Calculate the value of [n2]eq if [h2]eq = 2.0 m, [nh3]eq = 0.5 m, and kc = 2.n2(g) + 3 h2(g) ↔ 2 nh3(g)0.062 m62.5 m0.40 m0.016 m0.031 m

[ N₂(g) ]  = 0.016 M

Explanation:

N₂(g) + 3 H₂(g) ↔ 2 NH₃(g)

The equilibrium constant for the above reaction , can be written as the product of the concentration of product raised to the power of stoichiometric coefficients in a balanced equation of dissociation divided by the product of the concentration of reactant raised to the power of stoichiometric coefficients in the balanced equation of dissociation .  

Hence ,  

Kc = [ NH₃ (g) ]² / [ N₂(g) ]  [ H₂(g) ]³

From the question ,

[ NH₃ (g) ] = 0.5 M

[ N₂(g) ] = ?

[ H₂(g) ] = 2.0 M

Kc = 2

Now, putting it in the above equation ,  

Kc = [ NH₃ (g) ]² / [ N₂(g) ]  [ H₂(g) ]³

2 =  [ 0.5 M ]² / [ N₂(g) ]  [ 2.0 M ]³

[ N₂(g) ]  = 0.016 M .

14,16 18 reacts with group