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

Byju's Answer

Standard XII

Chemistry

Reaction Quotient

The equilibri...

Question

The equilibrium constant of the equation, $H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)$ is $0.44$ at $1259 K$ . The value of equilibrium constant of the equation, $H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)$ will be:

0.22

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

\frac{1}{0.44}

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

- \frac{1}{0.44}

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

0.44

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

Open in App

Solution

The correct option is B $\frac{1}{0.44}$
Given
Equilibrium constant is 0.44
Solution
$K c = \frac{[H_{2}] * [C O_{2}]}{[H_{2} O] * [C O]}$
As we can see reaction is reversed Therefore Equilibirium constant become 1/0.44
The correct option is B

Suggest Corrections

Similar questions

Q. One mole of

H_{2} O

and one mole of CO are taken in 10 L vessel and heated to 725 K. At equilibrium 40% of water reacts with CO according to the equation,

H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)

. What is the value of equilibrium constant

(K_{c})

for the reaction?

Q. From the given data of equilibrium constants of the following reactions,

C a O (s) + H_{2} (g) ⇌ C a (s) + H_{2} O (g); K = 60

C a O (s) + C O (g) ⇌ C a (s) + C O_{2} (g); K = 400

Calculate the equilibrium constant of the reaction,

C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

Q. If

CoO (s) + H_{2} (g) ⇌ C o (s) + H_{2} O (g), K_{1} = 60

;

C o O (s) + C O (g) ⇌ C o (s) + C O_{2}, K_{2}

= 180 then the equilibrium constant of the reaction

C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

will be-

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. From the given data of equilibrium constants of the following reactions,

C a O (s) + H_{2} (g) ⇌ C a (s) + H_{2} O (g); K = 60

C a O (s) + C O (g) ⇌ C a (s) + C O_{2} (g); K = 400

Calculate the equilibrium constant of the reaction,

C O_{2} (g) + H_{2} (g) ⇌ C O (g) + H_{2} O (g)

Join BYJU'S Learning Program

The equilibrium constant of the equation, H2O(g)+CO(g)⇌H2(g)+CO2(g) is 0.44 at 1259 K. The value of equilibrium constant of the equation, H2(g)+CO2(g)⇌H2O(g)+CO(g) will be:

The correct option is B 10.44GivenEquilibrium constant is 0.44SolutionKc=[H2]∗[CO2][H2O]∗[CO]As we can see reaction is reversed Therefore Equilibirium constant become 1/0.44The correct option is B

The equilibrium constant of the equation, $H_{2} O (g) + C O (g) ⇌ H_{2} (g) + C O_{2} (g)$ is $0.44$ at $1259 K$ . The value of equilibrium constant of the equation, $H_{2} (g) + C O_{2} (g) ⇌ H_{2} O (g) + C O (g)$ will be:

The correct option is B $\frac{1}{0.44}$
Given
Equilibrium constant is 0.44
Solution
$K c = \frac{[H_{2}] * [C O_{2}]}{[H_{2} O] * [C O]}$
As we can see reaction is reversed Therefore Equilibirium constant become 1/0.44
The correct option is B