dependent only, the sum of the changes from 1 to 3, 3 to 4 and then 4 to 2
must be the same as the change from 1 to 2. Therefore;
DS ¼ ST B À ST A ¼ S
T
B À S
B
À
Á þ S B À S
A
À
Á þ S
A À ST A
À
Á
¼ ST B abs þ S
B À S
A
À
Á À S
T
A abs
Because (S° B − S° A ) = 0 the equation reduces to
DS ¼ S
T
B abs À S
T
A abs
Here then lay the answer that Haber and others were seeking. We can
determine the free energy change and hence the equilibrium constant of a
chemical reaction by determining DH and then calculating DS if we can find
reversible path through absolute zero to the temperature of the reaction. In
T
Entropy
1
2
3
4
A →B
Fig. 2.4 The figure is a gross simplification. For representational purposes it has been
assumed that both A and B are (ideal) gases at temperature T, so that the horizontal
portions of the curves 1–3 and 4–2 represent fusions and evaporations. At absolute
zero, points 3 and 4 are coincident (S A ° = S B °). Because the point of coincidence makes
no difference to the calculation of DS it can occur anywhere along the S axis not
necessarily at the origin as my diagram shows. But it is counter intuitive to assume any
other position. It has also been assumed that S B
T > S A
T but there is no reason why this
should always be so and the argument is the same if the opposite is the case
64
D. Sheppard