You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Chemistry

Molality

Calculate the...

Question

Calculate the equilibrium constant $K_{p}$ for the reaction $C_{(s)} + C O_{2 (G)} ⇌ 2 c o_{(g)}$ at 1300K from the following data $C_{(s)} + 2 H_{2} O_{(g)} ⇌ C O_{2 (g)} + 2 H_{2 (g)} K_{p} (1300 K) = 3.9 a t m$
$H_{2 (g)} + C O_{2 (g)} ⇌ C O_{(g)} + H_{2} O_{(g)} K_{p} (1300 K) = 0.7 a t m$

0.49 atm

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

1.911 atm

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

0.7 atm

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

2.73 atm

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

Open in App

Solution

The correct option is B 1.911 atm
$K_{p} = \frac{[C O]^{2} [H_{2} O]^{2}}{[H_{2}]^{2} [C O_{2}]^{2}} \times \frac{[H_{2}]^{2} [C O_{2}]}{[H_{2} O]^{2}} = \frac{[C O]^{2}}{[C O]^{2}} = (0.7) 2 \times 3.9 = 1.911 a t m$

Suggest Corrections

Similar questions

Calculate the equilibrium constant ( $K_{p}$ ) for the reaction, $C (s) + C O_{2} (g) ⇌ 2 C O (g)$ , at $1300 K$ from the following data:
$C (s) + H_{2} O (g) ⇌ C O (g) + H_{2} (g); K_{p} (1300 K) = 3.9 a t m$

$H_{2} (g) + C O_{2} (g) ⇌ C O (g) + H_{2} O (g); K_{p} (1300 K) = 0.623 a t m$

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} 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. Considerthe following reactions,
I.

C (s) + \frac{1}{2} O_{2} (g) \to C O (g); Δ H_{1} = a

II.

C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); Δ H_{2} = b

III.

C (s) + C O_{2} (g) \to 2 C O (g); Δ H_{3} = c

Select the correct statement

Join BYJU'S Learning Program

Calculate the equilibrium constant Kp for the reaction C(s)+CO2(G)⇌2co(g) at 1300K from the following data C(s)+2H2O(g)⇌CO2(g)+2H2(g)Kp(1300K)=3.9atm H2(g)+CO2(g)⇌CO(g)+H2O(g)Kp(1300K)=0.7atm

The correct option is B 1.911 atmKp=[CO]2[H2O]2[H2]2[CO2]2×[H2]2[CO2][H2O]2=[CO]2[CO]2=(0.7)2×3.9=1.911atm

The correct option is B 1.911 atm
$K_{p} = \frac{[C O]^{2} [H_{2} O]^{2}}{[H_{2}]^{2} [C O_{2}]^{2}} \times \frac{[H_{2}]^{2} [C O_{2}]}{[H_{2} O]^{2}} = \frac{[C O]^{2}}{[C O]^{2}} = (0.7) 2 \times 3.9 = 1.911 a t m$