1 Gases
2 Microscopic Energies
3 First Law
4 2nd & 3rd Law
5 Phase transitions
6 Mixtures
7 Phase Diagrams
8 Equilibrium
9 Molecular Interactions
Thermochemistry
A specialized subset of the generic thermodynamics is thermochemistry. In thermochemistry, we generally are involved with the actual measurement of thermodynamic parameters and as such, it generally involves containers, etc. in the system. Additionally, we need to be able to compare our experimental results with each other so we need standards under which we can report our results. These are called standard conditions or standard state conditions.
Currently, the standard state of a substance is defined as it's pure form at whatever temperature we're working in at a pressure of 1 bar (100 kPa). Note that it is 25 years since atmospheres were used as a standard pressure, even though many text books still teach this.
Thus, for example, the standard state of water at 25ºC is liquid but the standard state of water at 150ºC is vapour.
Any values we need to report are also done as a common temperature of 25ºC (298.15K) but this is not a requirement for standard state. It is the commonly accepted temperature for reporting and tabulating results. Thus, any value you find in a thermodynamics table of values will most likely be valid at 25ºC alone. At other temperatures, these values may or may not be valid, depending on other circumstances.
One exception to the temperature rule is obviously enthalpies of vaporization. These are always valid only at the boiling point of the substance in question. Thus, the enthalpy of vaporization of water, at standard state, at 373K is
ΔvapHO(373K)
= +40.66 kJ/mol
At any other temperature, these values may be less accurate.
Other standard enthalpy changes relating to phase changes are fusion, a.k.a. melting, and subliming.
for example, the enthalpy change of fusion of water at 1 bar and 0.0ºC is
ΔfusHO(273K)
= +6.01 kJ/mol
the standard enthalpy change of sublimation of water is
now something we can calculate. Obviously, since enthalpy is a
state function, we can find the enthalpy change of sublimation by going
a different path. if water goes from solid to vapour then the
energy requirements should be the same as the sum of energies needed to
first melt the ice and then vaporize the liquid
| H2O(s) → H2O(ℓ)
ΔvapH |
= +40.66 kJ/mol |
| H2O(ℓ)
→ H2O(g)
ΔfusH |
= +6.01 kJ/mol |
| H2O(s) → H2O(g)
ΔvapH |
= +46.67 kJ/mol |
This all hinges on the enthalpy changes from the
tables still being valid at the temperatures involved.
Another consequence of enthalpy being a state function is that if we reverse the process, we change the sign of the enthalpy change.
ΔcondensationHO(373K)
= -DvapHO(373K)
= -40.66
kJ/mol
Notice that the subscript is on the delta symbol, not the enthalpy. this is to reaffirm that enthalpy is enthalpy. There are not different kinds of enthalpy. There are, however, different kinds of changes. This is a relatively new symbolism
The text book (Table 2.4 in Atkins, v.8) contains a list of different kinds of system changes and the corresponding enthalpy change symbols as recommended by IUPAC.
Enthalpy of chemical changes
In the case of a chemical reaction, we can express the standard enthalpy change as being the change when the reactants in their standard state react to become products in their standard state.
For example, the reaction of methane and oxygen to become carbon dioxide and water (combustion) is
CH4(g) + 2 O2(g)
→ CO2(g) + 2 H2O(ℓ)
ΔcHO
= -890
kJ/mol
This is the thermochemical equation, the combination of the chemical equation and the pertinent thermodynamic enthalpy change. The units of per mole refers to one mole of the reaction as written. The subscript c in the enthalpy symbol refers to the fact that this equation is actually a standard enthalpy of combustion of methane.
The standard combustion reaction fully oxidizes the compound to carbon dioxide and water if only C,H,and O are in the original compound. If there is N then to N2 (since Oxygen cannot oxidize it further). Actually, it could be any oxidizing reaction, not restricted to oxygen. For example, fluorine is an excellent oxidizing agent so
CH2S + 6 F2 → CF4 + 2 HF + SF6
is a combustion reaction where the methylthione is combusted in excess fluorine gas.
We can figure out the enthalpy change per mol of one of the reactants or products using the stoichiometric ratios as conversion factors.
