The following equilibrium constants were determined at 1120 K -2 CO g - rightleftharpoons C s - CO - 2 g - - K - P - 1-10-14 atm -1CO g - Cl - 2 g rightleftharpoons COCl - 2 g - - K - p - 2-6 -10-3atm-1What is the equilibrium constant K C for the following reaction at 1120 K- C s - CO 2 g -2 Cl 2 g - 2 COCl 2 g A- 5-51 - 106 M -1B- None of theseC- 3-31 - 10-11 M -1D- 5-5 - 1010 M -1

The correct option is C 3.31 × 10^ 11M^ 1C(s) + CO_2(g) ⇌ 2CO(g); K_p = 10^14atm 2CO(g) + 2Cl_2 (g) ⇌ 2COOCl_2 ; K_P_2 = (6 × 10^ 3)^2 atm^ 2 Add (1) and ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $3.31 \times 10^{- 11} M^{- 1}$
$2 C O (g)$ ⇌ $C (s) + C O_{2} (g)$ , $K_{P_{1}} = 10^{- 14} a t m^{- 1}$
Reverse this reaction will get, $C (s) + C O_{2} (g)$ ⇌ $2 C O (g)$ , $K_{p_{1}} = \frac{1}{10^{- 14}} a t m^{- 1} = 10^{14} a t m^{- 1}$ ...........(1)
$C O (g) + C l_{2} (g)$ ⇌ $C O C l_{2} (g), K_{p_{2}} = 6 \times 10^{- 3} a t m^{- 1}$
Multiply this equation by 2, we get
$2 C O (g) + 2 C l_{2} (g)$ ⇌ $2 C O C l_{2} (g), K_{p_{2}}^{2} = (6 \times 10^{- 3})^{2} = 36 \times 10^{- 6} a t m^{- 1}$ ................(2)
Adding (1) and (2) equation, we get
$C (s) + C O_{2}) (g) + 2 C l_{2} (g)$ ⇌ $2 C O C l_{2} (g)$
$K_{p} = K_{p_{1}} . K_{p_{2}}^{2}$
= $10^{14} \times 36 \times 10^{- 6} = 36 \times 10^{8}$
We know that $K_{p} = K_{c} \times (R T)^{Δ n_{g}} K_{c} = K_{p} \times (R T)^{- Δ n_{g}}$
For given reaction $Δ_{n_{g}}$ = -2 (number of gaseous molecules of products(2) - number of gaseous molecules of reactants(4) ), R = 0.0821 $L . a t m . K^{- 1} m o l^{- 1}$ , T = 1120 K
therefore $K_{c} = K_{p} (R T)^{2}$
$K_{c} = 36 \times 10^{8} \times (0.0821)^{2} (1120)^{2} = 3.31 \times 10^{- 11} M^{- 1}$

Suggest Corrections

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)

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)

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} {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)

Join BYJU'S Learning Program

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