The Δ H 0 f (KF,s) is – 563 kJ mol –1 . The ionization enthalpy of K(g) is 419 kJ mol –1 . and the enthalpy of sublimation of potassium is 88 kJ mol –1 . The electron affinity of F(g) is 322 kJ mol –1 and F–F bond enthalpy is 158 kJ mol –1 . Calculate the lattice enthalpy of KF(s).The given data are as follows :
(i) K(s) + ½F 2 (g) → KF(s) Δ H 0 f = – 563 kJ mol –1
(ii) K(g) → K + (g) + e – Δ H 0 Ioniz = 419 kJ mol –1
(iii) K(s) → K(g) Δ H 0 sub = 88 kJ mol –1
(iv) F(g) + e – → F – (g) Δ H 0 eg = – 322 kJ mol –1
(v) F 2 (g) → 2F(g) Δ H 0 diss = 158 kJ mol –1
(vi) K + (g) + F – (g) → KF(s) Δ H 0 L = ?
Text Solution
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(– 827 kJ mol –1 )
Equation (vi) can be generated by the following manipulations.
Eq. (i) – Eq. (ii) – Eq. (iii) – Eq. (iv) – ½ Eq. (v)
Carrying out the corresponding manipulations on Δ H 0 L , we get
Δ H 0 L = (– 563 – 419 – 88 + 322 – 79) kJ mol –1 = – 827 kJ/mole
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