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One "mole" of N(2) is mixed with three m...

One "mole" of `N_(2)` is mixed with three moles of `H_(2)` in a `4L` vessel. If `0.25% N_(2)` is coverted into `NH_(3)` by the reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`, calculate `K_(c)`. Also report `K_(c)` for
`1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)`

Text Solution

Verified by Experts

`{:(,N_(2)(g),+,3H_(2)(g),hArr,2NH_(3)(g)),("moles at t=0",1,,3,,0),("moles at equilibrium",(1-x),,(3-3x),,2x):}`
`:. K_(c )=((2x//V)^(2))/((1-x//V)(3-3x//V)^(3))=(4x^(2)V^(2))/((1-x)(3-3x)) ...(i)`
Now given `x=0.25//100 and V=4 L`
`:.` By equation (i), `K_(c )=1.49xx10^(-5) L^(2) "mol"^(-2)`
Now for `1//2 N_(2)+3//2 H_(2) hArr NH_(3)`
`K_(c_(1))=([NH_(3)])/([N_(2)]^(1//2)[H_(2)]^(3//2))`
`:. K_(c_(1))=sqrt((K_(c )))=3.86xx10^(-3) L "mol"^(-1)`
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