how to find equilibrium constant

Equilibrium Constant

equilibrium constants defined by volume fractions), the copolymer in the system without intercomponent interactions (χab, χ=0) can be formed only for KABKBA>1 (heterodyads prevail, a certain tendency of an alternating copolymer).

From: Polymer Science: A Comprehensive Reference , 2012

Chemical Thermodynamics

Robert T. Balmer , in Modern Engineering Thermodynamics, 2011

15.13 Rules for Chemical Equilibrium Constants

Tables of equilibrium constants are usually limited in size. This limitation can often be overcome by combining reactions available in a table to produce a desired reaction. For example, many tables include equilibrium constants for the water dissociation reaction ${\text{H}}_2\text{O}\leftrightarrows {\text{H}}_2+(1/2){\text{O}}_2$ . But what do you do if you need the reverse reaction for the formation of water ${\text{H}}_2+(1/2){\text{O}}_2\leftrightarrows {\text{H}}_2\text{O}$ ? Are the equilibrium constants for these two reactions the same? The following three chemical equilibrium constant rules can be used to determine the equilibrium constant for a reaction equation that is not listed in any table but can be constructed from reactions with known equilibrium constants that are listed in a table.

Equilibrium Constant Rule 1

Let K_{e 1} be the equilibrium constant for the reaction ${\nu }_{A}A+{\nu }_{B}B\leftrightarrows {\nu }_{C}C+{\nu }_{D}D$ and let K _e2 be the equilibrium constant for the reverse reaction ${\nu }_{C}C+{\nu }_{D}D\leftrightarrows {\nu }_{A}A+{\nu }_{B}B.$ Then, these two equilibrium constants are related as follows: K_{e 2} = 1/K_{e 1}.

Equilibrium Constant Rule 2

Let K _e1 be the equilibrium constant for the reaction ${\nu }_{A}A+{\nu }_{B}B\leftrightarrows {\nu }_{C}C+{\nu }_{D}D$ and let K_{e 2} be the equilibrium constant for the reaction $\alpha ({\nu }_{A}A+{\nu }_{B}B)\leftrightarrows \alpha ({\nu }_{C}C+{\nu }_{D}D)$ , where α is any constant. Then, these two equilibrium constants are related as follows: $K_{e2}={(K_{e1})}^{\alpha }$ .

Equilibrium Constant Rule 3

Let K _e1 be the equilibrium constant for the reaction ${\nu }_{A}A+{\nu }_{B}B\leftrightarrows {\nu }_{C}C+{\nu }_{D}D$ and let K _e2 be the equilibrium constant for a second reaction: ${\nu }_{E}E+{\nu }_{F}F\leftrightarrows {\nu }_{G}G+{\nu }_{H}H$ . Then, the equilibrium constant for a third reaction, formed by multiplying the first reaction by a constant α and adding it to the second reaction multiplied by a constant β,

$\alpha ({\nu }_{A}A+{\nu }_{B}B)+\beta ({\nu }_{E}E+{\nu }_{F}F)\leftrightarrows \alpha ({\nu }_{C}C+{\nu }_{D}D)+\beta ({\nu }_{G}G+{\nu }_{H}H)$

is

$K_{e3}={(K_{e1})}^{\alpha }{(K_{e2})}^{\beta }.$

Example 15.17

Determine the equilibrium constants for the following reactions at 5000. K.

a.

${\text{H}}_2+(1/2){\text{O}}_2\leftrightarrows {\text{H}}_2\text{O}$

b.

${\text{O}}_2+{\text{N}}_2\leftrightarrows 2\text{NO}$

c.

${\text{O}}_2+3.76{\text{N}}_2\leftrightarrows 2\text{O}+7.52\text{N}$

Solution

a.

Table C.17 gives equilibrium constant values for the spontaneous dissociation of water into hydrogen and water at 5000. K ( ${\text{H}}_2\text{O}\leftrightarrows {\text{H}}_2+(1/2){\text{O}}_2$ ) as K _e1 = 10^0.450 = 2.82. The formation of water from hydrogen and oxygen gases is simply the reverse reaction, so rule 1 can be used to determine its equilibrium constant as K _e2 = 1/K _e1 = 1/2.82 = 0.355.

b.

The reaction ${\text{O}}_2+{\text{N}}_2\leftrightarrows 2\text{NO}$ is the same as the reaction $2[(1/2){\text{O}}_2+(1/2){\text{N}}_2]\leftrightarrows 2\text{NO},$ which can be obtained from the reaction $(1/2){\text{O}}_2+(1/2){\text{N}}_2\leftrightarrows \text{NO}$ that appears in Table C.17 by multiplying it by α = 2. From Table C.17 at 5000. K, we find that the equilibrium constant for the reaction $(1/2){\text{O}}_2+(1/2){\text{N}}_2\leftrightarrows \text{NO}$ is K _e1 = 10^−0.298 = 0.504. Then, using rule 2, we can calculate the equilibrium constant for the reaction ${\text{O}}_2+{\text{N}}_2\leftrightarrows 2\text{NO}$ as K_{e 2} = (K_{e 1}) ^α .

c.

In Table C.17, we find the reactions ${\text{O}}_2\leftrightarrows 2\text{O}$ and ${\text{N}}_2\leftrightarrows 2\text{N}$ . If we multiply the second reaction by 3.76 and add it to the first reaction, we get the given reaction ${\text{O}}_2\text{+}3{\text{.76N}}_2\leftrightarrows \text{2O}\text{+}7\text{.52N}$ and we can use rule 3 with α = 1 and β = 3.76.

At 5000. K, the equilibrium constant for the reaction ${\text{O}}_2\leftrightarrows \text{2O}$ is K _e1 = 10^−1.719 = 52.4 and the equilibrium constant for the reaction ${\text{N}}_2\leftrightarrows 2\text{N}$ is K_{e 2} = 10^−0.570 = 0.269, then rule 3 gives the equilibrium constant for the combined reaction as K_{e 3} = (K_{e 1})¹(K_{e 2})^3.76 = (52.4)¹(0.296)^3.76 = 0.376.

Exercises

49.

Determine the equilibrium constant for the reaction ${\text{H}}_2+(1/2){\text{O}}_2\leftrightarrows {\text{H}}_2\text{O}$ at 2000. K. Answer: K_e = 3396.

50.

Determine the equilibrium constant for the reaction at 5000. K. Answer: K_e = 1/0.2535 = 3.945.

51.

Determine the equilibrium constant for the reaction ${\text{O}}_2+{\text{N}}_2\leftrightarrows \text{NO}+\text{O}+\text{N}$ at 5000. K. Answer: K_e = 1.890.

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Fundamentals of energy systems

Ibrahim Dincer , Yusuf Bicer , in Integrated Energy Systems for Multigeneration, 2020

2.2.3.1 Thermodynamic equilibrium

There are several possibilities concerning the spontaneity and equilibrium state of a chemically reacting system, as listed in Table 2.3. A reaction may progress spontaneously in the direction of the consumption of the reactants and formation of products. A reaction may also progress toward non-spontaneous and thus does not proceed in the forward direction. The reaction can also be at equilibrium, where the backward and forward process have an identical rate and there will be no net change.

Table 2.3. Criteria for spontaneity and equilibrium of chemical reactions

Spontaneity	ΔG	ΔS tot	ΔH	ΔS	T	ΔG	Spontaneity	ΔH	ΔS	T	ΔG	Spontaneity
Spontaneous (S)	<   0	>   0	<   0	>   0	any	<   0	S	>   0	<   0	any	>   0	NS
Nonspontaneous (NS)	>   0	<   0	<   0	<   0	low	<   0	S	>   0	<   0	low	>   0	NS
Equilibrium (E)	0	0	<   0	<   0	high	>   0	NS	>   0	>   0	high	<   0	S

Source: Dincer I, Zamfirescu C. 1.5 Thermodynamic aspects of energy. In: Dincer I, editor. Compr. Energy Syst., 1st ed., Cambridge, United States: Elsevier; 2018, p. 153–211. doi: 10.1016/B978-0-12-809597-3.00105-X.

The criteria of spontaneous, nonspontaneous and equilibrium can be characterized depending on reaction enthalpy (ΔH) and entropy (ΔS). Using these parameters, the Gibbs free energy change of the reaction can be calculated as follows:

(2.25) $\mathrm{\Delta }G=\mathrm{\Delta }H-T\mathrm{\Delta }S$

Hence, the total entropy change of the reaction (system and surroundings) at constant temperature and pressure can be defined as follows:

(2.26) $\mathrm{\Delta }{\mathit{S}}_{\text{total}}=-\mathrm{\Delta }G/\mathrm{T}$

In general, both the enthalpy and entropy of reactants and products increase with the temperature. Therefore, the temperature change of reaction enthalpy and entropy is very slow. Thus, the temperature is the crucial factor affecting the reaction equilibrium for most of the reactions. If a system is at equilibrium, then the Gibbs free energy of the system is zero.

The equilibrium constant can be expressed using Gibbs free energy change of the reaction in standard conditions (Δ G ⁰) as follows:

(2.27) ${\mathit{K}}_{\text{equilibrium}}=exp\left(-\frac{\mathrm{\Delta }{\mathit{G}}^0}{\mathit{RT}}\right)$

The equilibrium constant is useful to define the mixture composition of the reaction at the equilibrium state [1]:

•

When the equilibrium constant is K _equilibrium  >   10³, then the equilibrium mixture contains mostly reaction products.

•

When the equilibrium constant is K _equilibrium  <   10^{−   3}, then the equilibrium mixture contains mostly reactants.

•

When the equilibrium constant is 10³  > K _equilibrium  >   10^{−   3}, then the equilibrium mixture contains both reactants and products.

In relation to Le Châtelier's rule, if a reaction mixture at equilibrium is positioned upon a stress, the system reacts in the direction of relieving the stress.

