Problem 3 (2012-07-09)

14.78g PCl₅(g) and 10.15g O₂(g) are combined in a 1.500L vessel and reach equilibrium with POCl₃(g) and ClO(g). If K = 8.53x10^-9, find all equilibrium concentrations.

	2 PCl5(g) +	3 O2(g) ⇄	2 POCl3(g) +	4 ClO(g)
[ ]initial	(14.78g / 208.239g/mol) / 1.500L = 0.047317M	(10.15g / 31.998g/mol) / 1.500L = 0.21147M	0 M	0 M
Δ [ ]	– 2x M	– 3x	+ 2x	+ 4x M
[ ]equilibrium	(0.047317 – 2x) M	(0.21147 – 3x) M	2x M	4x M

Plugging in to the equilibrium constant expression...

Solving this directly would be rough, so let's try a simplifying approximation. Since the equilibrium is quite reactant-favored, we can assume that 2x is small compared to 0.047317 and 3x is small compared to 0.21147. We need to check this later, but that will simplify the equilibrium constant expression to:

x⁶ = 1.76376x10^-16

x = 0.002368

BEFORE WE GO ANY FARTHER, CHECK THE ASSUMPTION WE MADE!! 2x = 0.0047, this is (0.0047/0.047317)*100 = 9.9%. This is a little too high for this assumption to work well. Oops. Don't worry, I don't expect anyone to solve a 6^th order polynomial on an exam, these numbers are a little off because I made this problem up during class. For numbers that work, try using K = 8.53x10^-12. , then x = 7.489x10^-4 and the assumption is OK. If you'd like to know how to solve the original problem, you could use the method of successive approximations, this was how I had to treat these problems when I was an undergrad taking Gen Chem.

Exam tomorrow, if you have questions let me know, I should be online until at least 7 or 8pm tonight.