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For the reversible reaction , N(2)(g)+...

For the reversible reaction ,
`N_(2)(g)+3H_(2)(g)hArr2NH_(3)(g)`
at `500^(@)C` the value of `K_(p)` is 1.44xx10^(-5)` when partial pressure is measured in atmospheres .The corresponding value of `K_(c)` with concentration in mole `litre ^(-1)`, is :

A

`1.44xx10^(-5)//(0.082xx500)^(-2)`

B

`1.44xx10^(-5)//(8.314xx773)^(-2)`

C

`1.44xx10^(-5)//(0.082xx773)^(2)`

D

`1.44xx10^(-5)//(0.082xx773)^(-2)`

Text Solution

Verified by Experts

The correct Answer is:
D
`K_(c)=K_(p)(RT)^(-Deltan_(g))`
Since `Deltan_(g)=-2`
`therefore K_(c)=1.44xx10^(-5)//(0.082xx773)^(-2)`
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