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>>
View the other Key Equations and Concepts in this
chapter
Free Energy
>> Parts of this equation/concept include:
A.  G
= H
- TS,
Equation 13.12 |
>> Predicting Free Energy Qualitatively
Three thermodynamic parameters are put together in this equation,
so the relationships are very important, as is the meaning of each
variable. Table 13.1 reviews each of these variables.
Table 13.1 MEANING OF THERMODYNAMIC PARAMETERS
| Thermodynamic parameter |
Name |
Positive value |
Negative value |
Zero-Row |
 H |
enthalpy |
endodermic |
exothermic |
for H°f,
an element in its standard state |
| S |
entropy |
more random |
more ordered |
for S°, a perfect crystal at 0 |
| G |
Gibbs free energy |
nonspontaneous |
spontaneous |
equilibrium |
The other important thing to see qualitatively is the relationship
of the values to G. Remember
that both H and S
are needed to make any statement about G,
and the effect may be temperature dependent. Also remember that
"low" and "high" are relative temperatures. In some situations,
500 K is high; for others it is low.
Table 13.2 EFFECT OF H AND
S on G
| Sign of H |
Sign of S |
Sign of G |
| positive |
negative |
positive (nonspontaneous) at all temperatures |
| positive |
positive |
positive (nonspontaneous) at low temperatures
negative (spontaneous) at high temperatures |
| negative |
negative |
negative (spontaneous) at low temperatures
positive (nonspontaneous) at high temperatures |
| negative |
positive |
negative (spontaneous) at all temperatures |
>> Example 1
Under what temperature conditions are the following reactions
spontaneous?
- NH4Cl(s) + OH–(aq)
NH3(g) + H2O(l) + Cl–(aq)
endothermic
- 2 NO(g) + O2(g)
2 NO2(g) exothermic
- CaCO3(s)
CaO(s) + O2(g) endothermic
Solution:
- Because the reaction is endothermic, H
is positive. Entropy increases as the solid becomes a gas and
a liquid, so S is positive.
Since S favors a spontaneous
reaction but H does not,
to maximize S, this reaction
is spontaneous at high temperatures.
- Because this reaction is exothermic, the sign on H
is negative. Going from more moles of gas to fewer decreases
entropy, so S is negative.
The enthalpy encourages spontaneity; entropy discourages it.
To minimize entropy, this reaction will be spontaneous at low
temperatures.
- Because this reaction is endothermic, H
is positive. A solid turning to a gas results in an increase
in entropy, so S is positive.
While S is favorable for
a spontaneous reaction, H
is not. To maximize the effect of S,
high temperatures are required. Therefore this reaction will
be spontaneous at high temperatures.
>> Calculating Free Energy Quantitatively
The most common mistakes using this equation are with units. Typically,
the units of H are kJ/mol and
of S are J/mol•K. To add
these values together, the energy units must be the same. Thus either
H must be converted to joules
or S must be converted do kJ.
(Don't do both. You'll end up as you started!) In addition, S
requires that the temperature be in units of K. You may have to
use equations 11.13 and 13.5 to determine the values for H
and S. Be sure that all
the parameters—G, H,
and S—refer to the same
reaction. If the superscript is zero, the temperature is 298 K.
>> Example 2
What is the G° for the
following reaction? (Calculate H
and S from values in the appendix.)
Is it spontaneous?
NH4Cl(s) + OH–(aq)
NH3(g) + H2O(l) + Cl–(aq)
Solution:
Using the appendix to look up values of S° and H°f
for each reactant and product
| |
NH4Cl |
OH– |
NH3 |
H2O |
Cl– |
| H° (kJ/mol) |
–314.4 |
–230.0 |
–46.1 |
–285.8 |
–167.2 |
| S° (J/mol•K) |
94.6 |
–10.8 |
192.3 |
69.9 |
56.5 |
Using Equation 11.13 to determine H
H =  nHproducts
– mHreactants
H = [–46.1 + –285.8
+ –167.2] – [–314.4 + –230.0]
H = [–499.1] –
[–544.4]
H = +45.3 kJ/mol
Using Equation 13.5 to determine S
S = nSproducts
– mSreactants
S = [192.3 + 69.9 + 56.5]
– [94.6 + –10.8]
S = [318.7] – [83.8]
S = +234.9 J/mol•K
Before using Equation 13.12, the units of either H
or S must be changed. Changing
the units of H
| H |
= |
+45.3 kJ/mol |
 |
|
= |
+45,300 J/mol |
Use this value for H in Equation
13.12 with S. Because the
question asks for G°,
T = 298 K. The superscript indicates standard conditions,
which includes a temperature of 25 °C.
