Applications of the van’t Hoff equation
• Provided the reaction enthalpy,ΔrHθ, can be assumed to be
independent of temperature, eqn. 7.23b (or 9.26b in 7th edition)
illustrates that a plot of –lnK against 1/T should yield a straight line
of slope ΔrHθ/R.
• Example: The data below show the equilibrium constant measured
at different temperatures. Calculate the standard reaction enthalpy
for the system.
T/K
350
400
450
500
K
3.94x10-4 1.41x10-2
1.86x10-1
1.48
Solution:
1/T
-lnK
2.86x10-3
7.83
2.50x10-3
4.26
2.22x10-3
1.68
2.00x10-3
-0.39
Continued
• Self-test 7.5: The equilibrium constant of the reaction
2SO2(g) + O2(g) ↔ 2SO3(g)
is 4.0x1024 at 300K, 2.5x1010 at 500K, and 2.0x104 at 700K.
Estimate the reaction enthalpy at 500K.
Solution:
discussion:
1. Do we need a balanced reaction equation here?
2. What can be learned about the reaction based on the
information provided?
3. Will the enthalpy become different at 300K or 700K?
Calculate the value of K at different
temperatures
• The equilibrium constant at temperature T2 can be obtained in terms
of the known equilibrium constant K1 at T1.
• Since the standard reaction enthalpy is also a function of
temperature, when integrating the equation 9.26b from T1 to T2, we
need to assume that ΔrHө is constant within that interval.
1
K2
K
T2
d ln( K )


1

rH

d(
R
1
1
)
T
T1
• so
ln(K2) – ln(K1) =
 rH
R

(
1
T2

1
)
(7.24)
T1
• Equation 7.24 provides a non-calorimetric method of determining
standard reaction enthalpy. (Must keep in mind that the reaction
enthalpy is actually temperature-dependent!)
Example, The Haber reaction
N2(g) + 3H2(g) ↔ 2NH3(g)
At 298 K, the equilibrium constant K = 6.1x105. The standard enthalpy
of formation for NH3 equals -46.1 kJ mol-1. What is the equilibrium constant
at 500K?
Answer: First, calculate the standard reaction enthalpy, ΔrHө,
ΔrHө = 2*ΔfHө(NH3) - 3* ΔfHө(H2) - ΔfHө(N2)
= 2*(-46.1) – 3*0 - 1*0
= - 92.2 kJ mol-1
then
1
ln(K2) – ln(6.1*105) = 8 .3145
*(-92.2*1000 J mol-1) (1/500 – 1/298)
ln(K2) = -1.71
K2 = 0.18
•
Despite the decrease in equilibrium constant as a result of temperature
increase, yet in industrial production it is still operated at an elevated
temperature (kinetics vs thermodynamics)
Practical Applications of the Knowledge of the temperature
dependence of the equilibrium constant
(i) M(s) + 1/2O2(g) → MO(s)
(ii) 1/2C(s) + 1/2O2(g) → 1/2CO2(g)
(iii) C(s) + 1/2O2(g) → CO(g)
(iv) CO(g) + 1/2O2(g) → CO2(g)
• This is carried out based on two criteria
(1) Gibbs energy is a state quantity and thus
can be added or subtracted directly.
(2) When the standard reaction Gibbs energy
is negative, the forward reaction is
favored (i.e. K > 1).
Standard reaction Gibbs energy is sometimes
referred as Free Gibbs energy.
Equilibrium Electrochemistry
Thermodynamic functions of ions in solution
(10.1 & 2 of 7th edition or 5.9 in 8th edition)
• The standard enthalpy and Gibbs energy of ions are used in the
same way as those for neutral compounds.
• Cations cannot be prepared without their accompanying anions.
Thus the individual formation reactions are not measurable.
• Defining that hydrogen ion has zero standard enthalpy and Gibbs
energy of formation at ALL temperature.
ΔfHθ(H+, aq) = 0; ΔfGθ(H+, aq) = 0
• The standard Gibbs energy and enthalpy of formation for other ions
can be calculated in relative to the value of hydrogen ion.
• Consider: ½ H2(g) + ½ Cl2(g)  H+(aq) + Cl-(aq)
ΔrGθ = -131.23kJ mol-1 (this can be obtained from K)
ΔrGθ = ΔfGθ(H+, aq) + ΔfGθ(Cl-, aq) – ½ ΔfGθ(H2, g) + ½ ΔfGθ(Cl2, g)
= 0 + ΔfGθ(Cl-, aq) – 0 – 0
= ΔfGθ(Cl-, aq)
therefore the standard Gibbs energy of formation for Cl- ion can
be obtained from the standard Gibbs energy of reaction.
Standard Gibbs energy and enthalpy of formation of other
ions can be achieved through the same approach.
• Now, consider: Ag(s) + ½ Cl2(g)  Ag+(aq) + Cl-(aq)
ΔrHθ = ΔfHθ(Ag+, aq) + ΔfHθ(Cl-, aq) - 0 - ½ *0
ΔrGθ = ΔfGθ(Ag+, aq) + ΔfGθ(Cl-, aq) - 0 - ½ *0
(Once the standard reaction Gibbs energy is calculated, the
calculation of the equilibrium constant will be the same as discussed
for neutral solutions)
Thermodynamic cycles
(chapter 3.6, 8th edition)
• The sum of the Gibbs energy for all steps around a circle is ZERO!
• The Gibbs energy of formation of an ion includes contributions from
the dissociation, ionization, and hydration.
• Gibbs energies of solvation can be estimated from Max Born
equation.
2 2
 solv G

