Page 157 - Physical chemistry understanding our chemical world

124 ENERGY AND THE FIRST LAW OF THERMODYNAMICS

(ii) ionize the gaseous atoms of chlorine to form anions (for which the energy is the
electron affinity E (ea) ). Finally, we need to account for the way the sodium chloride
O
forms from elemental sodium and chlorine, so the cycle must also include H .
f
Worked Example 3.15 What is the lattice enthalpy H (lattice) of sodium chloride at
◦
25 C?
Strategy. (1) We start by compiling data from tables. (2) We construct an energy cycle.
(3) Conceptually, we equate two energies: we say the lattice enthalpy is the same as the
sum of a series of enthalpies that describe our converting solid NaCl first to the respective
elements and thence the respective gas-phase ions.
(1) We compile the enthalpies:

for sodium chloride
These energies are
1 O −1
huge. Much of this Na (s) + Cl 2(g) −−−→ NaCl (s) H f =−411.15 kJ mol
2
energy is incorporated
into the lattice; other- for the sodium
wise, the value of H O
f O −1
would be massive. Na (s) −−−→ Na (g) H (sublimation) = 107.32 kJ mol
+
Na (g) −−−→ Na (g) + e − I (Na) = 502.04 kJ mol −1
for the chlorine

1 · 1 O −1 −1
2 Cl (g) −−−→ Cl (g) 2 H BE = 121.68 kJ mol (i.e. half of 243.36 kJ mol )
Cl · − − E (ea) =−354.81 kJ mol −1
(g) + e −−−→ Cl (g)
(2) We construct the appropriate energy cycle; see Figure 3.8. For simplicity, it is usual
to draw the cycle with positive enthalpies going up and negative enthalpies going down.
(3) We obtain a value of H O , equating it to the energy needed
(lattice)
to convert solid NaCl to its elements and thence the gaseous ions. We
We multiply H O by
f construct the sum:
‘−1’ becausewecon-
sider the reverse pro- O O O 1 O
H (lattice) =− H + H (sublimation) + I (Na) + H BE + E (ea)
f
2
cess to formation. (We
travel in the opposite Inserting values:
direction to the arrow
representing H O in O
f H (lattice) =−(−411.153) + 107.32 + 502.04
Figure 3.8.)
+ 121.676 + (−354.81) kJ mol −1
so
H O = 787.38 kJ mol −1
(lattice)

SAQ 3.11 Calculate the enthalpy of formation H O for calcium fluoride.
f
Take H O =−2600 kJ mol −1 , H O = 178 kJ mol −1 ; I
+
(lattice) (sublimation) 1(Ca→Ca )