At 400K, in a 1.0L vessel `N_2O` is allowed to attain equilibrium `N_(2)O_(4(g))hArr 2NO_(2(g))`. At equilibrium the total pressure is 600 mm Hg, when `20%` of `N_2O_4` is disscociated. The `K_rho` value for the reaction is

Options

A. 50
B. 100
C. 150
D. 200

Answer

B

Explanation

`K_rho =(P^2NO_2)/(PN_2O_4)`
`underset(1)(N_(2)O_(4))=underset(0)(2NO_2)`
At equilibrium
Total moles `=0.4 +0.8=1.2`
`K_(rho)=(((0.4)/(1.2)xx600)^2)/(((0.8)/(1.2)xx600)^2)=100`