Affairs of State

Chm 1311 Lectures for 30 June 1999

State Variables	The "state" of the system is all those parameters needed to describe the system such that it is identifiably the same if they are all returned to their original values after having been changed in any way. Among them are intensive variables like temperature and pressure and extensive variables like energy and number of moles of component compounds.
	The important point about the conservation of state variables is that processes which change and then return a system to a given state can only take place with ZERO CHANGE in the original state variables. That lets us imagine multiple ways to play around with a system and even stop short of returning to our starting point! Then by the difference of the state variable in the endpoint of the truncated path vs. the starting point (the original state of the system), tells us immediately how the state variables must change (that difference exactly) along any path from where we quit back to the original state!
Non-State Variables	This is not to say that any variable must be a state variable. Only a few are so privileged. As a counter-example, consider the work done in climbing straight up a hill vs. zig-zagging back and forth on convenient "switchbacks." Although you'd arrive at the same altitude (a state variable) either way, you'd find vastly different work done trodding the different paths. So work is not a state variable.
Energy, E	Energy is a state variable, and so is chemical potential. So we can take advantage of them being so when trying to determine the energy ebb and flow as chemical reactions take place.
	We'll need to measure or predict such energies since all systems tend to try to find their most stable (lowest energy) condition, knowing the chemical potential of several possible outcomes of chemical reactions may help us identify the most likely of them! It will turn out that stability is more complex than just knowing the lowest energy. That'll be a necessary but not quite sufficient condition, but for the moment we'll imagine it to be an important touchstone.
	The physicists have already copyrighted E as energy, so us chemists need another symbol. So why don't we make it a form of energy most useful to bench chemists, eh? And one of the hallmarks of "bench" chemistry is that it is done under atmospheric pressure!
	"So what?", you ask. So consider a chemical reaction which produces a gas such as: CaCO3(s) + Heat CaO(s) + CO2(g) Where does the CO2 gas go? Dissipates into the atmosphere unless we trap it, of course, but it actually adds volume to the atmosphere in so doing! And that requires pushing back the atmosphere that had occupied the benchtop. That displaced air pushes back with, well, atmospheric pressure. Which means the CO2 is expanding against a force, and that takes work!
Enthalpy H	And from where comes the energy the CO2 requires to do that work? From the energy supplied to the process. So we'd actually have to supply more heat to expand the CO2 against the atmosphere than we would have had to if the decomposition had been carried out in a closed container. The chemist's energy that incorporates that added work is called Enthalpy, and it carries the symbol, H. It's a real convenience that tables of H don't have to be corrected for gas expansion (or compression...we could have reactions like HCl(g) + H2O H3O+ + Cl -(aq) which consume gas...and correspondingly collapse the atmosphere a little).
Reaction Enthalpies	Now we're all set to assign ethalpy consequences to chemical reactions. These will enable us to predict the heat consumed or evolved by chemical processes. Bear in mind that these heat transfers are not the consequence of temperature differences between the System (the reaction vessel of interest) and its Surroundings (the rest of the Universe). Indeed, the tables that enable us to look up enthalpies assume that both System and Surroundings are at the same temperature, viz., a standard 25°C (a temperature accessible to any terrestrial chemist).
	Instead, these enthalpies result from heat (evolved or consumed) as a consequence of the products having different chemical potentials from the reactants. Since energy is neither created nor destroyed, that difference has to show up as heat transfer between the System and its Surroundings.
Exo- Endo- thermic	Chemists have special names to distinguish reactions which produce heat (exothermic) from those which require it (endothermic). More Greek; exo- means outside of as in "crabs have an exoskeleton" while endo- is its opposite inside of as in "people have endoskeletons; where does that leave crabby people?" And which is which is designated by the sign of H. Exothermic reactions transfer heat to their surroundings, losing enthalpy in the process; so their change in enthalpy, D H < 0; so reactions with large negative enthalpies should be treated with great respect...they're probably fireworks. On the other hand, a D H > 0 implies that the System's enthalpy has risen by the amount of heat transferred in; expect to have to heat them.
	But what if D H = 0 ? Not surprisingly, chemists call those thermoneutral.
Using Enthalpies	Enthalpies like energies are extensive properties; the more stuff that reacts, the more heat is transferred. Duhhh. However, they are tied to specific reaction stoichiometries as well as the reactions themselves! So if we replicate the Hindenberg Disaster in the lab via H2(g) + ½O2(g) H2O(l) D H = - 285.9 kJ/mol the associated reaction enthalpy tells us we get almost 286,000 Joules blown out into the surroundings per mole of water formed. Strongly exothermic.
