Determine the standard enthalpy of formation of SrI2(s) using a Born–Haber cycle. Enthalpy of sublimation of Sr(s) = 164 kJ/mol 1st ionization energy of Sr(g) = 549 kJ/mol 2nd ionization energy of Sr(g) = 1064 kJ/mol Enthalpy of sublimation of I2(s) = 62.4 kJ/mol Bond dissociation energy of I2(g) = 152.55 kJ/mol 1st electron affinity of I(g) = –295.15 kJ/molLattice energy of SrI2(s) = –1959.75 kJ/mol

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mickymike92 mickymike92 · Answer 1 · 2021-01-23T07:42:54+01:00

Answer:

Enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol

Explanation:

Sr(s) -------> Sr(g) ΔHsublimation = +164 kJ/mol

Sr⁺(g) -------> Sr⁺(g) First Ionization energy, IE₁ = +549 kJ/mol

Sr⁺(g) -------> Sr²⁺(g) Second ionization energy, IE₂ = +1064 kJ/mol

I₂(s) --------> I₂(g) ΔHsublimation = +62.4 kJ/mol

I₂(g) -------> 2I(g) Bond dissociation energy, BE = +152.55 kJ/mol

2I(g) ----> 2I⁻(g) Electron affinity, EA = 2 (-295.15 kJ/mol) = -590.3 kJ/mol

Sr²⁺(g) + 2I⁻(g) ------> SrI₂(s) Lattice energy, LE = -1959.75 kJ/mol

Overall equation for the formation of SrCl₂ is given as:

Sr(s) + I₂(s) ----> SrI₂(s) ΔHformation = ?

Enthalpy of formation, ΔHformation = (ΔHsublimation of Sr(s) + IE₁ + IE₂ + ΔHsublimation of I₂(s) + BE + EA + LE)

ΔHformation = {164 + 549 + 1064 + 62.4 + 152.55 + (-590.3) + (-1959.75)} kJ/mol

Sr(s) + I₂(s) ----> SrI₂(s) ΔHformation = -558 kJ/mol

Therefore, enthalpy of formation of strontium chloride, SrCl = -558.1 kJ/mol