applications of the van’t hoff equation
DESCRIPTION
Applications of the van’t Hoff equation. Provided the reaction enthalpy, Δ r H θ , can be assumed to be independent of temperature, eqn. 7.23b ( or 9.26b in 7 th edition ) illustrates that a plot of –ln K against 1/T should yield a straight line of slope Δ r H θ /R . - PowerPoint PPT PresentationTRANSCRIPT
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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 2.86x10-3 2.50x10-3 2.22x10-3 2.00x10-3
-lnK 7.83 4.26 1.68 -0.39
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Continued
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• 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?
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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.
• 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!)
)()ln(T
dRHKd
K
K
T
T
r 12
1
2
1
1
1
)1
12
1(TTR
Hr
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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 ln(K2) – ln(6.1*105) = *(-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)
3145.81
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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.
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Equilibrium Electrochemistry
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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.
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• 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)
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Thermodynamic cycles(chapter 3.6, 8th edition)
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• 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.
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:
ri
Aisolv r
NezG
11
8 0
22
142
1086.6)/(
kJmolpmr
zGi
isolv
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• 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).
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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.
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• 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.
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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(SO4
2-) = 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 ?
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• 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
Solutions contain more than one types of electrolytes
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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;
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Experimental test of the Debye-Hückel limiting law
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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 = - 0.5576so γ = 0.277
Error = (0.47-0.277)/0.47 * 100% = 41%
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Extended Debye-Hückel law
•
B is an adjustable empirical parameter.
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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(0.907) = - (0.509|1*1|*0.011/2)/(1+ B*0.011/2)
B = - 1.67
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