Readings for this section:
Petrucci: Chapter 7
To begin the study of the transformation of energy in chemical (or other) processes, we need to first develop a few terms that we will use. These definitions must be very tightly specified and understood in order to properly follow some of the more complex logic in later pages.
One of the most common definitions for work is when a force f displaces an object by a distance Δx does work w = f × Δx. (also known as w = f × d). This is a mechanical work and serves as the basis for our definition of PV work.
Work can also be done by pushing back the atmosphere (piston in a cylinder). In this case, one pushes back against the atmospheric pressure P_{atm} and moves the piston a distance Δx. The force needed to push back the atmosphere is actually P_{atm}× A where A is the area of the piston. Hence, we can calculate the magnitude of the work done as
w = P_{atm}× A × Δx = P_{atm}
× ΔV.
We call this atmospheric work. The only thing left to decide is the sign
convention. (Described later) Since the system increased in volume and
hence, did the work of pushing back the atmosphere (positive value for ΔV),
it has used up energy so we can now write down a more complete definition for
PV work as:
w = P_{atm} × ΔV
In chemical situations, we are normally interested only in the atmospheric work done when gases are evolved or used up in the reaction.
both of these can be forms of work and of heat energy  KE = 1/2 mv^{2}^{
} PE = mgh lifting mass m by height h against acceleration due to gravity g = f × d force times distance. 
Work is organized energy, lifting a book, etc.
Heat (defined later) is random (disorganized) energy that is released or absorbed, often as a result of some work being done, e.g., sound (energy) dissipates through air, when a book drops and hits the floor and the floor gets warmer (molecules vibrate more rapidly) spreading out from the spot where the hit occurred. Thus, the work done to raise the book (increasing its potential energy) is converted to kinetic energy upon release and then to thermal energy (energy is released as heat) as the book strikes the floor.
One can also transfer energy into or out of a sample via lightenergy and
electrical work
(We include this type of work in our catchall term
w' ).
OR
OR
We need a few more definitions so we can discus thermodynamics more easily.
Note that the experimental reality may not quite match up with the definitions we use here. For example, An insulated system will still loose or gain heat because no amount of insulation is perfect. However, we can often consider the amount to be negligible as long as the rate of heat loss is very slow compared to the time scale of the experiment we have carried out. For example, it may take 30 minutes or so for the contents of a coffee cup to cool down to room temperature from the boiling point of water, so the cup is not a perfect insulator. However, we might still use a coffee cup as an insulated container for a thermodynamics experiment that we can complete in a few seconds. We will consider the coffee cup to be a perfectly insulated container in this experiment as long as the amount of heat lost over the short time of the experiment is quite small compared to the heat exchange involved in the experiment itself.
Energy transfer can be done in one of two ways:
There is a general sign convention that chemists use when describing q and w . When energy flows into the system as a result of heat or work, the sign is positive (the system gains energy). When energy flows out of a system, the sign is negative (the system looses energy. Both q and w refer, not to an amount of energy, but to an amount of energy transferred as a result of the process.
Note the sign convention
How do we measure the actual amount of heat contained in a material.
Can we even measure it.
For the initial part of this chapter, we will be looking at a specific type of cases where the system undergoing the change is held at constant pressure. This, in fact, represents the way the majority of reactions are done in our chemical experience.
At this point, we're looking at a black box. We can't really understand what's inside, we just define terms to try to understand it 
We define a term called enthalpy H
is the energy transferred between a system and the surroundings under constant
pressure. We cannot measure the absolute enthalpy of a system but we can
measure the change in enthalpy for a process. If the Initial and final
states have enthalpies
H_{i}
and
H_{f}, respectively then the
change in enthalpy for the process can be
defined as
We're familiar with state functions. For instance, the distance between Kingston and Toronto can be measured by a "straight" line on the map. It never changes. It is only a function of the position of the two cities. However, the distance driven to get from Kingston to Toronto depends on the route we take. Obviously, some routes are shorter than others. So, we cannot use a nonstate function (distance driven) to measure a state function (distance separating two cities) without some specific restrictions on the measurements. In this case, we would need to drive our vehicle in a straight line over the surface of the earth, ignoring roads lakes and hills (a "crow" could do it hence the term "as the crow flies" when referring to distances). 
This definition is only good if H is a state function, i.e., the change in enthalpy depends only on the initial and final states, not on the process itself.
If we simply measure the heat evolved (or absorbed) during a process, we will not have a measure of the change in enthalpy since, in general, q is not a state function. It depends on how the process occurs.
We need to restrict our process measurements to specific conditions in order to be able to measure enthalpy directly. In this case, if we maintain constant pressure and then simply measure the heat transferred as a result of the process q_{p} we will have a measure of the change in enthalpy. The subscript p refers to the fact that this heat was measured with the system held at constant pressure conditions.
Thus, we have simply
ΔH = q_{p}
at constant pressure
For chemical reactions we can write
$\Delta H\;=\;\sum_p H_p  \sum_r H_r$
where subscripts p and r refer to products and reactants, respectively. If ΔH is negative, then there is a net flow of heat from chemical system to surroundings (heat is released during the reaction). This kind of process is called Exothermic (EXO sounds like exit).
