Question

KI KI has a lattice energy of − 649 kJ/mol. −649 kJ/mol. Consider a generic salt, AB AB , where A 2 + A2+ has the same radius as K + , K+, and B 2 − B2− has the same radius as I − . I−. Estimate the lattice energy of the salt AB AB .

Answer 1

Answer:

The answer to this question is

The lattice energy of the salt AB is ≅ -2600 kJ/mol rounded to three significant figures

Explanation:

To solve this we list out the known variables and the required unkown, then we look for the appropriate relation between the known and the unknown thus

Lattice energy of KI = -649 kJ/mol

Latice energy of salt AB = unknown

Charge on A²⁺ =+2

Charge on B²⁻ = -2

radius of A²⁺ = radius of K⁺

radius of B²⁻ = radius  of I⁻

the Born-Lande equation for lattice energies is as follows

$E=-\frac{N_{A} Mz^{+}z^{-} e^{2} }{4\pi e_{0} r_{0} } (1-\frac{1}{n})$

$E=-\frac{N_{A} M e^{2} }{4\pi e_{0} } (1-\frac{1}{n})\frac{|z|^{+}|z|^{-}}{r_{0}}$

Where to the above question, the following terms are important

z+ =  positive ion charge

z− = negative ion charge

r₀ → distance between the positive and the negative ions;

Taking all variables of KI and AB as identical we have.

$E=-\frac{N_{A} M e^{2} }{4\pi e_{0} } (1-\frac{1}{n})\frac{|z|^{+}|z|^{-}}{r_{0}}$

$E=CONSTANT\frac{|z|^{+}|z|^{-}}{r_{0}}$

Where radius of A²⁺ = radius of K⁺ and

radius of B²⁻ = radius  of I⁻ = r₀

we have

$E_{AB} = E_{KI} (|z|^{+} |z|^{-} )$

As for AB z⁺ = +2 and z⁻ = -2 we have

$E_{AB} = E_{KI} (|+2|^{+} |-2|^{-} ) = 4E_{KI}$

= 4 ×(-649 kJ/mol) = -2596 kJ/mol

$E_{AB}$ ≅ -2600 kJ/mol rounded to three significant figures