Chem help please

Question

At a certain temperature, 0.4011 mol of N2 and 1.561 mol of H2 are placed in a 4.50 L container.

N2(g)+3H2(g)↽−−⇀2NH3(g)

At equilibrium, 0.1601 mol of N2 is present. Calculate the equilibrium constant, 𝐾c .

Answer 1

Convert mols to mols/liter:

0.4011 mol N₂ / 4.50 L = 0.0891 M N₂

1.561 mol H₂ / 4.50 L = 0.3469 M H₂

@equilibrium N₂ = 0.1601 mol / 4.50 L = 0.0356 M N₂

N₂(g) + 3H₂(g) <===> 2NH₃(g)

0.0891..0.3369............0...........Initial

-x........-3x..................+2x.........Change

0.0891-x..0.3369-3x...2x..........Equilibrium

0.0891-x = 0.03562 M

x = 0.0535 M

Summary @ equilibrium:

[N₂] = 0.0891 - 0.0535 = 0.0356 M

[H₂] = 0.3369 - (3x0.0535) = 0.176 M

[NH₃] = 2 x 0.0535 = 0.107 M

Keq = [NH₃]² / [N₂][H₂]³

Keq = (0.107)² / (0.0356)(0.176)³

Keq = 59.0