If ΔH is positive, then net flow of heat from surroundings to chemical system (heat absorbed). Endothermic (ENDO sounds like enter)
CH_{4}(g) + 2O_{2}(g) $\rightarrow$ CO_{2}(g) + 2H_{2}O(l)  ΔH = 890 kJ/mol (that's per mole of Rxn as written) (EXOTHERMIC) 
H_{2}(g) + I_{2}(g) $\rightarrow$ 2 HI(g)  ΔH = 52.2 kJ/mol (ENDOTHERMIC) 
One of the easiest ways to measure a heat transfer is to measure a change in
temperature and then calculate from that, the heat transferred. We use the
concept of heat capacity to do this. Heat
capacity of a substance describes the amount of heat that can be absorbed by the
substance for a given unit rise in temperature (unit = 1 K). It can be expressed
in general as :
The Heat capacity can be expressed as a function of the amount of material. The heat capacity per mole is called the molar heat capacity (units J K^{1}mol^{1}). The heat capacity per gram is called the specific heat capacity (or just specific heat) (units J K^{1}g^{1}). It can simply be the capacity of the unit object. For example, a calorimeter (made up of or containing several materials, container, insulation, water, etc.) will have a heat capacity unique to it.
It is not completely correct to talk of heat capacity in terms of q since q is not a state function. We are better off using the state functions ΔH rather than q.
We assume that no electrical or mechanical work w' is being done.
At constant pressure, we measure heat as ΔH = q_{p}, and hence the heat capacity we need to use is C_{p}, where we define
Thus, we can write (at constant P)
ΔH = C_{p}ΔT
The units of C_{p} are J/K or J/ºC (remember that ΔT is the same whether its measured in ºC or in K). Sometimes, we tabulate heat capacities in per mole or per gram values. If this is the case, we need to multiply the molar heat capacity by the number of moles or the gram heat capacity (specific heat) by the number of grams to get the total heat capacity. In these cases, we can think of modifying our equation for enthalpy to be
ΔH = nC_{p}ΔT (for molar heat capacity values)
OR
ΔH = mC_{p}ΔT (for specific heat values).
Obviously, these equations hold C_{p} to be independent of temperature. In reality, while that's a reasonable estimate for our purposes, it is not strictly correct. The heat capacity of any substance changes slightly with changes in temperature.
Now, Let's quickly review a few definitions we will need to use in this section.
A sample of 50. mL of a 0.20 M solution of HCl was mixed with 50. mL of 0.20 M NaOH in a coffee cup calorimeter. The initial temperature of both solutions was 22.2^{º}C. After mixing, the temperature rose to 23.5^{º}C. What is the enthalpy change for the neutralization reaction which occurred?
To Start, we need to define the system. Since this case involves a reaction in a liquid solution where the chemicals are intimately involved with the solvent (solvation in water), we cannot really separate the water from the chemicals. We will use the chemicals and water as the system. The insulated cup is the boundary between the system and the surroundings and we will assume that the cup itself absorbs no heat and that no heat passes through the cup to the surroundings.
We have thus, a two step process to deal with.
H_{3}O^{+}(aq) + OH^{}(aq) $\rightarrow$ 2 H_{2}O(l)
Total volume = 100. mL, dilute aqueous so we assume
density = 1.00 g/mL
mass of solution: m_{sol}
= 100. mL × 1.00 g/mL = 100. g
For the cooling process:
ΔT = T_{f}  T_{i} = 22.2^{º}C  23.5^{º}C = 1.3^{º}C (= 1.3 K)
The temperature was lowered from it's high value back down to its starting point. Thus, we expect a negative temperature change. Another way of viewing this: We removed heat from the system to return it to its starting point. Hence, q and ΔH are negative (the reaction is exothermic).
This temperature change is happening to the system.
q_{p} = C_{p}ΔT = mC_{m} ΔT (C_{m} is specific heat, i.e., heat capacity per gram)
NOTE: to figure out which equation you need simply look at the units of the quantities you have and figure out how to cancel out the undesired units by the correct combination of C with the values you have. In this case, we have
q_{p} = 4.18 J K^{1}g^{1}
× 100. g ×
1.3 K
= 540. J
# mol H_{3}O^{+}
= # mol HCl = 0.50 L × 0.20 mol/L = 0.010 mol.
ΔH' = q_{p} = 540
J
Note that the ΔH shown here is for the whole reaction amount. It is not a molar amount; hence, the prime on the ΔH. I use the prime here to distinguish the global quantity and no prime to show a molar quantity.
\[ \Delta H=\frac{540\;\mathrm{J}}{0.010\;\mathrm{mol}}=54.\;\mathrm{kJ}/\mathrm{mol}\;\mathrm{H}_3\mathrm{O}^+\]
We'll use the stoichiometry from the balanced chemical equation to convert from moles of H_{3}O^{+} to moles of reaction. According to the equation
H_{3}O^{+}(aq) + OH^{}(aq) $\rightarrow$ 2 H_{2}O(l),
there is one hydronium ion for every one reaction. So, we convert to moles of equation as follows:
\[\Delta H=\frac{540\;\mathrm{J}}{0.010\;\mathrm{mol}\;\mathrm{H}_3\mathrm{O}^+}\left \frac{1\;\mathrm{H}_3\mathrm{O}^+}{1\;\mathrm{equation}}\right=54.\;\mathrm{kJ}/\mathrm{mol}\;\mathrm{equation}\]
The value reported for the enthalpy change must always be accompanied by the balanced chemical reaction we used in the determination.
