Question

25 moles of NOCl were placed in a 2.50 L reaction chamber at 427ºC. After equilibrium was reached, 1.10 moles of NOCl remained. Calculate the equilibrium constant, Kc, for the reaction 2NOCl(g) ↔ 2NO(g) + Cl2(g).

Answer 1

2NOCl(g) ↔ 2NO(g) + Cl2(g)

intially 10 0 0

change -2x +2x +x

equilitium 10-2x 2x x

intially concentration of NOCl = 25 moles/2.5L = 10 moll/L

change concentration of NOCl = 1.1 moles/2.5L = 0.44 moll/L

2x=0.44 , x=0.22

equilitium [NOCl] = 10-2*0.22=9.56, [NO]=2*0.22=0.44, [Cl₂]=0.22

K_c= [NO]²[Cl₂] / [NOCl]² = (0.44)² * 0.22 / (9.56)² = 4.66*10^-4