For an exothermic reaction, ?rH? < 0, thus , suggesting that increasing the reaction temperature will reduce the equilibrium constant.

1

• For an exothermic reaction, ?rH? lt 0, thus
  , suggesting that increasing the
  reaction temperature will reduce the equilibrium
  constant.

2
Applications of the vant Hoff equation

• Provided the reaction enthalpy,?rH?, 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 ?rH?/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 2.86x10-3
  2.50x10-3 2.22x10-3 2.00x10-3
• -lnK 7.83 4.26 1.68 -0.39

3
Continued
4

• Example 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?

5
Calculate the value of K at different temperatures

• The equilibrium constant at a temperature T2 can
  be obtained in terms of the know 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 should assume
  that ?rH? is constant within that interval.
• so ln(K2) ln(K1)
  (7.24)
• 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

6

• 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, ?rH?,
• ?rH? 2?fH?(NH3) - 3 ?fH?(H2) -
  ?fH?(N2)
• 2(-46.1) 30 - 10
• - 92.2 kJ mol-1
• then ln(K2) ln(6.1105)
  (-92.21000 J mol-1) (1/500 1/298)
• ln(K2) -1.71
• K2 0.18

7
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)

8
Equilibrium Electrochemistry
9
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.
• ?fH?(H, aq) 0 ?fG?(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.

10

• Consider Ag(s) ½ Cl2(g) ? Ag(aq)
  Cl-(aq)
• ?rH? ?fH?(Ag, aq) ?fH?(Cl-, aq)
• ?rG? ?fG?(Ag, aq) ?fG?(Cl-, aq)
• (Once the standard reaction Gibbs
  energy is calculated, the calculation of the
  equilibrium constant will be the same as
  discussed for neutral solutions)
• Consider ½ H2(g) ½ Cl2(g) ? H(aq)
  Cl-(aq)
• ?rG? -131.23kJ mol-1
• ?rG? ?fG?(H, aq) ?fG?(Cl-, aq) ½ ?fG?(H2,
  g) ½ ?fG?(Cl2, g)
• 0 ?fG?(Cl-, aq) 0 0
• ?fG?(Cl-, aq)
• therefore the standard Gibbs energy of
  formation for Cl- ion can be obtained from the
  standard reaction Gibbs energy.
• Standard Gibbs energy and enthalpy
  of formation of other ions could be achieved
  through the same approach.

11
Thermodynamic cycles(chapter 3.6, 8th edition)
12

• 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 of hydrogen.
• Gibbs energies of solvation can be estimated from
  Max Born equation.
• where zi is the charge number, e is the
  elementary charge, NA is the Avogadros constant,
  e0 is the vacuum permittivity, er is the
  relative permittivity, ri is ions radius.
• ?solvG? is strongly negative for small, highly
  charged ions in media of high relative
  permittivity.
• For water at 25oC

13

• Example Estimate the value of ?solvG?(Br-, aq)
  - ?solvG? (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
• ?solvG?(Br-, aq)
  -103.96 kJ mol-1
• ?solvG?(Cl-, aq) -131.23 kJ mol-1
• So ?solvG?(Br-, aq) -
  ?solvG? (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 ?solvG?(Br-, aq) - ?solvG? (Cl-,
  aq) - (1/196 1/181)6.86104 kJ mol-1
• 
  29.00 kJ mol-1
• (The calculated result is slightly larger
  than the experimental value).

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Ion activities

• The activity relates to the molality b via
  a ? b/b?
• where ? is called the activity
  coefficient and b? equal 1mol kg-1.
• Now the chemical potential will be expressed by
  the following equation
• µ µ? RT ln(b/b?) RTln(?)
  µideal RTln(?)
• Consider an electrically neutral real solution of
  M X-,
• G µ µ- µideal µ-ideal RTln(?)
  RTln(?-)
• Gideal RT(??-)
• Since there is no experimental way to separate
  the product (??-) into contributions from the
  cations and anions, mean activity coefficient ??
  is introduced here to assign equal
  responsibility for nonideality to both kind of
  ions.

15

• The mean activity coefficient ?? is calculated
  as (??-)1/2
• The chemical potential for individual ions M and
  X- then becomes
• µ µideal RTln(??)
• µ- µideal RTln(??)
• For a general compound of the following form
  MpXq, the mean activity coefficient is expressed
  as
• ?? (?)p(?-)q1/s
  with s pq
• Debye-Hückel limiting law is employed to
  calculate the mean activity coefficient
• log(??) -zz-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 ?zi2(bi/b?)
  10.4
• where zi is the charge number of the ion I and
  bi is the molarlity of ion I.

16

• Example 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 its molecular formula, we can get
• b(Mg2) b(solution)
• b(Cl-)
  2b(solution)
• Z(Mg2) 2
• Z(Cl-) -1
• So I ½((2)2b (-1)2(2b)) ½(4b
  2b) 3b
• Fe2(SO4)3 From the molecular formula, we can
  get
• b(Fe3) 2b(solution)
• b(SO42-) 3b(solution)
• Z(Fe3) 3
• Z(SO42-) -2
• So I ½((3)2(2b) (-2)2(3b))
  ½(18b 12b) 15b

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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
  K3Fe(CN)6(aq), 0.040 mol kg-1 NaCl(aq), and
  0.03 mol kg-1 Ce(SO4)2 (aq).
• Solution
• I(K3Fe(CN)6) ½( 12(0.053) (3)20.05
  (-1)2(0.056))
• 0.45
• I(NaCl) ½(120.04 (-1)20.04) 0.04
• I(Ce(SO4)2) ½(420.03
  (-2)2(20.03)) 0.36
• So, I I(K3(Fe(CN)6) I(NaCl)
  I(Ce(SO4)2)
• 0.45 0.04 0.36 0.85

18
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. 10.3, 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 10.3.
• I ½(220.002
  (-1)2(20.002)) 30.002 0.006
• From equation 10.3, log(?) -
  21A(0.006)1/2
• 
  - 20.5090.0775
• 
  -0.0789
• ?
  0.834

19
Experimental test of the Debye-Hückel limiting law