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A mixture of N_(2)and H_(2) in the molare ratio 1:3 attains equlibrium when 50% of mixture has reacted. If P is the total pressue of the mixture, the partial pressure of NH_(3) formed is P//y. The value of y is

Answer»


Solution :`N_(2)+3H_(2)hArr2NH_(3)`
`{:("Initial mole",1,3,0),("At eq.",1-x,3-3x,2x):}`
`therefore` out of 4 mole of MIXTURE, 2 mole have REACTED and hence 2 mole are LEFT.
`therefore1-x+3-3x=2`
or `4x=2or x=0.5`
`therefore"Total mole atequlibrium"=1-x+3-3x+2x=4-2x=4-2xx0.5=3`
No. of mole of `NH_(3)` at equlibrium `2x=2xx0.5=1`
`thereforeP_(NH_(3))=1/3xxP=P/3thereforey=3`


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