G = H
– T S
G = (+45,300 J/mol) –
(298 K)(+234.9 J/mol•K)
G = +45,300 – 70,000
= –24,700 J/mol = –2.47 × 104 J/mol
in kJ
| –24,700 J/mol |
|
|
= |
–24.7 kJ/mol |
The negative sign indicates that the reaction is spontaneous
at this temperature.
Alternately, changing S to
units of kJ
| +234.9 J/mol |
|
|
= |
+0.2349 kJ/mol |
G = H
– T S
G = (+45.3 kJ/mol) –
(298 K)(+0.2349 kJ/mol)
G = –24.7 kJ/mol
The negative sign indicates that the reaction is spontaneous.
>> Example 3
What is the G° and at
what temperatures is the following reaction spontaneous?
H2O(l) + SO3(g)
H2SO4(aq)
Solutiong:
Using the appendix to get values of H°f
and S°,
| |
H2O |
SO3 |
H2SO4 |
| H°(kJ/mol) |
–285.8 |
–395.7 |
–909.3 |
| S°(J/mol•K) |
69.9 |
256.6 |
20.0 |
Using Equation 11.13,
H = nHprod
– mHreact
H = [–909.3] –
[–285.8 + –395.7] = [–909.3] – [–681.5]
= –227.8 kJ/mol
Using Equation 13.5,
S = nSprod
– mSreact
S = [20.0] – [69.9
+ 256.6] = [20.0] – [257.5] = –237.5 J/mol•K
Converting S to kJ,
| –237.5 J/mol•K |
|
|
= |
–0.2375 kJ/mol•K |
Using Equation 13.12 (recall that T = 298 K for G°),
G = H
– T S
G = –227.8 –
(298)(–0.2375) = –227.8 + 70.775 = –157.0 kJ/mol
To determine the temperatures at which this reaction is spontaneous,
use Equation 13.12, set G
= 0, and solve for temperature. Not only is G
= 0 equilibrium, but it is the turning point from spontaneous
to nonspontaneous reactions.
G = H
– T S
0 = –227.8 – T(–0.2375)
227.8 = 0.2375T
959.2 K = T
However, at this point, you have determined the temperature at
equilibrium, and the question asks at which temperature the reaction
is spontaneous. There are two ways to decide whether it is spontaneous
at higher or lower temperatures than the calculated one.
The calculation at 298 K indicates that it is spontaneous at
that temperature. Since it is lower than the equilibrium temperature,
the reaction is spontaneous at T < 959.2 K.
Alternately, the S value
is negative, which does not favor spontaneity. Consequently, this
reaction is spontaneous at low temperatures. Putting that together
with the equilibrium temperature, the reaction must be spontaneous
at T 959.2 K.
>> back
to the Top of the Page
| B. Calculating G°rxn
from G°f,
Equation 13.14 |
G°rxn = nG°f,products
– mG°f,reactants
Another way of calculating G
is from the G°f
values. These are listed in the same appendix as H
and S°. It is an equally valid way of calculating G.
Choosing this equation over Equation 13.12 is appropriate if G°f
values are given. Note that these values refer to formation reactions.
You can review formation reactions in Chapter 11.
>> Example 4
Using G°f
values, determine G°rxn
for the gaseous reaction
2 NO + O2
2 NO2
Values of G° are
G° of NO2
= +51.3 kJ/mol
G° of O2
= +0 kJ/mol
G° of NO = +86.6 kJ/mol
Solution:
This equation is solved in the same way as Equations 11.13 and
13.5.
G°rxn =
nG°f,products
– mG°f,reactants
G° = [2(51.3)] –
[2(86.6) + 0] = [102.6] – [173.2] = –70.6 kJ/mol
Significant figures for each value have been underlined except
the zero for oxygen which has infinite significant figures. Like
H°f, G°f
for an element at standard state is defined to be zero.
G° = [2(51.3)]
– [2(86.6) + 0] = [102.6] – [173.2]
= –70.6 kJ/mol = –71 kJ/mol
>> View
the other Key Equations and Concepts in this chapter
|
G
= 
