2.0 mol each of PCl5, PCl3, and Cl2 are placed in a 2.0L reaction vessel at...

2.0 mol each of PCl₅, PCl₃, and Cl₂ are placed in a 2.0L reaction vessel at 620K. K_c for the reaction PCl₅(g) ↔ PCl₃(g) + Cl₂(g) is 0.60 at the given temperature. What is the equilibrium concentration of Cl₂? 1.146 is incorrect.

Expert Solution

PCl₅(g) ↔ PCl₃(g) + Cl₂(g)

I 2 2 2

C +x -x -x

E 2+x 2-x 2-x

[PCl5] = 2+x/2

[PCl3] = 2-x/2

[Cl2] = 2-x/2

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

0.6 = (2-x)(2-x)/2(2+x)

0.6*2(2+x) = (2-x)(2-x)

x = 0.328

[Cl2] = 2-x/2 = 2-0.328/2 = 0.836M >>>>answer

queen_honey_blossom answered 2 years ago

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