The Foundations of Chemistry

Another interpretation of Hess’s Law lets us use tables of
H^0 fvalues to calculate the
enthalpy change for a reaction. Let us consider again the reaction of Example 15-8.

C 2 H 4 (g)H 2 O(
)88nC 2 H 5 OH(
)

A table of
H^0 f values (Appendix K) gives
H^0 fC 2 H 5 OH(
)277.7 kJ/mol,

H^0 fC 2 H 4 (g)52.3 kJ/mol, and
H^0 fH 2 O(
)285.8 kJ/mol. We may express this infor-
mation in the form of the following thermochemical equations.

H^0

2C(graphite)3H 2 (g)^12 O 2 (g)88nC 2 H 5 OH(
) 277.7 kJ/mol rxn (1)

2C(graphite)2H 2 (g)88nC 2 H 4 (g) 52.3 kJ/mol rxn (2)

H 2 (g)^12 O 2 (g)88nH 2 O(
) 285.8 kJ/mol rxn (3)

We may generate the equation for the desired net reaction by adding equation (1) to the
reverse of equations (2) and (3). The value of
H^0 for the desired reaction is then the
sum of the corresponding
H^0 values.

The capital Greek letter sigma () is

read “the sum of.” The nmeans that

the 
H^0 fvalue of each product and

reactant must be multiplied by its

coefficient, n,in the balanced equation.

The resulting values are then added.

15-8 Hess’s Law 607

Problem-Solving Tip:
Hf^0 Refers to Specific Reaction

The 
H^0 for the reaction in Example 15-8 is 44 kJ for each mole of C 2 H 5 OH(
)

formed. This reaction, however, does not involve formation of C 2 H 5 OH(
) from its

constituent elements; therefore, 
H^0 rxnis not
H^0 ffor C 2 H 5 OH(
). We have seen the

reaction for 
H^0 fof C 2 H 5 OH(
) in Example 15-6.

Similarly, the 
H^0 rxnfor

CO(g)^12 H 2 (g)88nCO 2 (g)

is not
H^0 ffor CO 2 (g).

We see that
H^0 for this reaction is given by

Hrxn^0 
H^0 (1)
H^0 (2)
H(^0 3)

or by
product reactants

H^0 rxn
H^0 fC 2 H 5 OH(
)[
H^0 fC 2 H 4 (g)
H^0 fH 2 O(
)]

In general terms this is a very useful form of Hess’s Law.

H^0 rxnn
H^0 f productsn
H^0 f reactants

The standard enthalpy change of a reaction is equal to the sum of the standard molar

enthalpies of formation of the products, each multiplied by its coefficient, n,in the

balanced equation,minus the corresponding sum of the standard molar enthalpies of

formation of the reactants.

H^0

2C(graphite)3H 2 (g)^12 O 2 (g)88nC 2 H 5 OH(
) 277.7 kJ/mol rxn (1)
C 2 H 4 (g)88n2C(graphite)2H 2 (g) 52.3 kJ/mol rxn (2)
H 2 O(
)88nH 2 (g)^12 O 2 (g) 285.8 kJ/mol rxn (3)
net rxn: C 2 H 4 (g)H 2 O(
)88nC 2 H 5 OH(
)
H^0 rxn44.2 kJ/mol rxn

o m^8 gb