If we had balanced the chemical equation differently, say, to
show one mole of water, it might look like:
1/2 H_{3}O^{+}(aq)
+ 1/2 OH^{}(aq) $\rightarrow$ H_{2}O(l)
In this balanced equation, there is one half a hydronium ion for every
equation thus, our conversion would be
\[\Delta H=\frac{540\;\mathrm{J}}{0.010\;\mathrm{mol}\;\mathrm{H}_3\mathrm{O}^+}\left \frac{1/2\;\mathrm{H}_3\mathrm{O}^+}{1\;\mathrm{equation}}\right=27.\;\mathrm{kJ}/\mathrm{mol}\;\mathrm{equation}\]
Our test calorimeter is not that accurate (significant heat loss to the surroundings). The true ΔH for the first reaction is 56.02 kJ/mol and is valid for the heat of neutralization of any strong acid with a strong base in water.
Here is an example where we are able to issolate the chemicals separately as the system and directly measure the heat removed from the system.
0.510 g of ethanol is burned in a flame calorimeter containing 1200 g of water. The water is initially at 22.46ºC and is warmed up to 25.52ºC as a result of the reaction. What is the ΔH for one mole of ethanol?
In this case, we can more easily separate the gas reaction mixture from the water in the calorimeter which absorbs (all) the heat from the reaction. Therefore, we can use the chemicals themselves as the system. In this case, the insulation in the calorimeter serves to isolate a small portion of the surroundings (the water) so we can measure its temperature change directly, while ignoring the rest of the surroundings.
C_{2}H_{5}OH(l) + 3 O_{2}(g) $\rightarrow$ 2CO_{2}(g) + 3 H_{2}O(l)
ΔT_{water} = 25.52ºC  22.46ºC = 3.06ºC = 3.06 K
We have a specified mass of water so we will start with that and look up and use a heat capacity which has grams (specific heat) to cancel out g and leave the desired units of energy.
\[q_\mathrm{surroundings}=\underset{\mathrm{net\;heat\;capacity}}{\underbrace{\frac{1200.\;\mathrm{g}\;\mathrm{H}_2\mathrm{O}}\;{\textstyle\left{\displaystyle\frac{4.18\;\mathrm{J}}{\mathrm{k}\;\mathrm{g}\;\mathrm{H}_2\mathrm{O}}}\right}}}\;\;\underset{\mathrm{Temp\;change}}{\underbrace{3.06\;\mathrm{K}}}=15.3\;\mathrm{kJ} \]
This is the total heat absorbed by the water (surroundings) and is the negative of the heat evolved by the system. Hence,
Note that the q of the surroundings does not need to have a specification of constant P or constant V. It just doesn't matter. Whatever energy is absorbed by the surroundings must have been given off by the system (First Law). Now, the system must be specified to be constant pressure if you need to equate the heat exchanged, q_{p}, to enthalpy change, ΔH, of the system. 
q_{p,system} = q_{surroundings} = 15.3 kJ
\[\frac{0.510\;\mathrm{g\;ethanol}}{\;}\left\frac{1\;\mathrm{mol}}{46.05\;\mathrm{g}}\right\;=0.0111 \;\mathrm{mol\;ethanol}\]
The other thing we need to do to calculate the enthalpy is to determine the heat per mole since enthalpies for a chemical reaction are always reported in units of kJ/mol of reaction.
\[\Delta H_\mathrm{system}\;=\;\frac{15.3\;\mathrm{kJ}}{0.0111\;\mathrm{mol}\;\mathrm{ethanol}}\;=\;1380\;\mathrm{kJ}/\mbox{mol ethanol}\]
Finally, we need to translate the enthalpy change into units of kJ/mol of equation as written, rather than kJ/mol of one particular chemical.
\[\Delta H\;=\;\frac{1380\;\mathrm{kJ}}{1\;\mathrm{mol}\;\mathrm{ethanol}}\left\frac{1\;\mathrm{ethanol}}{1\;\mathrm{equation}}\right\;=\;1380\;\mathrm{kJ}/\mathrm{mol}\]
It is important to note that this reaction is written for one mole of a reactant burned completely in oxygen. This type of reaction is called a combustion reaction and the ΔH is called a heat of combustion. We could simply tabulate ΔH_{comb} (ethanol) and know that the reaction is as written above.
Generally, we cannot use or report a enthalpy change for a chemical reaction unless we have also specified the chemical reaction. Had we written the reaction with different coefficients (still balanced) then the enthalpy change would have been different. Consider the reaction above written as follows.
