The following equilibrium constants were determined at 1120 K- 2COg - Cs - CO2g Kp1 - 10-14 atm-1COg - Cl2g - COCl2 Kp2 - 6 - 10-3 atm-1What is the equilibrium constant Kp for the following reaction at 1120 K- Cs - CO2g - 2Cl2g - 2COCl2g

2CO(g) ⇌ C(s) + CO_2(g) ; Kp_1 = 10^ 14 atm^ 1......(1) CO(g) + Cl_2(g) ⇌ COCl_2(g); Kp_2 = 6 × 10^ 3 atm^ 1.....(2) Multiply eqn (2) by (2), we get nbsp; ...

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Standard XII

Chemistry

Molality

The following...

Question

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

36 \times 10^{4} a t m^{- 1}

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3.6 \times 10^{9} a t m^{- 1}

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3.6 \times 10^{5} a t m^{- 1}

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3.6 \times 10^{10} a t m^{- 1}

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Solution

The correct option is B $3.6 \times 10^{9} a t m^{- 1}$
$2 C O (g) ⇌ C (s) + C O_{2} (g); K p_{1} = 10^{- 14} a t m^{- 1} . . . . . . (1)$
$C O (g) + C l_{2} (g) ⇌ C O C l_{2} (g); K p_{2} = 6 \times 10^{- 3} a t m^{- 1} . . . . . (2)$
Multiply eqn (2) by (2), we get
$2 C O (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)$ $K^{'} p_{2} = (6 \times 10^{- 3})^{2} a t m^{- 1}$ ...(3)
reverse eqn (1) we get
$C (O) + C O_{2} (g) ⇌ 2 C O$ $K p_{1}^{1} = \frac{1}{10^{- 14}} a t m^{- 1}$ ...(4)
add eqn (3) and (4)
$C (s) + C O_{2} (g) + 2 C l_{2} (g) ⇌ 2 C O C l_{2} (g)$
$K = K p_{2}^{1} \times K p_{1}^{1}$
$K = (6 \times 10^{- 3})^{2} \times \frac{1}{10^{- 14}}$
$, K = \frac{36 \times 10^{- 6}}{10^{- 14}}$
$K = 36 \times 10^{8}$
$K = 3.6 \times 10^{9} a t m^{- 1}$

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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} {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} (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. 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)

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 1120 K: 2CO(g)⇌C(s)+CO2(g) Kp1=10−14 atm−1 CO(g)+Cl2(g)⇌COCl2 Kp2=6×10−3 atm−1 What is the equilibrium constant Kp for the following reaction at 1120 K. C(s)+CO2(g)+2Cl2(g)⇌2COCl2(g)