Physical Chemistry Third Edition

318 7 Chemical Equilibrium

Reactions with Electrolyte Solutes

Reactions involving electrolyte solutes present some new aspects. Consider the reaction

NaOH(aq)+HCl(aq)
H 2 O+NaCl (7.3-5)

The solutes in this reaction are allstrong electrolytes, which means that they appear to

exist in solution in completely ionized or dissociated form. We can write the reaction

equation in ionic form:

Na++OH−+H++Cl− 
H 2 O+Na++Cl− (7.3-6)

We can also write thenet ionic equation, which omits the Na+and Cl−ions that occur

on both sides of the reaction equation:

OH−+H+
H 2 O (7.3-7)

Writing H+does not express any particular assumption about hydration of the hydro-

gen ions. Many aqueous hydrogen ions are apparently covalently bonded to a water

molecule, forming H 3 O+(thehydronium ion), but others are apparently bonded to

water dimers, producing H 5 O+ 2.^2 Others can be attached to trimers, etc. The symbol

H+stands for all of these species taken together.

The standard-state Gibbs energy change for the reaction of Eq. (7.3-7) could be

calculated from the Gibbs energy changes of formation of H 2 O, NaCl, NaOH, and HCl

if the data were available. However, actual tables include data for separate ions instead

of neutral electrolytes. We use the net ionic equation

∆G◦∆fG◦(H 2 O)−∆fG◦(H+)−∆fG◦(OH−) (7.3-8)

Gibbs energy changes of formation for separate ions cannot be determined experimen-

tally because of the near impossibility of adding ions of one charge to a system without

also adding ions of the opposite charge. An arbitrary choice has been made that assigns

zero value to the Gibbs energy change of formation, enthalpy change of formation,

and molar entropy of the hydrogen ion in its standard state in aqueous solution. This

convention is equivalent to assigning the entire Gibbs energy change of formation of

aqueous HCl to the Cl−ion or that of aqueous HNO 3 to the NO− 3 ion, and so on. All

other table values are made consistent with this convention.

EXAMPLE 7.8

Using values in Table A.8 of Appendix A, calculate∆H◦,∆G◦, and∆S◦for the reaction of

Eq. (7.3-7):

∆H◦∆fH◦(H 2 O)+(−1)∆fH◦(OH−)−∆fH◦(H+)

(− 285 .830 kJ mol−^1 )+(−1)(− 229 .994 kJ mol−^1 )+(−1)0

− 55 .836 kJ mol−^1

(^2) H.-P. Cheng and J. L. Krause,J. Chem. Phys., 107 , 8461 (1997).