Question

A 2.40 mole quantity of NOCl was initially in a 1.50 L reaction chamber at 400°C. After equilibrium was established, it was found that 32.0 percent of the NOCl had dissociated:

Answer 1

2NOCl --> 2NO + Cl2
2.4moles NOCl in 1.5L = 1.6M NOCl

32% dissociation
32% of 2.4 moles undergoes dissociation leaving 0.768 moles NOCl in ion form (dissociated).
0.768moles NOCl / 1.5L = 0.512M NOCl = 0.512M NO and 0.512M Cl2
0.512 = x for the ICE chart
......2[NOCl]....2[NO]......[Cl2]
.......1.6............0.............0
.....-2x..............+2x........+x
...1.6-2x...........2x...........x

[NOCl] = 1.6 - 2x = 1.6 - 1.024 = 0.576M
[NO] = +2x = 1.024M
[Cl2] = x = 0.512M

Kc = [NO]^2[Cl2]/[NOCl]^2
Kc = (1.024)^2(0.512) / (0.576)^2
Kc = 1.62