I am having some difficulties with Hess Law calculations. Even though I have performed them repeatedly, I get values that are inaccurate. Therefore, I must be doing something wrong. These calculations deal with the combustion of Hydrogen gas versus the combustion of the hydrocarbon methane. Since the water gas product, in a standard atmosphere, immediately goes to its lower energy state of a liquid, I am using the liquid water enthalpy value in both reactions.
I would appreciate the use of your Columbia University graduated Chemist brain.
This is what I have calculated:
Combusting the common hydrocarbon CH4 gas (methane), we get CO2 plus water. According to the Standard Enthalpy of Formation table, CH4 has a value of -74.81 kJ/mol, CO2 −393.509 kJ/mol and water (liquid) −285.8 kJ/mol.
CH4 + 2(O2) --> CO2 + 2(H2O) The enthalpy of the reactants is subtracted from the enthalpy of the products after accounting for the molar quantities.
The enthalpy of the reactants equals ONE mole of CH4 (ignoring the two moles of O2 gas because, according to Hess Law convention, elements in their standard state have an ARBITRARY VALUE OF ZERO) = -74.81 kJ
The enthalpy of the products equals One mole of CO2 + TWO moles of H2O = 965.509 kJ.
−393.509 kJ + [2(-285.8 kJ) MINUS [(-74.81 kJ) + (2 (ZERO))] = -890.299 kJ/mole
That was pretty straightforward. Now for the combustion of hydrogen gas.
In the following reaction (Thermochemical Properties of selected substances at 298 degrees Kelvin and 1 atmosphere of pressure.), oxygen gas + hydrogen gas = water gas. WATER gas has an enthalpy of formation of −241.818 kJ/mol.
That means that an Exothemic (energy releasing reaction) sent out (a certain amount of) kJ/mole of ENERGY from two gases.
When hydrogen gas combusts with oxygen gas, we get water gas, which quickly turns into to liquid in a standard atmosphere.
2(H2) + O2 --> 2(H2O) The enthalpy of the reactants is subtracted from the enthalpy of the products after accounting for the molar quantities.
The total enthalpy of the reactants is, according to Hess's Law, ZERO.
The enthalpy of the products equals one mole of water (liquid) −285.8 kJ/mole.
-285kj - [(2 times ZERO) + ZERO] = −285.8 kJ/mole
So, it appears that burning methane releases (890.299 kJ minus 285.8 kJ) 679.709 kJ/mole (about 76%) MORE ENERGY than burning hydrogen. This is inaccurate.
I did some checking and the more accurate figure for the energy given off by the combustion of one mole of methane is 802 kJ.
THE ENERGY FACTS:
"Since there are 500 moles of hydrogen gas in a kilogram, this means that burning a kilogram of hydrogen gas releases 500 times as much energy, or 121 MJ (million joules), assuming that the water comes out as a gas, as is usually the case in a combustion process."
"The energy given off by the combustion of one mole of methane turns out to be 802 kJ. The combustion of one kilogram of methane releases 50 MJ. Heavier hydrocarbons generally yield more energy per mole, but approximately the same energy on a per-kilogram basis. Gasoline, a mixture of hexane, heptane, octane, and various other hydrocarbons, yields about 44 MJ per kilogram."
NASA preferred Hydrogen over methane in their main space shuttle tank for a no bullshit thermodynamic higher energy release per unit mass reason. Yes, I know the fossil fuel industry peddles the "hydrogen is just an energy carrier, not a source" baloney 24/7. Bad mouthing Hydrogen as an energy source is right behind ethanol in the fossil fuel propaganda fun and games. But somehow they couldn't convince NASA of that .
So, what am I doing wrong that I can't make Hess's Law math come out right for the chemical reaction of hydrogen gas with oxygen gas?
For each bond you break through oxidation, burning H2 gas gives you more energy back than each bond you break in the oxidation of methane. The bond energy for H-H is 432 Kj/mole. The bond energy for the C-H bond in Methane (and all the single bonded alkanes +/- a bit) is 410 Kj/mole. So for an equal number of moles of H2, you have more energy stored in the Hydrogen than the methane. However, in the Gas phase, methane is more energy dense than hydrogen, because each molecule has 4 C-H bonds to break, whereas each molecule of H2 only has one bond to break. So it is almost 4X less efficient a store of energy in the gas phase, even though each individual bond holds greater energy.
This changes if you liquify the gas though. In this case, an equal amount of liquid hydrogen to liquid methane would contain more energy and release it on oxidation. For space applications, it is not that energy consumptive to keep these gases liquified, because space itself is so cold. On earth, this is a much bigger problem. So NASA for space uses Hydrogen, but on earth generally methane works better with a higher energy density in the gas phase.
Liquid Alkanes (basically pentane and up at normal earth temps) pack a lot more punch than both hydrogen and methane by volume, because they are liquid at normal earth temps. To keep Hydrogen gas liquified at earth surface temps would itself expend tremendous energy, so it is not a practical alternative on earth.
