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Characteristics of Equilibrium Constant

The equilibri...

Question

The equilibrium mixture for the reaction $C O (g) + H_{2} O_{(g)} ⇌ C O_{2 (g)} + H_{2 (g)}$ contains $2, 3, 4, 6$ moles respectively at $500$ K in a $2$ litre vessel. Now this mixture is transferred in a $4$ lit vessel. What is the new equilibrium constant?

A

4

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B

2

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C

8

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D

6

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Solution

The correct option is A $4$
${C O}_{(g)} + H_{2} O_{(g)} ⇌ {C O}_{(g)} + H_{2 (g)}$
2moles 3moles 4moles 6moles
conc. $\frac{2}{2} = 1$ $\frac{3}{2}$ $\frac{4}{2}$ $\frac{6}{2}$
$K_{C} = \frac{[{C O}_{2}] [H_{2}]}{[C O] [H_{2} O]} = \frac{2 \times 3}{1 \times 1.5} = 4$

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Similar questions

Q. The equilibrium constant for the reaction

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

is 3 at 500K. In a 2 litre vessel 60gm of water gas [equimolar mixture of CO(g) and

H_{2} (g)

] and 90 gm of steam is initially taken.

What is the equilibrium concentration of

H_{2} (g)

at equilibrium (mole/L)?

675

H_{2} (g)

C O_{2} (g)

C O (g)

H_{2} O (g)

K_{p}

for the reaction is

0.16

. If a mixture of

0.25

H_{2} (g)

0.25

C O_{2}

675

%

C O (g)

in equilibrium mixture is:

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.

675 K

H_{2} (g)

C O_{2} (g)

C O (g)

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0.16

. If a mixture of

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0.25

C O_{2}

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, mole percentage of

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Q. The equilibrium constant for the reaction,

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

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0.63

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