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For the reaction at 800 K N(2)(g)+3...

For the reaction at 800 K
`N_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)`
the ratio of `K_(p)` and `K_(c )` is :
(R = 0.082 L atm `mol^(-1)K^(-1)`)

A

`2.3xx10^(-4)`

B

`3.2xx10^(-6)`

C

`2.3xx10^(4)`

D

`3.2xx10^(6)`

Text Solution

Verified by Experts

The correct Answer is:
A
`K_(p)=K_(c )(RT)^(Delta n)`
or `K_(p)=K_(c )(RT)^(-2)`
`=(K_(c ))/((0.082xx800)^(2))`
or `(K_(p))/(K_(c ))=2.32xx10^(-4)`
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