| 
>>
View the other Key Equations and Concepts in this
chapter
Relationship to  G
and H
>> Parts of this equation/concept include:
A.  G°
= –RT ln K, Equation 15.19 |
>> Qualitative Relationships
The key to qualitative relationships is to understand what the
variables represent.  G°
is the free energy at standard state, which is 298 K and 100 kPa
(about 1 atm), so it is dependent on the type of reaction rather
than on reaction conditions. Recall that a negative value of G°
represents a spontaneous reaction, and a positive value is a nonspontaneous
reaction. The choices for equilibrium constant are greater than
or less than 1. (There are no negative values of K.) The
natural log of a fraction is negative and the natural log of a number
greater than 1 is positive. Therefore reactions that are spontaneous
at standard state have K > 1 and reactions that are nonspontaneous
at standard state have K < 1.
>> Example 1
Which reactions are spontaneous at standard state?
- 2 NOCl
2 NO + Cl2 KC
= 4.4 x 10–6 at 400 K
- C2H4 + H2
C2H6 KC
= 0.99 at 500 K
Solution:
That the K values given are not at standard state is not
relevant, since the equilibrium constant value is related to standard-state
conditions, not to nonstandard conditions. There is a relationship
for nonstandard conditions [G
= G° + RT ln Q],
but since the question does not refer to it, it is not needed.
- Since K is a fraction, G°
will be positive and the reaction is nonspontaneous.
- Since K is a fraction, G° will be positive and the reaction is nonspontaneous.
>> Quantitative Relationships
The key to doing quantitative relationships is units. Since G
is a measure of energy, the gas constant value used is R
= 8.314 J/mol•K. To make the units match this value, temperature
must be in units of Kelvin and G°
must be in units of joules. Most tables (including the one in the
textbook's appendix) list G°
with units of kJ. Therefore these values must be converted to joules.
The value of K does not have units, but usually atmospheres
and molarity apply.
>> Example 2
What is the G° for the following reactions?
- 2 NOCl
2 NO + Cl2 KC
= 4.4 x 10–6 at 400 K
- C2H4 + H2
C2H6 KC
= 0.99 at 500 K
Solution:
-
Using the equation G°
= –RT ln K, and including the values given
G° = –(8.314
J/mol•K)(400 K) ln(4.4 x 10–6)
G° = –(8.314
J/mol•K)(400 K)(–12.33)
G° = +41,004 J/mol
The significant figure rule for logs is that a the number
of significant figures in the number becomes the number of
decimal places in the answer (after the log function). Therefore
12.33 is the answer with the correct significant figures,
which is now four. Assuming all the zeros are significant
in the temperature value (3 sf), the answer with appropriate
significant figures is
G° = +4.10 x 104
J/mol or 41.0 kJ/mol
-
Using the equation
G° = –(8.314)(500)
ln(0.99)
G° = –(8.314)(500)(–0.01)
G° = 41.57 J/mol
Since two decimal places in the natural log leave only one
significant figure, the final answer is
G° = 4 x 101
J/mol or 4 x 10–2 kJ/mol
>> Example 3
What is the equilibrium constant for the following reactions?
- C2H4 + Cl2
CH2ClCH2Cl G°
= –148 kJ/mol at 500 K
- PbCl2
Pb2+ + 2 Cl– G°
= +28.1 kJ/mol at 25 °C
Solution:
-
Using the equation
G° = –RT
ln K
–148,000 = –(8.314)(500) ln K
35.6 = ln K
3 x 1015 = K
If you consider the significant-figure rule described in
the previous example, the reverse is that number of decimal
in the log term (35.6) becomes the number of significant figures
in the antilog.
-
Using the equation
G° = –RT
ln K
+28,100 = –(8.314)(298) ln K
–11.3 = ln K
1.19 x 10–4 = K
>> back
to the Top of the Page
The van't Hoff equation is used to determine the value of the equilibrium
constant with temperature changes. The equation is
| ln |
|
= |
| –H° |
|
| R |
|
 |
( |
|
– |
|
) |
Because the gas constant is used with an energy term (H),
the value R = 8.314 J/mol•K is appropriate.
Because R uses units of joules, H
should also be in units of joules, so that the units will cancel.
As usual, temperature must be in units of Kelvin.
>> Example 4
If the equilibrium constant is 1.5 x 10–8 at
298 K for a reaction with a H°
of +77.2 kJ/mol, what is the equilibrium constant at 400 K?
Solution:
| ln |
|
= |
| –H° |
|
| R |
|
|
( |
|
– |
|
) |
If the value of equilibrium constant is called K1,
then the values substituted into the equation are
K1 = 1.5 x 10–8
H° = +77,200 J/mol
(converted from kJ by 77.2 kJ/mol x (1000 J/1 kJ))
R = 8.314 J/mol•K
T1 = 298 K
T2 = 400 K
K2 = ?
Substituting these values into the equation,
| ln |
|
= |
|
|
( |
|
– |
|
) |
| ln |
|
= |
–9285.5 |
|
(2.5 x 10–3
– 3.36 x 10–3) |
| ln |
|
= |
–9285.5 |
|
(–8.56 x 10–4) |
K2 = 4.2 x 10–4
>> View
the other Key Equations and Concepts in this chapter
|
G°
= –RT ln K, Equation 15.19
