Chemistry, Third edition

You might be wondering why the calculation of H^ —fromH—f^ values by this

method works. The answer lies in the definition of standard enthalpy of formation,

which requires H—f^ values for elements in their reference states to be zero.

Although the standard enthalpies (i.e. energies) of compounds and elements cannot

be measured or calculated, the differencebetween the standard enthalpies of two sub-

stances is equal to the difference in the corresponding standard enthalpies of forma-

tion. This is the reason why equation (13.15) strongly resembles equation (13.2).

Standard enthalpies of formation and compound

stability

The value of H—f^ is an approximate measure of the stability of a substance relative to

the elements from which it is made. The standard enthalpies of formation of graphite,

diamond, water, ethyne (acetylene, C 2 H 2 ), ammonia and sodium chloride are shown

in Fig. 13.6. The reference states of elements define an energy baseline or ‘sea level’.

Compounds such as ethyne, for which H—f^ is positive, and which therefore possess a

greater enthalpy than their constituent elements, appear above ‘sea level’ and are called

endothermic compounds. Compounds such as water, ammonia and sodium chloride,

for which H—f^ is negative and which therefore possess a lower enthalpy than their

constituent elements, appear below ‘sea level’ and are called exothermic compounds.

SinceH—f^ values represent the difference in enthalpy between compounds and

the elements from which they are made, it is unhelpful to compare the H—f^ values of

STANDARD ENTHALPY OF FORMATION 229

Example 13.4 (continued)

H—^ f(H 2 O(l) 285.83 kJ mol^1

H—^ f(CuO(s) 157.3 kJ mol^1

H—^ f(NH 3 (g) 46.11 kJ mol^1

and remembering that the standard enthalpy of formation of the elements N 2 and

Cu is zero

H—^ [03(285.83)3(0)][3(157.3)2(46.11)]

293.4 kJ mol^1

Therefore the standard enthalpy change for reaction (A)is293.4 kJ mol^1.

Standard enthalpy change

Use the H^ —fdata in Table 13.2 to calculate the standard enthalpy change at 298 K of the

following reactions:

(i) The burning of two moles of carbon monoxide gas:

2CO(g)O 2 (g) 2CO 2 (g)

(ii)the reduction of one mole of iron(III) oxide by carbon monoxide:

Fe 2 O 3 (s)3CO(g) 2Fe(s)3CO 2 (g)

(iii)the decomposition of one mole of ammonia gas:

NH 3 (g)^1 ⁄ 2 N 2 (g)^3 ⁄ 2 H 2 (g)

Exercise 13H