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Chemical Energy
 Propane,
C3H8, natural gas, CH4, and phosphorous,
P4 react with oxygen O2, and these reactions release
energy in the form of heat and light. No doubt, according to the principle of
conservation of energy, energy is required to reverse the reactions. Thus,
energy stored in chemicals (compounds) and energy released or absorbed in
chemical reactions are called chemical energy, which also covers topics
such as bond energy, ionization potential, electron affinity, electronegativity,
lattice energy, etc.
For example, at standard conditions, the combustion of 1.0 mole hydrogen with
oxygen releases 285.8 kJ of energy. We represent the reaction.
H2(g) + 1/2 O2 ->
H2O (l), dH = -285.8 kJ/mol
where dH represent the heat (or enthalpy)
of reaction, and a negative value means that the heat is released. Usually,
dH is represented by DH in textbooks, but
using dH notation is much less work on Internet documents.
For the reverse reaction, 285.8 kJ/mol is required, and the sign for
dH value changes.
H2O (l) -> H2(g) + 1/2
O2, dH = +285.8 kJ/mol
| A Chemical Energy Level Diagram
|
------------H2(g) + 1/2O2
|
| |
286 kJ | | -286 kJ
| |
| ¯
------------ H2O
| We can also use an energy level diagram to show the
relative content of energy. The energy content of H2(g) + 0.5
O2 is 285.8 kJ higher than a mole of water, H2O.
Oil, gas, and food are often called energy by the news media, but more
precisely they are sources of (chemical) energy -- energy stored in chemicals
with a potential to be released in a chemical reaction. The released energy
performs work or causes physical and chemical changes.
It is obvious that the amount of energy released in a chemical reaction is
related to the amount of reactants. For example, when the amount is doubled, so
is the amount of energy released.
2 H2(g) + O2 -> 2 H2O (l),
dH = -571.6 kJ/mol Example 1 shows the
calculation when the amount of reactants is only a fraction of a mole.
Example 1
How much energy is release when a balloon containing 0.15 mole of hydrogen
is ignited in the air?
Solution
The amount released is 0.15 mol * 285.8 kJ/mol = 42.9
kJ
Discussion
The sudden release of energy causes an explosion.
Endothermic and Exothermic Reactions
A reaction that releases energy is called exothermic reaction. Energy
is released in the form of heat, light and (pressure-volume) work. For example,
when methane or propane is oxidized by O2, the heat released causes
the gas to expand (explosion in some cases); releasing heat & light and
doing work at the same time. In this case, the energy source came from chemical
reactions instead of at the expense of internal energy described in the previous module.
Endothermic reactions absorb energy, and in all cases, the energy is
supplied from another source, in the form of electrical energy, heat or light. Pressure-volume Work in Chemical Reactions
Many chemical reactions involve gases, and when a gas is formed, it displaces
other gases by pushing them out against a pressure. Work, defined in Newtonian
physics as a force times the distance along the force direction, is performed in
such an action. The work is called the pressure-volume (P-V) work, which is a
form of energy and it must be analyzed and its quantity included in chemical
energy calculations.
The SI units for pressure are N m-2 and that of the volume is
m3. Pressure times volume gives the unit of N m, which is the
definition of joule,
1 Pa * 1 m3 = 1 N m-2 m
= 1 N m
= 1 J Since 1 atm = 101300 Pa, and 1 L = 0.001
m3. Thus,
1 atm L = 101.3 J. The P-V work under constant pressure (P) is
simply the pressure times the change in volume dV.
w = - P dV This method applies to reactions that produce
gases, which are released into the atmosphere. When work is done by the system,
the work is negative, as the formula indicates. In reactions where gases are
consumed to produce liquid or solid, work is done on the system by its
environment. The work is positive.
In cases the pressure varies, an integral approach is required to evaluate
the pressure volume work.
w = - Ó d (P V)
= - Ó P dV - (the integral of) V dP
The negative sign is retained, because the work done by the system is
negative. However, the integral of P V work depends on the path, and we
will not get into the detail discussion at this stage.
Example 2
In the reaction to produce oxygen,
KClO3(s) = KCl(s) + 3/2 O2(g),
calculate the pressure-volume work done by 8.2 g of KClO3.
Solution
The molar mass of KClO3 is 123.5 g/mol and
8.2 g is 0.067 mol. Thus, the amount of oxygen produced is 0.10 (= 0.067*2/3)
mol. Apply the ideal gas law to the pressure volume work (P V), w have
P V = n R Tw = - D P V
= - D n R T
= - 0.10 mol*8.312 (J / (mol K))*298 K
= - 248 J
Discussion
The work done is due to the formation of gas
O2 which expands against the atmosphere of 1.0 atm or 101.3 kPa. The
volume changes of the solids are insignificant compared to that of the gas.
In case both pressure and volume change, and the work is the
difference of the pressure-volume product, DP V.
Enthalpy The enthalpy, usually represented by H is the
energy released in a chemical reaction under constant pressure, H =
qP. The enthalpy is a convenient property to evaluate for
reactions taking place at constant pressure. Enthalpy differ from internal
energy, E, defined in Energy as the energy input to a system at constant volume. The
energy released in a chemical reaction raises the internal energy, E, and
does work under constant pressure at the expense of energy stored in compounds.
Thus,
H = qP = E + P dV Of course, the enthalpy
change (dH) of a chemical reaction depends on the amount of reactants,
the temperature, and pressure. Under normal conditions, the ideal gas law can be applied to give reasonable results.
Like the internal energy, enthalpy is also a thermodynamic state function,
depending only on the initial and final states of the system, but not on the
rate of reaction.
Standard Enthalpy of Reaction In order to make the data useful for
scientific and engineering applications, there is a general agreement to report
and tabulate enthalpy changes for a mole of reaction at a standard temperature
and pressure. Such quantities are called the standard enthalpy of
reaction.
