In the following reaction started only with `A_(B), 2A_(B)(g) hArr3A_(2)(g)+A_(4)(g)` mole fraction of `A_(2)` is found to `0.36` at a total pressure of `100 atm` at equilibrium. The mole fraction of `A_(8)(g)` at equlibrium is `:`
A
`0.28`
B
`0.72`
C
`0.18`
D
None of these
Text Solution
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The correct Answer is:
1
`{:(,2A_(B),hArr,2A_(3),+,3A_(2),+,A_(4)),(t=0,2,,0,,0,,0),(t=t_(aq),2-2alpha,,2alpha,,3alpha,,alpha):}` `n_(r)=2+4alpha` given mole fraction of `A_(2)` is `=0.36`. `0.36=(3alpha)/(2+4alpha)` `alpha=0.46` Mole fraction of `A_(B)=(2-2alpha)/(2+4alpha)=(2-2xx0.46)/(2+4xx0.46)=0.28`
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