At 1000 K, the equilibrium constant, `K_(c)` for the reaction `2NOCl(g) hArr 2NO(g) +Cl_(2)(g)` is `4.0xx10^(-6) mol L^(-1)`. The `K_(p)` (in bar) at the same temperature is `(R=0.083" L bar K"^(-1) mol^(-4))`

Options

A. `3.32xx10^(-6)`
B. `3.32xx10^(4)`
C. `3.32xx10^(-4)`
D. `3.32xx10^(-3)`

Answer

C

Explanation

Given, `K_(c)=4xx10^(-6)" mol/L"`
`2NOCl_(g) hArr 2NO(g) +Cl_(2) (g)`
`Delta n`= product mole - reactant mole
=3-2=1
From `K_(p)` equation
`K_(p) =K_(c) xx (RT)^(Delta n)`
`=4xx10^(-6)" mol/L"xx (0.0831L" bar k"^(-1)mol^(-1)xx1000 K)^(2)`
`K_(p) =3.32xx10^(-4)` bar