The number of Schottky defects (n) present in an ionic compound containing `N` ions at temperature `T` is given by `n = Ne^(-E//2KT)`, where `E` is the energy required to create `n` Schottky defects and `K` is the Boltzmann constant, If the mole fraction of Schottky defect in `NaCl` crystal at `2900 K` is `X`, then calculate `-ln(x)`,
Given: `DeltaH` of Schottky defect `= 2 eV` and
`K = 1.38 xx 10^(-23) J K^(-1)`
`1 eV = 1.608 xx 10^(-19) J`

Answer

`-In X -= 4`

Explanation

The number of Schottky defects `(n)` is given as:
`n = Ne^(-E//2KT)`
Number of moles of ions `=` Number of moles of Schottky
defects `= (N)/(N_(A)`.
Mole fraction of defect `(X)` is given as:
`X = (n//N_(A))/(n//N_(A) + N//N_(A)) = ((n)/(n + N)) = ((n//N)/((n)/(N) + 1))`,
From Eq. (i)
`X = (e^(-E//2KT)/(e^(-E//2KT) + 1))`
(Since `E = DeltaH = 2 eV = 2 xx 1.608 xx 10^(-19) J`) ,
`:.` The value of `E//2KT` is
`= (2 xx 1.608 xx 10^(-19) J)/(2 xx 1.38 xx 10^(-23) J K^(-1) xx 2900 K)`
`-= 4`
`:. X = e^(-4//(e^(-4 + 1))` The value of `e = 2.7`
implies `(2.7)^(-4)/((2.7^(-4)) + 1) = (0.018)/(0.018 + 1)`
`:. X = (0.018)/(1.018) = 0.0176`
Taking `(-In)` of both sides
`:. -In X = -In (0.0176)`
implies `-In X -= 4`