	But you probably wouldn't have balanced the reaction that way; that ½ stoichiometric factor, although OK, would've annoyed. Instead, you're likely to have written 2 H2(g) + O2(g) 2 H2O(l) , D H = - 571.8 kJ/mol which requires that doubled enthalpy change since it states that each time the reaction runs, twice as much reactant turns into twice as much product as before. So now what does the "per mol" stand for? O2? No, because we could just as easily have had a reaction none of whose stoichiometric coefficients happened to be one. So the "per mol" refers to the reaction proceeding once as written, with all of the moles consumed and produced as specified by the stoichiometric coefficients! (If you must associate the "per mol" with a symbol there, make it the reaction arrow!)
Reverse Reactions	If instead we electrolyze water, i.e., run DC current through it to attract the few OH -(aq) to the anode and the few H3O +(aq) to the cathode, we can decompose the water back into its elements! (The primary reaction of a "Hydrogen Economy," if we can ever wean ourselves of fossil fuels.) 2 H2O(l) 2 H2(g) + O2(g) , D H = + 571.8 kJ/mol which reverses everything about the original reaction including the flow of energy which is now, of course, endothermic by exactly the same amount as the combustion reaction had been exothermic.
Chemical Algebra	Now comes the neat part. Because we can treat chemical reactions and their enthalpies just like algebraic equations, we can add and subtract them (subtraction means "reverse the reaction before adding") to our hearts content. And the exercise is not idle! It is the cornerstone of obtaining the enthalpies of reactions we haven't run from others involving their compounds which we have run or found in tables of enthalpies!
Standard States	The trick here is to agreeing upon some standard state for all of the atoms which go to make up all of the chemical compounds and then assigning a standard enthalpy to those atoms. The standard state of any atom in the Periodic Table is taken to be that allotrope (alternate form of the element like graphite vs. diamond vs. C60 for carbon) which is the most stable (OK, graphite) at a temperature of 25°C and a pressure of 1 atmosphere. And since we can set an arbitrary scale for enthalpies only once (differences not absolute values being all we can ever measure anyway), let's say the Enthalpy of every element is zero at standard state.
	Of course, we have to make sure we're talking about the phase of the element that's stable at standard state too. Thus, the versions of the 3 following halogens are all the ones which get awarded zero enthalpy: Cl2(gas), Br2(liquid), I2(solid). For sulfur, it'd be S8(s). And so on.
Enthalpies of Formation	Then with this ground floor of elements, we can refer enthalpies of all compounds to their component elements be using Enthalpies of Formation, DH f° , (from elements, of course) for all compounds. (The ° means "at standard state.") This is going to work because we can treat chemical equations like algebraic ones; enthalpies of formation add and subtract along with their compounds. So we can imagine, as did Hess, that making products from reactants involves mentally decomposing the reactants to their elements and then using those as tinkertoy parts to reconstruct the product compounds. That first mental step, decomposition of reactant molecules, is going to necessitate subtracting DH f° for the reactants since we're un-forming them but adding DH f° for the products since we are forming them!
	As an example, let's look at the key step in the production of sulfuric acid, oxidation of SO2 to SO3 in the presence of a catalyst [V2O5 which we'll ignore ...also, it's accepted practice to use S(s) in place of S8(s)] S(s) + (3/2)O2(g) SO3(g) , D H = D H f° (SO3) = - 395.2 kJ/mol SO2(g) S(s) + O2(g) , D H = - D H f°(SO2) = + 296.9 kJ/mol
add & cancel	SO2(g) + ½O2(g) SO3(g) , D H = D H f° (SO3) - D H f° (SO2) so that overall reaction enthalpy is ( - 359.2) - ( - 296.9) = - 98.3 kJ/mol.
	Tables of D H f° are found on page 209 and pages A40-A47 in your book. More are in the Handbook of Chemistry and Physics, various National Bureau of Standards publications, and Stull, Westrum, and Sinke, The Chemical Thermodynamics of Organic Compounds.
	So the general rule for using those tables is that the enthalpy of any reaction will be the sum of the enthalpies of formation of the products (each scaled by its stoichiometric coefficient, of course) minus the sum of the enthalpies of formation of the reactants (also so scaled). And skip all elements in their most stable form since their enthalpy is zero by choice. Dead simple.