A 0.313 mol sample of PCl5(g) is injected into an empty 3.25 L reaction vessel held at 250 ∘C. Calculate the concentrations of PCl5(g) and PCl3 (g) at equilibrium.

Question

Phosphorus pentachloride decomposes according to the chemical equation PCl 5 (g) ↽− −⇀ PCl 3 (g)+ Cl 2 (g)

𝐾 c =1.80 at 250∘C

Answer 1

This is the balanced chemical equation and equilibrium constant needed to create the ICE table and write the equilibrium constant expression to find equilibrium concentration values of phosphorous pentachloride and phosphorous trichloride:

PCl_5(g) <----> PCl_3(g) + Cl_2(g) K_c = 1.80 @ 250 ^oC

These are the initial conditions @ 250 ^∘C (temperature given to confirm that the K value can be used):

ICE tables are needed to arrange data and calculate equilibrium values for products and reactants given for reactions. When using an equilibrium-constant expression for an equilibrium concentration constant, concentration values in units of moles per liter need to be used.

0.313 mol PCl_5(g) with a volume of 3.25 L

initial M PCl_5(g) = 0.313 mol = 0.0963 M PCl_5(g)

3.25 L

PCl_5(g) <> PCl_3(g) + Cl_2(g)

initial 0.0963 M 0 0

change -x +x +x

equilibrium 0.0963 -x x x

K_{c @250}^o_C = 1.80 = [PCl₃] [Cl₂]

[PCl₅]

1.80 = x²

(0.0963 - x)

1.80 (0.0963 - x) = x²

0.173 - 1.80x = x²

0= x² + 1.80x - 0.173

±x = -b ± √(b²-4ac) when a=1; b=1.80; c=-0.173

2a

±x = -1.80 ± √(1.80²-[(4)(1)(-0.173)]

2(1)

±x = -1.80 ± √(3.24 + 0.693) ₌ -1.80 ± 1.98

2 2

+x = -1.80 + 1.98 = 0.0916 -x = -1.80 - 1.98 = -1.89

2 2

Calculate [PCl_5(g)]_eq & [PCl_3(g)]_eq when Δx = 0.0916

[PCl_5(g)]_eq= 0.0963 - x = 0.00466 M PCl₅

[PCl_3(g)]_eq = x = 0.0916 M PCl₃

*****Note: even though values are rounded to the correct number of significant figures in the explanation, full unrounded values are used for calculations until the final answers are computed and reported