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For the reaction: 2NOCl(g) hArr 2NO(g) +...

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)`

Text Solution

Verified by Experts

The correct Answer is:
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)`
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