Question

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Q1.At 1atp 90%pcl5 dissociated into pcl3 and cl2 then find Kp of the reaction?





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Answer 1

Answer :

Suppose we have 1 mole of $PCl_{5}$.

Number of moles dissociated =  ( 1×90) / 100

                                                 = 0.9

Number of moles left at equilibrium = 1- 0.9

0.9 moles of $PCl_{5}$ will dissociate to give 0.9 moles each of $PCl_{3}$ and $Cl_{2}$ .

Therefore , (see the attachment )

∴ Total number of moles at equilibrium = (1-0.9) + 0.9 + 0.9

                                                                =  1.9

Total pressure at equilibrium = 1 atm (given )

As we know ,

Partial pressure = Mole fraction × total pressure ,

we have, the partial pressures (see the attachment )

According to the law of mass action and the law of equilibrium ,

Kp = (see the formula in attachment )

Kp = ( 0.47× 0.47 ) / 0.05

    = 4.418

    = 4.5

Answer 2

Answer:

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