1/3 C_{2}H_{5}OH(l) + O_{2}(g) $\rightarrow$ 2/3 CO_{2}(g) + H_{2}O(l)
This way, we have specified the reaction per mole of H_{2}O produced. The value for ΔH for this reaction is 1/3 that for the reaction written for one mole of ethanol.
Thus, the enthalpy change for this reaction can be calculated, starting again from the enthalpy change for one mole of ethanol
\[\Delta H\;=\;\frac{1380\;\mathrm{kJ}}{1/3\;\mathrm{mol}\;\mathrm{ethanol}}\left\frac{1\;\mathrm{ethanol}}{1\;\mathrm{equation}}\right\;=\;460\;\mathrm{kJ}/\mathrm{mol}\]
This is the enthalpy change for the reaction written above with only one mole of water produced.
Reaction energies depend on the conditions under which they were measured. In order to be able to record energies and tabulate them in a way that others can make sense of them, we define a standard state. This standard state is then used to define the state of both the reactants (before reaction) and products (after reaction). Recall that since ΔH and ΔU do not depend on the path, only the initial and final state. Hence this is a useful thing to do.
Thermodynamic Standard state should not be confused with the standard conditions (STP)used for Ideal Gas calculations and tabulations. They aren't the same.
Standard State:
Pressure = 100 kPa (1 bar)*
Concentration = 1 mol/L, (solutions)
[Temperature = for tabulation purposes only, normally but not
exclusively 25 ºC ]
What are elements in their standard states? How do we decide which ones to use?
So how do we specify standard state in general? Normally, we just have to be careful to write the correct phase of the element or compound for the apropriate temperature. The following is a list of elements and compounds and the proper standard state at two different temperatures
substance  T=25ºC (298.15K)  T=100ºC  T=120ºC 
hydrogen  H_{2}(g)  H_{2}(g)  H_{2}(g) 
oxygen  O_{2}(g)  O_{2}(g)  O_{2}(g) 
carbon  C(s,graphite)  C(s,graphite)  C(s,graphite) 
water*  H_{2}O(l)  H_{2}O(l) or H_{2}O(g)  H_{2}O(g) 
methane  CH_{4}(g)  CH_{4}(g)  CH_{4}(g) 
pentane (bp=36ºC)  C_{5}H_{15}(l)  C_{5}H_{15}(g)  C_{5}H_{15}(g) 
methanol dissolved in water**  CH_{3}OH(aq)  CH_{3}OH(aq)  CH_{3}OH(g) 
* We see that since water exists in two different standard, stable forms at 100ºC and P=1 bar, that there are two different possible states we can indicate, both of which are 'standard' states at 100ºC. The problem here would be that the difference between them would be the difference in energy (enthalpy) of vaporization. Be careful of the states specified.
** The final entry in the table shows methanol, a molecule that can dissolve in water. At or below 100ºC, it is in standard state if the concentration is 1 mol/L. Any other concentration means not standard state. Note that above 100ºC, water doesn't exist in liquid form at standard conditions so methanol would only exist in gas form at p=1 bar along side the water, which would also have to be at 1 bar if it were to be claimed to be standard state.
We'll see later that a simpler definition for standard state is "activity = 1" (See NOTES 12.b)
Temperature is not part of the definition. It is merely the experimental temperature for which most of the results are tabulated.
We indicate Thermodynamic quantities measured at standard conditions by using the superscript zero (sometimes pronounced "not") as in ΔHº (delta H not)
eg. CH_{4}(g) + 2 O_{2}(g) $\rightarrow$ CO_{2}(g) + 2 H_{2}O(l) ΔHº = 890.4 kJ/mol
NOTE: When we see the superscript 'not' on the symbol ΔHº it means that all the reactants and products must be in their defined standard state at a defined pressure of 1 bar (and/or concentration of 1 mol/L for solutes).
Defined reaction types
Recall that the value we write down for the enthalpy of a reaction depends on how we have written the balanced chemical equation for that reaction. Hence, we either need to include the complete balanced chemical reaction with our values of ΔHº or we need to define a way of writing the equations so we don't have to actually specify them. The latter is the way we do it in many cases.
Two common defined reaction types and their associated enthalpy changes are given here.
H_{2}(g) + 1/2 O_{2}(g) $\rightarrow$ H_{2}O(l) ΔHº_{f} = 286. kJ/mol.
See how the standard form of hydrogen and oxygen is diatomic gas while that of water is molecular liquid. This is because the normal temperature of T=25ºC (and standard P=1 bar) is used unless otherwise specified. For example, had we been seeking the enthalpy change for the same reaction at a temperature T=120ºC (and P=1bar), we would still have calculated a standard enthalpy of formation but the standard states would be different.
H_{2}(g) + 1/2 O_{2}(g) $\rightarrow$ H_{2}O(g) ΔHº_{f} = 327. kJ/mol
Note that value for the standard enthalpy change at this temperature is not the same as at room temperature because we had to incorporate the enthalpy of vaporization of ~41 kJ/mol for water. This is still a 'standard enthalpy change' because all the chemicals are at standard state for T=120ºC and P = standard pressure of 1 bar.
