The sign
of DG
can be used to predict the spontaneity of a reaction at constant temperature and pressure:
·
If DG is
negative, the reaction or system is spontaneous.
This occurs when the reaction is exothermic, DH is negative. However, it can also occur if the reaction is endothermic
and there is a significant increase in entropy (i.e. TDS
is large and negative, resulting in a negative DG).
·
If DG is
positive, the reaction or system is not spontaneous and is energetically unfavourable.
In this case,
energy must be supplied for the reaction to progress forward. If energy is not
supplied, the reaction will proceed spontaneously in the reverse direction. This will happen if the reaction is highly endothermic (DH is positive)
and the entropic term (TDS) is relatively small. This can also happen if the reaction is exothermic (DH is negative) but the process results in significant increase in order (i.e. TDS is a negative number).
·
If DG is zero, the system or reaction is at equilibrium and there is no net reaction.
The change in enthalpy (DH) is equal in magnitude and opposite in sign to the entropy change term (TDS).
Note:
DG can only predict spontaneity at constant temperature and pressure.
EXAMPLE
Use the values of DH° and DS° to predict whether the following reaction is spontaneous at 25°C [4] :
N2(g)
+ 3 H2(g) →2 NH3(g)
DH° = -92.22 kJ
DS° = -198.75 J/K
Solution
It is clear from the signs of the below values
that this reaction is favored by enthalpy but not by entropy:
DH° = -92.22 kJ
(favorable)
DS° = -198.75 J/K
(unfavorable)
Before we can compare these terms to see which
is larger, we have to incorporate into our calculation the temperature at which the reaction is run:
TK =
25o C + 273.15 = 298.15 K
We then multiply the entropy of reaction by
the absolute temperature and subtract the T DSo term from the DHo term:
DGo = DHo - T DSo
= -92,220 J - (298.15 K x -198.75 J/K)
=
-92,220 J + 59,260 J
= -32,960 J
DGo = -32.96 kJ
According to this calculation,
the reaction should be spontaneous at 25°C.