For a reaction that diminishes the moles of gas:

•

For an upsurge of pressure formed by reduction of volume, the direction of the reaction is toward reducing the moles of gas.

•

An expansion of volume producing a reduction of pressure will shift the reaction toward an increase in the moles of gas.

A reaction can realize full transformation once all the reactants are spent and products formed in a stoichiometric amount. Most of the time, this is not the case for actual reactions due to a number of reasons. When a reaction has an equilibrium constant of approximately 1, there will not be a full conversion of the reactants into products since both reactants and products will be present in the reaction mixture. Furthermore, two or more reactions can take place concurrently, thus forming undesired products. The theoretical yield is the quantity of product that can be produced by a reaction based on the stoichiometry. The actual yield will be less than the theoretical yield. The percent yield is expressed as the ratio of actual yield to theoretical yield [1].

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MIXED SOLVENTS

Y.Y. Fialkov , V.L. Chumak , in Handbook of Solvents (Second Edition), Volume 1, 2014

8.1.4.4.1 Selective solvation. Resolvation

When component B is added to solution E in solvent A (E is neutral molecule or ion), resolvation process takes place:

[8.1.90] $\text{E}{\text{A}}_{\text{p}}+\text{q}\text{B}=\text{E}{\text{B}}_{\text{q}}+\text{p}\text{A}$

Equilibrium constant of this process in ideal solution (in molar parts of the components) equals to:

[8.1.91] ${\text{K}}_{\text{u}\text{s}}={\text{m}}_{\text{E}{\text{B}}_{\text{q}}}{\text{m}}_{\text{A}}^{\text{p}}/{\text{m}}_{\text{E}{\text{A}}_{\text{p}}}{\text{m}}_{\text{B}}^{\text{q}}$

where:

mi	a number of moles
p, q	stoichiometric coefficients of reaction

On the other hand, resolvation constant K_us equals to the ratio of equilibrium constants of solvation processes E + pA ↔ EA_p(I); E + qB ↔ EB_q(II):

[8.1.92] ${\text{K}}_{\text{u}\text{s}}={\text{K}}_{\text{I}\text{I}}/{\text{K}}_{\text{I}}$

The method of K_us determination, based on the differences between free energy values of electrolyte transfer from some standard solvent to A and B, respectively, leads to a high error.

If the concentration ratio of different solvative forms is expressed as α, the concentrations A and B are expressed as 1 – x_B and x_B, respectively, (x_B – molar part B), equation [8.1.91] may be presented in the form:

[8.1.91a] ${\text{K}}_{\text{u}\text{s}}=\alpha {\text{x}}_{\text{B}}^{\text{p}}/{\left(1-{\text{x}}_{\text{B}}\right)}^{\text{q}}$

or

[8.1.93] $\text{ln}{\text{K}}_{\text{u}\text{s}}=\text{ln}\text{a}+\text{p}{\text{lnx}}_{\text{B}}-\text{q}\text{ln}\left(\text{a}-{\text{x}}_{\text{B}}\right)$

The last equation permits to calculate value for the solvent of fixed composition and x_B determination at certain composition of the solvate complexes p and q and resolvation constant K_us.

It follows from the equations [8.1.91] and [8.1.91a]

[8.1.91b] ${\text{K}}_{\text{u}\text{s}}=\left[{\text{m}}_{\text{E}}^{\left(\text{q}\right)}\text{q}/\left({\text{m}}_{\text{E}}^0-{\text{m}}_{\text{E}}^{\text{q}}\right)\right]\left[{\text{x}}_0^{\left(\text{q}\right)}{\left(1-{\text{x}}_{\text{B}}^{\left(\text{q}\right)}\right)}^{\text{q}}\right]$

where:

m0 E	a number of moles E in solution
m(q) E	a number of moles E solvated by the solvent A

Hence

[8.1.91c] ${\text{K}}_{\text{u}\text{s}}={{\text{x}}^{\prime }}_{\text{B}}{\text{x}}_{\text{B}}^{\text{p}}/\left[\left(1-{{\text{x}}^{\prime }}_{\text{B}}\right){\left(1-{\text{x}}_{\text{B}}\right)}^{\text{q}}\right]$

where:

x'B	molar fraction of B in solvate shell

Equation [8.1.91c] developed for the ideal solution E–A–B permits us to establish the relationship between the composition of mixed solvent x_B and the solvate shell composition x'_B. For the special case of equimolar solvates, the expression of resolvation constant is written in the form:

[8.1.91d] ${\text{K}}_{\text{u}\text{s}}=\alpha {\text{x}}_{\text{B}}/\left(1-{\text{x}}_{\text{B}}\right)$

It follows that even in ideal solution of the simplest stoichiometry, α is not a linear function of the mixed solvent composition x_B. Dependence of isotherm a on the solvent composition at K_us = 1 is presented in Figure 8.1.9a.

Figure 8.1.9. Characteristics of solvate shells in the mixed solvent A-B: a – dependence a = EA/EB on composition of the mixed solvent; b – dependence of the solvate shell composition, x'_B, on composition x_B of the mixed solvent E–A–B.

Analytical correlation between the composition of solvate shell x_B and the mixed solvent composition can be developed:

[8.1.94] ${\text{K}}_{\text{u}\text{s}}=\left[{{\text{x}}^{\prime }}_{\text{B}}/\left(1-{{\text{x}}^{\prime }}_{\text{B}}\right)\right]\left[{\text{x}}_{\text{B}}^{\text{p}}/{\left(1-{\text{x}}_{\text{B}}\right)}^{\text{q}}\right]$

The dependence for K_us = 1 is given in Figure 8.1.9b. As it follows from the analysis of equation [8.1.94] and Figure 8.1.9, the compositions of solvate shell and mixed solvent are different.

A similar approach has been developed for calculation of solvate shell composition Na⁺ and I⁻ in the mixed solvent formed by the components with close values of permittivity. Such solvent selection permits us to eliminate the permittivity effect on solvation equilibrium. Resolvation constants have been determined from the calorimetric study. The composition of anions solvate complex has been determined from experimental data of electrolyte Bu₄NI assuming lack of the cation specific solvation. Experimental data are presented in Figure 8.1.10.

Figure 8.1.10. Selective solvation of NaI in mixed solvent cyanomethane–methanol (a) and DMFA–methanol (b, curve 1), DMFA – cyanomethane (b, curve 2): x_b – composition of mixed solvent; x'_b – composition of solvate shell in molar parts of the second component.

Padova ³⁶ has developed this approach to non-ideal solutions. He has proposed an equation based on electrostatic interaction which relates molar fractions of the components (x_B – in the mixed solvent and x'_B – in the solvate shell) to the activity coefficient of components of the binary solvent:

[8.1.95] $\alpha =\text{ln}\left[\left(1-{{\text{x}}^{\prime }}_{\text{B}}\right)/\left(1-{\text{x}}_{\text{B}}\right)\right]=\text{ln}{\gamma }_{\text{B}}^2$

Strengthening or weakening interaction (ion-dipole interaction or dipole-dipole interaction) of universal solvation leads to re-distribution of molecules in the mixed solvate and to the change of the composition of solvate shell in contrast to the composition of mixed solvent.

The method for determination of average filling of molecules' coordination sphere of dissolved substance by molecules of the mixed solvent (with one solvate-inert component) has been proposed. ³⁷ The local permittivity is related to average filling of molecules' coordination sphere expressed by the equation:

[8.1.96] ${\overline{\epsilon }}_{\text{p}}={\epsilon }_{\text{A}}{{\text{x}}^{\prime }}_{\text{A}}+{\epsilon }_{\text{B}}{{\text{x}}^{\prime }}_{\text{B}}$

where:

[8.1.97] ${{\text{x}}^{\prime }}_{\text{A}}={\text{z}}_{\text{A}}/\left({\text{z}}_{\text{A}}+{\text{z}}_{\text{B}}\right);{{\text{x}}^{\prime }}_{\text{B}}={\text{z}}_{\text{B}}/\left({\text{z}}_{\text{A}}+{\text{z}}_{\text{B}}\right)$

where:

zA, zB

average numbers of A and B molecules in the first solvate shell.

The last equations can be used for the development of the next expression permitting to calculate the relative content of B molecules in the solvate shell

[8.1.98] ${{\text{x}}^{\prime }}_{\text{B}}=\left({\epsilon }_{\text{p}}-{\epsilon }_{\text{A}}\right)/\left({\epsilon }_{\text{B}}-{\epsilon }_{\text{A}}\right)$

where:

ɛp	permittivity of binary solvent.
ɛA, ɛB	permittivities of components

Value x'_B can be found from the equation linking the location of maximum of absorption band of IR spectrum with refraction index and ɛ of the solution.

The data on selective solvation of 3-aminophthalimide by butanol from butanol–hexane mixture are presented in Figure 8.1.11. The data have been calculated from the equations presented above. Alcohol content in solvate shell has higher concentration than in solvent composition even at low concentration of alcohol in the solvent. For example, when molar fraction of n-butanol in the mixture was 7%, the relative molar fraction of n-butanol in the solvate shell of aminophthalimide was 30%. Resolvation process completes at n-butanol concentration in solution ≈ 90 %.

Figure 8.1.11. Selective solvation of 3-aminophthalimid (A) by n-butanol (B) from the mixed solvent hexane-n-butanol.

Mishustin ³⁷ proposed a strict and accurate method for selective solvation study. The method is based on data of free energy transfer of electrolyte from individual solvent A to mixed solvent A–B. The method takes into account non-ideality of the system, and allows calculation of the concentration of different solvate forms and their dependence on the mixed solvent composition.

An example of application of this method is in the work. ³⁹ Authors have calculated relative concentration of different solvate forms of Li⁺ in the mixed solvent ethylenediamine–DMSO and ethylenediamine–DMFA (Figure 8.1.12). Free energy of lithium transfer from DMSO (DMFA) in the mixed solvent has been calculated from the time of spin-lattice relaxation of kernel ⁷Li. The curves presented in Figure 8.1.12 depict quantitatively the selectivity of Li⁺ relative to ethylenediamine, which is more basic component in contrast to the second components of the mixed solvent, namely DMSO and DMFA.