Note also that we assumed the heat capacity of the products and reactants don't contribute significantly to the overall enthalpy change for this reaction under the temperature change from 25ºC to 120ºC. This is not necessarily good assumption as we'll see later (See Temperature Change and Enthalpy).
C_{2}H_{6}(g) + 7/2 O_{2}(g) $\rightarrow$ 2 CO_{2}(g) + 3 H_{2}O(l) ΔHº_{Comb} = 1560.4 kJ/mol
Since both of these types of reactions are now well defined, we can recreate the chemical equation exactly knowing only the compound of interest and the type of reaction. Hence, the values for the standard enthalpy changes for these defined types of reactions can be tabulated with need to only specify the chemical of interest. Thus, the following two specifications are complete because the reaction type and the chemical for each is specified so we can duplicate the appropriate balanced chemical equation that goes with each enthalpy change value.
ΔHº_{f} (H_{2}O)= 286. kJ/mol
ΔHº_{Comb} (C_{2}H_{6}) = 1560.4 kJ/mol
(since T was not specified for these two values, you must assume T=25ºC)
Since enthalpy change, ΔH, is a state function, we can go to products via several routes and the enthalpy change will still be the same.
Another way to say this is: the enthalpy change in a reaction does not depend on the reaction pathway. This allows us to define any path we choose to get from reactants to products and, as long as we keep track of the enthalpy changes for each step, we will be able to calculate the overall enthalpy change for the process.
Let's look at the example of ethane burning in excess oxygen to produce carbon dioxide and water. We have two possibilities: we make a series of reactions (1,2,3) that gets us to the final products and keep track of the enthalpy changes for each step; we simply burn the ethane in a calorimeter and measure the enthalpy change directly (labelled T). These two possibilities are diagrammed below where the vertical axis represents enthalpy.
We can consider the reaction to occur in one complete process [T] or in three distinct steps [1], [2] and [3]. In either case, the total enthalpy change should be the same.
[1] Ethane is heated to drive off H_{2}.
C_{2}H_{6} $\overset{\Delta}{\rightarrow}$ C_{2}H_{4} + H_{2} 
ΔH_{1}º = +136.2 kJ 
[2] Ethene is burned in excess O_{2}. C_{2}H_{4} + 3 O_{2} $\rightarrow$ 2 CO_{2} + 2 H_{2}O 
ΔH_{2}º = 1410.8 kJ 
[3] H_{2} is burned in excess O_{2}. H_{2} + 1/2 O_{2} $\rightarrow$ H_{2}O 
ΔH_{3}º = 285.8 kJ 
[T] overall reaction is sum [1]+[2]+[3] C_{2}H_{6} + 3.5 O_{2} $\rightarrow$ 2 CO_{2} + 3 H_{2}O 
ΔH_{T}º = 1560.4 kJ 
The example here may not seem to have been practical at first glance. After all, why set up three separate calorimetry experiments when we could have just done the overall experiment in one step? Well, it serves as a good demonstration of the principle. Additionally, we can use this idea to calculate enthalpies changes for reactions where direct measurement is impractical. Also, we can use this method to calculate the expected enthalpy change of a reaction using tabulated values without the need to do any experiments.
Example: Use the following standard enthalpies of combustion to calculate the standard enthalpy change for the formation of methane.
The standard formation reaction for methane is:
C(s) + 2 H_{2}(g) $\rightarrow$ CH_{4}(g) ΔH = ?
This will be our target reaction. We need to use the Standard Combustion Enthalpies listed here.
1) ΔHº_{Comb}(C) =
393.5 kJ
2) ΔHº_{Comb}(H_{2}) = 285.8
kJ
3) ΔHº_{Comb}(CH_{4}) = 890.4
kJ
Knowing the standard format for the combustion reaction (1 mol of reactant burns to completion in oxygen to produce fully oxidized products; CO_{2} and/or H_{2}O in this case) we can easily write down the three reactions.
1) C(s) + O_{2}(g) $\rightarrow$ CO_{2}(g)  ΔHº = 393.5 kJ/mol 
2) H_{2}(g) + 1/2 O_{2}(g) $\rightarrow$ H_{2}O(g)  ΔHº = 285.8 kJ/mol 
3) CH_{4}(g) + 2 O_{2}(g) $\rightarrow$ CO_{2}(g) + 2H_{2}O(l)  ΔHº = 890.4 kJ/mol 
Now lets add up the three reactions to give the target reaction 

1) C(s) + O_{2}(g) $\rightarrow$ CO_{2}(g)  ΔHº = 393.5 kJ/mol 
2) 2 ×[ H_{2}(g) + 1/2 O_{2}(g) $\rightarrow$ H_{2}O(g)  ΔHº = 285.8 kJ/mol] 
3) CO_{2}(g) + 2H_{2}O(l) $\rightarrow$ CH_{4}(g) + 2 O_{2}(g)  ΔHº = +890.4 kJ/mol 
C(s) + 2 H_{2}(g) $\rightarrow$ CH_{4}(g)  ΔHº = 74.7 kJ/mol 
To get to the desired final reaction, we needed to use equation 2 twice (needed to use up 2 hydrogens). Thus, we also have added the enthalpy for that reaction twice (note the square brackets extends around the equation and the enthalpy value. We also reversed the direction of equation 3 (multiplied all coefficients by 1, and hence multiplied the enthalpy change by 1) Whatever we do to the equation, we must also do the ΔHº value.
