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Handbook of Solvents - George Wypych - ChemTech - Ventech!

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9.4 Mixed solvent influence on the chemical equilibrium 535<br />

( ) ( [] [ ] )<br />

[ ] [ ] [ ]<br />

/<br />

K = EF E − EF F − EF<br />

us<br />

0 0<br />

[9.79]<br />

Inserting [9.76] in [9.77] and [9.79] in [9.78], we come to the equation relating the equilibrium<br />

constants <strong>of</strong> all considered processes:<br />

( )( )<br />

K = K 1+ E 1 + F<br />

[9.80]<br />

EF us A A<br />

A similar form <strong>of</strong> equation [9.80] has been proposed elsewhere. 25<br />

If it is impossible to neglect the solvent concentration (or the concentration <strong>of</strong> solvate<br />

component <strong>of</strong> the mixed solvent) in comparison with [E] 0 and[F] 0, equation [9.80] is written<br />

in the form:<br />

A<br />

S ( )( FA )<br />

K = K 1+ K 1 + K<br />

[9.81]<br />

EF us EA<br />

Let E be acidic (acceptor) reagent in reaction [9.68], F basic (donor) reagent. Then, if<br />

A is acidic solvent, one can neglect specific solvation <strong>of</strong> the reagent E. Thus equations for<br />

equilibrium constant <strong>of</strong> the process [9.68] can be written in the form<br />

A<br />

( ) ( )<br />

K = K 1+ K K = K 1 + K<br />

[9.82]<br />

EF us FS EF us FS<br />

If A is a basic component, equation [9.82] can be re-written to the form<br />

A<br />

( ) ( )<br />

K = K 1+ K K = K 1 + K<br />

[9.83]<br />

EF us EA EF us EA<br />

Equations [9.82] and [9.83] were developed 26-28 for the case <strong>of</strong> specific solvation.<br />

Thus for K EF calculation one must obtain the equilibrium constant <strong>of</strong> processes: [9.71]<br />

-K us, [9.69] - K EA and [9.70] - K FS from conductance measurements. The constant K EF is<br />

identified in literature as “calculated by taking into account the specific solvation”. 29 The<br />

value K EF characterizes only the universal solvation effect on the process <strong>of</strong><br />

heteromolecular associate formation. The approach cited above can be illustrated by equilibrium:<br />

(CH 3) 2SO + o-CH 3C 6H 4OH ↔ (CH 3) 2SO•CH 3C 6H 4OH [9.84]<br />

studied in binary mixed solvents:<br />

A) CCl 4 - hepthylchloride 30 formed by two solvate inert components [9.84]<br />

B) CCl 4 - nitromethane 31 formed by solvate inert component (CCl 4) and<br />

acceptor (nitromethane) component<br />

C) CCl 4 - ethyl acetate 32 formed by solvate inert component (CCl 4) and<br />

donor (ester) component;<br />

The isotherms lnK vs. 1/ε (298.15K) are presented in Figure 9.8. These dependencies<br />

(right lines 1,2,4,5) are required for calculation <strong>of</strong> equilibrium constants <strong>of</strong> the<br />

heteromolecular association process free from specific solvation effect. It can be seen from<br />

Figure 9.8 that the values lnK, regardless <strong>of</strong> solvent nature, lie on the same line 3, which describes<br />

the change <strong>of</strong> equilibrium constants <strong>of</strong> the process [9.84] in the universal solution<br />

CCl 4-heptylchloride.

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