`1` mol of nitrogen is mixed with `3` mol of hydrogen in a `4 L` container. If `0.25%` of nitrogen is converted to ammonia by the following reaction
`N_(2)(g)+3H_(2)(g) hArr 2NH_(3)(g)`
then calculate the equilibrium constant `K_(c)` in concentration units. What will be the value of `K_(c)` for the following equilibrium?
`1/2 N_(2)(g)+3/2 H_(2)(g) hArr NH_(3)(g)`

Explanation

`overset(0(N_2)+overset(3)(3H_(2)) Leftrightarrow overset(0)(2NH_(3))" Initial moles"`
`"(1-0.0025) "(3-3xx0.0025)" "(2 xx 0.0025)" Moles at equilibrium"`
`(0.9975)/(4)" "(2.9925)/(4)" "(0.005)/4 " Molar concentration at eqb".`
`K_(c)=([NH_(3)]^(2))/([N_2][H_2]^3)=((0.005)/(4))^(2)/((0.9975)/(4))((2.9925)/4)^3=1.481 xx 10^(-5) ("mol/L")^(-2)`.
Since the reaction `1/2 N_2+3/2H_2 Leftrightarrow NH_(3)` has been obtained by multiplying the equation `N_2+3H_2 Leftrightarrow 2NH_3" by "1/2`, the new equilibrium constant `=(1.481 xx 10^(-5))^(1/2)`
`K_c=3.84 xx 10^(-3) ("mole/litre")^(-1)`.