Calculate the equilibrium constant, Kc.

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.

Kc=??

Answer 1

N₂(g) + 3H₂(g) ==> 2NH₃(g)

0.4011.....1.501...............0..........Initial

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

0.4011-x....1.501-3x........2x........Equilibrium

To find x, we know 0.4011-x = 0.1401 mol, so x = 0.2610 moles

At equilibrium, we have the following concentrations:

[N₂] = 0.1401 mols / 4.50 L = 0.0311 M

[H_2] = 1.501 - (3 x 0.1401) = 1.501 - 0.4203 = 1.081 mol / 4.50 L = 0.240 M

[NH₃] = 2 x (0.1401) = 0.2802 mol / 4.5 L = 0.0623 M

Kc = [NH₃]² / [N₂][H₂]³

Kc = (0.0623)² / (0.0311)(0.240)³

Kc = 0.00388 / 0.0004299

Kc = 9.03

(be sure to check the math)