The following equilibrium constants were determined at 1120K- 2COg- Cs- CO 2 g- K p 1 - 10 -14 atm -1 COg-Cl 2 g- CO Cl 2 g- K p 2 - 6- 10 -3 atm -1What is the equilibrium constant K c for the following reaction at 1120K- Cs- CO 2 g-2Cl 2 g- 2CO Cl 2 g

The correct option is A 3.31× 10 ^ 11 M ^ 1 2CO(g)⇌ C(s)+ CO _ 2 (g); K _ p _ 1 = 10 ^ 14 atm ^ 1 so for reverse reaction: C(s)+ CO _ 2 ( ...

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Standard XII

Chemistry

Characteristics of Equilibrium Constant

The following...

Question

The following equilibrium constants were determined at $1120 K$ :
$2 C O (g) ⇌ C (s) + {C O}_{2} (g); K_{p_{1}} = 10^{- 14} {a t m}^{- 1}$
$C O (g) + C l_{2} (g) ⇌ C O {C l}_{2} (g); K_{p_{2}} = {6 \times 10}^{- 3} {a t m}^{- 1}$
What is the equilibrium constant $K_{c}$ for the following reaction at $1120 K$ ?
$C (s) + {C O}_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O {C l}_{2} (g)$

{3.31 \times 10}^{11} M^{- 1}

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{5.5 \times 10}^{10} M^{- 1}

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{5.51 \times 10}^{6} M^{- 1}

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None of these

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Solution

The correct option is A ${3.31 \times 10}^{11} M^{- 1}$
$2 C O (g) ⇌ C (s) + {C O}_{2} (g); K_{p_{1}} = 10^{- 14} {a t m}^{- 1}$ so for reverse reaction:
$C (s) + {C O}_{2} (g) ⇌ 2 C O (g); K_{p^{'}} = 1 / 10^{- 14} {a t m}^{- 1} = 10^{14}$
$C O (g) + C l_{2} (g) ⇌ C O {C l}_{2} (g); K_{p_{2}} = {6 \times 10}^{- 3} {a t m}^{- 1}$ so for
$2 C O (g) + 2 C l_{2} (g) ⇌ 2 C O {C l}_{2} (g); K_{p^{''}} = {6 \times 10}^{- 3}^{2} = 36 \times 10^{- 6}$ so for
$C (s) + {C O}_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O {C l}_{2} (g)$
$K_{p} = K_{p}^{'} \times K_{p}^{''} = 36 \times 10^{8}$ and $K_{c} = K_{p} R T = 3.31 \times 10^{11} .$

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Similar questions

Q. Given,

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}

The value of

K_{c}

for the following reaction at

1120 K

will be?

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} {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)

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)

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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The following equilibrium constants were determined at 1120K:2CO(g)⇌C(s)+CO2(g);Kp1=10−14atm−1CO(g)+Cl2(g)⇌COCl2(g);Kp2=6×10−3atm−1What is the equilibrium constant Kc for the following reaction at 1120K?C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g)