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 1.5 times the electron gain enthalpy of X(g). The first and second ionization enthalpies of the metal (M) are 1.2 and 2.8 times of the enthalpy of sublimation of M(s). The bond dissociation enthalpy of X2(g) is 0.8 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 -96 kcal/mol, then what is the enthalpy of sublimation (in kcal/mol) of the metal (M)?
Options:
Answer:
Solution:
Let S be the sublimation enthalpy of M.
Given the following relationships:
Ionization Energy 1: IE1 = 1.2S
Ionization Energy 2: IE2 = 2.8S
Dissociation energy of X₂: D(X2) = 0.8 × 1.2S = 0.96S
Lattice energy magnitude: U = 5D = 4.8S
Enthalpy of formation: Δ Hf = 1.5 × (-96) = -144 kcal
Using the Born-Haber cycle equation:
Δ Hf = S + IE1 + IE2 + D(X2) + 2(EA) - U
Substitute the given values and relationships into the equation:
Δ Hf = S + 1.2S + 2.8S + 0.96S + 2(-96) - 4.8S
Combine the terms involving S:
Δ Hf = (1 + 1.2 + 2.8 + 0.96 - 4.8)S - 192
Δ Hf = (1.16)S - 192
Now, substitute the value of Δ Hf:
-144 = 1.16S - 192
Solve for S:
1.16S = 192 - 144
1.16S = 48
S = 48/1.16
S ≈ 41.38 kcal/mol
This corresponds to option (a).
