + 2H2O (l) |
ΔH° = |
891 kJ |
o Lets look at this problem another way:
|
|
CH4 (g) |
+ 2O2 (g) |
> |
CO2 (g) |
+ 2H2O (l) |
|
ΔHf° |
75 kJ |
0 kJ |
|
394 kJ |
286 kJ |
|
x |
1 |
2 |
|
+1 |
+2 |
|
ΔH° = |
+75 kJ |
+0 kJ |
|
394 kJ |
572 kJ |
|
ΔH° = |
891 kJ |
||||
The (balanced) equation
for the reaction is written in the first row.
The standard (molar)
enthalpies of formation then are written in the second row for each component
of the reaction.
The third row contains a
set of multipliers where the magnitude of each multiplier is the stoichiometric
coefficient from the reaction equation and the signs are for each reactant
and + for each product.
The fourth row contains
terms that are obtained by multiplying together the entries in the second and
third rows.
When the terms in the
fourth row are added together, the result (fifth row) is the standard enthalpy
change for the reaction, our desired result.
- Equation 6.1 (p. 249): The procedure we have just used to obtain the standard
enthalpy of combustion of methane can be generalized so that it applies to
any reaction where the initial and final states correspond to standard
conditions. The standard enthalpy
change for a given reaction can be calculated by subtracting the
enthalpies of formation of the reactants from the enthalpies of formation
of the products. (These enthalpies of formation are calculated for the
actual numbers of moles of each component in the reaction.)
Mathematically, this statement can be written as Equation 6.1:
ΔH°reaction
= Σnp
ΔHf°(products) Σnr ΔHf°(reactants) (6.1)
In Equation 6.1, the
symbol, Σ (sigma), means to compute the sum of the terms,
the nps represent the stoichiometric coefficients of the products,
and the nrs represent the stoichiometric
coefficients of the reactants.
The five row procedure
that we just used is a systematic and compact way to apply Equation 6.1 to a
standard enthalpy change problem. We listed the molar enthalpies of formation (ΔHf°) in the second row for each component in the
reaction, and we listed the stoichiometric coefficients (the nps
and the nrs) in the third row (with signs for the nrs). Then we computed the individual terms of the
summation in row four (np ΔHf° and nr
ΔHf°) and did the final summation in row five to get
the desired answer.
- Key Concepts for Performing Enthalpy
Calculations: Your text
offers the following summary (found in the middle of p. 249):
- Sample Exercise 6.9 (pp. 249-51): Compute the standard enthalpy
change for the reaction of ammonia burning in air to produce nitrogen
dioxide and water. Refer to Table 6.2 for standard enthalpies of
formation.
The text spreads this
calculation over two full pages; well do it in five lines:
|
|
4NH3 (g) |
+ 7O2 (g) |
> |
4NO2 (g) |
+ 6H2O (l) |
|
ΔHf° |
46 kJ |
0 kJ |
|
34 kJ |
286 kJ |
|
x |
4 |
7 |
|
+4 |
+6 |
|
ΔH° = |
+184 kJ |
+0 kJ |
|
+136 kJ |
1716 kJ |
|
ΔH° = |
1396 kJ |
||||
- Sample Exercise 6.10 (p. 251): Compute the standard enthalpy
change for the reaction of aluminum metal with iron (III) oxide to form
aluminum oxide and iron metal. (This is commonly called the thermite
reaction.)
We will again do it in
five lines:
|
|
2Al (s) |
+ Fe2O3 (s) |
> |
Al2O3 (s) |
+ 2Fe (s) |
|
ΔHf° |
0 kJ |
826 kJ |
|
1676 kJ |
0 kJ |
|
x |
2 |
1 |
|
+1 |
+2 |
|
ΔH° = |
+0 kJ |
+826 kJ |
|
1676 kJ |
+0 kJ |
|
ΔH° = |
850. kJ |
||||
- Sample Exercise 6.11 (p. 252): Compare the standard enthalpy of
combustion of one gram of (liquid) methanol (CH3OH (l)) with that of one gram of
gasoline. Assume that gasoline is pure liquid octane (C8H18).