Figure 8.1.12. Dependence of concentration (in molar parts) of solvate forms Li(EDA)_mS_n on molar fraction, x_B, of the second component in the binary solvents ethylenediamine-DMSO (a) and ethylenediamine-DMFA (b): 1 – m = 4, n = 0; 2 – m = n = 2; 4 – m = 0, n = 4.

The following systems can serve as examples of the effect of composition of the mixed solvent on the solvate shell composition:

[Cr(NH)_S(H₂O)_m(DMSO)_n]³⁺–H₂O–DMSO ⁴⁰

[Be(H₂O)_m(EG)_n]SO₄–H₂O–ethylene glycol ⁴¹

[Be(H₂O)_m(HMPTA)_n]SO₄–H₂O–HMPTA ⁴²

[Ni(H₂O)_mS_n]C10₄–H₂O–S (where S is methanol, ethanol, propanol, DMSO) ⁴³

Data presented in Figure 8.1.13 contain information on the composition of solvate shell as a function of molar fraction of water in the mixed solvent H₂O–other solvents. ⁴³ Monograph ⁴⁴ contains collection of data on resolvation constants of the ions in the mixed solvents.

Figure 8.1.13. The composition of solvate shell Ni²⁺ (in molar parts of water) in the mixtures of solvents formed from water and propanol (1), ethanol (2), methanol (3) and DMSO (4). Data from Ref. 43.

The above presented dependencies of the composition of solvate shell on the mixed solvent composition as well as resolvation constants permit calculation of the solvate composition by varying solvent composition. The dependence of resolvation constants on the permittivity of the solvent is discussed in the example of the proton resolvation process.

In the mixed solvents water–non-aqueous solvent, in spite of its donor and polar properties, water is a preferred solvating agent. This generalization has some exceptions (solvation in systems Ag⁺–H₂O–acetonitrile, Cr³⁺–H₂O–DMSO, F–H₂O–ethylene glycol). ⁴⁵

Solvation energy of proton by donor solvents is very high. The regularities of the proton selective solvation and resolvation processes were studied in more detail in comparison with other ions.

Let us consider the changes in the system, when donor component is added to protic acid HA in solvent A. Anion solvation can be neglected, if both solvents have donor character. The interaction influences the proton resolvation process.

[8.1.99] $\text{H}{\text{A}}_{\text{p}}^{+}+\text{B}\leftrightarrow {\left[\text{H}{\text{A}}_{\text{p}-1}\text{B}\right]}^{+}\stackrel{+\text{B}}{\leftrightarrow }\cdots \leftrightarrow \text{H}{\text{B}}_{\text{p}}^{+}+\text{A}$

or

[8.1.99a] $\text{H}{\text{A}}_{\text{p}}^{+}+\text{q}\text{B}\leftrightarrow \cdots \leftrightarrow \text{H}{\text{B}}_{\text{q}}^{+}+\text{p}\text{A}$

The model related to eq. [8.1.99a] was evaluated, ⁴⁶ resulting in the supposition that the equilibrium of two forms of solvated proton ( $\text{H}{\text{A}}_{\text{p}}^{+}$ and $\text{H}{\text{B}}_{\text{q}}^{+}$ ) is important. Solvation stoichiometry was not considered. Both proton solvated forms are denoted as HA⁺ and HB⁺.

If B is the better donor component (it is a necessary requirement for equilibrium [8.1.99] shift to the right hand side of equation), the equation from work ⁴⁶ can be simplified to the form: ⁴⁷

[8.1.100] ${\text{K}}_{\text{a}}={\text{K}}_{\text{a}}^{\text{A}}+\left({\text{K}}_{\text{a}}^{\text{B}}-{\text{K}}_{\text{a}}^{\text{A}}\right)\left\{\left[{\text{K}}_{\text{u}\text{s}}{\text{x}}_{\text{B}}/\left(1-{\text{x}}_{\text{B}}\right)\right]/\left[1+{\text{K}}_{\text{u}\text{s}}{\text{x}}_{\text{B}}/\left(1-{\text{x}}_{\text{B}}\right)\right]\right\}$

where:

Ka	ionic association constant.
Ka, Ka	the ionic association constants of acid in individual solvents A and
Kus	a constant of resolvation process.
xb	molar fraction of B

For the calculation of resolvation constant, one must determine the experimental constant of HA association in the mixed solvent and determine independently K_a ^A and K.

When the resolvation process is completed at low concentration of the second component, the change of permittivity of mixed solvent A–B may be ignored. Thus, one may assume that ${\text{K}}_{\text{a}}^{\text{A}}$ and ${\text{K}}_{\text{a}}^{\text{B}}$ are constant and calculate K_us from the equation [8.1.100] in the form:

[8.1.101] $1/\left({\text{K}}_{\text{a}}-{\text{K}}_{\text{a}}^{\text{A}}\right)=1/\left({\text{K}}_{\text{a}}^{\text{B}}-{\text{K}}_{\text{a}}^{\text{A}}\right)+\left\{\left[\left(1-{\text{x}}_{\text{B}}\right)/{\text{x}}_{\text{B}}\right]\left[1/{\text{K}}_{\text{u}\text{s}}\left({\text{K}}_{\text{a}}^{\text{B}}-{\text{K}}_{\text{a}}^{\text{A}}\right)\right]\right\}$

Thus K_us is calculated as a slope ratio of coordinates: $1/\left({\text{K}}_{\text{a}}-{\text{K}}_{\text{a}}^{\text{A}}\right)=\left(1-{\text{x}}_{\text{B}}\right)/{\text{x}}_{\text{B}}$ and $\left({\text{K}}_{\text{a}}-{\text{K}}_{\text{a}}^{\text{A}}\right)$ remainder is obtained as Y-intercept. Dependence of eq. [8.1.101] is presented in Figure 8.1.14.

Figure 8.1.14. Dependence of 1/(K_a – K_a ^A) for CF3COOH on the mixed solvent composition DMSO (A)-dimethylalanine at different temperatures.

K_us for proton was calculated in a series of mixed solvents. It was shown ⁴⁷ that, when pyridine, dimethylalanine or diphenylamine (resolvating agents with decreasing donor numbers) are added to the solution of trifluoroacetic acid in DMSO, proton decreases consecutively and its values are equal to 2.7×10⁴; 4.2×10³ and 35.4, respectively.

Consideration of permittivity of the mixed solvent has allowed calculation of proton in whole concentration range of the mixed solvent DMSO–diphenylamine. The data have been approximated using equation [8.1.55].

$\text{ln}{\text{K}}_{\text{u}\text{s}}=-12.2+3400/\text{T}+453.7/\epsilon -75043/\epsilon \text{T}$

Equilibrium constants for exchange process of alcohol shell of solvates to water shell were calculated: ⁴⁸

[8.1.102] $\text{R}\text{O}{\text{H}}_2^{+}+{\text{H}}_2\text{O}\leftrightarrow {\text{H}}_3{\text{O}}^{+}+\text{R}\text{O}\text{H}$

The solutions of HCl and HOSO₃CH₃ in aliphatic alcohol (i.e., C_nH_2n+1OH) – normal alcohol C₁-C₅ and isomeric alcohol C₃-C₅ have been studied. If the components taking part in resolvation process are capable of H-bonding, the anion solvation by these components cannot be neglected. The differences in K_us values for both acids in different solvents may be explained as follows.

The dependence of K_us on permittivity and temperature is described by the equation:

[8.1.103] $\text{ln}\text{K}={\text{a}}_{00}+{\text{a}}_{01}/\text{T}+{\text{a}}_{02}/{\text{T}}^2+\left({\text{a}}_{10}+{\text{a}}_{11}/\text{T}\right)/\epsilon$

The coefficients of equation [8.1.103] are presented in Table 8.1.5.

Table 8.1.5. Coefficients of the equation [8.1.103] for constants of resolvation process

System	a00	a01×10−4	a02×10−4	a10	a11×10−3	±δ
HCl–n-alcohol	12.4	46.0	56.3	186.8	71.1	0.3
HCl–isomeric alcohol	13.1	5.9	−36.9	318.4	94.7	0.2
HOSO3CH3–n-alcohol	6.3	4.4	−5.4	177.7	60.9	0.3

The solvent effect in the following resolvation process was studied: ⁴⁹

[8.1.104] $\text{R}\text{O}{\text{H}}_2^{+}+\text{P}\text{y}\leftrightarrow \text{H}\text{P}{\text{y}}^{+}+\text{R}\text{O}\text{H}$

The solutions of trifluoroacetic acid in ethanol and methanol in the temperature range 273.15-323.15K were investigated. The dependence of K_us on permittivity and temperature is described by the equation:

$\text{ln}{\text{K}}_{\text{u}\text{s}}=9.44+1768/\text{T}+77.8/\epsilon -11076/\epsilon \text{T}$

Unlike the processes considered above, in the case of process [8.1.104] permittivity decrease leads to decreasing K_us. The explanation of the results based on covalent and electrostatic components of resolvation process enthalpy is given. ⁵⁰

The reaction [8.1.104] also has been studied for isodielectric mixtures of alcohol–chlorobenzene with ɛ = 20.2 (permittivity of pure n-propanol) and ɛ =17.1 (permittivity of pure n-butanol) to investigate the relative effect of universal and specific solvation on the resolvation process. The mixtures were prepared by adding chlorobenzene to methanol, ethanol, and C₁-C₃ alcohol. Alcohol is a solvate-active component in these isodielectric solvents. K_us. data are given in Table 8.1.6.

