Question

A 2.30 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

Calculate concentrations first

M = mol/L

then

[NOCl] = 2.3/1.5 = 1.5333

in equilibrium:

[NOCl] = 1.5333 - 2x

[NO] = 0+2x

[Cl2] = 0 + x

and

22% is dissociated so

0.88 of original is left

0.88*(1.5333) = 1.349304 M left in euqilibrium so

if

[NOCl] = 1.5333 - 2x is true

then

1.5333 - 2x = 1.349304

x = (1.349304-1.5333 )/-2 = 0.091998

then

[NO] = 0+2*0.091998 = 0.183996

[Cl2] = 0 + 0.091998 = 0.091998

then

Kc = [NO]^2[Cl2] / [NOCl]^2

Kc = (0.183996^2)(0.091998)/(1.349304^2) = 0.00171070661

Kc = 1.71*10^-3