The entropy of a substance increases as the temperature is increased. This is readily understood because the addition of heat to a substance results in an increase in molecular randomness. Conversely cooling a substance makes it more ordered and decreases its entropy.

At absolute zero, the entropy of a perfect crystalline substance may be taken as zero. This statement is sometimes called the Third Law of Thermodynamics.

On the basis of the third law, absolute entropies can be calculated from heat capacity data by extrapolating to absolute zero.

Example : 2 Hg (l) + O₂ (g) ® 2HgO (s)

The third law entropies are

S⁰ (Hg) = 18.5 Cal / ⁰K mole

S⁰ (O₂) = 49 Cal / ⁰K mole

S⁰ (HgO) = 17.2 Cal / ⁰K mole

D H⁰f (HgO) = - 21.68 K Cal / mole

D S⁰ = - 51.6 Cal / ⁰K

Now

D H⁰ = 2 D H⁰f

= - 43,400 Cal ( 2 moles )

\ D G⁰ = D H⁰ - TD S0

= - 43,400 - (298) (51.6)

= - 28,000 Cal

Since there is formation of two moles

D G⁰_f = -14.0 K Cal/mole

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[next chapter]

Index

10.1 Introduction
10.2 First Law of Thermodynamics
10.3 Enthalpy
10.4 Second Law of Thermodynamics
10.5 Gibb's Free Energy
10.6 Absolute Entropies

Chapter 11