The equilibrium constant for the reaction. CO ( g )+ H 2 O ( g ) leftharpoons CO 2( g )+ H 2( g ) is 5. How many moles of CO 2 must be added to 1 liter container already containing 3 mol each of CO and H 2 O to make 2 M equilibrium concentration CO?

Question

The equilibrium constant for the reaction. CO ( g )+ H 2 O ( g ) leftharpoons CO 2( g )+ H 2( g ) is 5. How many moles of CO 2 must be added to 1 liter container already containing 3 mol each of CO and H 2 O to make 2 M equilibrium concentration CO? (A) 15 (B) 19 (C) 5 (D) 20

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/b59cfdcf3ccc6a4da3c679c9e72dfe76-.png" /> 5=((x+1)/4)=20=x +1 ⇒ x=19

Q. The equilibrium constant for the reaction.
$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$ is $5.$ How many moles of $C O_{2}$ must be added to $1$ liter container already containing $3 m o l$ each of $CO$ and $H_{2} O$ to make $2 M$ equilibrium concentration $CO$ ?

15

19

5

20

$5 = \frac{( x + 1 )}{4} = 20 = x + 1$

$\Rightarrow x = 19$

Q. The equilibrium constant for the reaction. CO(g)+H2​O(g)⇌CO2​(g)+H2​(g) is 5. How many moles of CO2​ must be added to 1 liter container already containing 3mol each of CO and H2​O to make 2M equilibrium concentration CO?

15

19

5

20

5=4(x+1)​=20=x+1 ⇒x=19

Q. The equilibrium constant for the reaction.
$CO (g) + H_{2} O (g) ⇌ C O_{2} (g) + H_{2} (g)$ is $5.$ How many moles of $C O_{2}$ must be added to $1$ liter container already containing $3 m o l$ each of $CO$ and $H_{2} O$ to make $2 M$ equilibrium concentration $CO$ ?

$5 = \frac{( x + 1 )}{4} = 20 = x + 1$

$\Rightarrow x = 19$