 
zi e N A 
1 
1 

8 0 ri 
 r 
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.
• ΔsolvGθ is strongly negative for small, highly charged ions in media of
high relative permittivity.
• For water at 25oC:
2
 solv G


zi
( ri / pm )
 6 .86  10
4
kJmol
1

• Example (Self-test 10.2 7th edition): Estimate the value of
ΔsolvGө(Br-, aq) - ΔsolvGө (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:
ΔsolvGө(Br-, aq) = -103.96 kJ mol-1;
ΔsolvGө(Cl-, aq) = -131.23 kJ mol-1;
So
ΔsolvGө(Br-, aq) - ΔsolvGө (Cl-, aq) = -103.96 – ( 131.23)
= 27.27 kJ mol-1;
In order to apply the Born equation, 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 ΔsolvGө(Br-, aq) - ΔsolvGө (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).
Ion activity and mean activity coefficient
• 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 will be expressed by the following
equation:
μ =μө + RT ln (b/bө) + RTln (γ) = μideal + RTln (γ)
• Consider an electrically neutral solution of M+ X-,
G = μ+ + μ- = μ+ideal + μ-ideal + RTln(γ+) + RTln (γ-)
=Gideal + RTln (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 ions.
•
The mean activity coefficient γ is calculated as (γ+γ-)1/2
•
The chemical potential for individual ions M+ and X- then becomes:
μ+ = μideal + RTln (γ)
μ- = μideal + RTln (γ)
For a general compound of the form, MpXq, the mean activity coefficient is
expressed as:
γ = [(γ+)p(γ-)q]1/s with s = p+q
•
•
Debye-Hückel limiting law is employed to calculate the mean activity
coefficient:
log(γ) = -|z+z-| A I1/2 (5.69 in 8th edition)
A = 0.509 for aqueous solution at 25oC. I is the ionic strength,
which is calculated as the following:
I = ½ zi2(bi/bө) (5.70 in 8th edition)
where zi is the charge number and bi is the molality of the ion.
Example: Relate the ionic strength of (a) MgCl2, (b) Fe2(SO4)3
solutions to their molality, b.
Solution: To use the equation 5.70, we need to know the charge
number and the molality of each ion:
MgCl2: From molecular formula, we can get
b(Mg2+) = b(MgCl2); Z(Mg2+) = +2;
b(Cl-) = 2*b(MgCl2); Z(Cl-) = -1;
So
I = ½((2)2*b +(-1)2*(2b)) = ½(4b + 2b) = 3b;
For Fe2(SO4)3: From the molecular formula, we get
b(Fe3+) = 2*b(Fe2(SO4)3); with Z(Fe3+) = +3;
b(SO42-) = 3*b(Fe2(SO4)3); with Z(SO42-) = -2;
So
I = ½((3)2*(2b) +(-2)2*(3b)) = ½(18b + 12b) = 15b
* What is the unit of I ?
Solutions contain more than one types of
electrolytes
• The ionic strength of the solution equals the sum of the ionic
strength of each individual compound.
Example: Calculate the ionic strength of a solution that contains 0.050
mol kg-1 K3[Fe(CN)6](aq), 0.040 mol kg-1 NaCl(aq), and 0.03 mol kg-1
Ce(SO4)2 (aq).
Solution:
I (K3[Fe(CN)6]) = ½( 12*(0.05*3) + (-3)2*0.05)
= ½ (0.15 + 0.45) = 0.3;
I (NaCl) = ½(12*0.04 + (-1)2*0.04) = 0.04;
I (Ce(SO4)2) = ½(42*0.03 + (-2)2*(2*0.03)) =0.36;
So,
I = I(K3(Fe(CN)6]) + I(NaCl) + I(Ce(SO4)2)
= 0.3 + 0.04 + 0.36 = 0.7
Calculating the mean activity coefficient
Example: Calculate the ionic strength and the mean activity coefficient
of 2.0m mol kg-1 Ca(NO3)2 at 25 oC.
Solution: In order to calculate the mean activity coefficient with the eq.
5.70, one needs to know the ionic strength of the solution. Thus, the
right approach is first to get I and then plug I into the equation (5.70).
I = ½(22*0.002 + (-1)2*(2*0.002)) = 3*0.002 = 0.006;
From equation 5.70, log(γ±) = - |2*1|*A*(0.006)1/2;
= - 2*0.509*0.0775;
= -0.0789;
γ± = 0.834;
Experimental test of the Debye-Hückel
limiting law
Accuracy of the Debye-Hückel limiting law
Example: The mean activity coefficient in a 0.100 mol kg-1 MnCl2(aq)
solution is 0.47 at 25oC. What is the percentage error in the value
predicted by the Debye-Huckel limiting law?
Solution: First, calculate the ionic strength
I = ½(22*0.1 + 12*(2*0.1)) = 0.3
to calculate the mean activity coefficient.
log(γ) = -|2*1|A*(0.3)1/2;
= - 2*0.509*0.5477
= - 0.5576
so
γ = 0.277
Error = (0.47-0.277)/0.47 * 100%
= 41%
Extended Debye-Hückel law
•
log(   )  
A | z z | I
1  BI
1/2
1/2
B is an adjustable empirical
parameter.
Calculating parameter B
Example : The mean activity coefficient of NaCl in a diluted aqueous solution at
25oC is 0.907 (at 10.0mmol kg-1). Estimate the value of B in the extended
Debye-Huckel law.
Solution: First calculate the ionic strength
I = ½(12*0.01 + 12*0.01) = 0.01
From equation log(   )  
A | z z | I
1  BI
1/2
1/2
log(0.907) = - (0.509|1*1|*0.011/2)/(1+ B*0.011/2)
B = - 1.67