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Consider thr reaction where K(p)=0.497 a...

Consider thr reaction where `K_(p)=0.497 `at 500K
`PCl_(5)(g)hArrPCl_(3)(g)+Cl_(2)(g)`
If the htree gasses are mixed in a right container so that the partial pressure of each gas in initially 1 atm ,then which is correct observation ?

A

More `PCl_(5)` will be produced

B

More `PCl_(3)` will be proudced

C

Equlibrium will be established when `50%` reaction is complete

D

None of the above

Text Solution

Verified by Experts

The correct Answer is:
A
`Q_(p)=(pPCl_(3)xxpCl_(2))/(pPCl_(5))=(1atmxx1atm)/(1atm)=1atm`
Since, `Q_(p)gtK_(p),` the equlibrium shifts towards left.
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