Answer to Question #189149 in Organic Chemistry for chibi

Question

The two common chlorides of phosphorus, PCI₃ and PCI_5, both important in the production of other phosphorus compounds, coexist in equilibrium through the reaction

PCI₃(g) + Cl₂(g) → PCI₅(g)

At 250°C, an equilibrium mixture in a 2.50 L flask contains 0.105g PCI_{5, }0.220g PCI₃ and 2.12g Cl_2. What are the values of (a) Kc and (b) Kp for this reaction at 250°C?

Show solutions.

Answer 1

a) Kc = [PCl5]/[PCl3][Cl2]

with,

molarity = moles/Volume of solution (L)

moles = g/molar mass

we have,

[PCl5] = 0.105/208.24 x 25 = 2.02 x 10^-5 mols

[PCl3] = 0.220/137.333 x 25 = 6.41 x 10^-5 mols

[Cl2] = 2.12/70.906 x 25 = 1.20 x 10^-3 mols

Feed values in Kc equation,

Kc = (2.02 x 10^-5)/(6.41 x 10^-5)(1.20 x 10^-3) = 262.6105

b) Relation between Kc and Kp,

Kp = Kc(RT)^delta(n)

delta(n) = 1 - (1+1) = -1

R = 0.0821 L.atm/K.mol

T = 250 oC = 250 + 273 = 523 K

Feed values,

Kp = 262.6105 (0.0821 x 523)^-1 = 6.116