One mole of `N_(2)` (g) is mixed with 2 moles of `H_(2)` (g) in a 4 litre vessel. If 50% of `N_(2)` (g) is converted to `NH_(3)` (g) by the following reaction:
`N_(2)(g) + 3H_(2) (g) hArr 2NH_(3) (g)`
What will be the value of `K_(c )` for the following equilibrium ?
`NH_(3) (g) hArr (1)/(2) N_(2) (g) +(3)/(2) H_(2) (g)`

Options

A. 256
B. 16
C. `(1)/(16)`
D. None of these

Answer

C

Explanation

`{:(,N_(2)(g) , +, 3H_(2)(g) ,hArr , 2NH_(3)(g)), ("Initial moles", 1,,2,,0),("At equilibrium",1-x,,2-3x,,2x):}`
where x=0.5
`K_(c ) =((1//4)^(2))/( ((0.5)/(4)) ((0.5)/(4))^(2))=256`
Equilibrium constant for the reaction
When the reaction is reversed equilibrium constant becomes inverse of initial.
and when the reaction is halfed equilibrium constant becomes root of initial.
`NH_(3)(g) hArr (1)/(2) N_(2) (g) +(3)/(2) H_(2)(g)`
`K_( c)^(1) =(1)/(sqrt(K_(c )))=(1)/(sqrt(256))=(1)/(16)`