Question

Phosphorus pentachloride decomposes according to the chemical equation

PCl5(g)−⇀↽−PCl3(g)+Cl2(g) Kc=1.80 at 250

A 0.1414 mol sample of PCl5(g) is injected into an empty 2.00 L reaction vessel held at 250 ∘C.

Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Answer 1

Solution:-

Initial [PCl₅] = moles / volume = (0.1414 mol) / (2.00 L) = 0.0707 M

	PCl5 (g)	PCl3 (g)	Cl2 (g)
Intial	0.0707	0	0
Change	-x	​​+x	+x
Equilbrium	(0.0707 - x)	x	x

Now,

Kc = [PCl₃] [Cl₂] / [PCl₅]

1.80 = (x.x) / (0.0707 - x)

1.80 = x² / (0.0707 - x)

0.1273 - 1.80x = x²

^{​​​​​​}or,

x² + 1.80x - 0.1273 = 0 ................................(i)

Using quadratic formula,

±x = [(-b) ± √(b² - 4ac)] / 2a

From eq.(i) ,

a = 1 ; b = 1.80 ; c = -0.1273

±x = [(-1.80) ± √(1.80)² - 4 (1) (-0.1273)] / 2 (1)

±x = [(-1.80) ± √(3.34 + 0.509)] / 2  

±x = [(-1.80) ± 1.93623] / 2

x = +0.0681 or, x = -1.868

Using positive value of x,

x = 0.0681 M

Thus, at equilibrium,

[PCl_{3 (g)}] = x = 0.0681 M

[PCl₅ (g)] = (0.0707 - x) = (0.0707 - 0.0681) = 0.0026 M

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