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Introduction to Reversible and Irreversible Reactions

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Question

The following equilibrium constants were determined at $1120 K$ :
$2 C O (g) ⇌ C (s) + C O_{2} (g); K_{p 1} = 10^{- 14} {atm}^{- 1}$
$C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p 2} = 6 \times 10^{- 3} {atm}^{- 1}$
What is the equilibrium constant $K_{P}$ for the following reaction at same temperature?
$C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)$

A

6 \times 10^{8} {atm}^{- 1}

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B

36 \times 10^{11} {atm}^{- 1}

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C

36 \times 10^{8} {atm}^{- 1}

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D

6 \times 10^{11} {atm}^{- 1}

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Solution

The correct option is C $36 \times 10^{8} {atm}^{- 1}$
$2 C O (g) ⇌ C (s) + C O_{2} (g); K_{p 1} = 10^{- 14} {atm}^{- 1}$
Therefore,
$C (s) + C O_{2} (g) ⇌ 2 C O (g) . . . . . . (1)$
$K_{p 3} = 10^{14} atm$
$C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p 2} = 6 \times 10^{- 3} {atm}^{- 1}$
Therefore,
$2 C O (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g) . . . . . . (2)$
$K_{p 4} = (6 \times 10^{- 3})^{2} {atm}^{- 2}$
Add $(1)$ and $(2)$
$C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g);$
$K_{p} = 10^{14} \times 36 \times 10^{- 6}$
$\Rightarrow K_{p} = 36 \times 10^{8} {atm}^{- 1}$

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Similar questions

Q. The following equilibrium constants were determined at 1120 K:

2 C O (g) ⇌ C (s) + C O_{2} (g) K p_{1} = 10^{- 14} a t m^{- 1}

C O (g) + C l_{2} (g) ⇌ C O C l_{2} K p_{2} = 6 \times 10^{- 3} a t m^{- 1}

What is the equilibrium constant

K_{p}

for the following reaction at 1120 K.

C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)

Q. The following equilibrium constants were determined at

1120 K

2 C O (g) ⇌ C (s) + C O_{2} (g); K_{P_{1}} = 10^{- 14} a t m^{- 1} C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}

What is the equilibrium constant

K_{C}

for the following reaction at

1120 K

C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)

Q. The following equilibrium constants were determined at

1120 K

:

2 C O (g) ⇌ C (s) + {C O}_{2} (g); K_{p_{1}} = 10^{- 14} {a t m}^{- 1}

C O (g) + C l_{2} (g) ⇌ C O {C l}_{2} (g); K_{p_{2}} = {6 \times 10}^{- 3} {a t m}^{- 1}

What is the equilibrium constant

K_{c}

for the following reaction at

1120 K

?

C (s) + {C O}_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O {C l}_{2} (g)

2 C O (g) ⇌ C (s) + C O_{2} (g); K_{p_{1}} = 10^{- 14} a t m^{- 1}

1120 K

C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}

K_{c}

for the following reaction at

1120 K

C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)

The following equilibrium constants were determined at $1120 K$ :
$2 C O (g)$ ⇌ $C (s) + C O_{2} (g)$ ; $K_{p_{1}} = 10^{- 14} a t m^{- 1}$
$C O (g) + C l_{2} (g)$ ⇌ $C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}$

What is the equilibrium constant $K_{C}$ for the following reaction at $1120 K$ ;

$C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)$

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