Question

A 1.90 mole quantity of NOCl was initially in a 1.50 L reaction chamber at 400°C. After equilibrium was established, it was found that 22.0 percent of the NOCl had dissociated: 2NOCl(g) 2NO(g) + Cl2(g) Calculate the equilibrium constant Kc for the reaction.

Answer 1

20% of NOCl dissociate

1.9*20/100 = 0.38 moles dissociate

remaining 1.9-0.38 = 1.52 moles

[NOCl] = 1.52/1.5 = 1.013M

from the equation
0.38 moles of NOCl will produce 0.38 moles of NO

0.38 moles of NO in 1.5 L

[NO] = 0.38/1.5 = 0.26m

0.38 moles of NOCl produces the 0.19 moles of Cl2

[Cl2} = 0.19/1.5 = 0.126M

2NOCl(g) ----->2NO(g) + Cl2(g

Kc = [NO]²[Cl2]/[NOCl]²

    = 0.26*0.26*0.126/1.013 = 0.0084 >>> answer