CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) H = -802 kJ
2H2O(g) -> 2H2O(l) H = -88 Kj
(Note: under conditions of standard temperature and pressure the liquid state of water is the normal state. Thus, the gas would be expected to condense. This is an exothermic process under these conditions. In a related process, it should get warmer when it rains)
Combining these equations yields the following:
CH4(g)+2O2(g)+2H2O(g) -> CO2(g)+2H2O(g)+2H2O(l)
H = (-802) kJ + (-88) kJ= -890 kJ
Hess's Law
if a reaction is carried out in a series of steps, H for the reaction will be equal to the sum of the enthalpy changes for the individual steps
the overall enthalpy change for the process is independent of the number of steps or the particular nature of the path by which the reaction is carried out.
Thus we can use information tabulated for a relatively small number of reactions to calculate H for a large number of different reactions
Determining H for the reaction
is difficult because some CO2 is also typically produced.
However, complete oxidation of either C or CO to yield CO2 is experimentally pretty easy to do:
We can invert reaction number 2 (making it endothermic) and have CO(g) as a product. (This describes the decomposition of CO2 to produce CO and O2)
Thus, we now have two equations with known enthalpies of reaction: the first describes the combustion of carbon and oxygen to produce CO2 and the second describes how CO2 can be decomposed to produce carbon monoxide (and oxygen). We can combine these together to describe the production of carbon monoxide from the combustion of carbon and oxygen:
The overall reaction, going from left-hand side reactant(s) to the right-hand side product(s) would be:
We can algebraically subtract the one-half O2 from both sides to yield the following equation with the associated overall enthalpy:
Another way to look at the method of combining reactions would be as follows:
plus
gives:
H = (-393.5 kJ) + (283.0 kJ) = -110.5 kJ
canceling out identical compounds from the left and right hand sides of this reaction gives
Carbon occurs in two forms: graphite and diamond. The enthalpy of combustion of graphite is -393.5 kJ, and that of diamond is -395.4 kJ
C(graphite) + O2(g) -> CO2(g) H = -393.5 kJ
C(diamond) + O2(g) -> CO2(g) H = -395.4 kJ
Calculate H for the conversion of graphite to diamond
What we want is H for the reaction:
C(graphite) -> C(diamond)
C(graphite) + O2(g) -> CO2(g) H = -393.5 kJ
CO2(g) -> C(diamond) + O2(g) H = +395.4 kJ
C(graphite) -> C(diamond) H = +1.9 kJ
We can never expect to obtain more or less energy from a chemical reaction by changing the method of carrying out the reaction ("conservation of energy").
Another way of saying this is that the particular pathway chosen to arrive at the same reactants yields the same H for the overall reaction.
Consider the previous example of the combustion of methane to produce gaseous H2O and then the condensation of the gaseous H2O to the liquid state. How is this represented in an Enthalpy Diagram?
The key features are the following:
1. Each line represents a set of reactants or products for a balanced chemical reaction. When going from one line to another, the atoms must balance. For example, if we were to ask what is the enthalpy associated with the condensation of water we would have (from the above data):
The CO2(g) on both sides will cancel to yield:
2. The relative distance of each line must reflect the relative enthalpy difference (H) between the reactants/products. If the enthalpy change in going from reactants to products is negative, then the line for the products must be below the reactants. Furthermore, the length of the distance must be proportional. For example, the distance reflecting the enthalpy associated with the condensation of water (H = -88 kJ)is only about 10% as long as the distance between the reactants and products for the combustion of methane to CO2 and liquid water (H = -890 kJ)
