The reaction: PCl5(g) <-> PCl3(g) + Cl2(g) has Kc=0.0900. A 0.1000 mol sammple of PCl5 is...

The reaction: PCl₅(g) <-> PCl₃(g) + Cl₂(g) has K_c=0.0900. A 0.1000 mol sammple of PCl₅ is placed in an empty 1.00 flask and the above reaction is allowed to come to equilibrium at a certain temp. How manny moles of PCl_5, PCl₃ and Cl₂, respectively, are present at equilibrium?

Expert Solution

PCl₅(g) <-> PCl₃(g) + Cl₂(g)

I 0.1 0 0

C -x +x +x

E 0.1-x +x +x

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

0.09 = x*x/(0.1-x)

0.09*(0.1-x) = x^2

x = 0.06

[PCl3] = x = 0.06moles

[Cl2] = x = 0.06 moles

[PCl5] = 0.1-x = 0.1-0.06 = 0.04 moles

queen_honey_blossom answered 2 years ago

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