For the reaction : `2NO(g) + Cl_(2) (g) hArr 2NOCl (g), NO` and `Cl_(2)` are initially taken in mole ratio `2 : 1`. The total pressure at equilibrium is found to be `1 atm`. If mole of `NOCl` are one fourth of that of `Cl_(2)` at equilibrium, `K_(p)` for the reaction is:

Options

A. `13//36`
B. `13//256`
C. `13//512`
D. `13//128`

Answer

B

Explanation

`rho alpha n` at constant volume of temperature.
`2NO(g) + CL_(2) (g) hArr 2NOCl (g)`
`{:(2p,p,O,),(2p - 2x,p - x,2x,):}`
`p_(T) = 2p - 2x + p - x + 2x = 1`
`3p - x = 1` & `2x = (p - x)/(4)`
Solving `x = (1)/(26` & `p = (9)/(26)`
`k_(p) = (2(x)^(2))/((p - x) (2p - 2x)^(2))`