Page 47 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch04 Elementary Physical Chemistry

32 Elementary Physical Chemistry

The standard entropies of H 2 (g), O 2 (g) and H 2 O (l) at 25 Care
◦
respectively 130.7, 205.2, 70.0 kJ mol −1 , giving ∆S =(2 × 70.0 − 205.2 −
2 ×130.7)kJ = −326.6 kJ. [This is the correct way to calculate ∆S,since S
is a state function. Had we calculated ∆S from q/T , the result would have
been wrong. Why?]
The rule that ∆S has to be positive in an irreversible change applies
to an isolated system. In the above example, the system is not isolated.
To get around this difficulty, the standard procedure is to consider also
the entropy of the surroundings (the rest of the universe), and regard the
system and surroundings as an isolated system. Thus,

∆S total =∆S system +∆S surroundings (4.3)

4.4. The Entropy of the Surroundings
In treating the surrounding entropy, it is common practice to assume that
the surrounding changes reversibly even though the system may change
irreversibly. [The rationale is that the surrounding, being so enormous,
would change quasi-statically.]

Example 4.2. Let us return to the water reaction. It is found that at
◦
25 C and one atmosphere the heat released is −572 kJ mol −1 . As stated, the
heat released by the system is assumed to be absorbed by the surrounding
reversibly even though the system looses heat irreversibly.
The heat absorbed by the surrounding is q P = +572 kJ and thus
∆S surr = 572 kJ/298K = 1.92 kJ K −1 = 1920J K −1 . On the other hand,
the system entropy, as noted before, is ∆S sys = −327J K −1 and so ∆S tot
equals (−327 + 1920)J K −1 = 1539J K −1 and is positive, as it should be.
In general, at constant T ,∆S surr = q surr/T = −q sys/T and if P is also
constant, and if there is no work other than pV work,

∆S surr = −∆H sys /T (4.4)

Comment: ∆H can always be related to q, regardless of whether q
is reversible or irreversible. ∆S can only be equated to q/T if q is
reversible.