The vapor pressure of water is constant regardless of the amount of liquid
water present. In fact the amount of liquid does not affect the equilibrium,
and for this reason it is not included in the expression for the equilibrium
constant. It is a general rule that in heterogeneous systems, the activity of
pure solids and liquids is always 1. In our example of the H 2 O(l) ⇌ H 2 O(g)
equilibrium, K p = P H 2 O = a H 2 O . Another example is decomposition of
calcium carbonate to calcium oxide and carbon dioxide,
CaCO 3 s
ð Þ ⇌ CaO s
ð Þ + CO 2 g
ð Þ
(2.74)
The amount of calcium carbonate and calcium oxide does not affect the
position of equilibrium. Again, K p = P CO 2 = a CO 2 .
In many instances, nanoparticles are present or functioning in aqueous
solutions. A fraction of the particles may dissolve and result in a saturated
solution. The remaining solid will remain in equilibrium with the solution,
but the amount will not affect the position of equilibrium. For example,
barium sulfate nanoparticles may dissolve according to the reaction
BaSO 4 s
ð Þ !
H 2 O Ba
2+ aq
ð Þ + SO
2−
4 aq
ð Þ
(2.75)
The equilibrium constant in this case is known as the solubility produced,
K sp , and is given by Equation 2.76:
K sp = a Ba 2+ a SO 2−
4
(2.76)
2.6.3 The relationship between Gibbs energy
and the equilibrium constant
We can use our understanding of the laws of thermodynamics to obtain a
relationship between the Gibbs energy and the thermodynamic equilibrium constant. Starting with the definition of the Gibbs and the enthalpy
state functions (Equations 2.77 and 2.78),
G = H − TS
(2.77)
H = E + PV
(2.78)
we can write Equation 2.79:
G = E + PV − TS
(2.79)
Taking derivatives yields
dG = dE + PdV + VdP − TdS − SdT
(2.80)
CHAPTER 2: Thermodynamics and Nanoscience
52
water present. In fact the amount of liquid does not affect the equilibrium,
and for this reason it is not included in the expression for the equilibrium
constant. It is a general rule that in heterogeneous systems, the activity of
pure solids and liquids is always 1. In our example of the H 2 O(l) ⇌ H 2 O(g)
equilibrium, K p = P H 2 O = a H 2 O . Another example is decomposition of
calcium carbonate to calcium oxide and carbon dioxide,
CaCO 3 s
ð Þ ⇌ CaO s
ð Þ + CO 2 g
ð Þ
(2.74)
The amount of calcium carbonate and calcium oxide does not affect the
position of equilibrium. Again, K p = P CO 2 = a CO 2 .
In many instances, nanoparticles are present or functioning in aqueous
solutions. A fraction of the particles may dissolve and result in a saturated
solution. The remaining solid will remain in equilibrium with the solution,
but the amount will not affect the position of equilibrium. For example,
barium sulfate nanoparticles may dissolve according to the reaction
BaSO 4 s
ð Þ !
H 2 O Ba
2+ aq
ð Þ + SO
2−
4 aq
ð Þ
(2.75)
The equilibrium constant in this case is known as the solubility produced,
K sp , and is given by Equation 2.76:
K sp = a Ba 2+ a SO 2−
4
(2.76)
2.6.3 The relationship between Gibbs energy
and the equilibrium constant
We can use our understanding of the laws of thermodynamics to obtain a
relationship between the Gibbs energy and the thermodynamic equilibrium constant. Starting with the definition of the Gibbs and the enthalpy
state functions (Equations 2.77 and 2.78),
G = H − TS
(2.77)
H = E + PV
(2.78)
we can write Equation 2.79:
G = E + PV − TS
(2.79)
Taking derivatives yields
dG = dE + PdV + VdP − TdS − SdT
(2.80)
CHAPTER 2: Thermodynamics and Nanoscience
52