The following enthalpies of reaction have been obtained at a temperature of 298

Question

The following enthalpies of reaction have been obtained at a temperature of 298 K: Sublimation of solid potassium:K(s) rightarrow K(g) +64 kJ Ionization of gaseous potassium:K(g) rightarrow K^+(g) + e^- +419 kJ Dissociation of fluorine:F_2(g) rightarrow 2 F(g) + 478 kJ Electron attachment (affinity) to fluorine: F(g) + e^- rightarrow F (g)-349 kJ Reaction between potassium and fluorine: 2 K(s) + F2(g) rightarrow 2 KF(s) -874 kJ The lattice energy of solid potassium fluoride: KF(s) rightarrow K^+(g) + F^-(g) ? The lattice energy of KF(s) is: 64 kJ 128 kJ 197 kJ 262 kJ 437 kJ 810kJ 1486 kJ 1620 kJ None of the above; the correct lattice energy is_kJ

Explanation / Answer

Let's start by reversing (5) so we get KF(s) as a reactant:

(5) 2KF(s) ---> 2K(s) + F2(g) H = 874 kJ
Now get K(s) to K+(g) by adding (1) and (2):
K(s) ---> K+(g) + e- H = 64 + 419 = 483 kJ
Add these together to get:
2KF(s) ---> 2K+(g) + 2e- + F2(g) : H = 874 + 483 = 1357 kJ
Now let's convert F2 to 2F atom (half of equation 3 )
F2(g) ---> 2F(g) : H = 478 kJ
And add to get:
KF(s) ---> K+(g) + e- + F(g) : H = 478 + 1357 = 1835 kJ
Finally, we add equation to get:
KF(s) ---> K+(g) + F-(g) : H = 1835 + (-349) = 1486 kJ

answer is 1486 KJ

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