In a container equilibrium N2O4 (g) leftharpoons 2NO2 (g) is attained at 25°C. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N2O4 at this temperature will be:

Question

In a container equilibrium N2O4 (g) leftharpoons 2NO2 (g) is attained at 25°C. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N2O4 at this temperature will be: (A)  (1/3) (B) (1/2) (C) (2/3) (D) (1/4)

Tardigrade Admin · Accepted Answer

Kp=0.667 atm =(2/3) atm= (4α2/1-α2); P=(4α2/1-α2). (1/2) SO, (4α2/1-α2) =(4/3) ⇒ 3α2=1α2 So, α2 =(1/4)⇒ α=(1/2)

Q. In a container equilibrium $N_{2} O_{4} (g) ⇌ 2 N O_{2} (g)$ is attained at $25° C .$ The total equilibrium pressure in container is $380$ torr. If equilibrium constant of above equilibrium is $0.667$ atm, then degree of dissociation of $N_{2} O_{4}$ at this temperature will be:

$\frac{1}{3}$

$\frac{1}{2}$

$\frac{2}{3}$

$\frac{1}{4}$

$K_{p} = 0.667 a t m = \frac{2}{3} a t m = \frac{4 α ^{2}}{1 - α ^{2}}; P = \frac{4 α ^{2}}{1 - α ^{2}} . \frac{1}{2}$
$SO, \frac{4 α ^{2}}{1 - α ^{2}} = \frac{4}{3} \Rightarrow 3 α^{2} = 1 α^{2}$
$S o, α^{2} = \frac{1}{4} \Rightarrow α = \frac{1}{2}$

Q. In a container equilibrium N2​O4​(g)⇌2NO2​(g) is attained at 25°C. The total equilibrium pressure in container is 380 torr. If equilibrium constant of above equilibrium is 0.667 atm, then degree of dissociation of N2​O4​ at this temperature will be:

31​

21​

32​

41​