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16 mol of PCl(5)(g) is placed in 4 dm^(-...

16 mol of `PCl_(5)(g)` is placed in 4 `dm^(-3)` closed vessel. When the temperature is raised to 500 K, it decompses and at equilibrium, 1.2 mol of `PCl_(5)(g)` remains. What is `K_(c)` value for the decomposition of `PCl_(5)(g)` to `PCl_(3)(g)` and `Cl_(2)(g)` at 500K.

A

0.013

B

0.05

C

0.033

D

0.067

Text Solution

Verified by Experts

The correct Answer is:
C
`Kc = [PCl_(3)][Cl_(2)]/[PCl_(5)] = (0.4/2 xx 0.4/2)/(1.2/4) = 0.4^(2)/(4 xx 1.2) = 0.4/12 = 4(10 xx 12) = 0.033`
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