C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (l) ΔHm = –2219.2 kJ (1)
But boilers are almost always operated with flue temperatures near the combustion temperature, and maintained at over 100 oC to prevent water from condensing to liquid. Since water is produced as a gas, the standard thermodynamic equation (1), does not apply. If it were formed, liquid water would dissolve acidic flue gases like HCl and SO2 to make acidic solutions which corrode the system. These gases are normally removed by limestone "scrubbers", leaving just the water vapor. The white plumes over smokestacks are water droplets forming by condensation of water vapor. They often disappear as the droplets evaporate.
The precision of our definition of standard enthalpy guarantees that our calculations of heat energy will be sound, but it requires that we pay attention to each energy cost and source. We'll see that inattention to such details has led to misinformation and confusion. Since Higher Heating Value (HHV) may be used by some air quality management authorities, while Lower Heating Values are used by many engineers and European power facilities, it is important to know precisely what the terms mean and which is being used. The USDA reports the following heating costs apparently, but not explicitly, stated to be LHVs:
for 1 million Btu of usable heat
annual fuel costa
|Natural gas||1.03 million Btu/1000 ft3||80||0.82 million Btu/1000 ft3||1,220 ft3||$7/1000 ft3||$854|
|Propane||91,200 Btu/gal||79||72,000 Btu/gal||13.86 gal||$1.25/gal||$1,730|
|Fuel oil #2||138,800 Btu/gal||83||115,000 Btu/gal||8.68 gal||$1.40/gal||$1,220|
|Seasoned firewood||20 million Btu/cord||77||15.4 million Btu/cord||0.065 cord||$115/cord||$747|
|Electricity||3,413 Btu/kWh||98||3,340 Btu/kWh||299 kWh||$0.08/kWh||$2,390|
|Premium wood pellets||16.4 million Btu/ton||83||13.6 million Btu/ton||0.073 ton||$120/ton||$882|
a Based on 100 million Btu of energy for the heating season, a typical value for an average sized house.
There are many variables that affect total heating costs (furnace efficiencies, local energy costs, etc.), but one variable that can potentially be understood is the heat available from fuel combustion. Without precise definitions, even this may be obscured. To illustrate this point, let's examine what are known as the Higher Heating Value (HHV) and Lower Heating Value (LHV) are calculated for propane.
The standard heat of combustion of propane isC3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (l) ΔHm = -2219.2 kJ = HHV = ΔH1 (1)
This is the process which yields the HHV, because the fuel is burned and the combustion products are cooled to 25 oC, removing all heat resulting from the condensation of water in the process.
But we want the enthalpy for the reaction:C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (g) ΔHm = LHV = ΔH2 (2)
In this case, the water is not condensed, so some of the energy is not recovered, so the ΔHm is the Lower Heating Value. We can imagine reaction (2) occurring in two steps. First, reaction (1), then
H2O(l) → H2O(g) ΔHm = 44.0 kJ (3)We notice that equation (1) produces 4 mol of H2O (l), so we multiply equation (3) by 4, so that 4 moles of H2O (l) are consumed in equation (3a):
4 H2O(l) → 4 H2O(g) ΔHm = 4 x 44.0 kJ = 176.0 kJ (3a)
If we add equations (1) and (3a) as below, canceling the 4 mol H2O (l) that are produced (appear on the right) with the 4 mol H2O (l) which are consumed (appear on the left), we get equation (2).
C3H8(g) + 5 O2(g) → 3 CO2(g) +
4 H2O (l) ΔHm = –2219.2 kJ (1)
4 H2O(l) → 4 H2O(g) ΔHm = 4 mol x 44.0 kJ mol–1 = 176.0 kJ (3a)
C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (g) ΔHm = LHV = ΔH2=–2043.2 kJ (2)
Experimentally it is found that the enthalpy change for reaction (2) is the sum of the enthalpy changes for reactions (1) and (3a):
ΔH2 = –2219.2 kJ + (4 x 44.0 kJ) = –2043.2 kJ = ΔH1 + 4 x ΔH3
This value of ΔHm should be the Lower Heating Value. Let's see if it matches the USDA value of 91,200 BTU/gal in the table above, given that a gallon of propane is about 4.23 lb
Our value is not close to the USDA LHV, but it matches the Oak Ridge National Laboratory (ORNL) value for the HHV of 84 250 BTU/gal.
The Higher Heating Value is the enthalpy change for reaction (1), which includes the heat released when 4 mol of gaseous water from the combustion cool to 25 oC, so its value is more negative than the LHV by four times the heat of condensation of water (–2043.2 + 4 x (–-44) = –2219.2 kJ):
H2O(g) → H2O(l) ΔHm = –44 kJ (4)
Repeating the calculation for the heat liberated in equation (1) in BTU/gal, we get 91 500 BTU/gal, which is the HHV reported by ORNL. The USDA table above apparently reports the HHV, perhaps unknowingly. The variety of values found on the web for both HHV and LHV attests to the fact that much confusion can result from not being careful with the meaning and application of a standard enthalpy. Technically, the lower heating value of a fuel is defined as the amount of heat released in the combustion of the fuel to give products at 150°C.
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Acetylene (C2H2) cannot be prepared directly from its elements according to the equation2C(s) + H2(g) → C2H2(g) (1) Calculate ΔHm for this reaction from the following thermochemical equations, all of which can be determined experimentally: C(s) + O2(g) → CO2(g) ΔHm = –393.5 kJ (2a) H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ (2b) C2H2(g) + O2(g) → 2CO2(g) + H2O(l) ΔHm = –1299.8 kJ (2c)
Solution We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):
a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2a) by 2.
b) Since Eq. (1) has 1 mol H2 on the left, we leave Eq. (2b) unchanged.
c) Since Eq. (1) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of Eq. (2c) we write Eq. (2c) in reverse.We then have Thus the desired result is 2C(s) + H2(g) → C2H2(g) ΔHm = 227.0 kJ