Concise Physical Chemistry

c19 JWBS043-Rogers September 13, 2010 11:29 Printer Name: Yet to Come

306 CLASSICAL MOLECULAR MODELING

Elements

-21 kJ mol-1

-21 kJ mol-1

-21 kJ mol-1

FIGURE 19.1 Enthalpies of formation of “adjacent”n-alkanes.

as –42 kJ mol−^1. Provided that these enthalpies are transferable, the enthalpy of

formation ofn-pentane CH 3 (CH 2 ) 3 CH 3 , for example, is estimated as 2(−42)+

3(−21)=147 kJ mol−^1. The calculation agrees with the experimental value:

fH^298 (CH 3 (CH 2 )nCH 3 )=147 kJ mol−^1

to a precision of±1kJmol−^1. Although the agreement is not always this nice, the

procedure is very encouraging.

What of groups other than alkyl groups? Can thisgroup additivitystrategy

be extended to include the CH group and the C atom in branched alkanes like

2-methylbutane (isopentane) and 2,2-dimethylpropane (neopentane)? The answer is

yes, and this strategy has been extended to cover many functional groups in organic

molecules including oxygenated, nitrogenous, and halogenated hydrocarbons (Cohen

and Benson, 1993).

19.2 BOND ENTHALPIES

Molecular enthalpy can be segregated in other ways. One familiar way involves

associating an enthalpy with each bond in the molecule. For example, inserting a

CH 2 group into ann-alkane requires breaking the backbone of the alkane molecule

at a cost of one C C bond followed by forming two C C bonds and increasing the

number of C H bonds in the molecule by two:

R CH 2 CH 2 R→R CH 2 CH 2 CH 2 R

The sum total of the process is a gain of one C C bond and two C H bonds. If we

associate –348 kJ mol−^1 with the C C bond and –413 kJ mol−^1 with the C H bond

relative to the isolated atoms C and H, the insertion entails a change in total bond

enthalpies of

C C+2(C H)=− 348 +2(−413)=−1174 kJ mol−^1

Why such a big number? This big number arises because theoretical folks find

it convenient to use a reference state that is different from the thermodynamicist’s