1 mole of H 2 O and CO are taken in a 10 liter container at equilibrium point 40 - of H 2 O reacts with CO to form H 2 O - CO 2 Calculate the equilibrium constant for the reaction-

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1 mole of H 2...

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1 mole of H₂O and CO are taken in a 10 liter container at equilibrium point 40% of H₂O reacts with CO to form H₂O & CO₂Calculate the equilibrium constant for the reaction.

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1

H_{2} O

1

C O

10

1259 K

40 % H_{2} O

C O

H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)

H_{2} O

H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)

(K_{c})

H_{2} O

C O

10

725 K

40

H_{2} O_{(g)} + C O_{(g)} ⇌ H_{2 (g)} + C O_{2 (g)}

C O_{(g)} + H_{2} O_{(g)} ⇌ C O_{2 (g)} + H_{2 (g)}

2.2

H_{2} O

2.0

([C O_{2}] = [H_{2}] = 0.2986 M, [C O] = [H_{2} O] = 0.2014 M)

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,

Calculate the equilibrium constant for the reaction.

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