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`K_(p)=0.04 atm` at `899 K` for the equilibrium shown below. What is the equilibrium concentration of `C_(2)H_(6)` when it is placed in a flask at `4.0 atm` pressure and allowed to come to equilibrium?
`C_(2)H_(6)(g) hArr C_(2)H_(4)(g)+H_(2)(g)`

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Correct Answer - `[C_(2)H_(6)]_(eq) = 3.62 atm`
Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
Now, according to the reaction,
`{:(,C_(2)H_(2(g)),harr,C_(2)H_(2(g)),+,H_(2(g))),("Initial conc.",4.0"atm",,0,,0),("At equilibrium",4.0-p,,p,,p):}`
We can write,
`(p_(C_(2)H_(4)) xx p_(H_(2)))/(p_(C_(2)H_(6))) = K_(p)`
`rArr (pxxp)/(4.0-p)= 0.04`
`rArr p^(2) + 0.16 - 0.04`
`rArr p^(2) + 0.16 - 0.04 p`
`rArr p^(2)+ 0.04 p - 0.16 = 0`
Now, `p = (-0.04 +-sqrt((0.04)^(2) - 4 xx 1 xx (-0.16)))/(2 xx 1)`
`= (-0.04 +- 0.80)/(2)`
`= (0.76)/(2)` , (Taking positive value)
Hence, at equilibrium
`[C_(2)H_(6)] - 4 - p = 4 - 0.38`
`= 3.62 "atm"`

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