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

For the reversible reaction,
`N_(2)(g) +3H_(2)(g) hArr 2NH_(3)(g)` at `500^(@)C`, the value of `K_(p)` is `1.44 xx 10^(-5)` when partial pressure is measured in atmosphere. The corresponding value of `K_(c )`, with concentration in mole litre`""^(-1)` , is -

A

`(1.44 xx 10^(-5))/((0.082 xx 500)^(-2))`

B

`(1.44 xx 10^(-5))/((8.314 xx 773)^(-2))`

C

`(1.44 xx 10^(-5))/((8.314 xx 500)^(-2))`

D

`(1.44xx 10^(-5))/((0.082xx 773)^(-2))`

Text Solution

Verified by Experts

The correct Answer is:
D

`Deltan_(g)=n_(p)-n_(r )" " {(n_(p)="number of moles of products"),(n_(r )="number of moles of reactants"):}`
`Deltan_(g) = 2-4`
` = -2`

`K_(c ) = (K_(p))/((RT)^(Deltan))" " ("R in atom " K^(-1)"mol"^(-1))`.
`K_(c )= (1.44 xx 10^(-5))/( (0.082 xx 773)^(-2))`
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