In handbooks and textbooks, the standard enthalpy change is represented by
Ho. For simplicity, we use dHo to
represent the changes of standard enthalpy in our discussion to avoid (very)
slow loading of the delta onto your computer.
Example 3
The standard enthalpy for the combustion of methane is 890.4 kJ per mole,
CH4(g) + 2 O2(g) -> CO2(g) + 2
H2O(g), dHo = -890.4
kJ/mol calculate the standard enthalpy change when 1.0 cubic meter of
natural gas is burned converting to gaseous products.
Solution
When 1.0 mol or 22.4 L of CH4, at 273K and
1 atm, is oxidized completely, the standard enthalpy change is 890.4 kJ. One
cubic meter is 1000 L (/22.4 = 44.6 mol). Thus, the standard enthalpy of change
is,
dH = 44.6 mol * 890.4 kJ/mol = 39712 kJ or 39 million joules.
A problem can be made up using any of the following standard
enthalpy of reactions. These are given here to illustrate the type of reactions
and the representation of enthalpy of reactions.
2 H(g) -> H2(g)
dHo = -436 kJ/mol
2 O(g) -> O2(g)
dHo = -249 kJ/mol
H2O(l)
-> H2O(g) dHo = 44
kJ/mol at 298 K
H2O(l) -> H2O(g)
dH = 41 kJ/mol at 373 K, non-standard condition
Mg(s) +
S(s) -> MgS(s) dHo = -598
kJ/mol
2 H(g) + O(g) -> H2O(g)
dHo = -847 kJ/mol
Cu(s) +
1/2O2(g) -> CuO(s)
dHo = -157
kJ/mol
1/2N2(g) + O2(g) ->
NO2(g) dHo = 34
kJ/mol
Mg(s) + 1/2O2(g) -> MgO(s)
dHo = -602 kJ/mol
2 P(s) + 3
Cl2(g) -> 2 PCl3(s)
dHo = -640 kJ/mol
2 P(s) + 5 Cl2(g) -> 2
PCl5(s) dHo = -880
kJ/mol
C(graphite) + 2 O(g) -> CO2(g)
dHo = -643 kJ/mol
C(graphite) + O2(g)
-> CO2(g) dHo = -394
kJ/mol
C(graphite) + 2 H2(g) -> CH4(g)
dHo = -75 kJ/mol
2 Al(s) +
Fe2O3(s) -> Al2O3(s) + 2Fe(s)
dHo = -850 kJ/mol
As we
shall see, the application of Hess Law will make these data very useful. For example, applying
Hess law using a few of these reactions enable us to calculate the heat of
combustion of methane to form liquid water (as opposed to gaseous water) and
carbon dioxide,
CH4 + 2 O2 -> 2 H2O(l) +
CO2(g) dH = -980 kJ/mol.
Enthalpy
is an important topic in thermodynamics. Various methods have been devised for
the accurate measurement of heat of reaction under constant pressure or under
constant volume. This link gives a more advanced treatment on enthalpy.
Standard Enthalpy of Formation, dHf
When the standard enthalpy is for a reaction that forms a compound from its
basic elements also at the standard state, the standard enthalpy of reaction is
called the standard enthalpy of formation, represented by
dHof. Unless specified, the temperature is 298 K.
| Table of dHof
|
| Compound
| dHof
|
| MgS
| -598 kJ/mol
|
| CuO
| -157
|
| PCl3
| -320
|
| PCl5
| -440
|
| H2O
| -286
|
| NO2
| + 34
|
| MgO
| -602
|
| CO2
| -394
|
| CH4
| -75
| In the above list, some of the equations lead to
the formation of a compound from its elements at their standard state. These
equations and their enthalpy of formation are given below:
Mg(s) + S(s) -> MgS(s)
dHof = -598 kJ/mol
P(s) +
3/2 Cl2(g) -> PCl3(g)
dHof = -320 kJ/mol
P(s) +
5/2 Cl2(g) -> PCl5(g)
dHof = - 440
kJ/mol
H2(g) + 1/2 O2(g) ->
H2O(g) dHof =
-286 kJ/mol
1/2N2(g) + O2(g)
-> NO2(g)
dHof = + 34 kJ/mol
Cu(s) +
1/2 O2(g) -> CuO(s)
dHof = -157 kJ/mol
Mg(s) +
1/2 O2(g) -> MgO(s)
dHof = -602 kJ/mol
C(graphite) +
O2(g) -> CO2(g)
dHof = -394 kJ/mol
C(graphite) + 4
H2(g) -> CH4(g)
dHof = -75 kJ/mol
In all the above equations of reaction, the right hand side has only one
product and that its coefficient is 1. A general rule is to consider standard
enthalpy of formation of all elements at the standard condition to be zero.
Then, there is no need to write the complete equation for the tabulation of the
standard enthalpy of formation. The above list can be simplified to give the
table shown here.
A simple application of the standard enthalpy of formation is illustrated by
Example 4.
Example 4
For NH3, dHf = -46.1 kJ/mol. Estimate energy
released when 10 g of N2 react with excess of H2 to form
ammonia.
Solution
Ten grams of nitrogen is less than 1 mol, and we carry
out the calculation in the following manner: 1 mol N2 - 46.1 kJ
10 g N2 ---------- ---------- = - 32.9 kJ
28.1 g N2 0.5 mol N2
Thus, 32.9 kJ is released when 10 g of N2 is consumed.
Standard enthalpies of formation and standard entropies are important thermodynamic
data, and this link gives an extensive table of values for some key
compounds.
The principle of conservation of energy leads to the formulation of the Hess law. It's application makes the enthalpy of reaction and
standard enthalpy of formation very useful. |