The Standard Enthalpy of formation is defined as the enthalpy change of a standard formation reaction. We saw the definition of the reaction type earlier. Now let's explore how we can use these enthalpy change values.
Here are a few reactions that can be defined as standard formation reactions and whose enthalpy changes are useable as standard enthalpies of formation.
C(s,graphite) + O_{2}(g) $\rightarrow$ CO_{2}  ΔH_{f}º (CO_{2}) = 393.5 kJ/mol 
H_{2}(g) + 1/2 O_{2} $\rightarrow$ H_{2}O(l)  ΔH_{f}º (H_{2}O) = 285.8 kJ/mol 
C(graphite) + 2H_{2}(g) $\rightarrow$ CH_{4}(g)  ΔH_{f}º (CH_{4}) = 74.7 kJ/mol 
Note that the phase of standard state carbon is solid graphite, not just solid. That's because there are more than one solid phase of carbon, for example, diamond is solid carbon. In the cases where there are more than one allotrope of an element, the most stable one is defined to be the standard state.
Question: What will be the standard enthalpy of formation of an element in its standard state [like O_{2}(g)]?
Answer: Zero. Consider the reaction that is being described in the question. It would look like this: O_{2}(g) $\rightarrow$ O_{2}(g). Clearly there is no reaction as the product and the reactant are the same so no enthalpy change has actually occurred. We can generalize this discussion to say: the standard enthalpy of formation of any element in its standard state is zero.
Now let's use these with Hess' law to determine the reaction enthalpy for the following reaction.
CH_{4}(g) + 2 O_{2}(g) $\rightarrow$ CO_{2}(g) + 2H_{2}O(l) ΔH_{1}º (combustion of methane)
According to Hess' law, we could chose to break down the reactants to their constituent elements in their standard state and then reconstruct them into the products.
CH_{4}(g) + 2 O_{2}(g) $\rightarrow$ C(s) + 2H_{2}(g) + 2O_{2}(g)  ΔH_{2}º 
C(s) + 2H_{2}(g) + 2O_{2}(g) $\rightarrow$ CO_{2}(g) + 2H_{2}O(l)  ΔH_{3}º 
From Hess' law, we have, ΔH_{1}º = ΔH_{2}º + ΔH_{3}º
Lets explore ΔH_{2}º + ΔH_{3}º a bit further.
In step 2, we broke the reactants (methane and oxygen) into their constituent elements at standard state. This is the reverse of the formation of these reactants so the total enthalpy change is simply the negative of the sum of the standard enthalpy of formation for the reactants (multiplied, of course by the approprite coefficients to give us the correct stoichiometry of the desired reaction). Thus,
$\Delta H_2^o$ =  [sum of Std enthalpies of formation of reactants].
\[\Delta H_2^o\;=\;\sum \Delta H_f^o (react) \]
In step 3, we took elements in their standard state and formed the products. Thus, the total enthalpy change is simply the sum of the standard enthalpy of formation for the products (multiplied, of course by the approprite coefficients to give us the correct stoichiometry of the desired reaction). Thus,
$\Delta H_2^o$ = + sum of Std enthalpies of formation of products
\[\Delta H_2^o\;=\;\sum \Delta H_f^o (prod) \]
So now, we can write:
For the combustion of methane we're dealing with here, we can write
ΔH_{1}º = [ΔH_{f}º (CO_{2}) + 2 ΔH_{f}º (H_{2}O)]  [ΔH_{f}º (CH_{4}) + 2 ΔH_{f}º (O_{2})]
We can look up the values for Heat of formation in any standard reference book.
ΔH_{1}^{º} = [(393.5 kJ) + 2 (285.8 kJ)]  [(74.7 kJ) + 2 (0)]
ΔH_{1}^{º} = 890.4 kJ
In general, we can write:
\[\Delta H^o\;=\;\sum \Delta H_f^o (prod) \;\;\sum \Delta H_f^o (react) \] 
A more completely mathematically correct equation is written as follows 
\[\Delta H^o\;=\;\sum_{p=1}^{p} n_p \Delta H_f^o (p) \;\;\sum_{r=1}^{r} n_r \Delta H_f^o (r) \] 
Note that the coefficients n used here are unitless because they represent mole ratios, not numbers of moles. This allows the final entalpy change values to have units of kJ/mol, as they must have. Note that if one or more of the reactants or products are not in their standard state, we simply can remove the superscript 'not' from the final enthalpy change symbol and the same equation will still work.
This procedure is merely a special application of Hess' Law whereby we are adding the enthalpies of formation in a way that gives us the overall enthalpy change.
