1.25mole Nocl was placed in a2.50l reaction chamber at 427c. After equilibrium was reached .1.10moles of Nocl remained. Calculate the equilibrium constant Kc for the reaction. 2Nocl(g)=2No(g)+cl2(g)

Kc → 5.58×10⁻⁴

Explanation:

Equilibrium reaction is:

2NOCl (g) ⇄  2NO (g) + Cl₂(g)

Initially we have 1.25 moles of NOCl

After the equilibrium, we have 1.10 moles. So, during the process:

(1.25 mol - 1.1 mol) = 0.15 moles have reacted.

As ratio are 2:2, and 2:1, 0.15 moles of NO and (0.15 /2) = 0.075 moles of chlorine, were produced in the equilibrium.

Finally in equilibrium we have: 1.10 moles of NOCl, 0.15 moles of NO and 0.075 moles of Cl₂. But these amount are not molar, so we need molar concentration in order to determine Kc:

1.10 mol /2.50L = 0.44 M

0.15 mol / /2.50L = 0.06 M

0.075 mol /2.50L = 0.03 M

Let's make expression for Kc → [Cl₂] . [NO]² / [NOCl]²

Kc = (0.03 . 0.06²) / 0.44² → 5.58×10⁻⁴

i believe the answer is three water molecules