CK-12-Chemistry Intermediate

http://www.ck12.org Chapter 20. Entropy and Free Energy

0=∆H°-T∆S°

T=

∆H◦

∆S◦

=

177 .8 kJ/mol

0 .1605 kJ/K·mol

=1108 K= 835 ◦C

At any temperature higher than 835°C, the value of∆G° will be negative, and the decomposition reaction will be
spontaneous.

FIGURE 20.5

This lime kiln in Utah was used to produce

quicklime (calcium oxide), an important

ingredient in mortar and cement.

Recall that we are assuming the values for∆H and∆S are independent of temperature. Because they actually vary
slightly at non-standard temperatures, the point at which the sign of∆G° switches from being positive to negative
(835°C) is an approximation. Also, it is not correct to assume that absolutely no products are formed below 835°C
and decomposition suddenly begins once that temperature is reached. Rather, at lower temperatures, the reactants are
simply favored to some extent over the products when the reaction is at equilibrium. When this reaction is performed,
the amount of products can be detected by monitoring the pressure of the CO 2 gas that is produced. Above about
700°C, measurable amounts of CO 2 are produced. The pressure of CO 2 at equilibrium gradually increases with
increasing temperature. Above 835°C, the pressure of CO 2 at equilibrium begins to exceed 1 atm, the standard-state
pressure. This is an indication that the products of the reaction are now favored above that temperature. When
quicklime is manufactured, the CO 2 is constantly removed from the reaction mixture as it is produced. This causes
the reaction to be driven toward the products according to Le Châtelier’s principle.

Changes of State

At the temperature at which a change of state occurs, the two states are in equilibrium with one another. For an
ice-water system, equilibrium takes place at 0°C, so∆G° is equal to 0 at that temperature. The heat of fusion of
water is known to be equal to 6.01 kJ/mol, and so the Gibbs free energy equation can be solved for the entropy
change that occurs during the melting of ice. The symbol∆Sf usrepresents the entropy change during the melting
process, while Tfis the freezing point of water.

0 =∆Hf us- Tf∆Sf us

∆Sfus=

∆Hfus

Tf

=

6 .01 kJ/mol

273 K

= 0 .0220 kJ/K·mol= 22 .0 J/K·mol

The entropy change is positive as the solid state changes into the liquid state. A similar calculation can be performed
for the vaporization of liquid to gas.