c04 JWBS043-Rogers September 13, 2010 11:24 Printer Name: Yet to Come
58 THERMOCHEMISTRYH=U+nRTunder the ideal gas assumption. For the formation of one mole
of water,n=−^32fU^298 =fH^298 −nRT=fH^298 +^32 RT=− 285 ,600 J mol−^1 +^32 [8.31(298)]∼=− 281 .9kJmol−^1For comparison with theoretical calculations, one often needs to know the
thermodynamic properties of molecules in the gaseous state—for example,
fH^298 (H 2 O(g)). This is handled by adding the heat of vaporization of water to
fH^298 (H 2 O(l)) to obtainfH^298 (H 2 O(g))=− 285. 6 + 44. 0 =− 241 .6kJmol−^1However, water vapor is not in itsstandard state.4.3 STANDARD STATES
One can burn a diamond, C(dia), in an oxygen bomb calorimeter. When this is done,
the measured enthalpy of formation of CO 2 (g) is about 2 kJ mol−^1 more negative
than thefH^298 (CO 2 (g)) found when C(gr), carbon in the standard state, is burned.
The difference is not in the CO 2 (g) produced, but in the crystalline form of diamond,
which is not the standard state for carbon. Since the path from C to CO 2 (g) is about
2kJmol−^1 longer in the diamond combustion, the starting point C(dia) must have
been about 2 kJ mol−^1 higher in enthalpy than C(gr). Differences like this lead us
to define the standard state of all elements as thestableform at 1.000 atm pressure.
By the nature of enthalpy and energy (thermodynamic properties), we can set any
arbitrary point to zero as a reference point. Hence we define the enthalpy of formation
ofany element in its standard state as zero at all temperatures. This definition
works because elements are not converted from one to another in ordinary chemical
reactions.
Our previous observation thatfH^298 (C(dia))=0 but is about 2 kJ mol−^1 sug-
gests that, given this small enthalpy change, it might be possible to convert common
graphite into the nonstandard state of diamond. Indeed it is. Production of small
industrial diamonds for cutting tools is commercially feasible.4.4 MOLECULAR ENTHALPIES OF FORMATION
According to the first law of thermodynamics, the enthalpy of formation of a molecule
can be determined even if the formation reaction from its elements cannot be carried
out in an actual laboratory experiment. This is illustrated by the simple example of
carbon monoxide CO(g). We know from painful examples that this poisonous gas is
produced by incomplete combustion of hydrocarbon fuels, but the simple controlled