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Consider the reaction where K_(p)=0.497at 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 ? |
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Answer» <P>More `PCl_(5)` will be produced `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. |
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