Let's try a few examples to get some practice with these calculations:
What is the standard enthalpy change for
4 NH_{3}(g) + 5 O_{2}(g) $\rightarrow$ 4NO(g) + 6 H_{2}O(l)?
ΔHº = $\sum$ΔH_{f}º (prod)
 $\sum$ΔH_{f}º (reactants)
ΔHº = [4 ΔH_{f}º (NO) + 6 ΔH_{f}º (H_{2}O)]
 [4 ΔH_{f}º (NH_{3}) +
5 ΔH_{f}º (O_{2})]
ΔHº = [4 (90.25 kJ/mol) + 6 (285.83 kJ/mol)]
 [4 (46.11 kJ/mol) +
5 (0)]
ΔHº = 1169.54 kJ/mol
Now here's another one.
B_{5}H_{9}(g) reacts exothermically with O_{2} to form B_{2}O_{3}(s) and water. What is the standard enthalpy change for the reaction of 1 mol of B_{5}H_{9}(g)?
balance this  B_{5}H_{9}(g) + O_{2}(g) $\rightarrow$  B_{2}O_{3}(s) + H_{2}O(l) 
for 1mol B_{5}H_{9}  1 B_{5}H_{9}(g) + 6 O_{2}(g) $\rightarrow$  5/2 B_{2}O_{3}(s) + 9/2 H_{2}O(l) 
ΔHº = ΣΔH_{f}º(prod)

ΣΔH_{f}º(reactants)
ΔHº = [5/2 ΔH_{f}º (B_{2}O_{3}) +
9/2 ΔH_{f}º (H_{2}O)]
 [ΔH_{f}º (B_{5}H_{9})
+ 6 ΔH_{f}º (O_{2})]
ΔHº = [5/2 (1272.77) + 9/2 (285.83)]
 [(73.2) + 6 (0)]
ΔHº = 4541.4 kJ/mol
1 mol of benzene burns in air at standard state conditions and gives off 3267 kJ of heat. What is the enthalpy of formation of C_{6}H_{6}?
C_{6}H_{6}(l) + 7.5O_{2}(g) $\rightarrow$ 6 CO_{2}(g) + 3 H_{2}O(l)
ΔHº_{Comb}  = 3267 kJ/mol 
ΔHº_{Comb}  \[=\; \sum_p \Delta H^o_f(p)  \sum_r \Delta H^o_f(r)\] 
3267 kJ/mol  = [6 ΔH_{f}º (CO_{2},g) + 3 ΔH_{f}º (H_{2}O,l)]  [ΔH_{f}º (C_{6}H_{6},l) + 7.5 ΔH_{f}º(O_{2},g)] 
Substitute in values from data table.  
3267 kJ/mol  = [6 (393.5) + 3 (285.83)]  [ΔH_{f}º(C_{6}H_{6},l) + 7.5 (0)] 
Solve for the only unknown:
ΔH_{f}º (C_{6}H_{6},l) = 49 kJ/mol
The standard enthalpy data we find tabulated in thermodynamic tables is all for data at 25ºC. This data may not be exactly the information we need. What if a reaction doesn't actually happen at standard 25ºC. We have two options: 1, we can assume that there is no significant difference in enthalpy change or, 2, we can calculate the effect.
In cases where there is only small temperature changes the former option may not be too bad. However, in some cases, especially where the temperature of reaction is far from 25ºC we should take the latter option.
Consider a reaction between chemicals A and B to produce C and D at some temperature T.
We could calculate the ΔH for the process 2 using standard thermodynamic data. Steps 1 and 2 involve temperature changes and we can calculate the ΔH for these steps if we know the heat capacities of the compounds involved. According to Hess' law, the overall enthalpy change for the reaction at temperature T is the sum of the steps 1, 2 and 3.
Step 1:  This is simply a temperature change and we can
calculate the enthalpy change using the heat capacities of A and B. ΔH_{1}
= {aCº_{P}(A) + bCº_{P}(B)}ΔT_{1}
(298  T) 

Step 2:  ΔH_{2}
= ΔHº_{298} (calculate from whatever means possible, for
example you could use ΔH?_{f}
values.) 

Step 3:  Like step 1, this is simply a temperature
change. Calculate the enthalpy change using the heat capacities of C and
D. Note that the process is the reverse direction of step 1 and the sign of
the temperature change is opposite. ΔHº_{3} = {cCº_{P}(C) + dCº_{P}(D)}ΔT_{3} (T  298) NOTE: ΔT_{1}= ΔT_{3 } 

Step T: 

\[\Delta C_p^o\;=\;\sum_{products} C_p(products)\sum_{reactants} C_p(reactants)\]
The change in heat capacity (capacity of products  capacity of reactants) is a actually slightly dependent on temperature and hence this formulation is not valid if the temperature T is very different from standard temp. (298 K). In most cases, the temperature effect is not extremely large but that is not general.
Like the general concepts involved in Hess' Law, we simply add up the heats of the processes involved in order to determine the overall heat. This is merely another special application of Hess' Law.
Internal energy of the system is represented by the symbol U. It is a more general description of the energy of the system than enthalpy. We can never measure the absolute energy of a system but we can measure changes in the internal energy of the system using the state function
ΔU = U_{final}  U_{initial}.