Table 8.1.6. Equilibrium constants of the process [8.1.104] in isodielectric solvents

Solvents with ɛ = 20.2	Kus×10−5	Solvents with ɛ = 17.1	Kus×10−5
Methanol + 23.5% chlorobenzene	3.9	Methanol + 32.8% chlorobenzene	1.4
Ethanol + 13% chlorobenzene	4.9	Ethanol + 24% chlorobenzene	2.0
n-Propanol	7.3	n-Butanol	4.9

Insignificant increase of K_us in n-butanol (or n-propanol) solution in comparison to methanol is due to relaxation of the proton–alcohol bond, when the distance of ion–dipole interaction increases.

The change of donor property of the solvate-active component is not significant. The equations relating K_us to ɛ permit to divide free energy of resolvation process into the components. Corresponding data are presented in Table 8.1.7.

Table 8.1.7. The components of free energy (kJ mol⁻¹) of proton resolvation process at 298.15K

Solvent	ɛ	σΔGel in process
		[8.1.102]		[8.1.104]
		HCl	HOSO3CH3	CF3COOH
Methanol	32.6	3.9	2.0	−3.1
Ethanol	24.3	5.3	2.7	−4.1
n-Propanol	20.1	6.4	3.3	−5.0
n-Butanol	17.1	7.5	3.9	−5.9
n-Pentanol	14.4	8.9	4.6	–
-δΔGcov		8.2	17.7	−38.1
$-\delta \Delta {\text{G}}_{\text{E}=1}^{\text{el}}$		128	66.0	−100.6

In contrast to the processes considered earlier, the vacuum electrostatic component of resolvation process has high value whereas –δΔG^el values are comparable with the covalent component, –δΔG^cov.

δΔG values according to [8.1.51] are equal to:

$\delta \Delta \text{G}={\left(\Delta {\text{G}}_{\text{H}{\text{B}}^{+}}+\Delta {\text{G}}_{\text{A}}-\Delta {\text{G}}_{\text{H}{\text{A}}^{+}}+\Delta {\text{G}}_{\text{B}}\right)}_{\text{s}\text{o}\text{l}}$

For small additions of B (to component A), (ΔG_A)_sol = 0 and (ΔG_HA)_sol = 0, then

$\delta \Delta \text{G}={\left(\Delta {\text{G}}_{\text{H}{\text{B}}^{+}}+\Delta {\text{G}}_{\text{B}}\right)}_{\text{s}\text{o}\text{l}}$

Solvation energy of complex HB+ by solvent A is small because coordination vacancies of the proton are saturated to a considerable extent. Therefore the interaction energy between A and B influences significantly the value of σΔG. That is why, the mixed solvents (alcohol–water and alcohol–pyridine, for instance) are different because of the proton resolvation process. This can be explained in terms of higher energy of heteromolecular association for the alcohol–water in comparison with alcohol–pyridine.

The concept of solvent effect on the proton resolvation process was confirmed by quantum chemical calculations. ⁵¹ Above phenomena determine the dependence of resolvation constant on physical and chemical properties.

Let the resolvation process proceeds at substantial abundance of the component A in mixed solvent and initial concentrations HA (HA⁺) and B to be equal. The output of the process can be calculated from the equation similar to equation [8.1.66]. The large value of K_us in all considered processes of proton resolvation indicates the effect of permittivity change on the yield of complex HB+ formation. The output of resolvated proton in process [8.1.104] proceeding in methanol equals 100%, whereas in the same process in low polarity solvent (e.g., methanol–hexane), with abundance of the second component, the equilibrium is shifted to the left, resulting in solvate output of less than 0.1%. K_us values in single alcohol solvents are large, thus the output of reaction does not depend on solvent exchange.

The process H · DMSO⁺ + B ↔ HB⁺ + DMSO may be considered as an example of the effect of chemical properties of B on the output of the reaction [8.1.99]. The output of HB⁺ equals to 98%, if pyridine is included in the process at initial concentration of 0.1M. Use of diphenylamine, having lower donor properties, decreases the output to 60%. The output differs even more at smaller concentrations of component, such as 10⁻³ M, which gives yields of 83 and 33%, respectively.

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Gasification Theory and Modeling of Gasifiers

Prabir Basu , in Biomass Gasification and Pyrolysis, 2010

Gibbs Free Energy

Gibbs free energy, G, is an important thermodynamic function. Its change in terms of a change in entropy, ΔS, and enthalpy, ΔH, is written as

(5.34) $\Delta G=\Delta H-T\Delta S$

The change in enthalpy or entropy for a reaction system is computed by finding the enthalpy or entropy changes of individual gases in the system. It is explained in Example 5.2. An alternative approach uses the empirical equations given by Probstein and Hicks (2006). It expresses the Gibbs function (Eq. 5.35) and the enthalpy of formation (Eq. 5.36) in terms of temperature, T, the heat of formation at the reference state at 1 atmosphere and 298 K, and a number of empirical coefficients, a′, b′, and so forth.

(5.35) $\begin{array}{ll}\Delta G_{\prime f,T}^0=\hfill & \Delta h_{298}^0-a^{\prime }T\ln \left(T\right)-b^{\prime }{T}^2-\left(\frac{c^{\prime }}{2}\right)T^3-\left(\frac{d^{\prime }}{3}\right)T^4\hfill \\ \hfill & +\left(\frac{e^{\prime }}{2T}\right)+f^{\prime }+g^{\prime }T\text{kJ}/\text{mol}\hfill \end{array}$

(5.36) $\Delta H_{\prime f,T}^0=\Delta h_{298}^0+a^{\prime }T+b^{\prime }{T}^2+c^{\prime }{T}^3+d^{\prime }{T}^4+\left(\frac{e^{\prime }}{T}\right)+f^{\prime }\text{kJ}/\text{mol}$

The values of the empirical coefficients for some common gases are given in Table 5.5.

TABLE 5.5. Heat of Combustion, Gibbs Free Energy, and Heat of Formation at 298 K, 1 Atm, and Empirical Coefficients from Eqs. 5.35 and 5.36

Product	HHV(kJ/mol)	Δ fG 298(kJ/mol)	Δ fH 298(kJ/mol)	Empirical Coefficients
				a′	b′	c′	d′	e′	f′	g′
C	393.5	0	0
CO	283	−137.3	−110.5	5.619 × 10−3	−1.19 × 10−5	6.383 × 10−9	−1.846 × 10−12	−4.891 × 102	0.868	−6.131 × 10−2
CO2	0	−394.4	−393.5	−1.949 × 10−2	3.122 × 10−5	−2.448 × 10−8	6.946 × 10−12	−4.891 × 102	5.27	−0.1207
CH4	890.3	−50.8	−74.8	−4.62 × 10−2	1.13 × 10−5	1.319 × 10−8	−6.647 × 10−12	−4.891 × 102	14.11	0.2234
C2H4	1411	68.1	52.3	−7.281 × 10−2	5.802 × 10−5	−1.861 × 10−8	5.648 × 10−13	−9.782 × 102	20.32	−0.4076
CH3OH	763.9	−161.6	−201.2	−5.834 × 10−2	2.07 × 10−5	1.491 × 10−8	−9.614 × 10 −12	−4.891 × 10 2	16.88	−0.2467
H2O (steam)	0	−228.6	−241.8	−8.95 × 10−3	−3.672 × 10−6	5.209 × 10−9	−1.478 × 10−12	0	2.868	−0.0172
H2O (water)	0	−237.2	−285.8
O2	0	0	0
H2	285.8	0	0

Source: Adapted from Probstein and Hicks, 2006, pp. 55, 61.

The equilibrium constant of a reaction occurring at a temperature T may be known using the value of Gibbs free energy.

(5.37) $K_e=\exp \left(-\frac{\Delta G}{RT}\right)$

Here, ΔG is the standard Gibbs function of reaction or free energy change for the reaction, R is the universal gas constant, and T is the gas temperature.

Example 5.2

Find the equilibrium constant at 2000 K for the reaction

${\text{CO}}_2\to \text{CO}+½{\mathrm{O}}_2$

Solution

Enthalpy change is written by taking the values for it from the NIST-JANAF thermochemical tables (Chase, 1998) for 2000 K:

$\begin{array}{lll}\Delta H\hfill & =\hfill & {\left(h_f^0+\Delta h\right)}_{\text{CO}}+{\left(h_f^0+\Delta h\right)}_{{\mathrm{O}}_2}-{\left(h_f^0+\Delta h\right)}_{{\text{CO}}_2}\hfill \\ \hfill & =\hfill & 1\text{mol}\left(-110,527+56,744\right)\mathrm{J}/\text{mol}+1/2\text{mol}\left(0+59,175\right)\mathrm{J}/\text{mol}\hfill \\ \hfill & \hfill & -1\text{mol}\left(-393,522+91,439\right)\mathrm{J}/\text{mol}=277,887\mathrm{J}\hfill \end{array}$

The change in entropy, ΔS, is written in the same way as for taking the values of entropy change from the NIST-JANAF tables (see list that follows on page 140).

$\begin{array}{c}\Delta S=1\times S_{\text{CO}}+½\times S_{{\mathrm{O}}_2}-1\times S_{{\text{CO}}_2}\\ =\left(1\text{mol}\times 258.71\mathrm{J}/\text{mol K}\right)+\left(1/2\text{mol}\times 268.74\mathrm{J}/\text{mol K}\right)\\ -\left(1\text{mol}\times 309.29\mathrm{J}/\text{mol K}\right)\\ =83.79\mathrm{J}/\mathrm{K}\end{array}$

From Eq. (5.34), the change in the Gibbs free energy can be written as

$\begin{array}{c}\Delta G=\Delta H-T\Delta S\\ =277.887\text{kJ}-\left(2,000\mathrm{K}\times 83.79\mathrm{J}/\mathrm{K}\right)=110.307\text{kJ}\end{array}$

The equilibrium constant is calculated using Eq. (5.37):

(5.38) $K_{2000K}=e^{-\frac{\Delta G}{RT}}=e^{-\left(\frac{110.307}{0.008314*2000}\right)}=0.001315$

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Syngas

Mariano Martín Martín , in Industrial Chemical Process Analysis and Design, 2016

5.4.1.2.1 Thermodynamics of the reaction

Ammonia production is an exothermic reaction and thus low temperatures favor conversion. However, the kinetics sets the lower bound of the operating temperature. On the other hand, there is a decrease in the number of moles as the reaction progresses. Therefore, the higher the pressure, the larger the conversion. The ideal operation would be isothermal, since high conversions could be achieved. However, removing the energy generated as the reaction advances is a technical challenge and the design of the reactors look for an optimal operating line. The typical pressures and temperatures are 100–1000   atm and 400–600°C (Vancini, 1961). In Table 5.4 we see some values of the equilibrium constant as a function of both pressure and temperature. Alternatively, in the literature we can find correlations ( Eqs. 5.12 and 5.13) to compute the equilibrium constant either for atmospheric pressure, or as a function of it; the parameters b and u can be seen in Table 5.5.

