Download First, there are several issues regarding this course need to be

10.1 Thermodynamic functions of formation (continued)
In this lecture we will continue to study the thermodynamic functions of
formations of ions. We have just learned that the Gibbs energy of formation of ions
received contributions from several factors including dissociation, ionization and
hydration of hydrogen ion. At the end of the last lecture, we introduced Max Born
equation, which can be used to calculate the standard Gibb energy of salvation.
The standard solvation Gibbs energy ∆solvGө can be calculated by Born equation:
z i2 e 2 N A 
1
1 − 
10.1
∆ solv G
=−
8πε 0 ri 
εr 
where zi is the charge number, e is the elementary charge, NA is the Avogadro’s constant,
ε0 is the vacuum permittivity, εr is the relative permittivity, ri is ion’s radius.
θ
At 25 oC, when the solvent is water,
∆ solv G θ
=−
z i2
6.86 × 10 4 kJmol −1
(ri / pm)
(
)
10.2
Example 10.1c : Estimate the value of ∆solvGө(Br-, aq) - ∆solvGө (Cl-, aq) from the
experimental data and from the Max Born equation.
Solution: To calculate the difference of their experimental measurement, use the data
provided in Table 2.6:
∆solvGө(Br-, aq) = -103.96 kJ mol-1;
∆solvGө(Cl-, aq) = -131.23 kJ mol-1;
So
∆solvGө(Br-, aq) - ∆solvGө (Cl-, aq) = -103.96 – ( 131.23)
= 27.27 kJ mol-1;
In order to apply the Born equation (10.2), we need to know the radius of the
corresponding ions. These numbers can be obtained from Table 23.3
r(Br-) = 196 pm; r(Cl-) = 181 pm;
thus
∆solvGө(Br-, aq) - ∆solvGө (Cl-, aq) = - (1/196 – 1/181)*6.86*104 kJ mol-1
= 29.00 kJ mol-1
(The calculated result is slightly larger than the experimental value).
Quite often, we do not have to go through the above process in order to know the
standard Gibbs energy of formation or standard reaction Gibbs energy, which could be
employed for further calculations such as calculating equilibrium constant. Check the
following example:
Example 10.1d: Calculate the molar solubility of lead (II) sulfide at 25oC where the
standard Gibbs energy of formation is -98.7 kJ mol-1.
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Solution: The molar solubility of PbS(s) can be represented by the molality of S2- ,
b(s2-)in the aqueous solution.
The molality has the following relationship with activity α = γb/bө;
Assume this is a diluted solution, where γ = 1.0; so α = b/bө;
Because ln(K) = - ∆rGө/(RT)
We need to know the reaction Gibbs energy of the following process:
PbS(s)
↔ Pb2+(aq) + S2-(aq)
∆rGө = ∆fGө(Pb2+) + ∆fGө(S2-) - ∆fGө(PbS(s))
= 85.8 kJ mol-1 + (-24.43 kJ mol-1 ) - (-98.7 kJ mol-1)
= 160.07 kJ mol-1
so K = 8.7 x 10-29
From the above reaction equation, we get b(Pb2+) = b(s2-) (due to their 1:1
stiochiometric relationship)
Thus, (b/bө)2 = K
b = 9.3 x 10-15 mol kg-1;
(From the final result, we see that PbS barely dissolves in water, thus our assumption
that this is a diluted solution is all right.)
10.2 Ion activities
Recall what we just reviewed in the above section that the activity relates to the molality
b via
α = γ * b/bө
where γ is called the activity coefficient and bө equal 1mol kg-1.
Now the chemical potential can be expressed by the following equation:
µ =µө + RT ln(b/bө) + RTln(γ) = µideal + RTln(γ)
Consider an electrically neutral real solution of M+ X-,
G = µ+ + µ- = µ+ideal + µ-ideal + RTln(γ+) + RTln(γ-)
=Gideal + RT(γ+γ-)
Since there is no experimental way to separate the product (γ+γ-) into
contributions from the cations and anions, mean activity coefficient γ± is introduced
here to assign equal responsibility for nonideality to both kind of ions:
γ± = (γ+γ-)1/2
The chemical potential for individual ions M+ and X- becomes:
µ+ = µideal + RTln(γ±)
µ- = µideal + RTln(γ±)
For a general compound of the following form: MpXq
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The mean activity coefficient is expressed as:
γ± = (γ+pγ-q)1/s
s = p+q
The nonideality is induced by strong interactions between charges. To calculate
the mean activity coefficient , in 1923, Debye and Huckel devised the Debye-Huckel
limiting law
log(γ±) = -|z+z-|AI1/2
10.3
A = 0.509 for aqueous solution at 25oC. I is the ionic strength, which is calculated as
the following:
I = ½ Σzi2(bi/bө)
10.4
where bi is the charge number of ion I and bi is the molality.
To better understand the usage of the above two equations (10.3 and 10.4), we
study the following examples:
Example 10.2a: Relate the ionic strength of (a) MgCl2, (b) Fe2(SO4)3 solutions to
their molalities, b.
Solution: To use the equation 10.4, we need to know the charge numbers and the
molalities of the ions:
MgCl2:
From molecular formula, we can get b(Mg2+) = b(solution);
b(Cl-) = 2*b(solution);
Z(Mg2+) = 2;
Z(Cl-) = 1;
So
I = ½((2)2*b +(1)2*(2b)) = ½(4b + 2b) = 3b;
Fe2(SO4)3:
So
From the molecular formula, we can get b(Fe3+) = 2*b(solution);
b(SO42-) = 3*b(solution);
Z(Fe3+) = 3;
Z(SO42-) = 2;
I = ½((3)2*(2b) +(2)2*(3b)) = ½(18b + 12b) = 15b;
In the case the solution contains more than one types of electrolytes, the ionic
strength of the solution equals the sum of the ionic strength of each individual compound,
such as:
Example 10.2b: Calculate the ionic strength of a solution that contains 0.050 mol kgin K3(Fe(CN)6](aq), 0.040 mol kg-1 in NaCl(aq), and 0.03 mol kg-1 in Ce(SO4)2 (aq).
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3
Solution: Using the same approach as employed in example 10.2a to calculate the ionic
strength for the three compounds, respectively.
I(K3(Fe(CN)6]) = ½( 12*(0.05*3) + (3)2*0.05 + 12*(0.05*6)) = 0.45;
I(NaCl) = ½(12*0.04 + 12*0.04) = 0.04;
I(Ce(SO4)2 = ½(42*0.03 + 22*(2*0.03)) =0.36;
So, I = I(K3(Fe(CN)6]) + I(NaCl) + I(Ce(SO4)2
= 0.45 + 0.04 + 0.36 = 0.85
Example 10.2c: Calculate the ionic strength and the mean activity coefficient of
2.0m mol kg-1 Ca(NO3) at 25 oC.
Solution: In order to calculate the mean activity coefficient with eq. 10.3, one needs to
know the ionic strength of the solution. Thus, the right approach is to get I first and then
plug I into equation 10.3.
I = ½(22*0.002 + 12*(2*0.002)) = 3*0.002 = 0.006;
From equation 10.3, log(γ) = - |2*1|*A*(0.006)1/2;
= - 2*0.509*0.0775;
= -0.0789;
γ = 0.834;
While the Debye-Huckel limiting law provides us an approach of
calculating the mean activity coefficients. However, this law only work
within limit concentration range, more specifically at lower concentration
range
Discussing Fig. 10.5
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