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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 produced

C

Equilibrium 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=(""_(p)PCl_3xx""_(p)Cl_2)/(""_(p)PCl_5)=(1"atm"xx1"atm")/(1"atm")=1"atm"`
Since `Q_p gt K_p` , the equilibrium shifts towards left.
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