The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction PCl5(g) arrow PCl3(g)+Cl2(.g.) . Assuming ideal behavior of all gases, calculate the density of equilibrium mixture of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.5)

Question

The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction PCl5(g) arrow PCl3(g)+Cl2(.g.) . Assuming ideal behavior of all gases, calculate the density of equilibrium mixture of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.5) (A) 4.30 (B) 3.54 (C) 2.64 (D) 1.56

Tardigrade Admin · Accepted Answer

beginarraycccc PCl 5( g ) leftharpoons PCl 3( g )+ Cl 2( g ) t =0 a 0 0 t eq 0.6 a 0.4 a 0.4 a endarray nTotal=1.4a ∴ Ma v=(208 a/1.4 a) =148.57 ∴ Pm i x=(PMa v/R T) =(1 × 148.57/0.0821 × 420)=4.30

Q. The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction

$PC l_{5} (g) \to PC l_{3} (g) + C l_{2} (g)$ . Assuming ideal behavior of all gases, calculate the density of equilibrium mixture of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.5)

4.30

3.54

2.64

1.56

$t = 0 t_{e q} PC l_{5} (g) a 0.6 a ⇌ PC l_{3} (g) + C l_{2} (g) 0 0.4 a 0 0.4 a$

$n_{T o t a l} = 1.4 a$

$∴ M_{a v} = \frac{208 a}{1.4 a} = 148.57 ∴ P_{mi x} = \frac{P _{M_{a v}}}{RT} = \frac{1 \times 148.57}{0.0821 \times 420} = 4.30$

Q. The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction PCl5​(g)→PCl3​(g)+Cl2​(g) . Assuming ideal behavior of all gases, calculate the density of equilibrium mixture of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.5)