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Equilibrium Constant and Standard Free Energy Change

One mole of H...

Question

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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Solution

The given reaction is:

Therefore, the equilibrium constant for the reaction,

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Similar questions

H_{2} O

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

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

What is the value of equilibrium constant

(K_{c})

for the reaction.

H_{2} O

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

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

. What is the value of equilibrium constant

(K_{c})

for the reaction?

H_{2} O

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

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

. What is the value of equilibrium constant

(K_{c})

for the reaction?

H_{2} O

and one mole of

C O

10

litre vessel and heated to

725 K

. At equilibrium,

40

percent of water (by mass) reacts with carbon monoxide according to the equation:

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

Calculate the equilibrium constant for the reaction.

1

H_{2} O

1

C O

are heated in a

10

litre vessel at

1259 K

40 % H_{2} O

C O

according to the following equations,

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

;
Calculate equilibrium constant for the reaction.

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