Concise Physical Chemistry

c04 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come

MOLECULAR ENTHALPIES OF FORMATION 59

-393.5

-283.0

FIGURE 4.1 Combustion of C(gr) and CO(g).

experiment C(gr)+^12 O 2 (g) in a limited supply of O 2 (g) gives a mixture of products.

It is not a “clean” reaction. The reaction

CO(g)+^12 O 2 (g)→CO 2 (g)

isclean, however, and gives well-definedqpandqV. We find that the enthalpy change

is
rH^298 =− 283 .0kJmol−^1 for this reaction (Fig. 4.1). Thus we have two paths

connecting the same initial and final thermodynamic states. By a thermochemical

principle equivalent to the first law of thermodynamics known asHess’s law,the

enthalpy change over both paths must be the same. The third leg of the triangle

must be

fH^298 (CO(g))=− 393. 5 −(− 283 .0)=− 110 .5kJmol−^1

This method of indirect determination of
fH^298 is capable of wide extension.

For example, knowing
fH^298 (H 2 O(l)) and
fH^298 (CO 2 (g)), one can determine


fH^298 of a hydrocarbon like methane by burning it in a flame calorimeter and

setting up a state diagram that is a little more complicated than the triangle just

described but that works on the same principle. The combustion reaction

CH 4 (g)+2O 2 (g)→CO 2 (g)+2H 2 O(l) 
rH^298 =−890 kJ

has the same final state as

2H 2 (g)+C(gr)+2O 2 (g)→CO 2 (g)+2H 2 O(l) 
rH^298 =−966 kJ

In Fig. 4.2, the thermodynamicstateof the products of combustion of methane,

CO 2 (g)+2H 2 O(l) is reproduced by burning 2 mol of hydrogen and 1 mol of C(gr).

The heats of combustion are arranged in an enthalpy diagram so as to make everything

come out even except for one missing leg of the quadrangle, that of methane. The

thermodynamicstateof methane is found by difference; it is 76 kJ mol−^1 below

that of the elements. Formation of methane from its elements would give off 76 kJ

per mole of methane produced; therefore
fH^298 (methane)=−76 kJ mol−^1 .(The

tabulated NIST value is−74.9 kJ mol−^1 .)