Since we can never actually measure either U_{initial} nor U_{final} we measure, instead, the energy transferred into or out of the system in the form of heat and work to obtain.
ΔU = q + w ( = heat transferred + work done)
This is, in fact, the mathematical expression of the First Law of
Thermodynamics
and is completely general, not relying on any of the restrictions we
placed on enthalpy measurements. We can measure work and heat under any
conditions to get the change in internal energy because it turns out that
while neither
q
nor
w are state functions, their sum is.
Example:
If 15 kJ of work is done by the surroundings on the system and the system looses 10 kJ of energy as heat to the surroundings then
ΔU  = 10 kJ + 15 kJ 
= +5 kJ 
Chemical reactions often involve changes in Volume (exp. gases)
C_{3}H_{8}(g) + 5 O_{2}(g)  $\rightarrow$ 3 CO_{2}(g) + 4 H_{2}O(g) 
6 mol gas  7 mol gas $\Rightarrow$ increase in volume at const. P 
If this reaction is to happen, it needs to push back the rest of universe (atmosphere) to make room for itself. Let's put the reaction system into a cylindrical aparatus like that pictured at the beginning of this unit. As gas molecules are produced, the system expands to accomodate them and keep the pressure (inside and out) constantly the same. The amount of work done as the system expanded against constant external pressure can be calculated as:
work = force (a.k.a. pressure × area) × displacement (a.k.a. ΔX).
or w = P × A × ΔX or just P × ΔV
to fully define w, we need to recognise that if the system expands (does the work), the system lost energy so by definition, w is negative when volume expands (ΔV positive). So we can remove the magnitude formulation and put in a negative sign.
w =  P_{surr }ΔV (gas expanding against an external pressure P_{surr}.)
ΔU = q + w = q  P_{surr } ΔV
or q_{p} = ΔU + P_{surr } ΔV
in this case, we can use the ideal gas law to get a more useful equation as long as the gas molecules all behave ideally.
q_{p} = ΔU + Δn_{g}RT. which is also written as ΔH = ΔU + Δn_{g}RT.
We can use this formulation when the only volume increase comes from the fact that the number of moles of gas is different in products than in reactants. Note that the parameter ΔU is a value that is tied to the balanced chemical equation (just as we did for enthalpy changes). Thus, the value we used for ΔU and the value we used for Δn_{g} must be both for the same balanced chemical equation.
In a closed container (no volume change) there would be no work so
ΔV = 0 so:
ΔU = q_{v}.
subscript v refers to constant volume conditions.
Enthalpy is now better defined as
H = U + PV
Changes in enthalpy are
ΔH = ΔU + ΔPV since we normally work in a fixed pressure (atmosphere) situation, we can rewrite this as
ΔH = ΔU + PΔV.
To do these measurements under general conditions, we need to expand a few previous definitions. First, lets look at heat capacity.
In many substances, the two heat capacities differ considerably. Consider one mole of ideal gas. We have
ΔH  ΔU = Δ(PV) = RΔT
.
Therefore
C_{p}  C_{v} = R for one mole of ideal gas
Solids and liquids often have values of
C_{p} and
C_{v} that don't differ much. On the other hand there is no
consistent rule. Take for example water and benzene; for water,
C_{p}  C_{v} = 0.075R,
while for benzene we have C_{p}  C_{v} = 5.1R.
let's do an example where we look at the PV work done.
When 2.00 mol of SO_{2}(g) react completely with 1.00 mol O_{2}(g) to form 2.00 mol of SO_{3}(g) at 25ºC and constant pressure of 1.00 atm, 198 kJ of energy is released as heat. Calculate ΔU and ΔH for this reaction.
2 SO_{2}(g) + 1 O_{2}(g) $\rightarrow$ 2 SO_{3}(g)
The heat released is q_{p} since it was measured under constant pressure conditions. so we can quickly say
ΔH' = 198 kJ (negative sign since the heat is released, i.e., exothermic)
ΔH = 198 kJ/mol (trivial in this case since the molar quantities given exactly match the numbers in the equation)
ΔU' = q_{p} + w = q_{p}  PΔV (total PV work)
or (since we have the molar values)
ΔU = ΔH  PΔV (remember, this is the PV work per mole which just happens to equal the total PV work in this case)
Since we have constant P and T conditions, the only change in PΔV will be due to change in numbers of moles of gas molecules. We can use the stoichiometry from the equation as written to determine PΔV :
ΔU = ΔH  Δn_{g}RT  or  ΔH = ΔU + Δn_{g}RT 
Note: Either of these two equations above are valid for any situation where the T and P are constant but where the number of moles of gas molecules changes because of stoichiometry. Pick one and learn it. 
Δn = (#moles product gas  #moles reactant gas) per mole of equation
Δn = 2  3 = 1.
Note that the coefficients used here to calculate Δn are really unitless as they represent ratios of moles, not actual measures of moles.
ΔU  = 198 kJ/mol + (1 × 8.31451 J K^{1}mol^{1} × 298 K) 
= 198 kJ/mol + 2.48 kJ/mol  
= 196 kJ/mol 