We will first compute the
standard molar enthalpies of the two fuels, then
convert to enthalpies per gram for the comparison. The calculation for (one
mole of) methanol is:
|
|
CH3OH (l) |
+ 3/2O2 (g) |
> |
CO2 (g) |
+ 2H2O (l) |
|
ΔHf° |
239 kJ |
0 kJ |
|
394 kJ |
286 kJ |
|
x |
1 |
3/2 |
|
+1 |
+2 |
|
ΔH° = |
+239 kJ |
+0 kJ |
|
394 kJ |
572 kJ |
|
ΔH° = |
727. kJ |
||||
This result is the heat of
combustion of methanol in units of kJ/mol. To convert this to kJ/g, we divide
by the molar mass of methanol (32.0 g/mol). Thus the desired result is:
ΔH° = (727. kJ/mol)/(32.0 g/mol) = 22.7 kJ/g
The calculation for (one
mole of) octane (C8H18) is:
|
|
C8H18 (l) |
+ 25/2O2 (g) |
> |
8CO2 (g) |
+ 9H2O (l) |
|
ΔHf° |
269 kJ |
0 kJ |
|
394 kJ |
286 kJ |
|
x |
1 |
25/2 |
|
+8 |
+9 |
|
ΔH° = |
+269 kJ |
+0 kJ |
|
3152 kJ |
2574 kJ |
|
ΔH° = |
5457. kJ |
||||
This result is the heat of
combustion of octane in units of kJ/mol. To convert this to kJ/g, we divide by
the molar mass of octane (114.2 g/mol). Thus the desired result is:
ΔH° = (5457 kJ/mol)/(114.2 g/mol) = 47.8 kJ/g
(Since our data are
precise to 3 significant figures, we express this result to 3 significant
figures.) We have just demonstrated that the enthalpy of combustion of gasoline
(octane) is slightly more than twice that of methanol.
6.5
Present Sources of Energy
- But First, a Little History:
o Solar
Power The Ultimate Source of All Energy: All the energy we have used throughout human history comes from
stellar sources. Even our nuclear fuels were forged in stellar interiors. Most
of the energy we use today was captured from the sun by photosynthesis in prehistoric
times and laid down where it would be transformed into fossil fuels like
petroleum, natural gas, and coal. Even wind and water power are driven by solar
energy. Wind arises from solar energy heating the atmosphere and setting it
into motion. Water power is driven through solar heating that evaporates water
(mostly from the sea) which ultimately falls as rain and which can be channeled
to flow downhill through electric generators.
o The
Pre-Industrial Age: Biomass,
mostly wood, was the primary source of energy, and it was used mostly for
heating. There was also some exploitation of wind power (e. g., windmill-driven
pumps in
o The
Industrial Revolution: The
Industrial Revolution began in
o Petroleum: The first oil well went into production in 1859 to
supply kerosene for lighting when demand for lamp oil outstripped the supply of
oils derived from animal fat. Gasoline was an unwanted by-product until the
advent of automobiles. Another initially unwanted by-product was natural gas.
It was typically flared (burned) at the wellhead for lack of any system of gas
pipelines. In those days, gas for street and domestic lighting was made by coal
gasification in plants located near the urban areas they supplied.
o Sources
of Energy in the
- Petroleum and Natural Gas: Petroleum and Natural Gas are fossil fuels
consisting largely of hydrocarbons. With the singular exception of methane
(CH4), they feature carbon carbon bonds with chain lengths
that can be quite long. Table 6.3 contains the names and formulas of the
hydrocarbons with between 1 and 8 carbons:
o Natural
Gas: Natural Gas is largely
methane, but it also contains significant quantities of ethane, propane, and
butane, all of which are gases at room temperature. It is usually found in
combination with petroleum.
o Petroleum:
Petroleum (literally rock-oil)
is largely a mixture of liquid hydrocarbons ranging from pentane (C5H12)
to hydrocarbons containing 25 or more carbon atoms. Before petroleum becomes
really useful, it needs to be separated into its various fractions, by a
process called fractional distillation. These petroleum fractions and their
uses are listed in Table 6.4:
In the early days of the
petroleum industry, fractional distillation was the extent of petroleum
refining. The kerosene would be packaged and sold, but a substantial proportion
of the other fractions would have no market and would have to be discarded.
The advent of the
electric light and the automobile changed all that. Demand for kerosene
plummeted while demand for gasoline skyrocketed. The process of catalytic
cracking was developed to break kerosene and higher fractions down to
four-carbon (C4) compounds, and the process of reforming was
developed to react pairs of C4 molecules together to produce
desirable branched isomers of octane (C8H18).
(We will conclude Section 6.5 in class on November
19.)