Chemistry - Thermochemistry Question with Solution | TestHub
For an ionic solid MX2, where X is monovalent, the enthalpy of formation of the solid from M(s) and X2(g) is 1.2 times the electron gain enthalpy of X(g). The first and second ionization enthalpies of the metal (M) are 1.0 and 2.0 times of the enthalpy of sublimation of M(s). The bond dissociation enthalpy of X2(g) is 0.5 times the first ionization enthalpy of metal and it is also equal to one-fifth of the magnitude of lattice enthalpy of MX2. If the electron gain enthalpy of X(g) is -100 kcal/mol, then what is the enthalpy of sublimation (in kcal/mol) of the metal (M)?
Options:
Answer:
Solution:
Let S = sublimation enthalpy of M. Then IE1 = S, IE2 = 2S, D(X2) = 0.5S, U = 5D = 2.5S, and ΔHf = 1.2 × (-100) = -120 kcal. Born-Haber: ΔHf = S + IE1 + IE2 + D + 2(EA) - U = (1 + 1 + 2 + 0.5 - 2.5)S + 2(-100) = 2S - 200. So -120 = 2S - 200, giving S = 40 kcal/mol. This is option (b).
