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538 15 SOLIDS
neighbouring centres, A is the Madelung constant, ε0 is the vacuum per-
mittivity and d∗ is taken to be 34.5 pm.
Taking the ionic radii from Table 15C.2 on page 661 as rCa+ ≈ rK+ =
138 pm and rCl− = 181 pm, d = (138 + 181) pm = 319 pm.
Ep,min = −
NA∣zAzB∣e2
4πε0d
(1 − d
∗
d
)A
= (6.0221 × 1023mol−1) × (1.6022 × 10−19 C)2
4π × (8.8542 × 10−12 J−1 C2m−1) × (319 pm)
× (1 − 34.5 pm
319 pm
) × 1.763 = −685 kJmol−1
�us the lattice enthalpy of CaCl(s) is ∆HL = 685 kJmol−1 .
�e relevant Born–Haber cycle is shown in Fig. 15.8; all of the enthalpy
changes quoted in the diagram are in kJmol−1.�e �rst ionisation energy
of Ca, +589.7 kJmol−1, is taken from Table 8B.4 in the Resource section.
It follows from the cycle that ∆fH−○ = 176 + 589.7 + 122 − 349 − 685 =
−146 kJmol−1
CaCl(s)
Ca(s)+ 12Cl2(g)
Ca(g)+ 12Cl2(g)
Ca+(g)+ 12Cl2(g)
Ca+(g)+Cl(g)
Ca+(g) + Cl– (g)
∆fH−○
176
590
122 −349
−∆HL = −685
Figure 15.8
(b) �e disproportionation reaction forCaCl(s) is 2CaCl(s)→Ca(s)+CaCl2(s).
Using data from Table 2C.5 in the Resource section and using the enthalpy
of formation of CaCl(s) calculated in part (a),
∆rH−○ = ∆fH−○(Ca(s)) + ∆fH−○(CaCl2(s)) − 2∆fH−○(CaCl(s))
= [0 − 795.8 − 2(−146)] kJmol−1 = −504 kJmol−1
Hence the disproportionation reaction is strongly exothermic. �e en-
tropy change for this reaction is expected to be close to zero since there
is no change in the number of species or their physical state, so it can
reasonably by concluded that ∆rG−○ will also be negative, thus favouring
the products of the reaction.�us, it is plausible that CaCl is thermody-
namically unstable with respect to disproportionation.