Question

At 500 K, a 10.0 L equilibrium mixture contains 0.424 mol N2, 1.272 mol H2, and 1.152 mol NH3. The mixture is quickly chilled to a temperature at which the NH3 liquefies, and the NH3(l) is completely removed. The 10.0 L gaseous mixture is then returned to 500 K, and equilibrium is re-established.

How many moles of NH3(g) will be present in the new equilibrium mixture?
N2(g)+3H2(g)→2NH3(g)

Express your answer to three significant figures and include the appropriate units.

Answer 1

You didnt provide the value of Kc for the reaction at that temperature. So i am gonna solve upto that part where substuting Kc value gives the answer.

N₂ + 3H₂ = 2NH₃

conc. of N2 at = 0.424/10 = 0.0424

conc. of H2 = 1.272/10 = 0.1272

conc of NH3 = 1.152/10 = 0.1152

The above values are at equillibrium 1.

Now NH3 is completely removed so another equillibrium will be reached

I 0.0424 0.1272 0

C -x -3x +2x

E 0.042-x 0.1272-3x 2x

Kc = [NH₃]²/[N₂][H₂]³

Kc = 4x²/(0.042-x)(0.1272-3x)³

^{Use can use this link to solve the cubic equation :} http://www.1728.org/cubic.htm