Physical Chemistry , 1st ed.



    0.204 mol

You should be able to verify the value of using any of the other three sub-

stances in the reaction.

b.At equilibrium, (^) rxnGequals zero. Why? Because that’s one way to define
equilibrium: the instantaneous change in the Gibbs free energy is zero when
the reaction is at equilibrium. This is what the equality means in equations 4.9.
c. (^) rxn G°, on the other hand, is not zero. (^) rxnG° (note the ° sign) is the dif-
ference in the Gibbs free energy when reactants and products are in their
standard state of pressure and concentration. (^) rxnG° is related to the value of
the equilibrium constant by equation 5.10:
(^) rxnG°RTln K
Given a temperature of 120°C (393 K) and an equilibrium constant value of
4.00, we can substitute:
(^) rxnG°8.314 
mo

J

l K

(393 K)(ln 4.00)

Evaluating:

(^) rxnG°4530 J/mol
Because our equilibrium constant has been defined in terms of partial pres-
sures, we will have to convert to those values if some other unit of amount is
used, such as moles or grams. The following example illustrates a more com-
plex problem.
Example 5.6
Molecular iodine dissociates into atomic iodine at relatively moderate tem-
peratures. At 1000 K, for a 1.00-L system that has 6.00  10 ^3 moles of I 2
present initially, the final equilibrium pressure is 0.750 atm. Determine the
equilibrium amounts of I 2 and atomic I, calculate the equilibrium constant,
and determine if the relevant equilibrium is
I 2 (g) 2I (g)
Assume ideal-gas behavior under these conditions. Use atm as the standard
unit for pressure.
Solution
Since this example is a bit more complicated, let us map out a strategy before
we begin. We assume that some of the molecular iodine will dissociate—call
the amount x—and the amount of atomic iodine, given by the stoichiometry
of the reaction, will be  2 x. In a volume of 1.00 L at 1000 K, we can use the
ideal gas law to determine partial pressures. We have to constrain any possi-
ble answer to the fact that pI 2 pImust equal 0.750 atm.
We can construct a chart for this example:
Amount I 2 2I
Initial 6.00  10 ^3 mol 0 mol
Equilibrium 6.00  10 ^3 x  2 x

QP

JQPJ

0.102 mol 0.306 mol



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5.3 Chemical Equilibrium 127