Phosphorus Penta chloride when heated in a sealed tube at 700 K, it un

1.	Phosphorus Penta chloride when heated in a sealed tube at 700 K, it undergoes decomposition as, PCl5(g)⇌PCl3(g)+Cl2(g);KP=38 atm. vapour density of the mixture is 74.25.1) Percentage dissociation of PCl5 may be given as :-
Answer» EXPLANATION. Phosphorus Penta chloride is heated in a sealed tube at 700k. ⇒ PCI₅(g) ⇆ PCI₃(g) + CI₂(g). ⇒ Kp = 38 atm. vapor density of a mixture = 74.25. As we know that, Some concept. Relationship between Degree of Dissociation (α) and vapor density. For dissociation reaction, ⇒ Aₙ(g) ⇆ nA(g). Degree of Dissociation(α) = No of moles dissociated/Total no of moles taken. $\sf \implies \dfrac{D - d}{d(n - 1)} = \dfrac{M_{t} - M_{o}}{M_{o}(n - 1)}$ M(t) = theoretical (calculated) MOLECULAR weight. M(o) = observed (experimental) molecular weight. D = theoretical vapor density. d = OBSERVES vapor density. n = number of moles of product formed from 1 mole reactant. Density of gas mixture = PM/RT. As we know that, vapor density = Molecular weight/2. ⇒ 74.25 = molecular weight/2. ⇒ Molecular weight = 148.5. Molecular weight of PCI₅ = 31 + 35.5 x 5 = 208.5. ⇒ M(t) = PCI₅ = 208.5. ⇒ M(o) = 148.5. Put the value in equation, we get. ⇒ α = 208.5 - 148.5/148.5(2 - 1). ⇒ α = 60/148.5 ⇒ α = 0.404. Percentage dissociation = α x 100. Percentage dissociation = 0.404 x 100 = 40.4. MORE INFORMATION. LE CHATELIER'S PRINCIPAL. (1) = Increase of reactant concentration = shift forward. (2) = Decreases of reactant concentration = shift backward. (3) = INCREASES in pressure = Form more moles to less moles. (4) = Decreases of pressure = Form less moles to more moles.

Phosphorus Penta chloride when heated in a sealed tube at 700 K, it undergoes decomposition as, PCl5(g)⇌PCl3(g)+Cl2(g);KP=38 atm. vapour density of the mixture is 74.25.1) Percentage dissociation of PCl5 may be given as :-

Answer»

EXPLANATION.

Phosphorus Penta chloride is heated in a sealed tube at 700k.

⇒ PCI₅(g) ⇆ PCI₃(g) + CI₂(g).

⇒ Kp = 38 atm.

vapor density of a mixture = 74.25.

As we know that,

Some concept.

Relationship between Degree of Dissociation (α) and vapor density.

For dissociation reaction,

⇒ Aₙ(g) ⇆ nA(g).

Degree of Dissociation(α) = No of moles dissociated/Total no of moles taken.

$\sf \implies \dfrac{D - d}{d(n - 1)} = \dfrac{M_{t} - M_{o}}{M_{o}(n - 1)}$

M(t) = theoretical (calculated) MOLECULAR weight.

M(o) = observed (experimental) molecular weight.

D = theoretical vapor density.

d = OBSERVES vapor density.

n = number of moles of product formed from 1 mole reactant.

Density of gas mixture = PM/RT.

As we know that,

vapor density = Molecular weight/2.

⇒ 74.25 = molecular weight/2.

⇒ Molecular weight = 148.5.

Molecular weight of PCI₅ = 31 + 35.5 x 5 = 208.5.

⇒ M(t) = PCI₅ = 208.5.

⇒ M(o) = 148.5.

Put the value in equation, we get.

⇒ α = 208.5 - 148.5/148.5(2 - 1).

⇒ α = 60/148.5

⇒ α = 0.404.

Percentage dissociation = α x 100.

Percentage dissociation = 0.404 x 100 = 40.4.

MORE INFORMATION.

LE CHATELIER'S PRINCIPAL.

(1) = Increase of reactant concentration = shift forward.

(2) = Decreases of reactant concentration = shift backward.

(3) = INCREASES in pressure = Form more moles to less moles.

(4) = Decreases of pressure = Form less moles to more moles.