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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?

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प्रश्न

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?

\[\ce{C2H6 (g) ⇌ C2H4 (g) + H2 (g)}\]

संख्यात्मक
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उत्तर

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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Law of Chemical Equilibrium and Equilibrium Constant
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अध्याय 6: Equilibrium - EXERCISES [पृष्ठ २३३]

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एनसीईआरटी Chemistry Part 1 and 2 [English] Class 11
अध्याय 6 Equilibrium
EXERCISES | Q 7.17 | पृष्ठ २३३

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