Consider the reaction where `K_(p)=0.497` at 500 K.
`PCl_(5)(g)hArr PCl_(3)(g)+Cl_(2)(g)`
If the three gases are mixed in a rigid container so that the partial pressure of each gas is initially 1 atm, which is true ?

Options

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

Answer

A

Explanation

`PCl_(5)hArr PCl_(3)+Cl_(2)`
`K_(p)=0.497`
`Q=(P_(PCl_(3))xx P_(Cl_(2)))/(P_(PCl_(5)))=(1xx1)/(1)=1`
Since `Q gt K_(p)`, the reaction will proceed in backward direction. Hence, more `PCl_(5)` will be produced.