You can't just apply Hess' Law without dealing with the Phase Change problems here between Liquids and Gases.
Note: This assumes both Hydrogen Gas and Methane Gas operate according to the combined gas laws PV=nRT. Both gases do operate very closely to that law except at very low temps or very high pressures.
Although a certain member of the peanut gallery will vociferously disagree (and hurl ridicule and ad hominem arrogant insults along with bullshit about Hess Law "beauty") at what I am about to say, you have confirmed my view that Hess's Law belongs in the dust bin of thermodynamics history.
When Hess's Law was formulated, bond energies were not known. It was not until many, many years later that Linus Pauling's table of electronegativity was published. This aided in the knowledge of chemical reactions in regard to the positioning of molecular bonds between adjacent reactants and atomic orbitals in what physical chemistry now knows about the energy required to break bonds and the energy released (in exothermic reactions) when they are reformed.
As you pointed out in so many words, Bond energies, not Hess's Law, is the only accurate way, when gas phase changes are involved, to calculate energy density AND how much energy is released in a chemical reaction.
The following graphic illustrates the bond energy values, NOT AVAILABLE when using Hess's Law, whether phase changes are involved or not, that accurately reflect the high energy density
Although you mentioned that in certain states, methane is more energy dense that hydrogen, as a rule, that's a technicality that doesn't justify methane over hydrogen. For example, a very long chained hydrocarbon is much more energy dense than a short chained one.
HOWEVER, the refining process, using a massive amount of heat, probably a greater amount than that required to keep hydrogen cool prior to combustion, produces short chains and VOCs at the top of the cracking tower.
BEFORE you get to the point where those short chains are massaged chemically and thermally to get long chains so your internal combustion engine will actually run, you have gone through one hell of a lot of energy per unit mass. This makes the CH4 (and every other hydrocarbon liquid or gas) perceived advantage in energy density at certain gaseous states as insufficient to prefer it to Hydrogen gas as an energy source, regardless of the energy required to keep hydrogen gas at a low temperature.
Hess's Law confuses this issue, not just in regard to gases involving phase changes, but in arbitrarily assigning ZERO enthalpy of formation values to ALL elements in their ground state, be they gas or liquid. This fails to account for, among other things, catalytic energy of activation lowering action based on element or compound electronegativity differences, to the unwarranted perceived advantage of fossil fuels.
We KNOW the bond energies. We DON'T NEED Hess's Law any more to estimate energy density. As you pointed out, energy density varies by temperature and pressure. As I point out very often, the fossil fuel industry people peddle the energy density data according to Hess's Law without absolutely any regard for the vast amount of energy they expend to package their product. They will ENDLESSLY go into detail about all the energy difficulties of packaging hydrogen for combustion, but are silent as DEATH about the much greater energy expended to prepare gasoline or methane for the market.
I know you and I disagree on this. You often mention the high energy density of fossil fuels as the reason we still use them instead of a hydrogen plus solar plus wind EV run world. I simply ask you to remember the energy cost of packaging fossil fuels in addition to the energy cost of cleaning up after the pollution they expel. Hydrogen has always been more efficient in energy and cheaper in dollars, if fossil fuels were not subsidized.
One more thing that Hess's Law does not account for, but bond energies (with some detailed internal combustion physical chemistry) do account for, is the energy density of oxygen carrying fuels like ethanol. Measurements based on external combustion ignore some thermodynamic realities that molecular electronegativity evidence. When you are oxidizing a substance through combustion, the efficiency of that reaction is contingent on the availability of oxygen. Nobody wants to talk about that elephant in the fossil fuel energy density room (except for maybe Palloy with a thumb down following ridicule and huffing and puffing
There is simply NO WAY for a hydrocarbon to combust as efficiently and cleanly as an oxygen carrying fuel like ethanol in an atmosphere of 21 to 23% oxygen. As you know, the formation of incomplete combustion products in a combustion chamber is caused by the lack of sufficient oxygen reactant.
Gasoline never even releases, in the time it has to combust in the chamber, all of its energy density that SHOULD BE USEFUL ENERGY, never mind the waste heat, because of the lack of oxygen. Incomplete combustion is the NORM in gasoline. YET, Charles Hall deliberately ignored that in his ERoEI calculations with his erroneous assumption of complete combustion. Simply put, that was thermodynamic mendacity to make gasoline look better than ethanol.
That is one of the reasons that it is CRAP compared with ethanol. The other reason is that ethanol complete combustion reduces waste heat and engine friction.
Although I have never been able to convince you of the superiority of ethanol over gasoline or diesel, I hope that you consider agreeing with me that bond energy math is preferable to Hess's Law in computing energy density in elements and chemical compounds in a standard atmosphere, as well as the only truly accurate way to measure energy release in exothermic reactions and energy absorption in endothermic reactions.
If you don't agree, that's okay. Thanks again for the info.