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For the reaction N(2) (g) + 3 H(2) (g) h...

For the reaction `N_(2) (g) + 3 H_(2) (g) hArr 2 NH_(3) (g) " at " 400 K , K_(p) = 41`
Find the value of `K_(p)` for the following reaction :
` 1/2 N_(2) (g) + 3/2 H_(2) hArr NH_(3) (g) `

A

`6.4`

B

`0.02`

C

50

D

`4.6`

Text Solution

Verified by Experts

The correct Answer is:
A

`K_(p)=([p_(NH_(3))])/([P_(N_(2))][P_(H_(2))]^(3))=41`
`impliesP_(NH_(3))=sqrt(K_(p)[P_(N_(2))][P_(H_(2))]^(3))" "...(i)`
`K'_(p)=([P_(NH_(3))])/([P_(N_(2))]^(1//2)[P_(H_(2))]^(3//2))" "...(ii)`
Putting (i) in (ii), we get
`K'_(p)=sqrt(K_(p)[P_(N_(2))][P_(H_(2))]^(3))=sqrt(41)=6.4.`
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