Calculate the lattice energy from the following data(given 1 eV = 23.0 kcal mol-1) i. ΔfH⊖(KI)=−78.0 kcal mol−1ii. IE1 of K=4.0eV iii. ΔdissH⊖I2=28.0 kcal mol−1iv. ΔsubH⊖(K)=20.0 kcal mol−1v. EA of I=−70.0 kcal mol−1vi. ΔsubH⊖ of I2=14.0 kcal mol−1
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a
+14.1 kcal mol−1
b
−14.1 kcal mol−1
c
−141 kcal mol−1
d
+141 kcal mol−1
answer is C.
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Detailed Solution
Using Born-Haber cycle for KI. (All values are in k cal mol-1)∴ΔfH⊖=ΔsubH⊙+IE1+12ΔsubH⊙+12ΔdissH⊖+EA+ΔuH⊖−78=20.0+23.0×4.0+12×14.0+12×28.0−70.0+ΔuH⊖∴ΔuH⊙=−141 kcal mol−1