C 100 k 373 k mol j 1088 mol j 40626 s h t on

C 100 K 373
K mol
J
108.8
mol
J
40626
S
H
T 0
on vaporizati
on vaporizati
 





, the boiling point of H
2
O
*note the cancellation of units

18-12
e.g Consider the decomposition of… CaCO
3
(s)  CaO (s) + CO
2
(g)
 H
0
system
= +178 kJ - endothermic
 S
0
system
= +161 J/K - positive… gas is formed.
Assuming that the reaction entropy and the enthalpy are independent of
temperature, calculate the temperature at which CaCO
3
starts decomposing.
Check the signs of the enthalpy and the entropy:
Solution –  G =  H
system
– T  S
system
 G should be negative to start decomposition. The minimum
temperature is when  G = 0.
Therefore…  H
system
= T  S
system
C 37 8 K 110 1
K
J
61 1
kJ
J
10 kJ 78 1
S
H
T 0
3
ion decomposit
ion decomposit
 





Temperature must be greater than 837
0
C, (T > 837
0
C), to start decomposition.

18-13
Standard Free Energy Change,  G
0
 G is a state function.
 G
0
is the standard free energy change when reactants and products are in their
standard states.
 G
0
f
is the standard free energy of formation of a compound. Many of these are
tabulated at 25
0
C.
 G
0
f
of elements = 0
 G
0
reaction
=  n  G
0
f
(products) –  n  G
0
f
(reactants)
e.g. Calculate  G
0
for the reaction… 2 NO (g) + O
2
(g)  2NO
2
(g)
 G
0
f
(kJ/mol) 86.7 0 51.8
 G
0
reaction
=  n  G
0
f
(products) –  n  G
0
f
(reactants)
= [(2 mol)(51.8 kJ/mol)] – [(2 mol)(86.7 kJ/mol)]
= – 69.8 kJ (negative – the reaction is spontaneous)
 G
0
reaction
can also be calculated from  H
0
and  S
0
of the reaction.
e.g. Calculate  G
0
reaction
:
2NO(g) + O
2
(g)  2NO
2
(g)
Given:  H
0
= -114.1kJ and  S
0
= -146.2 JK
-1

18-14
e.g. Given the following:
C (s, diamond) + O
2
(g)  CO
2
(g)  G
0
reaction
= – 397 kJ
C (s, graphite) + O
2
(g)  CO
2
(g)  G
0
reaction
= – 394 kJ
Calculate the  G
0
reaction
for… C (s, diamond)  C (s, graphite)
C (s, diamond) + O
2
(g)  CO
2
(g)  G
0
reaction
= – 397 kJ
CO
2
(g)  C (s, graphite) + O
2
(g)  G
0
reaction
= +394 kJ
Add: C (s, diamond)  C (s, graphite)  G
0
reaction
= – 3 kJ
 G
0
reaction
is negative… Therefore, diamond should spontaneously change to
graphite. In other words, the reaction is feasible, but it
is very slow.
e.g. Is it feasible to make ethyl alcohol by the following reaction?
C
2
H
4
(g) + H
2
O (l)  C
2
H
5
OH
ethylene water alcohol
Given:  G
0
f
(kJ/mol) 68 – 237 – 125
 G
0
reaction
=  n  G
0
f
(products) –  n  G
0
f
(reactants)
= ( – 175 kJ) – (68 kJ – 237 kJ) = – 6 kJ
Thus… negative, thus the process is spontaneous at 25
0
C.
e.g. Hydrogen peroxide decomposes according to the reaction…
2 H
2
O
2
(l)  2 H
2
O (l) + O
2
(g)  G
0
reaction
= – 234 kJ
Since  G
0
reaction
is negative, one can conclude that H
2
O
2
(l) has the tendency to
decompose into H
2
O (l) and O
2
(g).
Indeed, prolonged storage of H
2
O
2
had shown that decomposition does
occur within a certain noticeable amount of time. Beware, when H
2
O
2
(l)
goes on sale in the supermarket. It might have lost some of its potency.

18-15
Kinetics and Thermodynamics: The Sign of  G
0
f
1) When  G
0
f
is negative, it means that, under standard conditions, the
compound can spontaneously form from its constituent elements.