Page 180 - Physical chemistry understanding our chemical world

INTRODUCING THE GIBBS FUNCTION 147

The sign of G

Equation (4.24) in Justification Box 4.2 shows clearly that a pro-
The Gibbs energy must
cess only occurs spontaneously within a system if the change in
Gibbs function is negative, even if the sign of S (system) is slightly be negative if a change
occurs spontaneously.
negative or if H (system) is slightly positive. Analysing the reaction
in terms of our new variable G represents a great advance: pre-
viously, we could predict spontaneity if we knew that S (total) was
positive – which we now realize is not necessarily a useful crite-
We see the analytical
rion, since we rarely know a value for S (surroundings) . It is clear
power of G when we
from Equation (4.21) that all three variables, G, H and S, each realize how its value
relate to the system alone, so we can calculate the value of G by does not depend on
looking up values of S and H from tables, and without needing the thermodynamic
to consider the surroundings in a quantitative way. properties of the sur-
A process occurring in a system is spontaneous if G is nega- roundings, but only on
tive, and it is not spontaneous if G is positive, regardless of the the system.
sign of S (system) . The size of G (which is negative) is maxi-
mized for those processes and reactions for which S is positive
and which are exothermic, with a negative value of H. A process occurring in a
system is spontaneous
Worked Example 4.8 Methanol (IV) can be prepared in the if G is negative, and is
gas phase by reacting carbon monoxide with hydrogen, according not spontaneous when
to Equation (4.25). Is the reaction feasible at 298 K if H O = G is positive.
−1
O
−90.7kJ mol −1 and S =−219 J K −1 mol ?
(4.25)
CO (g) + 2H 2(g) −−−→ CH 3 OH (g)
H
OH
H
H
(IV)

O
O
O
We shall use Equation (4.21), G = H − T S . Inserting
values (and remembering to convert from kJ to J): The Gibbs function is
a function of state,
−1
−1
O
G = (−90 700 J mol ) − (298 K ×−219 J K −1 mol ) so values of G O
obtained with the van’t
O
G = (−90 700 + 65 262) Jmol −1 Hoff isotherm (see
p. 162) and routes
O
G =−25.4kJ mol −1 such as Hess’s law
cycles are identical.
This value of G O is negative, so the reaction will indeed be
spontaneous in this example. This is an example of where