The standard heat of formation of PI3(s) is -24.7 kJ/mol and the PI bond energy

Question

The standard heat of formation of PI3(s) is -24.7 kJ/mol and the PI bond energy in this molecule is 184 kJ/mol. The standard heat of formation of P(g) is 334 kJ/mol and that of I2(g) is 62 kJ/mol. The I2 bond energy is 151 kJ/mol.tion of CaBr2 is -675 kJ/mol. The first ionization energy of Ca is 590 kJ/mol and its second ionization energy is 1145 kJ/mol. The heat of sublimation of Ca[Ca(s)Ca(g)] is 178 kJ/mol. The bond energy of Br2 is 193 kJ/mol, the heat of vaporization of Br2(l) is 31 kJ/mol, and the electron affinity of Br is -325 kJ/mol.

I need to calculate the heat of sublimation of PI3 in kJ/mol, but I have no idea where to start. Could someone show their work so I know where to go from here? Please help! Thank you in advance for your assistance and time.

Explanation / Answer

Sublimation if    solid to gas directly

PI3(s) ---> PI3 (g)   is equation for sublimation

Heat of sublimation = dH of reaction = dH PI3 ( g) - PI3 (s)

thus we nee to first find PI3( g) enthalphy of formation

the reaction is   P (g) + ( 3/2) I2 ( g) ----> PI3 (g)

Enthaphy of reaction = Bond energies of bonds broken - bond energies of bonds formed

                       =   3/2 BE ( I-I)   - 3 BE ( P-I)   = ( 3/2) 151 - 3( 184)   = -325.5 KJ

Now   for reaction   dH reaction = dH products - dH reactants

        - 325.5 = dH ( PI3 g) - dH ( P) - 3/2 dH ( I2)

-325.5 = dH ( PI3 g)   - 334 - ( 3/2) ( 62)

dH ( PI3 g) = 101.5 KJ /mol

Now heta of sublimation dH = dH ( PI3 g) - dH ( PI3 s) = 101.5 - ( -24.7) = 126.2 KJ/mol

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