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What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M? 2ICl(g)↽−−⇀IX2(g)+ClX2(g); KC = 0.14 - Chemistry

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प्रश्न

What is the equilibrium concentration of each of the substances in the equilibrium when the initial concentration of ICl was 0.78 M?

\[\ce{2 ICl(g) ⇌  I2(g) + Cl2(g)}\]; KC = 0.14

संख्यात्मक

उत्तर

The given reaction is:  

  2 ICl(g) I2(g) + Cl2(g)
Initial conc. 0.78 M   0   0
At equilibrium (0.78 - 2x) M   x M   x M

Now we can write, `(["I"_2]["Cl"_2])/["ICl"]^2 = "K"_"C"`

`=> (x xx x)/(0.78  - 2x)^2 = 0.14`

`=> x^2/(0.78 - 2x)^2` = 0.14

`=> x/(0.78 - 2x) = 0.374`

`=> x= 0.292 - 0.748  x`

`=> 1.748  x = 0.292`

⇒ x = 0.167

Hence, at equilibrium,

[ICl] = [I2] = 0.167 M[ICl]

= (0.78 - 2 × 0.167)M

= 0.446 M

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Law of Chemical Equilibrium and Equilibrium Constant
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पाठ 7: Equilibrium - EXERCISES [पृष्ठ २३३]

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एनसीईआरटी Chemistry - Part 1 and 2 [English] Class 11
पाठ 7 Equilibrium
EXERCISES | Q 7.16 | पृष्ठ २३३

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