Generally, we assume the energy of mixing, dissolving, etc. is insignificant compared to the energy of the reaction itself. The biggest deviation from this assumption is for ionic solutions where solvation energies are substance
A more general expression for this calculation is needed.
Consider the reaction: a A + b B → c C + d D
the enthalpy change for this reaction is
Or more generally,
where the symbol ν is the "frequency" (the stoichiometric coefficient) of the given substance in the equation.
Hess' Law
The standard enthalpy change of an overall chemical reaction is the sum of the standard enthalpies of the individual reactions into which a reaction may be divided.
Thus, if we can add up two or more chemical equations to achieve an overall equation we can add up the enthalpies for the individual equations to get the enthalpy of the overall equation.
Example:
Calculate the standard change in enthalpy for the following reaction:
| N2(g) + 2O2(g) → 2NO2(g) |
Using the following two reactions:
| N2(g) + O2(g) → 2NO(g) | ΔrH |
| NO2(g) → NO(g) + 1/2 O2(g) | ΔrH |
We will need to reverse the second equation and double the amount (multiply all the coefficients by –2) so that we are producing 2 NO2 and hence, the enthalpy change of our new equation will be multiplied by –2 as well.
| N2(g) + O2(g) → 2NO(g) | ΔrH |
| 2NO(g) + O2(g) → 2NO2(g) | ΔrH |
| N2(g) + 2O2(g) → 2NO2(g) | ΔrH |
Standard enthalpies of formation
Obviously, Hess' law is a very useful way to
determine enthalpies of reactions without actually performing
the reactions. We merely need to have on hand the
enthalpies of similar reactions that can be added together such
that the result is the desired final reaction.
unfortunately, this could involve a huge number of tabulated
reactions.
We choose often to use a more artificial path, whereby we can
reduce the amount of tabulated data we need.
We start by defining a new equation type
Formation equation: any chemical reaction that produces exactly one mole of one product from the products constituent elements in their standard states.
Now, we can imagine any reaction we are interested in as being composed of the complete dissociation of the reactants into their elements (reverse of formation) followed by the recombination of these elements into the products (formation of each of the products)
For example, the reaction
2 H2(g) + C(s,graphite) → CH4(g)
is a formation reaction for methane since only one mole of methane is formed (per mole of reaction) and the reactants are the elements hydrogen and carbon in their respective standard states.
The enthalpy change for this reaction is called
the enthalpy of formation of methane Δf
HO(CH4)
sometimes also called Δf
HO(CH4).
so now, if we have a reaction that involves methane, say, the
combustion of methane, we can use this formation reaction (among
others) to get the overall enthalpy of reaction.
Example,
What is the enthalpy of combustion of methane at 25ºC, given the enthalpy of formation of methane is –74.81 kJ/mol, of water is –187.78 kJ/mol and the enthalpy of formation of carbon dioxide is –393.51 kJ/mol?
The combustion reaction for methane is
| CH4(g) + 2 O2(g) → CO2(g) + 2H2O(g) | |
| first, break up the reactants into elements | |
| CH4(g) → 2 H2(g) + C(s,graphite) | ΔrH |
| then form new compounds from the elements (O2 is already an element) | |
| O2(g) + C(s,graphite) → CO2(g) | ΔrH |
| 2 H2(g) + O2(g) → 2 H2O(ℓ) | ΔrH |
| CH4(g) + 2 O2(g) → CO2(g) + 2H2O(g) | ΔrH |
a more general formula for this process that eliminates the need to write out the equations like this is
Now, a more succinct way to calculate the enthalpy of the reaction is
| ΔrH ΔrH ΔrH |
Enthalpy change is Temperature dependent
|
|
Consider a pure substance heated from temperature T1 to T2. we can express the enthalpy of the new temperature as
Now consider a reaction enthalpy at
T1. We can
express the new enthalpy of that reaction at that
T2 by
considering each component of the chemical reaction. The reactants
go fromT2
to T1 and the
products go from T1
to
T2
to complete Hess' cycle. The total enthalpy change for this
round-a-bout process would be
If we can assume constant heat capacities then this equation simplifies to
where
. This is known as Kirchoff's Law.