Table 5.4. Values for the Equilibrium Constant

	Temperature (°C)
P (atm)	325	350	400	450	500
1.0	0.0401	0.0266	0.0129	0.0060	0.0038
100			0.0137	0.0072	0.0040
300				0.0088	0.0049
600				0.0130	0.0065
1000				0.0233	0.0099

Table 5.5. Values for Parameters b and u

P (atm)	b	u
300	1.256×10−4	2.206
600	1.0856×10−4	3.059
1000	2.6833	4.473

(5.12) ${\log }_{10}(K_{\mathrm{p}})=\frac{2250.322}{T}-0.85430-1.51049{\log }_{10}T-2.58987\times {10}^{-4}T+1.48961\times {10}^{-7}{T}^2$

(5.13) ${\log }_{10}(K_{\mathrm{p}})=\frac{2074.8}{T}-2.4943{\log }_{10}T-b·T+1.8564\times {10}^{-7}{T}^2+uT[=]\mathrm{K}$

The equilibrium constant is a pseudoconstant since it is computed with partial pressures instead of fugacities. Thus, for the stoichiometry presented, the equilibrium is as follows Reklaitis (1983):

(5.14) $K_{\mathrm{p}}=\frac{P_{{\mathrm{NH}}_3}}{P_{{\mathrm{N}}_2}^{0.5}{P}_{{\mathrm{H}}_2}^{1.5}}{K}_{\mathrm{f}}=\frac{f_{{\mathrm{NH}}_3}}{f_{{\mathrm{N}}_2}^{0.5}{f}_{{\mathrm{H}}_2}^{1.5}}$

(5.15) $f_{{\mathrm{NH}}_3}=u_{{\mathrm{NH}}_3}{P}_{{\mathrm{NH}}_3}=u_{{\mathrm{NH}}_3}P{x}_{{\mathrm{NH}}_3}$

(5.16) $K_{\mathrm{f}}=\frac{f_{{\mathrm{NH}}_3}}{f_{{\mathrm{N}}_2}^{0.5}{f}_{{\mathrm{H}}_2}^{1.5}}=\frac{{\nu }_{{\mathrm{NH}}_3}}{{\nu }_{{\mathrm{N}}_2}^{0.5}{\nu }_{{\mathrm{H}}_2}^{1.5}}\frac{x_{{\mathrm{NH}}_3}}{x_{{\mathrm{N}}_2}^{0.5}{x}_{{\mathrm{H}}_2}^{1.5}}=K_{\mathrm{u}}{K}_{\mathrm{p}}\Rightarrow K_{\mathrm{p}}=\frac{K_{\mathrm{f}}}{K_{\mathrm{u}}}$

Example 5.5

Compute the ammonia fraction at equilibrium for an operating pressure equal to 1   atm. The reaction for the production of ammonia is given as follows:

$0.5{\mathrm{N}}_2+1{\mathrm{.5H}}_2\leftarrow \to {\mathrm{NH}}_3$

The feed enters the reaction in stoichiometric proportions. Assume that no inerts accompany the feed and that the correlation for the equilibrium constant is given by the following equation:

$\log (K_{\mathrm{p}})=\frac{2250.322}{T}-0.85340-1.51049\log (T)-2.58987\times {10}^{-4}T+1.48961\times {10}^{-7}{T}^2$

Solution

For this example, we use the following nomenclature: T=temperature, P _T=P, and the molar fraction for the species involved is detoned as x, y and z for NH₃, N₂ and H₂ respectively. Therefore:

$\begin{array}{l}{P}_{{\mathrm{NH}}_3}=x{P}_{\mathrm{T}};P_{{\mathrm{N}}_2}=y{P}_{\mathrm{T}};P_{{\mathrm{H}}_2}=z{P}_{\mathrm{T}};\hfill \\ K_{\mathrm{p}}=\frac{P_{{\mathrm{NH}}_3}}{P_{{\mathrm{N}}_2}^{0.5}·P_{{\mathrm{H}}_2}^{1.5}}=\frac{x·P_{\mathrm{T}}}{{(y·P_{\mathrm{T}})}^{0.5}{(z·P_{\mathrm{T}})}^{1.5}}=\frac{1}{P_{\mathrm{T}}}\frac{x}{y^{0.5}·z^{1.5}}\hfill \\ K_{\mathrm{p}}·P_{\mathrm{total}}=\frac{x}{y^{0.5}{z}^{1.5}}\hfill \end{array}$

$\begin{array}{l}z=3·y\hfill \\ x+y+z=1\hfill \\ x+y+3·y=1\hfill \\ x+4·y=1\Rightarrow y=\frac{1-x}{4}\hfill \\ z=3·y=\frac{3}{4}(1-x)\hfill \end{array}$

Thus:

$K_{\mathrm{p}}=\frac{x}{P_{\mathrm{T}}{\left(\frac{1-x}{4}\right)}^{0.5}{\left(\frac{3}{4}(1-x)\right)}^{1.5}}=\frac{4^2}{3^{1.5}}\frac{x}{P_{\mathrm{T}}{(1-x)}^2}=3.0792\frac{x}{P_{\mathrm{T}}{(1-x)}^2}$

We can solve the equation for x as follows:

$x=\frac{\left(2+\frac{3.0792}{K_{\mathrm{p}}·P_{\mathrm{T}}}\right)-\sqrt{\left({\left(2+\frac{3.0792}{K_{\mathrm{p}}·P_{\mathrm{T}}}\right)}^2-4\right)}}{2}$

We compute, for different temperatures, the fraction of ammonia given by the variable x (Table 5E5.1).

Table 5E5.1. Equilibrium Constant Values

P (atm)	T (K)	K p	x
1	473	0.59515055	0.14221511
1	523	0.17715127	0.051733
1	573	0.06429944	0.02005278
1	623	0.02717032	0.00867146
1	673	0.01293599	0.00416616
1	723	0.00677637	0.00219106

Assuming that the correlation is valid for different pressures, Fig. 5E5.1 presents the ammonia fraction of the product gas as a function of the pressure and temperature. As predicted by Le Chatelier's principle, the conversion increases with the pressure and for lower temperatures. As described above, the optimal operation in the reaction would be isothermal, since high conversion could be obtained in one stage. Adiabatic operation leads us to find the limit given by the equilibrium line. The reader can also use the values for K _p given in Table 5.4 to compute the ammonia fraction obtained.

Figure 5E5.1. Effect of pressure and temperature on the fraction of ammonia produced.

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Energy Fundamentals

Ibrahim Dincer , Calin Zamfirescu , in Comprehensive Energy Systems, 2018

1.5.5 Thermodynamic Equilibrium

The concept of equilibrium has been previously described in Sections 2 and 3. Here we will invoke the first and second laws of thermodynamics simultaneously to analyze the conditions required for a closed system to reach equilibrium. Equivalently, the results of this analysis apply for a closed system that departs from equilibrium. There are in total five distinct cases to be considered, depending on the process constraints imposed upon the closed system.

The first situation is when the closed system is constrained to fixed volume and fixed entropy together. The system is free to displace from equilibrium by changing other parameters than entropy and volume, for example, species concentration. In this situation the principle of minimum internal energy applies, which states that if a closed system evolves at constant entropy and volume then the internal energy of the system is maximum when the system reaches thermodynamic equilibrium. Mathematically, the following relationships are valid:

(109) ${\left(\frac{\partial U}{\partial X_i}\right)}_{S,V}=0\mathrm{and}{\left(\frac{{\partial }^{2}U}{\partial X_i^2}\right)}_{S,V}>0,i=1\dots n$

where U(S, V, X _i), i=1…n is the internal energy of a homogeneous closed thermodynamic system with n+2 degrees of freedom and X _i are mass fractions of each system component.

The second case reflects the minimum enthalpy principle, which states that a closed thermodynamic system that is restricted to evolve at constant entropy and pressure has a minimum enthalpy at thermodynamic equilibrium. Mathematically, the principle is expressed as follows:

(110) ${\left(\frac{\partial H}{\partial X_i}\right)}_{S,P}=0\mathrm{and}{\left(\frac{{\partial }^{2}U}{\partial X_i^2}\right)}_{S,P}>0,i=1\dots n$

The third case refers to a closed system restricted to evolving at constant temperature and volume when the principle of minimum Helmholtz energy $\mathcal{A}$ applies. The Helmholtz free energy of a thermodynamic system is defined as follows:

(111) $\mathcal{A}=U-TS$

The principle of minimum Helmholtz energy states that a closed thermodynamic system approaching equilibrium at constant temperature and volume tends to a state of minimum Helmholtz free energy. Mathematically this principle translates as follows:

(112) ${\left(\frac{\partial \mathcal{A}}{\partial X_i}\right)}_{T,V}=0\mathrm{and}{\left(\frac{{\partial }^2\mathcal{A}}{\partial X_i^2}\right)}_{T,V}>0,i=1\dots n$

The fourth case is when the closed system is restricted to fixed temperature and pressure, in which case the principle of minimum Gibbs free energy holds. A closed system approaching equilibrium at constant temperature and pressure tends to a state of minimum Gibbs free energy. Mathematically one can write for this case:

(113) ${\left(\frac{\partial G}{\partial X_i}\right)}_{T,P}=0\mathrm{and}{\left(\frac{{\partial }^{2}G}{\partial X_i^2}\right)}_{T,P}>0,i=1\dots n$

In addition, a fifth case does exist when the closed system is isolated (it has an impermeable boundary). The principle of maximum entropy holds in this case, which states that an isolated system with constrained constant internal energy will tend to a state of maximum entropy while approaching equilibrium. Mathematically, this principle is written as follows:

(114) ${\left(\frac{\partial S}{\partial X_i}\right)}_U=0\mathrm{and}{\left(\frac{{\partial }^{2}S}{\partial X_i^2}\right)}_U<0,i=1\dots n$

where S=S(U, X _i), i=1…n.

