Chemistry, Third edition

KEY POINTS ABOUT ENTHALPY CHANGES 221

shows that standard enthalpy change for reaction (13.3) is the sumof the enthalpy

changes for reactions (13.4) and (13.5):

H—^ (13.3)H—^ (13.4)H^ —(13.5)

(62)(9)53 kJ mol^1

Generalizing, if a reaction A E is carried out in several stages,

1234

A B C D E

then the enthalpy change for the overall reaction,

H(A E)

is the sum of the individual enthalpy changes:

H(A E)H(A B)H(B C)H(C D)H(D E)

or,

HoverallH 1 H 2 H 3 H 4

as illustrated in Fig. 13.2(b).

Hess’s law: a geographical analogy

Cardiff is 155 miles from London. Suppose that instead of travelling direct from

Cardiff to London, you travelled first from Cardiff to Edinburgh (455 miles) and

then from Edinburgh to London (413 miles) as in Fig. 13.3. Whichever route you

select, Cardiff is still only 155 miles from London. Similarly, whatever the chemical

route taken in Fig. 13.2(a), the energy differencebetween I 2 (s)H 2 (g) and 2HI(g)

remains at 53 kJmol^1.

Use of Hess’s law to calculate enthalpy changes

Hess’s law allows us to work out a Hvalue for a reaction without carrying out that

reaction in the laboratory. This is achieved by algebraically manipulating (adding,

subtracting, or multiplying by a factor) the chemical equations of related reactions

so that they produce the chemical equation for the reaction whose enthalpy change

is required. The value for His then obtained by applying the same operations to

the enthalpy changes of the related reactions. This procedure is illustrated in

Example 13.2.

Fig. 13.3A geographical

analogy of Hess’s law.

Example 13.2

From the thermochemical equations

H 2 (g)Cl 2 (g) 2HCl(g) H—^ (13.6)=184.6 kJ mol^1 (13.6)

2NH 3 (g) 3H 2 (g)N 2 (g) H—^ (13.7)=92.2 kJ mol^1 (13.7)

(^1) ⁄
2 N 2 (g)2H 2 (g)
(^1) ⁄
2 Cl 2 (g) NH 4 Cl(s) H
—
(13.8)=314.4 kJ mol
 (^1) (13.8)
calculate the enthalpy change for the reaction
HCl(g)NH 3 (g) NH 4 Cl(s) H—^ (13.9)=? (13.9)