Chemistry - Thermochemistry Question with Solution | TestHub
The standard molar enthalpies of formation of IF3(g) and IF5(g) are -470 kJ and -847 kJ, respectively. Valence shell electron-pair repulsion theory predicts that IF5(g) is square pyramidal in shape in which all I-F bonds are equivalent while IF3(g) is T-shaped (based on trigonal-bipyramidal geometry) in which I-F bonds are of different lengths. It is observed that the axial I-F bonds in IF3 are equivalent to the I-F bonds in IF5. Calculate the equatorial I-F bond strength (in kJ/mol) in IF3. Some other details given are: I2(s) → I2(g); ΔH = 62 kJ. F2(g) → 2F(g); ΔH = 155 kJ. I2(g) → 2I(g); ΔH = 149 kJ.
Answer:
Solution:
The enthalpy required to obtain one I(g) from I₂(s) is 62 + 149/2 = 105.5 kJ. For each F(g), it is 77.5 kJ.
For IF₅:
The atomization energy is 105.5 + 5(77.5) + 847 = 1340 kJ.
Therefore, each I–F bond energy is 1340/5 = 268 kJ/mol.
For IF₃:
The atomization energy is 105.5 + 3(77.5) + 470 = 808 kJ.
Assuming two axial bonds have the same energy as in IF₅, their combined energy is 2(268) = 536 kJ.
Thus, the equatorial bond energy is 808 - 536 = 272 kJ/mol.
The answer is 0272.
