1.

At constant temperature and 1 atm pressure, PCl5 dissociate 2% then at what pressure PCl5 dissociate4% :- (Use PCl_(5)(g) hArrPCl_(3)(g) + Cl_(2)(g))

Answer»

`1/8 ` atm
`1/(16) atm`
`(1)/(2) atm`
`1/4 atm`

Solution :`{:(PCL5 , hArr,PCl 3, + ,CL2),( 1,,0,,0),(1-alpha , ,alpha ,, alpha ):}`
`KP=( alpha^(2))/((1-alpha ))xx[(p)/((1+alpha))]^(1)`
`KP= (alpha^(2) P)/(1-alpha ^(2)) ~~- alpha ^(2)P `
` alpha_(1) ^(2)P_(1)= alpha _(2) ^(2)P_(2) `
`( 0.02 ) xx 1 = ( 0.04 ) ^(2).Pa`
`impliesP2= 0.25atm `
`=(1)/(4) atm`


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