Consider a mixture of chemical species. The infinitesimal Gibbs energy variation caused by a differential change in species concentrations, $\text{d}{n}_{\text{i}}$ , can be calculated using chemical potentials ${\mu }_{\text{i}}$ , defined for the component "i" of the system according to the equation

(115) ${\mu }_i=\frac{\partial G}{\partial n_{\mathrm{i}}},i=1\dots n$

Thenceforth,

(116) $\mathrm{d}G=\sum {\mu }_i\mathrm{d}{n}_i$

If a mixture of gases obeys the ideal gas law, then the following equation expresses the variation of chemical potential with pressure

(117) ${\mu }_i\left(P_i\right)-{\mu }_i\left(P_0\right)=RT\ln \left(\frac{P_i}{P_0}\right)$

where P ₀ is a reference pressure while P _i is the partial pressure of component "i."

A similar equation does exist for ideal solutions, where partial pressure is replaced by molar concentration, namely:

(118) ${\mu }_i\left(c_i\right)-{\mu }_i\left(c_0\right)=RT\ln \left(\frac{c_i}{c_0}\right)$

where c ₀ represents the concentration at reference conditions.

For real chemical systems, in order to account for nonideal conditions and still keep the same form for the relationship describing the variation of chemical potential, the thermodynamic activity "a _i " is defined by:

(119) ${\mu }_i\left(a_i\right)-{\mu }_i\left(a_i=1\right)=RT\ln \left(a_i\right)$

where the chemical potential for a _i=1 must be defined and it is known as the standard state thermodynamic potential, denoted by ${\mu }_i^0={\mu }_i\left(a_i=1\right)$ .

The thermodynamic activity has the unit of measure similar to the unit of concentration. Thermodynamic activity is often expressed through a dimensionless activity coefficient f that quantifies the abatement from ideal conditions of a component (species) of a chemical system through the equation

(120) $a_i=c_i{f}_i$

According to the definitions for chemical potential and thermodynamic activity given in Eqs. (116), (118), and (119), the Gibbs energy of a chemical reaction has the general expression given as follows:

(121) $\mathrm{\Delta }G=\sum \left(n_i{\mu }_i^0\right)+RT\ln \left(\Pi {\left(cf\right)}_i^{n_i}\right)$

where n _i is the stoichiometric number that is positive for the products (P) and negative for the reactants (R), the term $\sum \left(n_i{\mu }_i^0\right)$ can be expressed as the sum of products minus the sum of reactants, where reactants are the chemical species that are consumed, and the products represent the species that are generated.

Therefore, one has:

(122) $\sum \left(n_i{\mu }_i^0\right)=\sum _{\mathrm{P}}\left(n_i{\mu }_i^0\right)-\sum _{\mathrm{R}}\left(n_i{\mu }_i^0\right)=\mathrm{\Delta }{G}^0$

where ΔG ⁰ is the standard Gibbs free energy of the reaction.

From Eqs. (120) and (121) one obtains the following:

(123) $\mathrm{\Delta }G=\mathrm{\Delta }{G}^0+RT\ln \left(K_{\mathrm{eq}}\right)$

where

(124) $K_{\mathrm{eq}}=\Pi {\left(cf\right)}_i^{n_i}$

is known as the equilibrium constant of the reaction.

A potential diagram is useful to describe chemical equilibrium, and even more generally, thermodynamic equilibrium. Similar to the case represented previously as shown in Fig. 15, a potential surface (pathway) of any process can be described as shown in Fig. 29. In any general chemical reaction there must be two potential surfaces, one corresponding to reactants and another for products. Moving along the reaction coordinate, the potential energy of the system decreases and reaches a minimum in #2. An energy barrier 2-3 of magnitude ΔG _act has to be overcome in order for the system to move on the product's potential surface, toward the right. If an electric field is applied by polarization of the electrode (anode for the case shown in the figure), the whole potential surface characterizing the thermodynamic system of the reaction products displaces at lower energy with a magnitude proportional to the polarization potential, namely FΔE _act, where ΔE _act represents the activation overpotential. Consequently, the activation energy of the reaction under the influence of polarization decreases as it can evolve along the path $1-2-\overline{3}-\overline{4}$ .

Fig. 29. Potential surfaces and stable equilibrium states for a general chemical reaction.

The following possibilities do exist regarding the spontaneity and equilibrium state of a chemically reacting system, as given in Table 8. The reaction can proceed spontaneously (S) forward toward consumption of the reactants and generation of products. The reaction can be nonspontaneous (NS) and therefore does not proceed in a forward direction. The reaction is at equilibrium €, therefore the backward and forward processes have the same rate and there will be no net change.

Table 8. Criteria for spontaneity and equilibrium of chemical reactions

Spontaneity	ΔG	ΔStot	ΔH	ΔS	T	ΔG	Spontaneity	ΔH	ΔS	T	ΔG	Spontaneity
Spontaneous (S)	<0	>0	<0	>0	any	<0	S	>0	<0	any	>0	NS
Nonspontaneous (NS)	>0	<0	<0	<0	low	<0	S	>0	<0	low	>0	NS
Equilibrium (E)	0	0	<0	<0	high	>0	NS	>0	>0	high	<0	S

Those criteria (S, NS, E) can be categorized based on reaction enthalpy (ΔH) and entropy (ΔS). Those parameters allow for calculation of the reaction Gibbs free energy, ΔG=ΔH−TΔS and further of the total entropy change of the reaction (the thermodynamic system at constant $T$ and $P$ ) and its surroundings, as follows:

(125) $\mathrm{\Delta }{S}_{\mathrm{tot}}=-\mathrm{\Delta }G/T$

In general both the enthalpy and entropy of reactants and products increase with the temperature. Consequently, the temperature variation of reaction enthalpy and entropy are very slow. Therefore, in most cases, temperature is the key parameter that influences the reaction equilibrium. As given in Table 8, if a system is at equilibrium, then its Gibbs free energy is zero. From Eq. (125) the following approximate estimation of the temperature at equilibrium can be derived, by imposing ΔG=0, namely:

(126) $T_{\mathrm{eq}}\cong \frac{\mathrm{\Delta }G}{\mathrm{\Delta }S}$

The equilibrium constant is defined based on Gibbs free energy of the reaction in standard conditions, ΔG ⁰, as follows:

(127) $K_{\mathrm{eq}}=\exp \left(-\frac{\mathrm{\Delta }{G}^0}{RT}\right)$

Furthermore, the equilibrium constant can help determine the reaction mixture composition at the equilibrium as follows:

•

If equilibrium constant is K _eq>10³, then the equilibrium mixture comprises mainly reaction products.

•

If equilibrium constant is K _eq<10⁻³, then the equilibrium mixture comprises mainly reactants.

•

If equilibrium constant is 10³>K _eq>10⁻³, then the equilibrium mixture comprises both reactants and products.

According to Le Châtelier's principle, if a reaction mixture at equilibrium is placed upon a stress the system reacts in the direction that relieves the stress. For a reaction that reduces the moles of gas, this means that:

•

An increase of pressure produced by shrinking the volume moves the reaction direction toward reducing the moles of gas.

•

An expansion of volume that produces a decrease of pressure will move the reaction toward an increase of the mole of gas.

Similarly to the reaction quotient, the notion of extent of reaction – denoted as ξ – is very useful for calculating the equilibrium conditions. The extent of reaction is a parameter between 0 and some positive value that depends on the reaction stoichiometry. When the reaction quotient is smaller than 1, the extent of reaction is near zero, and the total free energy of reaction mixture decreases when the reaction proceeds spontaneously in the forward direction. The total Gibbs energy of the reaction mixture can be defined as the sum of moles of chemical species "i" multiplied by the respective molar specific Gibbs free energy at constant temperature and pressure. Therefore, the extent of a reaction is defined implicitly as follows:

(128) $\mathrm{\Delta }G={\left(\frac{\partial G}{\partial \xi }\right)}_{T,P}$

where G _i is the molar specific free energy of species i (reactant of product) and n _i is the number of moles of species "i."

Fig. 30 shows a plot of the total Gibbs free energy of the reaction mixture versus the extent of reaction. At equilibrium, one must have dG=0 with a minimum as suggested in the figure. Therefore, Gibbs free energy minimization will determine the reaction mixture composition at the equilibrium, in terms of the number of moles n _i.

Fig. 30. Gibbs free energy variation vs. the extent of reaction.

A reaction achieves full conversion when all the reactants are consumed and products generated in a stoichiometric amount. This is not the case with actual reaction for several reasons. When a reaction has an equilibrium constant around 1 there will not be a full conversion of the reactants into products because both reactants and products will be present in the reaction mixture. In addition, two or more reactions can occur simultaneously therefore creating undesired products. Also several types of losses occur with purification and separation of the products and the presence of impurities. The theoretical yield is the amount of product that can be generated by a reaction according to the stoichiometry. The actual yield will be smaller than the theoretical yield. The percent yield is defined as the ratio of actual to theoretical yield.

