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A quantity of PCl5 was heated in a 10 dm...

A quantity of `PCl_5` was heated in a `10 dm^3` vessel at `250^@C`. At equilibrium, the vessel contains 0.1 mole of `PCl_5` and 0.2 mole of `Cl_2`.The equilibrium constant for the reaction is

A

`0.05 `

B

`0.02 `

C

`0.03 `

D

`0.04 `

Text Solution

Verified by Experts

The correct Answer is:
D
`{:(PCl_5, `K_c = ([PCl_3][Cl_2])/([PCl_5]) = (0.02 xx 0.02)/(0.01) = 0.04`
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