Question

For the reaction: `2NOCl(g) hArr 2NO(g) +Cl_(2)(g), K_(c)` at `427^(@)C` is `3xx10^(-6) L mol^(-1)`. The value of `K_(p)` is
A. `7.50xx10^(-5)`
B. `2.50xx10^(-5)`
C. `2.50xx10^(-4)`
D. `1.72xx10^(-4)`

Answer 1

Correct Answer - D
`2NOCl_((g)) = 2NO_((g)) + Cl_(2(g))`
`K_(p) = K_(c)(RT)^(Delta n)`
`K_(p) = 3xx10^(-6)(0.0821xx700)`
`= 172.41xx10^(-6) = 1.72xx10^(-4)`