lattice energy of KCl = 699 kJ/mol first ionization energy of K = 418.7 kJ/mol e
ID: 858549 • Letter: L
Question
lattice energy of KCl = 699 kJ/mol
first ionization energy of K = 418.7 kJ/mol
electron affinity of Cl = 349 kJ/mol
bond energy of Cl-Cl = 242.7 kJ/mol
enthalpy of formation of KCl = -435.87 kJ/mol
What is the enthalpy of sublimation for K, in kJ/mol?
I know the enthalpy of formation is equal to the sum of the other numbers listed above, but I'm having trouble making equations.
Explanation / Answer
?H1 = ?H?f(KCl) the enthalpy of formation of potssium chloride: K(s) + 1/2Cl2(g) ==> KCl(s) ?Hf(KCl) = -435.87 kJ mol-1 ?H2 = ?H?atom(K) the enthalpy of atomisation of sodium: K(s) ==>K(g) ?Hatom(K) = x kJ mol-1 ?H3 = ?H?atom(Cl2) the enthalpy of atomisation of chlorine: Cl2(g) ==> 2Cl(g) ?Hatom(Cl) = +242.7 kJ mol-1/2 = 121.35 kJ/mol ?H4 = ?H?el.affin.(Cl) the 1st electron affinity of chlorine: Cl(g) + e- ==> Cl-(g) ?Helec.affin.(Cl) = -349 kJ/mol ?H5 = ?H?1st IE(K) the 1st ionization enthalpy of potassium : K(g) ==> K+(g) + e- ?H1st IE(K) = + 418.7 kJ/mol ?H6 = ?H?LE(KCl) the lattice energy of potassium chloride: K+(g) + Cl-(g) ==> KCl(s) ?HLE(kCl) = -699 kJ/mol This shows how to present and solve a Hess's Law cycle for the formation of an ionic metal chloride and is an example of the so-called Born-Haber Cycle. ?H1 = ?H2 + ?H3 + ?H4 + ?H5 + ?H6 keeping all values : ?Hatom(K) = x = ?H2 = 72.08kJ/mol
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