The Foundations of Chemistry

modynamics. It is different from the reference state for
H^0 f(see Section 15-7). The
absolute entropies, S^0298 , of various substances under standard conditions are tabulated in
Appendix K.
The standard entropy change, S^0 ,of a reaction can be determined from the absolute
entropies of reactants and products. The relationship is analogous to Hess’s Law.

S^0 rxnnS^0 productsnS^0 reactants

S^0 values are tabulated in units of J/molKrather than the larger units involving kilojoules
that are used for enthalpy changes. The “mol” term in the units for a substancerefers to
a mole of the substance, whereas for a reactionit refers to a mole of reaction. Each term
in the sums on the right-hand side of the equation has the units



The result is usually abbreviated as J/molK, or sometimes even as J/K. As before, we
will usually omit units in intermediate steps and then apply appropriate units to the result.

EXAMPLE 15-15 Calculation of S^0 rxn

Use the values of standard molar entropies in Appendix K to calculate the entropy change at
25°C and one atmosphere pressure for the reaction of hydrazine with hydrogen peroxide. This
explosive reaction has been used for rocket propulsion. Do you think the reaction is sponta-
neous? The balanced equation for the reaction is

N 2 H 4 (
)2H 2 O 2 (
)88nN 2 (g)4H 2 O(g) 
H^0 rxn642.2 kJ/mol reaction

Plan

We use the equation for standard entropy change to calculate
S^0 rxnfrom the tabulated values
of standard molar entropies, S^0298 , for the substances in the reaction.

Solution

We can list the S^0298 values that we obtain from Appendix K for each substance:

N 2 H 4 (
)H 2 O 2 (
)N 2 (g) H 2 O(g)

S^0 , J/molK: 121.2 109.6 191.5 188.7

S^0 rxnnS^0 productsnS^0 reactants

[S^0 N 2 (g) 4 S^0 H 2 O(g)][S^0 N 2 H 4 (
) 2 S^0 H 2 O 2 (
)]

[1 (191.5)4 (188.7)][1 (121.2)2 (109.6)]

S^0 rxn605.9 J/molK

The “mol” designation for
S^0 rxnrefers to a mole of reaction, that is, one mole of N 2 H 4 (
),
two moles of H 2 O 2 (
), and so on. Although it may not appear to be, 605.9 J/molK is a rela-
tively large value of
S^0 sys. The positive entropy change favors spontaneity. This reaction is
also exothermic (
H^0 is negative). As we shall see, this reaction mustbe spontaneous, because
both factors are favorable: the reaction is exothermic (
H^0 rxnis negative) and the disorder of
the system increases (
S^0 rxnis positive).

You should now work Exercise 92.

J



(mol rxn)K

J



(mol substance)K

mol substance



mol rxn

The nmeans that each S^0 value must

be multiplied by the appropriate

coefficient, n,from the balanced

equation. These values are then added.

15-14 Entropy, S 625

Small booster rockets adjust the

course of a satellite in orbit. Some of

these small rockets are powered by

the N 2 H 4 –H 2 O 2 reaction.

See the Saunders Interactive

General Chemistry CD-ROM,

Screen 20.5, Calculating 
Sfor a

Chemical Reaction.