Question

Phosphorus pentachloride decomposes according to the chemical equation

PCL_{5} <-> PCL_{3} + CL_{2}

A 0.382 mol sample of PCl5(g) is injected into an empty 4.20 L reaction vessel held at 250 °C. Calculate the concentrations of PCl5(g) and PCl3(g) at equilibrium.

Please help, I have tried this problem three times and I still do not seem to be getting the correct answers. I would like the explanation as well, please. I will rate your answer!

Thanks in advance.

Answer #1

Kc = [PCl3]¹ x [Cl2]¹ / [PCl5]¹

for this rxn.

and we know that

[PCl5] initial = 0.382mol / 4.20L = 0.091M

and if we say X mols / L of PCl5 were consumed to reach
equilibrium

then AT equilibrium we have this concentration of PCl5
remaining

[PCl5] = 0.091M - X

and because of the 1:1:1 ratios in the balanced equation, we've
produced

[PCl3] = X

[O2] = X

so that..

Kc = [PCl3]¹ x [Cl2]¹ / [PCl5]¹ = X x X / (0.091M - X)

Kc = X² / (0.091M - X)

and now we sub in Kc and solve

1.80 = X² / 0.091

nd now we sub in Kc and solve

1.80 = X² / (0.091 - X)

0.119 - 1.80X = X²

X² + 1.80X - 0.119= 0

solve however you like. I'll use the quadratic formula

X = (-b ±√ (b² - 4ac) ] / (2a)

X = (-1.80 ±√ (1.80² - (4) x (1) x (-0.119)) ] / (2 x 1)

X = (-1.80 ± 1.93) / 2

**X = 0.065M = [PCl3] = [Cl2]**

the other root (-1.80 - 1.93) gives a negative concentration so
is not meaningful.

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