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Energy Conversion

Murat Ozturk , Ibrahim Dincer , in Comprehensive Energy Systems, 2018

4.10.3.3 Stoichiometric Method of Gasification Cycles

The equilibrium constant is usually utilized to define the composition of gasification outputs. This constant is, whereas, only available to the stable state, and hence unsuitable to the examination of transient composition of gaseous in process, which is constantly unstable. The chemical cycle in the large-scale gasification chamber cannot be exactly described by using the kinetics and equilibrium states. The combination of gases generated by gasification process is a very clear measure of the chemical state in the gasification chamber.

In this study, the stoichiometric approach is accepted to investigate the coal gasification reaction structure. Therefore, the hypothetical chemical definition is improved depending on the ultimate analysis results of source feedstock. The gasification reaction can be defined as follows:

(12) ${\mathrm{CH}}_m{\mathrm{O}}_n+\alpha {\mathrm{O}}_2+\beta {\mathrm{H}}_2\mathrm{O}\to \gamma {\mathrm{H}}_2+\delta \mathrm{CO}+\epsilon {\mathrm{CO}}_2+\kappa {\mathrm{CH}}_4+\zeta {\mathrm{C}}_2{\mathrm{H}}_4+\theta {\mathrm{C}}_2{\mathrm{H}}_6+\lambda {\mathrm{CH}}_{m\prime }{\mathrm{O}}_{n\prime }$

where CH _m O _n and CH _m′O _n′ are coal sources and tar, respectively [32]. The elemental balance equations for C, H, and O in Eq. (12) can be defined as follows:

(13) $1=\delta +\epsilon +\kappa +2\zeta +2\theta +\lambda$

(14) $m+2\beta =2\gamma +4\kappa +4\zeta +6\theta +m\prime \lambda$

(15) $n+2\alpha +\beta =\delta +2\epsilon +n\prime \lambda$

The total moles of produced gas in Eq. (12) can be defined as follows:

(16) $\Phi =\gamma +\delta +\epsilon +\kappa +\zeta +\theta$

When the concentrations of H₂, CO, CO₂, CH₄, C₂H₄, and C₂H₆ (dry and N₂ free) are represented by p, q, r, s, t, and u, respectively, the mole numbers of all gases can be defined as follows:

(17) $H_2:\gamma =p\mathrm{\Phi }$

(18) $\text{CO}:\delta =q\mathrm{\Phi }$

(19) ${\mathrm{CO}}_2:\epsilon =r\Phi$

(20) ${\mathrm{CH}}_4:\kappa =s\Phi$

(21) ${\mathrm{C}}_2{\mathrm{H}}_4:\zeta =t\Phi$

(22) ${\mathrm{C}}_2{\mathrm{H}}_6:\theta =u\Phi$

(23) $\lambda =v\Phi$

The reference state should be essential to explain the reaction process given in Eq. (12).

(24) ${\mathrm{CH}}_m{\mathrm{O}}_n+0.5(1-n){\mathrm{O}}_2\to 0.5m{\mathrm{H}}_2+\mathrm{CO}$

(25) $\alpha =0.5(1-n)+{\mathrm{O}}_{\mathrm{ex}}$

In the O_ex>0 case when z mole of CH₄ is created, the formula can be written as follows:

(26) ${\mathrm{CH}}_m{\mathrm{O}}_n+\left\{0.5(1-n)+O_{\mathrm{ex}}\right\}{O}_2+(y-x-z){\mathrm{H}}_2\mathrm{O}\to (0.5m+y-x-3z){\mathrm{H}}_2+\left(1-2{\mathrm{O}}_{\mathrm{ex}}-y+x-z\right)\mathrm{CO}+\left(2{\mathrm{O}}_{\mathrm{ex}}+y-x\right){\mathrm{CO}}_2+z{\mathrm{CH}}_4$

In the O_ex<0 case, the formula can be defined as follows:

(27) ${\mathrm{CH}}_m{\mathrm{O}}_n+\left\{0.5(1-n)+{\mathrm{O}}_{\mathrm{ex}}\right\}{\mathrm{O}}_2+\left(-2{\mathrm{O}}_{\mathrm{ex}}+y-z\right){\mathrm{H}}_2\mathrm{O}\to \left(0.5m-2{\mathrm{O}}_{\mathrm{ex}}+y-3z\right){\mathrm{H}}_2+(1-y-z)\mathrm{CO}+y{\mathrm{CO}}_2+z{\mathrm{CH}}_4$

From the two formulas given above, the previously defined coefficients of each component can be defined numerically as illustrated in Table 6.

Table 6. Coefficients of all components in gasifier based on O_ex value

	Oex≥0	Oex≤0
α	0.5(1−n)+Oex	0.5(1−n)+Oex
β	y−x−z−2w−2u−(1−n′)v	−2Oex+y−z−2w−2u−(1−n′)v
γ	0.5m+y−x−3z−4w−5u−{(1−n′)+0.5m′}v	0.5m−2Oex+y−4w−5u−{(1−n′)+0.5m′}v
δ	1−2Oex−y+x−z−2w−2u−v	1−y−I−2w−2u−v
ε	2Oex+y−x	y
κ	Z	z
ξ	w	w
θ	u	u
λ	v	v

The heat of gasification reaction, h _r (kcal/mol-coal) can be defined as follows:

(28) $h_{\mathrm{r}}=\gamma h_{{\mathrm{H}}_2}+\delta h_{\mathrm{CO}}+y{h}_{{\mathrm{CO}}_2}+\kappa h_{{\mathrm{CH}}_4}-h_{\mathrm{coal}}$

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Reactions of 17- and 19-Electron Organometallic Complexes

Shouheng Sun , Dwight A. Sweigart , in Advances in Organometallic Chemistry, 1996

IV MIGRATORY INSERTION AND ISOMERIZATION REACTIONS

The equilibrium constant for the CO insertion shown in Eq. (36) (L   =   CO, PR₃) is greatly increased when the metal is oxidized from Fe(II) to Fe(III). ¹⁴⁹ Electrochemical studies show that K _eq is ca. 10¹¹ times greater for the Fe(III) analogues, and that this is largely due to an increase in the forward rate constant. This oxidatively induced insertion is not catalytic because the acyl product is more easily oxidized than is the reactant. The carbonylation reaction in Eq. (37), however, is subject to redox catalysis. ^{1, 150} Thus, K _eq for Eq. (37) is quite large, but the reaction does not progress detectably after 5   days under CO at 0   °C. The addition of a few mole percent of an oxidizing agent such as Ag⁺ or [Cp₂Fe]⁺ causes complete conversion to product within 2   minutes. The overall insertion rate

(36) $\left[\mathrm{CpFe}\left(\mathrm{C}\mathrm{O}\right)\left(\mathrm{L}\right)\mathrm{Me}\right]+\mathrm{MeCN}\to \left[\mathrm{CpFe}\left(\mathrm{L}\right)\left(\mathrm{MeCN}\right)\left(\mathrm{COMe}\right)\right]$

(37) $\left[\mathrm{CpFe}\left(\mathrm{C}\mathrm{O}\right)\left({\mathrm{P}\mathrm{P}\mathrm{h}}_3\right)\mathrm{Me}\right]+\mathrm{C}\mathrm{O}\to \left[\mathrm{CpFe}\left(\mathrm{C}\mathrm{O}\right)\left({\mathrm{P}\mathrm{P}\mathrm{h}}_3\right)\left(\mathrm{COMe}\right)\right]$

increases by a factor of ca. 10⁷ upon oxidation. The use of resolved [CpFe(CO)(PPh₃)Me] leads to a racemic acyl product, a fact attributed to configurational instability of the 17-electron [CpFe(CO)(PPh₃)Me]⁺ intermediate. The use of trityl cation as the oxidizing agent to initiate the catalytic carbonylation of [CpFe(CO)₂R] has also been described. ¹⁵¹ The alkyl to acyl migration in the 17-electron [CpFe(CO)(L)R]⁺ occurs in two steps, and after some debate, there seems to be general agreement ^{150, 152, 153} that the mechanism consists of nucleophilic attack followed by migration and not vice versa. This is not surprising, because the latter mechanism requires a 15-electron intermediate. Reductive activation can also be used to induce alkyl to acyl migrations, in which case there is good evidence that migration occurs directly from the initially formed 19-electron intermediate. ^{154, 155}

An interesting case of reductively induced hydride migration is summarized in Scheme 9. ¹⁵⁶ Chemical oxidation of [Cp*Fe(dppe)H] affords the 17-electron monocation, which reversibly binds CO at –80   °C. Spectroscopic studies established that [Cp*Fe(dppe)(CO)H]⁺ is a genuine 19-electron complex, which undergoes hydride migration to an endo site of the C₅Me₅ ring when reduced with cobaltocene.

Scheme 9.

The reductive elimination of ethane from [Cp*Rh(PPh₃)Me₂] was found to be increased by a factor of at least 3   ×   10⁹ upon oxidation to the 17-electron radical cation. ¹⁵⁷ The absence of solvent effects on the rate was interpretated to indicate direct elimination from the cation to give ethane and a 15-electron intermediate, which is rapidly trapped by solvent.

