Concise Physical Chemistry

c11 JWBS043-Rogers September 13, 2010 11:26 Printer Name: Yet to Come

PROBLEMS AND EXERCISES 179

107.3

495.8

Na

Na+ 1/2 I^2

I

I-

NaI

∆fH

62.4

-295.3

-287.8

-658

FIGURE 11.13 The Born–Haber cycle for NaI.

path the process takes because enthalpy is a state function and is path-independent.

Vaporization takes up 107.3 kJ mol−^1 and ionization takes up 495.8 kJ mol−^1 (both

endothermic). These enthalpy changes are shown as vertical lines for the left-hand

side of Fig. 11.13.

Next, we sublime 0.5 mol of I 2 (endothermic 62.4 kJ mol−^1 ) and add an electron

(electron affinity: exothermic, – 232.9 kJ mol−^1 ).

1

2 I^2 (s)→I(g)

I(g)+e−→I−

These two steps are shown in the top middle panel in Fig. 11.13. The enthalpy of

formation of NaI is shown as the slanted line at the bottom of the diagram and after

all this is done, we have the difference between Na and I in the state of their gaseous

ions and the crystal NaI in its standard state. The formation of NaI from the gaseous

ions is shown by the downward dotted arrow in Fig. 11.13. Adding all the enthalpies,

one gets−658 kJ mol−^1. The reverse of this arrow is the vaporization of NaI into its

gaseous ions. This is the lattice energy of the crystal, (^) latticeH(NaI)=658 kJ mol−^1.
Problem 11.1
Calculate the surface area of a unit volume of liquid confined within
(a) a sphere and
(b) a cube
Which is larger and what is the percentage difference?
Problem 11.2
Water is unique among common liquids because it has a surface tension of
72 .0mNm−^1 , nearly double the average of the other common liquids listed in the