equilibrium problem- college level general inorganic 2

Question

At a certain temperature, 0.4011 mol of N2 and 1.501 mol of H2 are placed in a 4.50 L container.

N2(g)+3H2(g)↽−−⇀2NH3(g)

At equilibrium, 0.1401 mol of N2 is present. Calculate the equilibrium constant, 𝐾c.

Answer 1

N₂(g) + 3H₂(g) <==> 2NH3(g)

0.0891...0.3336..............0.........Initial

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

0.0891-x...0.3336-3x.....2x........Equilibrium

At equilibrium, 0.1401 mol N2 are present = 0.1401 mol / 4.5 L = 0.0311 M. This is then equal to x.

Equilibrium concentrations:

N2 = 0.0311 M

H2 = 0.3336 - 0.0934 = 0.240 M

NH3 = 2x = 0.0622 M

K = [NH3]² / [N2][H2]³ = (0.0622)² / (0.0311)(0.240)³

K = 8.999 = 9.0

Answer 2

Since you have .1401 moles of N₂ at equilibrium, the amount of N₂ reacted is .4011-.1401 = .261 moles. The amount of H₂ at equilibrium will be 1.501 - 3(.261) = .718 moles. The amount of NH₃ at equilibrium is 2(.261)= .522 moles.

The equilibrium concentrations are the moles divided by 4.5 liters.

The Equilibrium constant expression is K_C = C_NH3²/(C_H2³C_N2)