First in-class problem from 2012-07-05

2.63M NOCl₂(g) → NO(g) + Cl₂(g) K=0.118, find all [ ]_eq

	NOCl2(g) ⇄	NO(g) +	Cl2(g)
[ ]initial	2.63 M	0 M	0 M
Δ [ ]	- x M	+ x M	+ x M
[ ]equilibrium	(2.63 – x) M	x M	x M

Plugging in to the equilibrium constant expression...

Because the value of K is not exceptionally large or exceptionally small, this one will probably require working through the quadratic. Breaking it down in steps...

(x)(x) = 0.118(2.63-x)

x² = 0.31034 – 0.118x

x² + 0.118x + (-0.31034) = 0

Plugging in to the quadratic formula:

NOTE: The other root is negative, it doesn't work in this problem.

[NOCl₂]_eq= 2.63 – 0.50 = 2.13M

[NO]_eq = x = 0.501M

[Cl₂]_eq= x = 0.501M