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One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation,

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प्रश्न

One mole of H2O and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium, 40% of water (by mass) reacts with CO according to the equation, 

\[\ce{H2O (g) + CO (g) ⇌ H2 (g) + CO2 (g)}\] 

Calculate the equilibrium constant for the reaction.

संख्यात्मक
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उत्तर

The given reaction is:

  H2O(g) + CO(g) H2(g) CO2(g)
Initial conc. `1/10`M   `1/10`M   0 0
At equilibrium `(1 - 0.4)/10`M   `(1 - 0.4)/10`M   `0.4/10`M `0.4/10`M
  = 0.06 M   = 0.06 M   = 0.04 M = 0.04 M

Therefore, the equilibrium constant for the reaction,

`"K"_"c" = (["H"_2]["CO"_2])/(["H"_2"O"]["CO"])`

`= (0.04 xx 0.04)/(0.06 xx 0.06)`

= 0.444 (approximately)

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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 7.14 | पृष्ठ २३३

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