Given- 2COg - Cs - CO2g- Kp1 - 10-14atm-1 at 1120 K COg - Cl2g - COCl2g- Kp2 - 6- 10-3 atm-1The value of Kc for the following reaction at 1120 K will be- Cs - CO2g - 2Cl2g - 2COCl2g

The correct option is C 3.31× 10^11M^ 1 (1) 2CO(g) ⇌ C(s) + CO_2(g); K_p_1 = 10^ 14atm^ 1 at 1120 K (2) CO(g) + Cl_2(g) ⇌ COCl_2(g); K_p_2 ...

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Byju's Answer

Standard XII

Chemistry

Characteristics of Equilibrium Constant

Given, 2COg...

Question

Given, $2 C O (g) ⇌ C (s) + C O_{2} (g); K_{p_{1}} = 10^{- 14} a t m^{- 1}$ at $1120 K$
$C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}$
The value of $K_{c}$ for the following reaction at $1120 K$ will be?
$C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)$

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

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

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

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4.35 \times 10^{8} M^{- 1}

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Solution

The correct option is C $3.31 \times 10^{11} M^{- 1}$
(1) $2 C O (g) ⇌ C (s) + C O_{2} (g); K_{p_{1}} = 10^{- 14} a t m^{- 1}$ at $1120 K$
(2) $C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}$
Multiply equation (2) with 2.
(3) $2 C O (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g); K_{p_{3}} = (6 \times 10^{- 3} a t m^{- 1})^{2} = 3.6 \times 10^{- 5} a t m^{- 2}$
Subtract equation (1) from equation (3).
(4) $C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g); K_{p_{4}} = \frac{3.6 \times 10^{- 5} a t m^{- 2}}{10^{- 14} a t m^{- 1}} = 3.6 \times 10^{9} a t m^{- 1}$
For equation (4) $Δ n = 2 - (1 + 2) = - 1$
$K_{c} = K_{p} (R T)^{- Δ n}$
$K_{c} = 3.6 \times 10^{9} a t m^{- 1} \times (0.08206 L a t m / m o l K \times 1120 K)^{- (- 1)}$
The value of $K_{c}$ for the following reaction at $1120 K$ will be
$3.31 \times 10^{11} M^{- 1}$

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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} (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 $1120 K$ ;

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

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} {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)

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Given, 2CO(g)⇌C(s)+CO2(g);Kp1=10−14atm−1 at 1120 KCO(g)+Cl2(g)⇌COCl2(g);Kp2=6×10−3atm−1The value of Kc for the following reaction at 1120 K will be?C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g)