There are many examples of redox-induced structural rearrangements in organometallic complexes; the reader is referred to a review by Connelly. ² The simplest rearrangement is cis/trans or fac/mer isomerization, usually induced by oxidation. Thus, cis-[Mn(CO)₂(dppe)₂]⁺ slowly converts to the trans isomer with a rate constant of 10^–5  s^–1 at room temperature. Upon oxidation, the cis   →   trans isomerization increases in rate by 7 powers of 10. ¹⁵⁸ The process is not catalytic, so that stoichiometric oxidation followed by reduction is required to synthetically utilize the increased reactivity of the radicals in the conversion of 18-electron cis-[Mn(CO)₂(dppe)₂]⁺ to trans-[Mn(CO)₂(dppe)₂]⁺. An example of oxidatively induced fac   →   mer isomerization is given in Scheme 10. ¹⁵⁹ The fac⁰  →   mer⁰ reaction for the neutral 18-electron isomer is slow, with k ₂  =   2   ×   10^–4  s^–1, K ₂  =   4. The reaction fac⁺  →   mer⁺ is much faster and also has a larger equilibrium constant: k ₁  =   1   ×   10^–1  s^–1, K ₁  =   640. In this case E ⁰ for the mer⁺/mer couple is about 0.15   V negative of that for fac⁺/fac, which means that the cross reaction is slightly unfavored (K ₃  =   1/160). Nevertheless, the fac⁺  →   mer⁺ conversion is rapid enough so that the cross reaction provides an effective pathway for the catalytic conversion fac⁰  →   mer⁰. An analogous study has been reported concerning fac/mer isomerization in [M(CO)₃(η ³-P₂P')] [M   =   Cr, Mo, W; P₂P'   =   Ph₂PCH₂CH₂P(Ph)CH₂CH₂PPh₂]. ¹⁶⁰

Scheme 10.

Upon oxidation, the vinylidene complex [(C₆Me₆)Cr(CO)₂{C   =   C(Si Me₃)₂}] undergoes rapid isomerization to the alkyne analogue, shown as V ⁺  → A ⁺ in Scheme 11. ^{161, 162} While isomerizations in the direction alkyne   →   vinylidene are not rare when hydrogen migration is involved, the chemistry depicted in Scheme 11 is rare because the migrating group is SiMe₃ and unique because it is redox promoted in the direction vinylidene   →   alkyne. The conversion A  → V is favored in the neutral complexes, and this isomerization is catalyzed by V ⁺.

Scheme 11.

Another interesting oxidatively promoted transformation is illustrated in Fig. 9. The syn-facial bimetallic complex (η ⁴,η ⁶-dimethylnaphthalene) Mn₂(CO)₅ (31) contains a fairly strong metal–metal bond. ¹⁶³ The addition of a trace of oxidant [Cp₂Fe]⁺ to 31 under an atmosphere of CO results in the catalytic conversion to the zwitterionic 32. Complex 32 retains the syn-facial structure, which is made possible by a large bending of the diene (η ⁴) plane from the η ⁶-ring. ¹⁶⁴ Significantly, 32 can be converted back to 31 by the stoichiometric addition of the reducing agent cobaltocene or by addition of trimethylamine oxide.

Fig. 9. Redox promoted interconversion between syn-facial (η ⁴,η ⁶-dimethylnaphthalene) dimanganese carbonyl complexes.

Most redox-induced isomerizations are initiated by oxidation. Reductions that lead to isomerization are much less common. Two examples are Eqs. (38) and (39), both of which are accomplished by initial reduction with sodium amalgam followed by stoichiometric oxidation. ^{165, 166} The neu-

(38) $fac-{\left[\mathrm{M}\mathrm{n}\left(\mathrm{C}\mathrm{O}\right){\left({\mathrm{CNBu}}^{\mathrm{t}}\right)}_3\left(\mathrm{bipy}\right)\right]}^{+}\to \mathrm{m}\mathrm{e}\mathrm{r}-{\left[\mathrm{M}\mathrm{n}\left(\mathrm{C}\mathrm{O}\right){\left({\mathrm{CNBu}}^{\mathrm{t}}\right)}_3\left(\mathrm{bipy}\right)\right]}^{+}$

(39) $cis,cis‐{\left[\mathrm{M}\mathrm{n}\left({\mathrm{C}\mathrm{O}}_2\right){\left({\mathrm{CNBu}}^{\mathrm{t}}\right)}_2\left(\mathrm{bipy}\right)\right]}^{+}\to cis,\mathit{trans}‐{\left[\mathrm{M}\mathrm{n}\left({\mathrm{C}\mathrm{O}}_2\right){\left({\mathrm{CNBu}}^{\mathrm{t}}\right)}_2\left(\mathrm{bipy}\right)\right]}^{+}$

tral radicals involved in this process have been shown to be of the 18   + δ variety. The isomerization of [CpCo(1,5-COT)] ^z to [CpCo(1,3-COT)] ^z is fairly slow in the 18-electron z  =   0 complex, but is fast in the 19-electron anion (z  =   –l). ¹⁶⁷ Reductively initiated ETC catalysis has been used for the synthesis of the isomers of [(CF₃)₂C₆Co₂{P(OMe)₃}₂] and other "flyover" organometallic clusters. ¹⁶⁸

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Salinity Effect and Ion Exchange

James J. Sheng Ph. D. , in Modern Chemical Enhanced Oil Recovery, 2011

3.3.4 Calculation of Mass Action Constant at Different Temperatures

Changes of equilibrium constants with temperature are usually described with the van't Hoff equation (Atkins and de Paula, 2006).

(3.28) $\frac{\mathrm{d}\ln \mathrm{K}}{\text{dT}}=\frac{\Delta {\mathrm{H}}_{\mathrm{r}}}{{\text{RT}}^2},$

where ΔH_r is the reaction enthalpy, or the heat lost or gained by the chemical system. For exothermal reactions, ΔH_r is negative (the system loses energy and heats up), and for endothermal reactions, ΔH_r is positive (the system cools). R is the gas constant (8.314 J/K/mol). At 25°C, the value of the reaction enthalpy,

, is calculated from the formation enthalpies.

This equation shows that K increases with temperature for positive ΔH_r and decreases with temperature for negative ΔH_r. Usually,

is constant within the range of a few tenths of degrees. Therefore, the preceding equation can be integrated to give two temperatures:

(3.29) $\log {\mathrm{K}}_{{\mathrm{T}}_1}-\log {\mathrm{K}}_{{\mathrm{T}}_2}=\frac{-\Delta {\mathrm{H}}_{\mathrm{r}}^0}{2.303\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right).$

Some values of

are provided by Dria et al. (1988).

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Adsorption Equilibrium Relationships, Isotherm Expressions, Their Determinations, and Predictions

Chi Tien , in Introduction to Adsorption, 2019

3.6 Is There Equivalence Between Adsorption Isotherm and Ion Exchange Equilibrium Constant?

Previously, we stated that the materials presented in this volume may be readily applied to ion exchange processes as well. Specifically, the solute uptake mechanisms and the rate of uptake in adsorption and those of ion exchange are essentially the same. Accordingly, the materials given in Chapters 4 and 5 4 5 can be applied to both types of operations.

In extending adsorption results, especially adsorption performance models to ion exchange, the question: What is the equivalent adsorption isotherm arises? More broadly, is it correct to treat ion exchange equilibrium data as single solute isotherm of liquid solutions?

Consider the following simple ion exchange process

${\mathrm{A}}^{+}+\mathrm{HR}\iff {\mathrm{H}}^{+}+\mathrm{AR}$

The ion exchange equilibrium constant K ^o (or more properly, the apparent equilibrium constant) is commonly defined as

(3.101a) $K^{\mathrm{o}}=\frac{\left(c_{\mathrm{t}}-c_{\mathrm{A}}\right)q_{\mathrm{A}}}{c_{\mathrm{A}}\left(q_{\mathrm{t}}-q_{\mathrm{A}}\right)}$

where c _A is the ionic concentration of A in the solution and q _A, the concentration of A⁺ in the resin phase. ⁹ c _t is the total concentration of all the ionic species of the solution and q _t is the total ionic species concentration in the resin phase are

$c_{\mathrm{t}}=c_{{\mathrm{A}}^{+}}+c_{{\mathrm{H}}^{+}}$

(3.101b) $q_{\mathrm{t}}=q_{{\mathrm{A}}^{+}}+q_{{\mathrm{H}}^{+}}$

From Eq. (3.100), q _A may be expressed as a function of c _A, or

(3.102) $q_{\mathrm{A}}=\frac{K^{\mathrm{o}}{q}_{\mathrm{t}}{c}_{\mathrm{A}}}{\left(c_{\mathrm{t}}-c_{\mathrm{A}}\right)+K^{\mathrm{o}}{c}_{\mathrm{A}}}=\frac{K^{\mathrm{o}}{q}_{\mathrm{t}}{c}_{\mathrm{A}}}{c_{\mathrm{t}}+\left(K^{\mathrm{o}}-1\right)c_{\mathrm{A}}}=\frac{\left(\frac{K^{\mathrm{o}}{q}_{\mathrm{t}}}{c_{\mathrm{t}}}\right)c_{\mathrm{A}}}{1+\frac{K^{\mathrm{o}}-1}{c_{\mathrm{t}}}{c}_{\mathrm{A}}}$

For adsorption of dilute liquid solution, the single solute isotherm is

(3.103) $q_{\mathrm{A}}=\frac{a{c}_{\mathrm{A}}}{1+{\mathit{bc}}_{\mathrm{A}}}$

By comparing Eq. (3.101a) and Eq. (3.103), the ion exchange equilibrium relationship may appear similar to the adsorption isotherm of the Langmuir type. Specifically, the quantities of K ^o q _t/c _t and (K ^o  −   1)/c _t may be viewed as a and b of the Langmuir equation.

This similarity, however, does not lead to an equivalence between the two expressions, as the quantity K ^o is known to be not constant, but a function of total ion concentration of the solution, c _t (Bauman and Eichhorn, 1947); while, for a given adsorbate-adsorbed system obeying the Langmuir equation, the equilibrium relationship is described by a single set of "a" and "b" values. For ion exchange, if the Langmuir equation is to be used to describe its equilibrium, the Langmuir parameter, a and b, as functions of c _t must be established.

With this understanding, the appropriateness of expressing ion exchange equilibrium with adsorption isotherm expression may be assessed. The data used for establishing the "isotherm expression" should be obtained under specified total ion concentration, c _t, and the isotherm parameters given as functions of c _t. While it is certainly possible to correlate ion exchange equilibrium data without considering the requirement of constant c _t, such expressions are purely empirical, and more importantly, the correlations developed are of uncertain accuracy.

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how to find equilibrium constant

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