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.

Calculate the heat of sublimation of PI3

Answer 1

Sol :-

Given

ΔH⁰_f of PI₃ (s) = - 24.7 KJ/mol

BDE of P-I bond = 184 KJ/mol

ΔH⁰_f of P(g) = 334 KJ/mol

ΔH⁰_f of I₂ (g) = 62 KJ/mol and

BDE of I-I bond = 151 KJ/mol

Step.1 : Calculation of ΔH⁰_rxn :-

2 P (g) + 3 I₂ (g) -------> 2 PI₃ (g) , ΔH⁰_rxn = ?

ΔH⁰_rxn = BDE of Reactants - BDE of products

ΔH⁰_rxn = [3 x BDE of  I-I bond ] - [ 2 x 3 x BDE of P-I bond ]

ΔH⁰_rxn = [ 3 x 151 KJ/mol ] - [ 6 x 184 KJ/mol ]

ΔH⁰_rxn = 453 KJ - 1104 KJ

ΔH⁰_rxn = - 651 KJ

Step.2 : Calculation of ΔH⁰_f of PI₃ (g) :-

2 P (g) + 3 I₂ (g) -------> 2 PI₃ (g) , ΔH⁰_rxn = - 651 KJ

We know

ΔH⁰_rxn =   ΔH⁰_f of Products -   ΔH⁰_f of Reactants

- 651 KJ = [ 2 x ΔH⁰_f of PI₃ (g) ] - [ 2 x ΔH⁰_f of P (g) + 3 x ΔH⁰_f of I₂(g) ]

- 651 KJ = 2 x ΔH⁰_f of PI₃ (g) - [ 2 x 334 KJ + 3 x 62 KJ ]

- 651 KJ = 2 x ΔH⁰_f of PI₃ (g) - [ 668 KJ + 186 KJ ]

- 651 KJ = 2 x ΔH⁰_f of PI₃ (g) - 854 KJ

2 x ΔH⁰_f of PI₃ (g) = 854 KJ - 651 KJ

2 x ΔH⁰_f of PI₃ (g) = 203 KJ

ΔH⁰_f of PI₃ (g) = 203 KJ / 2

ΔH⁰_f of PI₃ (g) = 101.5 KJ

Step.3 : Calculation of ΔH⁰_sub :-

PI₃ (s) ---------> PI₃ (g) , ΔH⁰_sub = ?

ΔH⁰_sub =   ΔH⁰_f of Products -   ΔH⁰_f of Reactants

ΔH⁰_sub = [ ΔH⁰_f of PI₃ (g) ] - [ ΔH⁰_f of PI₃ (s) ]

ΔH⁰_sub = [ 101.5 KJ/mol ] - [ - 24.7 KJ / mol]

ΔH⁰_sub = 101.5 KJ/mol + 24.7 KJ/mol

ΔH⁰_sub = 126.2 KJ/mol

Hence Heat of sublimation of PI₃ = 126.2 KJ/mol