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Question
Kp = 0.04 atm at 899 K for the equilibrium shown below. What is the equilibrium concentration of C2H6 when it is placed in a flask at 4.0 atm pressure and allowed to come to equilibrium?
\[\ce{C2H6 (g) ⇌ C2H4 (g) + H2 (g)}\]
Solution
Let p be the pressure exerted by ethene and hydrogen gas (each) at equilibrium.
Now, according to the reaction,
| C2H6(g) | ↔ | C2H4(g) | + | H2(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"`
`=> ("p" xx "p")/(4.0 - "p") = 0.04`
`=> "p"^2 + 0.16 - 0.04 "p"`
`=> "p"^2 + 0.04 "p" - 0.16 = 0`
Now `"p" = (- 0.04 +- sqrt((0.04)^2 - 4xx1xx (-0.16)))/(2xx1)`
`= (-0.04 +- 0.80)/2`
`= 0.76/2` (Taking positive value)
= 0.38
Hence, at equilibrium,
`["C"_2"H"_6] - 4 - "p"`
= 4 - 0.38
= 3.62 atm
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