Thermochemistry
Thermochemistry is the application of thermodynamics to
chemical systems. This field of study is particularly concerned
with the heat effects that accompany chemical reactions, the
formation and dilution of solutions, and phase changes.
The heat of a reaction is defined as the heat transferred between
the system (the reaction) and the surroundings when a known
amount of reactants react to form a known amount of products at
either a constant temperature and pressure:
T, P
reactants ------------> products
qp = DH
or a constant temperature and volume:
T, V
reactants ------------> products
qV = ?
Since most reactions are carried out at the constant ambient
pressure, we are usually concerned with the enthalpy of reaction
and this is what is typically meant when speaking of a heat of
reaction.
15.1
There are between 10 and 20 million known compounds that can
actually or hypothetically react with each other in an astronomical
number of ways and it is therefore literally impossible to catalog
all the possible heats of reaction. To get around this problem we
define for each substance a standard reaction and tabulate its
associated heat of reaction. These reactions and their associated
heats of reaction can then be used to calculate the heats of other
reactions.
This standard reaction is known as the standard formation reaction
of some substance and is defined as the balanced reaction in which
1 mole of that substance (the only product) is formed from the
elements which comprise it (the reactants) in their elemental
standard states at 1 bar of pressure and some temperature (usually,
but not necessarily 25.0 oC).
In older tables the standard pressure was defined as 1 atm, rather
than 1 bar. How significant is this change in the definition of
standard pressure?
For example, the standard formation reaction for the interhalogen,
iodine monobromide, is:
298 K, 1 bar
1/2 I2 (s) + 1/2 Br2 (l) ------------> 1 IBr (s)
The heat associated with the standard formation reaction is known
as the standard heat of formation for IBr (s) and is:
DHof, 298 K [IBr (s)] = - 10.5 kJ/mole of IBr (s)
The superscript, o, indicates standard state conditions, e.g., that the
pressure is 1 bar, while the subscript, f, indicates that this is a
formation reaction.
15.2
This is a page taken from the National Technical and Information
Service (NTIS) formerly the National Bureau of Standards (NBS)
publication The NBS Tables of Chemical Thermodynamic
Properties, Wagman. et. al., eds., Journal of Physical and Chemical
Reference Data, Vol 11 (Supp. No. 2), 1982.
Substance
Molar mass
Formual and Description State
ICI
-1
g mol
o
0K
o
DfH 0
298.15 K (25 C) and 0.1 MPa (1 bar)
o
o
o
o
o
H -H 0
S
DfH
DfG
-1
-1
kJ mol
-1
kJ mol
Cp
-1
J mol K
g
ao
162.3574
162.3574
162.3574
19.41
---
17.78
---13.8
-5.46
-17.1
-8.8
9.548
-----
247.551
--152.7
35.56
-----
ICl
ICl2
g
ao
162.3574
197.8104
992.9
---
997.5
---
---161.0
-----
-----
-----
ICl3
cr
233.2634
---
-89.5
-22.29
---
167.4
---
ao
cr
g
ao
289.2618
206.8314
206.8314
206.8314
--49.79
-----
-137.7
-10.5
40.84
---
-116.3
--3.69
---
----9.904
---
221.3
--258.773
---
----36.44
---
in CCl4:x
+
-
I2Cl
IBr
Note:
ao implies a 1 molal aqueous solution in which the
substance is not ionized, ai implies that it is ionized
cr implies the crystalline or solid state
What do you think g stands for?
Note further that the data has energy units of either kJ or J and that the
standard pressure is 1 bar.
Finally note the units on S and Cp are different from the units on DH and
o
DG , etc.
o
o
Why are there dashes, ---, for some of the entries? For which
thermodynamic variable is the greatest amount of data present?
15.3
The reactants in the formation reaction are the elements in their
elemental standard state, i.e., the most stable form of the
element at a pressure of 1 bar and the specified temperature
(again this is usually, but not necessarily 25 oC).
At 1 bar and 25 oC only a few elements are gases. All the Nobel
gases are monatomic gases:
He (g), Ne (g), Ar (g), Kr (g), Rn (g)
while the remaining elemental gases are all diatomic:
H2 (g), N2 (g), O2 (g), F2 (g), Cl2 (g)
Even fewer elements are liquids at 1 bar and 25 oC:
Br2 (l), Hg (l), and perhaps Fr (l) ?
All the remaining elements are solids at 1 bar and 25 oC. Many
solid elements can occur in several different crystal phases or
allotropes at 1 bar and some specified temperature, say 25 oC,
and only the most stable of these crystalline phases represents
the elemental standard state. For example elemental sulfur can
adopt 10 different allotropic forms, but it is the rhombohedral
crystalline phase that is stable at 1 bar and 25 oC and is therefore
the elemental standard state of sulfur at this temperature.
What do you think the elemental standard state of carbon is at
25 oC, diamond or graphite?
What do you think the elemental standard state of astatine, At, is
at 25 oC?
15.4
With this definition it should be clear that the standard enthalpies of
formation of the elements in their standard states are zero:
298 K, 1 bar
Ni (s) ----------> Ni (s)
DHof, 298 K [Ni (s)] = 0
The standard enthalpy of formation of copper(II) sulfate
pentahydrate, CuSO4x5H2O (s), at 25 oC is:
DHof, 298 K [CuSO4x5H2O (s)] = - 2279.7 kJ / mole
Write the balanced standard formation reaction associated with this
value.
15.5
Standard enthalpies of formation tabulated at some temperature
can be used to calculate the standard enthalpy change for some
reaction at that same temperature. The procedure is to sum the
molar standard enthalpies of formation of each product scaled by
the coefficients in the balanced reaction for each product and from
this sum over the products subtract a similar sum over the
reactants, where the molar standard enthalpy of formation of each
reactant is also scaled by the coefficient in the balanced reaction
for that reactant. This verbose procedure is best illustrated by an
example:
Fossil fuels contain sulfur and when these fuels are burned one of
the products is sulfur dioxide, SO2 (g). An important reaction in
the formation of acid rain involves the reaction of sulfur dioxide,
with molecular oxygen, O2 (g), to form sulfur trioxide, SO3 (g),
which can subsequently react with water droplets, H2O (l) to form
sulfuric acid, H2SO4 (aq):
298 K, 1 bar
2 SO2 (g) + O2 (g) -----------> 2 SO3 (g)
The standard enthalpy change for this reaction at 25 oC can be
calculated from the standard enthalpies of formation of all the
reactants and products as:
DHo298 K = +(2 mols SO3) DHof, 298 K [SO3 (g)]
- (2 mols SO2) DHof, 298 K [SO2 (g)]
- (1 mol O2) DHof, 298 K [O2 (g)]
15.6
Substituting tabulated values of the standard enthalpies of
formation gives:
DHo298 K = +(2 mols SO3) (- 395.72 kJ / mole SO3)
- (2 mols SO2) (- 296.830 kJ / mole SO2)
- (1 mol O2) ( 0 kJ / mole O2)
= - 197.78 kJ
Note that this result can be viewed as having units of moles in the
denominator, since we can interpret the result as:
= - 197.78 kJ / 2 moles of SO3 formed
or
= - 197.78 kJ / 2 moles of SO2 reacted
or
= - 197.78 kJ / 1 moles of O2 reacted
Using data from the National Technical and Information Service
(NTIS) calculate the standard enthalpy change, ΔHo, at 25 oC for
the combustion of 1 mole of octane, C8H18 (l) in the presence of
excess oxygen?
C8H18 (l) + 25/2 O2 (g) -------------> 8 CO2 (g) + 9 H2O (l)
Also calculate the standard internal energy change, ΔEo, for this
reaction?
15.7
The use of standard enthalpies of formation to calculate the standard
enthalpy change for some reaction is a specific example of a more general
procedure known as Hess’s law (it’s not really a law) that states that if you
have a set of reactions that add to give a net reaction, then thermodynamic
values for this set of reactions add in the same way to give the
corresponding thermodynamic value for the net reaction. Hess’s law is
usually applied to reaction enthalpies.
The direct conversion of graphite to diamond is not easily carried out (or we
would all be doing it) and hence it is also not easy to measure the enthalpy
change associated with this phase change. It is, however, fairly straight
forward to burn (combust) either graphite or diamond (there may be some
tear shedding as we part with our diamond) in the presence of exess oxygen
and measure the associated heats of combustion:
C (gr) + O2 (g) -------> CO2 (g)
DHo298 K = - 393.5 kJ
C (d) + O2 (g) -------> CO2 (g)
DHo298 K = - 395.4 kJ
Note that, if we reverse the 2nd reaction, the two reactions will add to give a
reaction describing the conversion of graphite to diamond:
C (gr) + O2 (g) -------> CO2 (g)
CO2 (g) -------> C (d) + O2 (g)
----------------------------------------C (gr) ---------> C (d)
DHo298 K = - 393.5 kJ
DH 298 K = - (- 395.4 kJ)
-------------------------------DHo298 K = + 1.9 kJ
o
Note that when adding reactions, equal numbers of moles of the same
species in the same phase that occur in both reactants and products cancel.
Any reactants or products that don’t cancel end up as reactants or products
in the net reaction. Note also that reversing a reaction, changes the sign on
the enthalpy change.
15.8
When solid metallic copper is dropped into concentrated nitric acid,
copious noxious dark brown fumes of nitrogen dioxide are produced:
3 Cu(s) + 8 HNO3(aq) + O2(g) ---> 3 Cu(NO3)2(aq) + 2 NO2(g)
+ 4 H2O(l)
The reaction of the solid copper with the nitric acid actually initally
produces nitric oxide:
3 Cu (s) + 8 HNO3 (aq) --> 3 Cu(NO3)2 (aq) + 2 NO (g) + 4 H2O (l)
which immediately reacts with molecular oxygen in the air to produce the
nitrogen dioxide:
NO (g) + 1/2 O2 (g) -----> NO2 (g)
DHo298 K = - 57.07 kJ / mole
In addition to the above data, use the following standard enthalpies of
formation:
DHof,298 K [H+ (aq)]  0
by definition
DHo,f,298 K [NO3- (aq)] = - 205.0 kJ / mole
DH
o
f,298 K
[Cu2+ (aq)] = + 64.77 kJ / mole
DHof,298 K [NO (g)] = + 90.25 kJ / mole
DHof,298 K [H2O (l)] = - 285.83 kJ / mole
to calculate the heat of the overall reaction initially described in this
problem. Note that you may have to scale all the coefficients in one or
more of these reactions by some constant scaling factor and, if you do this,
the enthalpy change for that reaction is also scaled by the same factor.
15.9
Use the following data:
DHo298 K, combustion [1-hexene] = - 4003 kJ / mole
DHo298 K, combustion [hexane] = - 4163.12 kJ / mole
to determine the enthalpy change for the hydrogenation of 1-hexene
to hexane:
CH3(CH2)3CH=CH2 (l) + H2 (g) -------> CH3(CH2)4CH3 (l)
You may need to look some data up in your textbook.
15.10